TCS

Analysis of TCS Placement Papers

Quantitative Aptitude/Logical Reasoning

Corporate Corporat e & International Relations

Amrita Vishwa Vidyapeetham


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TABLE OF CONTENTS Introduction 1. Number System

Questions taken from students’ forum Theory, Examples & Explanations

2. Percentages


3. Averages


4. Problems on Ages


5. Profit & Loss


6. Ratio Proportion


7. Time, Speed & Distance


8. Time & Work


9. Heights & Distance


10. Sets & Matrices


11. Functions


12. Sequence & Series

Theory, Examples & Explanations

13. Statistics


14. Clocks & Calendar


15. Geometry


16. Logical Reasoning


17. Data Interpretation


18. Miscellaneous

Questions taken from students’ forum

Formula Booklet

2

Corporate & International Relations



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The book “ Analysis of TCS Placement Papers ” is an attempt to provide you with a mirror to previous years’ papers. Only practicing with questions is not enough because of its unpredictability. Learning concepts can make you tackle all sorts of problems with ease. The Review of topics is designed to familiarize you with the mathematical skills and concepts likely to be tested. This material includes many definitions and examples with solutions. Note, however, this review is not intended to be comprehensive. It is assumed that certain basic concepts are common for all examinees. examinees. Emphasis is, therefore, placed on the more important skills, concepts, and definitions, and on those particular areas that are frequently asked. If any of the topics seem especially unfamiliar, we encourage you to consult appropriate mathematics texts for a more detailed treatment of those topics.

All the Best

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Tata Consultancy Service Quantitative Aptitude

Number System (Base Conversion, Logarithms, Prime Numbers) Questions taken from students’ forum 1. If log 0.317=________ and log 0.318=________ 0.318=________ , then find the value of log 0.319. 2. The base 5 representation of the decimal number 2048 is 3. Which is the largest prime number that can be stored in a 9 -bit memory? 4. What is the largest prime number that can be stored in an 8-bit memory? 5. In which base system, the decimal number 194 is equal to 1234? 6. Find the value of the 678 to the base 7. 7. Represent 222 in base 5 number system 8. What is the max possible 3 digit prime number? 9. Which is the power of 2 n?

a. 2068

b.2048

c.2668

10.The 10. The number 362 in decimal system is given by (1362) x in the X system of numbers. Find the value of X? 11.A 11. A sales person multiplied a number by 3 and get the answer is 3, instead of that number divided by 3. What is the answer that he actually has to get? 12.What 12. What is the largest prime number stored in 6 bit memory? 13.What 13. What is the largest prime number stored in 7 bit memory?

Theory, Examples & Explanations Number Theory Finding Remainders of a product (derivative of remainder theorem) If ‘a1‘is divided by ‘n’, the remainder is ‘r 1’ and if ‘a2’ is divided by n, the remainder is r 2. Then if a1+a2 is divided by n, the remainder will be r1 + r2 If a1 – a2 is divided by n, the remainder will be r 1 – r2 If a1 × a2 is divided by n, the remainder will be r 1 × r2

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Back to Table of Contents



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Ex. If 21 is divided by 5, the remainder is 1 and if 12 is divided by 5, the remainder is 2. Then if (21 + 12 = 33) is divided by 5, the remainder will be 3 (1 + 2). If 9(21 – 9(21 – 12) is divided by 5, the remainder will be 1 – 1 – 2 = – = – 1. But if the divisor is 5, – 5, – 1 is nothing but 4. 9 = 5 × 1 + 4. So, if 9 is divided by 5, the remainder r emainder is 4 and 9 can be written as 9 = 5 × 2 – 1. So here – here – 1 is the remainder. So – So – 1 is equivalent to 4 if the divisor is 5. Similarly – 2 is equivalent to 3. If 252(21 × 12) is divisible by 5, the remainder will be (1 × 2 = 2). If two numbers ‘a1’ and ‘a2‘ are exactly divisible by n. Then their sum, difference and product is also exactly divisible by n. i.e., If ‘a1’ and ‘a2’ are divisible by n, then

a1 + a2 is also divisible by n a1 – a2 is also divisible by n and If a 1 × a2 is also divisible by n. Ex.1

12 is divisible by 3 and 21 is divisible by 3.

Sol.

So, 12 + 21 = 33, 12 – 21 = – = – 9 and 12 × 21 = 252 all are divisible by 3.

Finding Remainders of powers with the help of Remainder theorem: 25

Ex.2

What is the remainder if 7 is divided by 6?

Sol.

If 7 is divided by 6, the remainder is 1. So if 7 25 is divided by 6, the remainder is 1 25 (because 725 = 7× 7 × 7… 25 times. 25

So remainder = 1 × 1 × 1…. 25 times = 1 ). Ex.3

What is the remainder, if 3 63 is divided by 14.

Sol.

If 33 is divided by 14, the remainder is – is – 1. So 3 63 can be written as (3 3)21. 21

So the remainder is ( – ( – 1) = – 1. If the divisor is 14, the remainder – remainder – 1 means 13. (14 – (14 – 1 = 13) By pattern method 33

Ex.4

Find remainder when 4

is divided by 7.

Sol.

If 41 is divided by 7, the remainder is 4. (4 1 = 4 = 7 × 0 + 4) 2

2

If 4 is divided by 7, the remainder is 2 (4 = 16 = 7 × 2 + 2) If 43 is divided by 7, the remainder is 1 (4 3 = 42 × 4, so the Remainder = 4× 2 = 8 = 1) If 44 is divided by 7, the remainder is 4 (4 4 = 43 × 4, so the Remainder = 1× 4 = 4) The remainders of the powers of 4 repeats after every 3 rd power. So, as in the case of finding the last digit, since the remainders are repeating after every 3 rd power, the remainder of 433 is equal to the remainder of 43 ( since 33 is exact multiple of 3) = 1. (OR) If 43 is divided by 7, the remainder is 1. So 4 33 = (43)11 is divided by 7, the remainder is 1 1 Back to Table of Contents

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Application of Binomial Theorem in Finding Remainders The binomial expansion of any expression of the form n

n

n

n

n – 1

(a + b) = C0 a + C1 a

1

n

× b + C2 × a

n – 2

2

n

1

n – 1

× b  ..….  Cn – 1 × a × b

n

+ Cn × b

n

Where nC0, nC1, nC2,…are all called the binomial coefficients. coefficients. In general, There are some fundamental conclusions that are helpful if remembered, i.e. a. There are (n + 1) terms. b. The first term of the expansion has only a. c. The last term of the expansion has only b. d. All the other (n – (n – 1) terms contain both a and b. n

n

n

e. If (a + b) is divided by a then the remainder will be b such that b < a. Ex.5

What is the remainder if 7 25 is divided by 6?

Sol.

(7) can be written (6 + 1) . So, in the binomial expansion, all the first 25 terms will have 6 in it. The 26 term is (1) .

25

25

th

25

Hence, the expansion can be written 6x + 1. 6x denotes the sum of all the first 25 terms. Since each of them is divisible by 6, their sum is also divisible divisible by 6, and therefore, can be written 6x, where x is any natural number. So, 6x + 1 when divided by 6 leaves the remainder 1. (OR) When 7 divided by 6, 6, the remainder is 1. So when 7 25 is divided by 6, the remainder will be 1 25 = 1.

Wilson’s Theorem

If n is a prime number, (n – 1)! + 1 is divisible by n. Let take n = 5 Then (n – (n – 1)! + 1 = 4! + 1 = 24 + 1 = 25 which is divisible by 5. Similarly if n = 7 (n – (n – 1)! + 1 = 6! + 1 = 720 + 1 = 721 which is divisible by 7. Corollary 2

p

If (2p + 1) is a prime number (p!) + ( – – 1) is divisible by 2p + 1. e.g

If p = 3 , 2p + 1 = 7 is a prime number (p!)2 + ( – – 1)p = (3!)2 + ( – – 1)3 = 36 – 36 – 1 = 35 is divisible by (2p + 1) = 7.

Property p

If “a” is natural number and P is prime number then (a – a) is divisible by P.

e.g.

6

If 2 31 is divided by 31 what is the remainder?





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(231/31) = (2 31-2+2)/31 = So remainder = 2 Fermat’s Theorem

If p is a prime number and N is prime to p, then (N) p-1 – 1 is a multiple of p. Corollary Since p is prime, p – p – 1 is an even number except when p = 2. Therefore (N

(p-1/2)

(p-1/2)

+1)(N

-1) = M(p)

Hence either N (p-1/2)+1 or N(p-1/2)-1 is a multiple of p, that is N (p-1/2) = kp+1 or kp-1, where, k is some positive integer. Base Rule and Conversion This system utilizes only two digits namely 0 & 1 i.e. the base of a binary number system is two. e.g. 1101 2 is a binary number, to find the decimal value of the binary number, powers of 2 are used as weights in a binary system and is as follows: 3

1×2 =8 1 × 22 = 4 1

0×2 =0 1 × 20 = 1 3

2

1

0

Thus, the decimal value of 11012 is 1 × 2 + 1 × 2 + 0 × 2 + 1 × 2 = 13. Conversion from decimal to other bases We will study only four types of Base systems, 1. Binary system (0, 1) 2. Octal system (0, 1, 2, 3, 4, 5, 6, 7). 3. Decimal system (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) 4. Hexa-decimal Hexa-decimal system (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C D, E, F) where A = 10, B = 11 … F = 15. Let us understand the procedure with the help of an example Ex.6

Convert 35710 to the corresponding binary number.

Sol.

To do this conversion, you need to divide repeatedly by 2, keeping track of the remainders as you go. Watch below: As you can see, after dividing repeatedly by 2, we end up with these remainders:


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Quants Funda

These remainders tell us what the binary number is! Read the numbers outside the division block, starting from bottom and wrapping your way around the right-hand side and moving upwards. Thus, (357)10 convert to (1011001 (101100101) 01) 2. This method of conversion will work for converting to any non-decimal base. Just don't forget to include the first digit on the left corner, which is an indicator of the base. You can convert from base-ten (decimal) to any other base.

Conversion from other bases to Decimal We write a number in decimal base as 2

1

345 = 300 + 40 + 5 = 3 × 10 + 4 × 10 + 5 × 10

0

Similarly, when a number is converted from any base to the decimal base then we write the number in that base in the expanded form and the result is the number in decimal form. Ex.7

Convert (1101)2 to decimal base

Sol.

(1101) 2 = 1 × 2 + 1 x 2 + 0 × 2 + 1 × 2

3

2

1

0

= 8 + 4 + 1 = 13 So (1101) 2 = (13) 10 Ex.8

Convert the octal no 3456 in to decimal number.

Sol.

3456 = 6 + 5 × 8 + 4 × 8 + 3 × 8

2

3

= 6 + 40 + 256 + 1536 = (1838) 10 Ex.9

Convert (1838)10 to octal.

Sol. = (3456) 8 Ex.10

What is the product product of highest 3 digit number & highest 2 digit number number of base 3 system? (1) (21000) 3

8

(2) (22200) 3

(3) (21222) 3


(4) (21201) 3

(5) None




Sol.

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The highest 3 digit & 2 digit numbers are 222 & 22 222 = 2 + 2 × 3 + 2 × 3 2 = 26 22 = 2 + 2 × 3 = 8 ∴

Product = 26 × 8 = 208

Convert back to base (21201)3 Answer: (4) 29

29

Ex.11

What is the remainder, if 24

Sol.

an + bn is always divisible by a + b, if is odd. 29

+ 34 is divided by 29?

29

24 + 34 is always divisible by 24 + 34 = 58. So, it is always divisible by 29. So, the remainder is 0. Ex.12

What is the remainder, if 12 243 is divided by 10?

Sol.

12

243

The remainder repeats after every 4 th power. 3

So, the required answer is the remainder of 12 is divided by 10. i.e. 8 Ex.13

What is the value of (FBA) 16 in binary system?

Sol.

A = 10, B = 11, F = 15 4

Since 2 = 16, While converting each digit of the decimal, can be written as 4 digit binary no: A = 1010, B = 1011, F = 1111 (FBA)10 = (111110111010) 2 Ex.14

Convert (721)8 to binary.

Sol.

Since 23 = 8, write each digit of octal no. as 3 digits of binary which gives equivalent value. 7 = 111, 2 = 010, 1 = 001 ∴

(721) 8 = (111 010 001) 2


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CYCLICITY At times there are questions that require the students to find the units digit in case of the numbers occurring in powers. If anyone asks you to find the unit digit of 3 3, you will easily calculate it also you can calculate for 35 but if any one ask you the unit digit of 173 99, it will be hard to calculate easily. But it’s very simple if we understand that the units digit of a product is determined by whatever is the digit at the units place irrespective of the number of digits digits.. E.g. 5 × 5 ends in 5 & 625 × 625 also ends in 5. Now let’s examine the pattern that a number generates when it occurs in See the last digit of different numbers.

Unit Digit Chart Power

From the above table we can conclude that the unit digit of a number repeats after an interval of 1, 2 or 4. Precisely we can say that the universal cyclicity of all the numbers is 4 i.e. after 4 all the numbers start repeating their unit digits.

Therefore, to calculate the unit digit for any exponent of a given number we have to follow the following steps Step 1: Divide the exponent of the given number by 4 and calculate the remainder. Step 2: The unit digit of the number is same as the unit digit of the number raise to the power of calculated remainder. Step 3: If the remainder is zero, then the unit digit will be same as the unit digit of N4.


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Quants Funda

Let us consider an example Ex.1

Find the last digit of (173) 99.

Sol.

We notice that the exponent is 99. On dividing, 99 by 4 we get 24 as the quotient & 3 as the remainder. Now these 24 pairs of 4 each do not affect the no. at the units place so, (173)

99

≈ (173) 3. Now, the number at the units place is 3 3 =

27. Factors A factor is a number that divides another number completely. e.g. Factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24. Number of Factors a

b

c

If we have a number, N = p × q × r Where p, q, and r are prime numbers and a, b, and c are the no. of times each

prime number occurs , then the number of factors of n is found by (a + 1) (b + 1)(c Example: 4

2

Find the number of factors of 2 × 3 . Number of factors = (4 + 1) (2 + 1) = 5(3) = 15

Number of Ways of Expressing a Given Number as a Product P roduct of Two Factors When a number is having even number of factors then it can be written as a product of two numbers in (a+1)(b+1)(c+1)/2 ways. But if a number have odd number of factors then it can be written as a product of two different numbers in [(a+1)(b+1)(c+1)-1]/2 ways and can be written as a product of two numbers (different or similar) in [(a+1)(b+1)(c+1)+1]/2 ways. Examples: 1. 148 can be expressed as a product of two factors in 6/2 or 3 ways. {Because (p + 1) (q + 1) (r + 1) in the case of 148 is equal to 6}. 2. 144 (2 4.32) can be written as a product of two different numbers in [(4 + 1)( 2 + 1) −1]/2 i.e. 7 ways Sum of the factors of a number: p

q

r

If a number N is written in the form of N = a .b .c ,where a, b & c are prime numbers and p, q & r are positive integers, then the sum of all the factors of the number are given by the formula Sum of factors = Back to Table of Contents

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Factorial Factorial is defined for any positive integer. It is denoted by L or !. Thus “Factorial n” is written as n! Or n! is defined as the product of all the integers from 1 to n. Thus n! = 1.2.3. …. n. (n! = n(n – 1)!) Finding the Highest power of the t he number dividing a Factorial Ex.2

Find the largest power of 3 that can divide 95! 95! without leaving any any remainder.OR remainder.OR Find the largest power of of 3 contained in 95!.

Sol.

First look at the detailed explanation and then look at a simpler method for solving the problem. When we write 95! in its full form, we have 95 × 94 × 93 ….. × 3 × 2 × 1. When we divide 95! 95! by a power 3, we have these 95 numbers in the numerator. The denominator denominator will have all 3’s. The 95 numbers in the numerator have 31 multiples of 3 which are 3, 6, 9….90, 93.Corresponding to each of these multiplies we can have a 3 in the denominator which will divide the 31

numerator completely without leaving any remainder, i.e. 3 can definitely divide 95! Further every multiple of 9, i.e. 9, 18, 27, etc. after canceling out a 3 above, will still have one more 3 left. Hence for every multiple of 9 in the numerator, we have an additional 3 in the denominator. There are 10 multiples of 9 in 95 i.e. 9, 18….81, 90. So we can take 10 more 3’s in the denominator. 3

Similarly, for every multiple of 3 we can take an additional 3 in the denominator. Since there are 3 multiples of 27 in 91 (they are 27, 54 and 81), we can have thre e more 3’s in the denominator. 4

Next, corresponding to every multiple of 3 i.e. 81 we can have one more 3 in the denominator. Since there is one multiple of 81 in 95, we can have one additional 3 in the denominator. Hence the total number of 3’s we can hav have e in the denominator is 31 + 10 + 3 + 1, i.e., 45. So 3

45

is the largest power of

3 that can divide 95! without leaving any remainder. The same can be done in the following manner also. Divide 95 by 3 you get a quotient of 31. Divide this 31 by 3 we get a quotient of 10. Divide this 10 by 3 we get a quotient of 3. Divide this quotient of 3 once again by 3 we get a quotient of 1. Since we cannot divide the quotient any more by 3 we stop here. Add all the quotients, i.e. 31 + 10 + 3 + 1 which gives 45 which is the highest power of 3.

Add all the quotients 31 + 10 + 3 + 1, which give 45. {Note that this type of a division where the quotient of one step is taken as the dividend in the subsequent step is called “Successive Division”. In general, in successive division, the divisor need not be the same (as it is here). Here, the number 95

is being successively divided by 3. Please note that this method is applicable only if the number whose largest power is to be Back to Table of Contents

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found out is a prime number. number. If the number is not a prime number, then we have to write the number as the product of relative primes, find the largest power of each of the factors separately first. Then the smallest, among the largest powers of all these relative factors of the given number will give the largest power required. Ex.3

Find the largest power of 12 that can divide 200!

Sol.

Here we cannot apply Successive Division method because 12 is not a prime number. Resolve 12 into a set of prime factors. We know that 12 can be written as 3 × 4. So, we will find out the largest power of 3 that can divide 200! and the largest power of 4 that can divide 200! and take the LOWER of the two as the largest power of 12 that can divide 200!. To find out the highest power of 4, since 4 itself is not a prime number, we cannot directly apply the successive division method. We first have to find out the highest power of 2 that can divide 200!. Since two 2’s taken together will give us a 4, half the power of 2 will give the highest power of 4 that can divide 200!. We find that 197 is the largest power of 2 that can divide 200!. Half this figure-98-will be the largest power of 4 that can divide 200!. Since the largest power of 3 and 4 that can divide 200! are 97 an 98 respectively, the smaller of the two, i.e., 97 will be the largest power of 12 that can divide 200! without leaving any remainder. 34

34

Ex.4

What is the last digit of 2 × 3 × 4

Sol.

Given = (24) 34

34

n

Last digit of 4 is 6, if n is even. Answer 6 Ex.5

What is the right most non zero digit of (270)

Sol.

The required answer is the last digit of 7 270.

270

Last digit of 7 powers repeat after every 4. So, the last digit of 7

270

2

is the last digit of 7 = 9.

Ex.6

How many factors do 1296 have?

Sol.

1296 = 4 × 324 = 4 × 4 × 81 4

4

=2 ×3

Number of factors = (4 + 1) (4 + 1) = 25. Ex.7

If x is the sum of all the factors of 3128 and y is the no of factors of x and z is the number of ways of writing ‘y’ as a

product of two numbers, then z = ? Sol.

3128 = 4 × 782 = 4 × 2 × 391 3

= 2 × 17 × 23

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Quants Funda

= 15 × (17 + 1) (23 + 1) =3×5×9 ×2×8×3 4

4

=2 ×3 ×5 y = (4 + 1) (4 + 1) (1 + 1) =2×5

2

z = 1/2 { (1 + 1) (2 + 1) } = 3 Ex.8

How many cofactors are there for 240, which are less than 240?

Sol.

240

= 16 × 15

= 24 × 3 × 5 Number of co primes to N, which are less than N

if N = ab × bq × - - - - (a, b, - - - - are Prime no.s) = 64 Ex.9

What is the sum of all the co primes to 748? Which are less than N?

Sol.

748

= 4 × 187

2

= 2 × 11 × 17

Sum of all the co primes to N. which are less than N is N/2 (number of co primes to N, which are less than N. Sum =

320 = 119680

Ex.10

In how many ways 5544 can be written as a product product of 2 co primes?

Sol.

If N = ap × bq × - - - -, where a, b, - - - - are prime numbers N can be written as a product of two co primes in 2 5544

n-1

ways, where n is the number of prime factors to N.

= 11 × 504

= 11 × 9 × 56 = 11 × 9 × 8 × 7 = 23 × 32 × 7 × 11 Answer: = 2 4-1 = 23 = 8. (Because, 2, 3, 7 & 11 are four different prime factors). Ex.11

If n! have 35 zeroes at the end. What is the least value ‘n’ will take?

(1) 110

14

(2) 120

(3) 130


(4) 140

(5) 145




Sol.

Quants Funda

Since the number of zeroes are 35, 5 35 should exactly divide n! by trail & error, take n = 140.

So, there are 34 zeroes. The answer should be 145. Answer: (5) Ex.12

‘N’ is a five digit number. The last digit of N

(1) 2 Sol.

(2) 3

(3) 7

(4) 8

35

is 2. What is the last digit of N?

(5) Cannot be determined

The last digit repeats after every 4 th power. Since the last digit of N

35

is 2

The last digit of N 3 is 2, which is possible only for 8. Answer: (4) Ex.13

What is the right most non zero digit in 40 40/2020

Sol.

= The required answer is the last digit of 2

60

=6

Percentages Questions taken from students’ forum 1. The si size ze of a prog program ram is N. And the mem memory ory occ occupi upied ed by the pro progr gram am is is give given n by by M =

. If If the the size size of the pro progra gram m is

increased by 1% then how much memory now occupied? 2. Given Y =

. Find the % change in Y if N is increased by 1%?

3. If A, B and C are the mechanisms used separately to reduce the wastage of fuel by 30%, 20% and 10%. What will be the fuel economy if they were used combined (in %)? 4. What percent of 60 is 12? 5. If the length of a rectangle is increased by 30% and the width is decreased by 20%, then the area is increased by_________%?

(a) 10%

(b) 5%

(c) 4%

(d) 20%

(e) 25% Back to Table of Contents

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Quants Funda

Theory, Examples & Explanations Percentages What is the meaning of percent? The terms percent means “for every hundred”. A fraction whose denominator is 100 is called ercentage and the numerator of the fraction is called the rate percent. percent. Thus, when we say a man made a profit of 20 percent we mean to say that he gained Rs. 20 for every hundred rupees he invested in the business, i.e. 20/100 rupees for each R upee. The abbreviation of percent is p.c. and it is generally denoted by %. Ex.1

84% of a particular total is 630 marks. What is 90% equal to? (1) 750

Sol.

(2) 675

The required answer is

(3) 450

(4) 550

(5) None of these

= 675.

Answer: (2) Ex.2

Two numbers numbers are greater than the third number by 25% 25% and and 20% 20% respectively. respectively. What What percent percent of first number number is the second number? (1) 92%

Sol.

(2) 94 %

(3) 96 %

(4) 98 %

(5) None of these

Assume the third number is 100. So the first number is 125 and the second number is 120. So the required answer is

= 96%.

Answer: (3) Ex.3

A is earning 20% more than B, who earns 20% 20% less than C. By what what percent A earns earns more or less than C?

Sol.

From (1) and (2) we have

i.e. A is 4% less than C. (b) Multiplying Factor While dealing with %age increase or decrease picture the following scale in your mind with reference as 100% ( = 1) in the center. So we can say that multiplying factor (M.F) of 10% increase is 1.1 and that of 15% decrease is 0.85. An increase by x% implies impl ies the value lies on the right hand side of 100% & vice versa. Let’s start with a number X Back to Table of Contents

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Quants Funda

1. X increased by 10% would become X + 0.1 X = 1.1X 2. X increased by 1% would become X + 0.01 X = 1.01X 3. X increased by 0.1 % would become X + 0.001 X = 1.001X 4. X decreased by 10% would become X – X – 0.1X = 0.9X 5. X decreased by 1% would become X – X – 0.01 X = 0.99X 6. X decreased by 0.1% would become X – 0.001 X = 0.999X 7. X increased by 200% would become X + 2X = 3X 8. X decreased by 300% would become X – 3X = − 2X Ex.4

Coconut oil is now being sold at Rs. 27 per kg. During last month its cost was Rs. 24 per kg. Find by how much percent a family should reduce its consumption, to keep the expenditure the same.

Sol.

Assume the consumption last year is 1 kg, and then it cost Rs. 24. But now for Rs.24, only

kg of oil will come.

So the % reduction in consumption = (c) Successive Percentage change: The population of a city increases by 10% in one year and again increases by 10% in the next year, then what is the net increase in the population in two years. The very common answer is 20% which is wrong. Why? Let us see If Original population = P st

After 1 year = After 2nd year = I.e increases by 21% of the original value. This successive change in the percentage can be calculated in the shortcut way as explained below: Let us consider a product of two quantities A = a x b. If a & b change (increase or decrease) by a certain percentage say x & y respectively, then the overall percentage change in their product is given by the formula Ex.5

If the volume of a milk and water solution solution is increased by 25% by pouring pouring only water. By what percentage percentage does does the concentration of milk reduce?

Sol.

Assume initially, there is 100 lts of solution, out of which x lts is milk. So the concentration of milk is

. Now it is

So the percentage decrease = Back to Table of Contents

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Quants Funda

This formula also holds true if there are successive changes as in the case of population increase or decrease. But care has to be taken when there are either more than 2 successive changes or there is a product of more than 2 quantities as in the case of volume. In these cases we have to apply the same formula twice. (i) If there is successive increase of x% and y%, then the net change will be (ii) If there is successive discount of x% and y%, then the total discount will be (a) If there is x% increase and then x% decrease, then the net change (b) If the values are different, then net change Ex.6

If A is increased increased by 20% and and B is decreased decreased by 20%, then then both the quantities quantities will be equal. equal. What What percentage percentage of B is A.

Sol.

If A is increased by 20%, it will be come 1.2 A. and B is decreased by 20%, it will become 0.8 B. It is given 1.2 A = 0.8 B,

A = B.

A = 0.66 B. A is 66.66% of B. Ex.7

If A is 3 times to B then B is what percentage of A.

Sol.

A = 3B

Ex.8

If ‘x’ is increased by 20% & 2 5% successively, then its value increases by 30. What is the value of X?

Sol.

If X is increased by 20%, it will become 1.2X or X If it is again increased by 25%, its value becomes =

It is given that 30.

Ex.9

A man spends spends 30% 30% of his salary salary for food and 20% of the remaining remaining on rent and 20% of the remaining remaining on other expenses. If he saves Rs. 8960, what is his salary?

Sol.

Let his salary be ‘K’. If he spent 30% on food, he will be left with 70% of K i.e. 0.7K or

on this, he spends

20% on rent and 80 left with 80%. i.e After spending 20% of this on other expenses he will be left with 80% of this. i.e.

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It is given as Rs.8960 K = 20000 Ex.10

Last year an an employee used to save 40% 40% of his his salary. But But now his salary is increased by 50% and expenses also increases by 20%. What is his percentage savings now?

Sol.

Assume ‘X’ was the salary of the employee last year. Since his savings were 40%, his expenses were 60% i,e, 0.6X. At present, his salary is 1.5X and expenses are 1.2(0.6X) = 0.72X Savings = 1.5X – 1.5X – 0.72X = 0.78X % Savings =

Ex.11

A man earns X% on on the first Rs. Rs. 2000 and Y% Y% on the rest of income. If he earns earns Rs. 700 from Rs.4000 Rs.4000 and Rs. 900 900 from Rs. 5000 income; find X%.

Sol.. Sol

We can form two equations from the above information as

Equa Eq uattio ion n (1 (1) 3 – Eq Equ uat atio ion n (2) (2) 2

X =15%

Averages Questions taken from students’ forum 1. In Mad Madra rass, tem temp per erat atur ure e at at noo noon n var variies ac acco cord rdin ing g to to

, wh where t is is ela elaps psed ed ti time me.. Fin Find d ho how mu much te temp mper erat atu ure

more or less in 4pm to 9pm. 2. Low temperature at the night in a city is 1/3 m ore than ½ hinge as higher temperature in a day. Sum of the low temp and highest temp is 100C. What is the low temperature? 3. Hansie made the following amounts in seven games of cricket in India: Rs.10, Rs.15, Rs.21, Rs.12, Rs.18, Rs.19 and Rs.17 (all figures in cores of Rs). Find his average earnings. 4. A truck contains 150 small packages, some weighing 1 kg each and some weighing 2 kg each. How many packages weighing 2 kg each are in the truck if the total weight of all the packages is 264 kg? Back to Table of Contents

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Quants Funda

Theory, Examples & Explanations Averages Definition Simple Average (or Mean) is defined as the ratio of sum of the quantities to the number of quantities. By definition, Average = Putting in symbols, Here

represent the n values of quantity under consideration and

is the mean. Average or Mean is said to be a

measure of central tendency. Ex.1

If a person with age 45 joins a group of 5 persons with an average age age of 39 years. What will be the new average average age of the group?

Sol.

Total age will be 45 + 5× 39 = 240. And there will be 6 persons now. So the average will be 240/6 = 40.

(or)

Since 45 is 6 more than 39, by joining the new person, the total will increase by 6 and so the average will increase by 1. So, the average is 39 + 1 = 40.

Ex.2

Two students with marks 50 and 54 leave class VIII VIII A and and move to class VIII B. As a result the average average marks of the class VIII A fall from 48 to 46. How many students were there initially in the class VIII A?

Sol.

The average of all the students of class VIII A is 46, excluding these two students. They have 4 and 8 marks more than 46. So with the addition of these two students, 12 marks are adding more, and hence the average is increasing 2. There should be 6 students in that class including these two. This is the initial number of students,

Ex.3

The average average of x successive successive natural natural numbers numbers is N. If the next natural number is included included in the group, the average average increases by:-

Sol.

(1) Depends on x

(2) Depends on the starting number of the series

(3) Both (1) and (2)

(4)

(5) None of these

The average of consecutive numbers is the middle number. If one more number is added to the list, the middle number moves 0.5 towards right. So the answer is (4).


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Weighted Mean If some body asks you to calculate the combined average marks of both the sections of class X A and X B, when both sections have 60% and 70% average marks respectively? Then your answer will be 65% but this is wrong as you do not know the total number of students in each sections. So to calculate weighted average we have to know the number of students in both the sections. Let N 1, N2, N3, …. Nn be the weights attached to variable values X 1, X2, X3,…….. Xn respectively. Then the weighted arithmetic mean, usually denoted by For any two different quantities taken in different ratios. The weighted average is just like a see-saw. More the ratio of a quantity more will be the inclination of the average from mid value towards the value with more ratios. Ex.4

The average marks of 30 students students in a section of class X are 20 while that that of 20 20 students of second section is 30. Find the average marks for the entire class X?

Sol.

We can do the question by using both the Simple average & weighted average method. = By the weighted mean method, Average =

= 24

=24

Real Facts about average 1. If each number is increased / decreased by a certain quantity n, then the mean also increases or decreases by the same quantity. 2. If each number is multiplied/ divided by a certain quantity n, then the mean also gets multiplied or divided by the same quantity. 3. If the same value is added to half of the quantities and same value is subtracted from other half quantities then there will not be any change in the final value of the average. Average Speed Average Speed = If d1, d2 are the distances covered at speeds v 1 and v2 respectively and the time taken are t 1 and t2 respectively, then the average speed over the entire distance (X 1 +X2) is given by If both the distances are equal i.e. d 1 = d2 = d then, Average Speed = But if the time taken are equal i.e. t 1 = t2 = t then, Average Speed = Ex.5

The average of 10 consecutive numbers starting from 21 is:

Sol.

The average is simply the middle number, which is the average of 5

Ex.6

There are two classes A and B., each has 20 students. The average average weight of class A is 38 and that of class B is 40. X

th

th & 6 no. i.e, 25 & 26 i.e. 25.5.

and Y are two students of classes A and B respectively. If they interchange their classes, and then the average weight of both the classes will be equal. If weight of x is 30 kg, what is the weight of Y ? Back to Table of Contents

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Sol.

Quants Funda

Total weight of class A = 38 × 20, and class B = 40 × 20, if X & Y are interchanged, then the total ages of both the classes are equal. 38 × 20 – 20 – x + y = 40 × 20 – y + x . 2(y – 2(y – x) = 2 × 20, y = x + 20 = 50 (OR) Since both the classes have same number of students, after interchange, the average of each class will be 39. Since the average of class ‘A’ is increasing by 1, the total should increase by 20. So, x must be replaced by ‘y’, who must be 20 years elder to ‘x’. So, y must be 50 years old..

Ex.7

The average weight of 10 apples apples is 0.4 0.4 kg. If the heaviest and and lightest apples are are taken out, the average average is 0.41 kg. If the lightest apple weights 0.2 kg, what is the weight of heaviest apple?

Sol.

Total weight of the apples is 0.4 × 10 = 4 kg. Weight of apples except heaviest & lightest = 0.41 × 8 = 3.28 kg Heaviest + lightest = 4 – 4 – 3.28 = 0.72 kg. It is given lightest = 0.2 kg. Heaviest is 0.72 – 0.72 – 0.2 = 0.52 kg.

Ex.8

While finding the average of ‘9’ consecutive numbers starting from X; a student interchanged the digits of second

number by mistake and got the average which is 8 more than the actual. What is X? Sol.

Since the average is 8 more than the actual, the second no will increase by 72 (9 × 8) by interchanging the digits. If ab is the second no, then 10a + b + 72 = 10b + a,

9(b – 9(b – a) = 72. ∴ b – a = 8.

The possible number ab is 19. Since the second no is 19. The first no is 18.

X = 18

Ex.9

There are 30 consecutive consecutive numbers. What is the difference between the averages averages of first and and last 10 numbers?

Sol.

The average of first 10 numbers is the average of 5 th & 6th no. Where as the average of last 10 numbers is the average th

th

th

th

of 25 & 26 no. Since all are consecutive numbers, 25 number is 20 more than the 5 number. We can say that the average of last 10 nos is 20 more than the average of first 10 nos. So, the required answer is 20. Instructions for next 3 examples: There are 60 students in a class. These students are divided into three groups A, B, C of 15, 20 & 25 students each. The groups A & C are combined to form group D. Ex.10

What is the average weight of the students in group group D? D? 1) More than the average weight of A. 2) More than the average weight of C. 3) Less than the average weight of C. 4) Less than the average weight of B. 5) Cannot be determined.

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Sol.

Quants Funda

We know only the no of students in each class, but we don’t know the average weight of any class, we can’t find the answer.

Ex.11

If one student from group group A is shifted to group B, which of of the following is necessarily necessarily true? true? 1) The average weight of both groups increases 2) The average weight of both groups decreases. 3) The average weight of class cl ass remains the same. 4) The average weight of group A decreases and that of group B increases. 5) None of these

Sol.

Options (1) & (2) are not possible. Average of both cannot increase or decrease. Option (4) can be eliminated because we are not sure, whether the average of A increases & B decreases or A decreases & B increases or both remains unchanged. It depends on the weight of the student, who shifted from A to B. Option (3) is always true because even the student shifts from one group to other. The average weight of the whole class does not change. Answer: (3)

Ex.12

If all the students students of the class have the same weight, which of the following following is false? false? 1) The average weight of all the four groups is same. 2) Total weight of A & C is twice that of B. 3) The average weight of D is greater than that of A. 4) The average weight of class remains same even the students shifts from one group to other. 5) None of these. these.

Sol.

Since each student has same weight (1) & (4) are right. Since the number of students in group A & C together is 15 + 25 = 40. Where as in B, there are only 20 students. So option (2) is also correct. But (3) is false, because, the average weight of each group is same, since all the students have same weight. Answer: (3)

Problems on ages (Word Problems) Questions taken from students’ forum 1. A power unit in bank of the river of 750 meters width. A cable is made from power unit to power a plant opposite to that of the river and 1500mts away from the power unit. The cost of the cable below water is Rs. 15/- per meter and cost of cable on the bank is Rs.12/- per meter. Find the total of laying the cable. 2. Given two pencils cost 8 cents. Find the cost of 5 pencils?

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3. One year ago Pandit was three times his sister's age. Next year he will be only twice her age. How old will Pandit be after five years? 4. There are two trees in a lawn. One grows at a rate 3/5 of the other in 4 years. If the total growth of trees is 8 ft. What is the height of the smaller tree after 2 years? 5. The sum of the digits of a two digit number is 8. When 18 is added to the number, the digits are reversed. Find the number? 6. Given the cost of 1 pencil, 2 pens and 4 erasers is Rs.22 and the cost of 5 pencils, 4 pens and 2 erasers is Rs.32.How much will 3 pencils, 3 pens and 3 erasers erasers cost? 7. Father’s age is 5 times his son's age. Four years back the father was 9 times older than son. Find the fathers' present age.

Theory, Examples & Explanations Linear Equations /Word Problems/Problems on ages General Theory of Equations Polynomial equation: A polynomial which is equal to zero is called a polynomial equation. For example, 2x + 5 = 0, x 2 – 2x + 5 = 0, 2

2x – 5x2 + 1 = 0 etc. are polynomial equations. Linear Equation: A linear equation is 1st degree equation. Its general form is ax + b = 0 Quadratic Equation: A quadratic equation is 2nd degree equation. The general form of a quadratic equation is 2

ax + bx + c = 0. Root of a polynomial equation: If f (x) = 0 is a polynomial equation and f (α ( α) = 0, then α is called a root of the polynomial equation f (x) = 0. Theorem: Every equation f (x) = 0 of nth degree has exactly n roots. If f(x) is linear the number of roots will be one, if it is quadratic, then number of roots is two. If we draw the graph of the expression y = x 2 – 2x + 1, then we would note that the graph cuts the x -axis at two points. This is because it has two real roots. Factor Theorem: If α If α is a root of equation f (x) = 0, then the polynomial f (x) is exactly divisible by x – α (i.e., remainder is zero). For example, x

2

– 5x + 6 = 0 is divisible by x – 2 because 2 is the root of the given equation. Remainder Theorem: If we divide polynomial f(x) by (x – (x – α ) then f( α ) is the remainder.

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For example, f(x) = x 2 – 5x + 7 = 0 is divided by x – 2 the remainder is f(2) where f(2) = 2 2 – 5 × 2 + 7 = 1. Linear Equation A linear equation is 1 st degree equation. It has only one root. Its general form is a x + b = 0 and root is – . LINEAR EQUATION OF ONE AND TWO VARIABLES Linear Equation: Equation: The equation which when reduced to its simplest form contains only the first power of the variable is called linear equation or simple equation. In other words, a linear equation in one variable is an equation of the type ax + b = 0, or ax = c, where a, b, c are constants (real numbers), a ≠ 0 and x is a variable. v ariable. We know that the solution of ax + b = 0, a ≠ 0 is x = - . We say x = -

is a root of the equation equation ax + b = 0.

Linear Equation in two variables: variables : A general linear equation in two variables x and y is usually written in the following forms: (i) ax + by + c = 0 where a ≠ 0, b ≠ 0; a, b, c are real constants (ii) ax + by = d where a ≠ 0, b ≠ 0; a, b, d are real constants (iii)

Any pair of values of x and y which satisfies ax + by + c = 0 is called its solution.

(iv)

A pair of linear equations in two variables say x a nd y is said to form a system of simultaneous linear equations in

two variables. The general form of a system of linear equations in two variables x and y is a1x + b1y + c1 = 0 a2x + b2y + c2 = 0 Set of Equations

Consistent System (means have solution)

Inconsistent System (means have no solution)

Dependent System (has infinitely many solutions) Condition

Independent System (Only one/Unique Solution) Condition

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Word Problems Questions based on word problems can be solved with the help of linear questions: Examples: 1.

Ten years ago, the average age of a family of four members was 25 years. Two children having been born, the average age of the family is the same today. What is the age of the youngest child if they differ in age by 2 years? (A) 3 years

Sol.

(B) 2 years

(C) 4 years

(D) 8 years

(E) 5 years

Present total age = 25 × 6 = 150; Total age of 4 members 10 years ago = 25 × 4 = 100. Total age of 4 members now = 100 + 40 = 140 Sum of ages of children = 150 – 150 – 140 = 10 years Let x be the age of youngest child. x + x + 2 = 10 x = 4. Answer: (C)

2.

If the average age of three persons is 20 years. With the addition of one more person, average age of all four becomes 21 years. Find the age of the fourth member. (A) 24 years

(B) 22 years

(C) 21 years

(D) 25 years

(E) None of

these Sol.

Sum of ages of three members = 20 × 3 = 60 years. Let’s the age of fourth members = X Now 60 + X = 21 × 4 = 84. So age of fourth member (X) = 24 years. Answer: (A)

Quadratic Equation Quadratic Equation in “x” is one in which the highest power of “x” is 2. The equation is generally satisfied by two values of “x”, but these values may be equal to each other. The quadratic form is generally represented by ax 2 + bx + c = 0 where a ≠ 0, and a, b, c are constants. e.g. x2 – 6x + 4 = 0 2

3x + 7x – 7x – 2 = 0 A quadratic equation in one variable has two and only two roots, which are

and


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Nature of Roots The term (b2 – 4ac) is called the discriminant of the quadratic equation ax2 + bx + c and is denoted by D. Rules: 1. if D > 0, x 1 & x2 are real and unequal. 2. if D = 0, x 1 & x2 are real and equal. 3. if D is a perfect square, x 1 & x2 are rational and unequal. 4. if D < 0, x 1 & x2 are imaginary, unequal, and conjugates of each other. If x1 and x2 are the two roots of ax 2 + bx + c = 0, then sum of roots = x1 + x2 = – b/a and product of roots = x1 x2 = c/a. Formation of equation from roots: 1. If x1 and x2 are the two roots, then (x – (x – x1) (x – (x – x2) = 0 is the required equation. 2. If (x 1 + x2) and x1 × x2 are given then equation is x 2 – (x1 + x2) x + x 1 x2 = 0. 2

x – Sx + P = 0 where S = sum of roots, P = product of roots.

Profit & Loss Questions taken from students’ forum 1. A shopkeeper bought a watch for Rs.400 and sold it for Rs.500.What is his profit percentage? 2. The most economical prices among the following prices is: (a) 10 kilo for Rs.160

(b) 2 kilo for Rs.30

(d) 20 kilo for Rs.340

(e) 8 kilo for Rs.130

(c) 4 kilo for Rs.70

Theory, Examples & Explanations Profit, Loss & Discount Definition Cost Price:

C.P. is the price at which one buys anything

Selling Price:

S.P. is the price at which one sells anything Back to Table of Contents

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Profit/Loss:

Quants Funda

This is the difference between between the selling price and the cost cost price. If the difference is positive it is called the

profit and if negative it is called as loss. Profit/Loss %:

This is the profit/loss as a percentage of the C.P.

Margin:

Normally is in % terms only. This is the profit as a percentage of S.P.

Marked Price:

This is the price of the product as displayed on the label.

Discount:

This is the reduction given on the marked price before selling it to a customer. If the trader wants to make a

loss he can offer a discount on the cost price as well Markup:

This is the increment on the cost price before being sold to a customer.

Formulae • Gain = (S.P. (S.P. – – C.P.), Loss = (C.P. – (C.P. – S.P.) • Gain % = (Gain × 100)/C. P, Loss % = (Loss × 100)/C. P. • Given the cost & the gain percent, S.P. S.P . = (100 + gain %) × C. P. / 100 • Given the cost & the loss percent, S.P. = (100 – (100 – loss %) × C. P. / 100 • Given the M.P. & the discount, C.P. = (100 – (100 – Discount %) x M.P / (100 + gain %) • Given the M.P. & the discount, C.P. = (100. (100. – – Discount %) x M.P / (100 (100 – – loss %) Ex.1

A person sells 36 oranges for one rupee and suffers suffers a loss of 4%. Find how many many oranges oranges per rupee to be sold to have a gain of 8%?

Sol.

Let ‘X’ is the cost price of each orange. Since he is giving 36 apples for one rupee, the selling price of an orange is 1/36 rupee. Since he got 4% loss, the selling price of each orange is 0.96X = 1/36 To get 8% gain he has to sell it for 1.08X= (1.08/0.96)*(1/36) (1.08/0.96)*(1/36) = 1/32 rupee. So for one rupee, he has to give 32 oranges to get a gain of 8%.

Discount You always come across different offers attracting the customers such as “Buy 1 get 2 Free” or “Buy 3 get 5 Free” or “SALE 50% + 40%”.

Can you calculate the discount offered to you? Most of us are not aware about the offer given to us. The percentage of the discount offered in the first case is not 200% but it is 66.66% only. The discount is always on the number of items sold, not on the number of items purchased. In case of successive discounts we can treat the problem as the problem of successive percentage percentage change and can use the formula Net discount = E.g.: 40% + 30% discount =

28


= (70 (70 – – 12) % = 58%.




Quants Funda

Markup Price It is also known as list price or Tag price which is written on the item. The markup price written is always greater than the actual C.P of the item and the percentage rise in the markup price is on the C.P of the item. Percentage increase in the Markup price = Ex.2

The price of a trouser is marked 50% 50% more than its cost price price and a discount discount of 25% is offered on the marked price price of the trouser by the t he shopkeeper. Find the percentage of profit/loss.

Sol.

M.P = 1.5 C.P S.P = 0.75× 1.5 C.P = 1.125 C.P So profit percentage = 12.5 %.

Ex.3

After allowing allowing a discount of 11.11%, 11.11%, a trader still still makes a profit profit of 14.28%. At how much percent percent above the cost price does he mark on his goods?

Sol.

Discount of 11.11% means a discount of 1/9 and 14.28% means 1/7 So selling price = So MP = CP. So it is

= 28.56% more than the CP.

Two different articles sold at same selling selli ng price Overall % loss = Ex.4

where x is the percent profit or loss on the transaction.

Each of the two horses is sold for Rs. 1875. The first one is sold sold at 25% profit and the other one at 25% loss. What is the % loss or gain in this t his deal?

Sol.

It is loss of

= 6.25 % loss.

Ex.5

What is the total loss or gain (in rupees) in the above example?

Sol.

Since he got 6.25 % loss means (1/16)

th

loss.

So, his selling price should be (15/16) th of the CP th

So loss is (1/15) of the SP = (1/15)(1875+1875) = Rs. 250. Faulty Balance Sometimes traders may sell their products at the rate at which they purchased or even less than the actual cost incurred to them. Even in this transaction they make profit by cheating on volume. If the weighing balance of a shopkeeper reads 1000 grams for every 900 grams, grams , what is the profit or loss the shopkeeper is making? On the other hand if the faulty balance reads 900 grams for every 1000 grams, grams , is he still making profit? If not why? Ex.6

Instead of a meter scale, a cloth cloth merchant merchant uses a 120 120 cm scale while buying, but uses uses an 80 cm cm scale while selling the same cloth. If he offers a discount of 20% on the cash payment, what is his overall profit percentage? Back to Table of Contents

29




Sol.

Quants Funda

When the merchant is buying he is using a scale of 120 cm instead of 100 cm thus multiplying factor for him in this transaction = 120/100 = 6/5 …….(1) When selling the cloth the merchant is measuring 80 cm for every 100 cm , so multiplying factor of this transaction is = 100/80 = 5/4 = …….(2) For the discount offered by the merchant the multiplying factor = 80/100 = 4/5 ……. (3) Net profit =

.

Hence making a profit of 20% in the whole transaction Ex.7

Mr. A purchased purchased an article article and sold it to Mr. B at 20% profit. profit. Mr. B sold it to Mr. C at at 20% profit. If If Mr. C paid Rs. 2880 to Mr. B, what is the profit in rupees earned by Mr. A?

Sol.

Assume that Mr. A purchased the article for Rs. X, Then he sold it to Mr. B for Rs. 1.2X and Mr. B sold it to Mr. C for R s. 1.2 × 1.2X = 1.44X. This is given as Rs. 2880. So X = Rs. 2000. Profit earned by A is 20% of X = Rs. 400.

Ex.8

A cloth merchant allows 25% discount discount on on a saree and and still makes 20% profit. By selling a saree, saree, he gained Rs. 160. 160. What is the marked price of that saree?

Sol.

He makes a profit of 20% and it is given as Rs. 160. 20% of C.P = 160 So C.P = Rs. 800. So S.P = 800 + 160 = Rs. 960. And 0.75 M.P = 960 (since he is giving 25% discount) M.P = Rs. 1280

Ex.9

A man purchased some chocolates chocolates at at 80 per Rs 100 and same same number of chocolates chocolates of other type at 120 per per Rs. 100. He sold each chocolate per 1 rupee each, what is his profit/loss percentage?

Sol.

The first type of chocolate costs Rs. 100/80 rupees = 5/4 rupee. The second type costs Rs. 100/120 = 5/6 rupee Since he purchased both the chocolates in equal number, the Average cost per chocolate is But he is selling each chocolate at Re. 1. So he gets a loss of


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Ratio Proportion Questions taken from students’ forum 1. In a mixture, R and S in the ratio of 2:1. In order to make S to 25% of the mixture, how much part of R should be added? 2. What number should be added to or subtracted from each term of the ratio 17: 24 so that ratio becomes equal to 1: 2. 3. In a fraction, if 1 is added to both the numerator and the denominator, the fraction becomes 1/2. Instead if numerator is subtracted from the denominator, the fraction becomes 3/4. Find the fraction. 4. Rs.1260 is divided between A, B and C in the ratio 2:3:4. Find the C's share?

Theory, Examples & Explanations Ratio Proportion Concept If a and b (b ≠ 0) are two quantities of the same kind, then Ratio is the relation which one quantity bears to another of the same kind in magnitude. Now in two quantities a and b the fraction a/b is called the ratio of a to b. It is usually expressed as a: b, a and b are said to be the terms of the ratio. The former (numerator) ‘a’ is called the Antecedent of the ratio and latter (denominator) ‘b’ is called consequent consequent.. Basics of Ratio (i) As ratio is a relation between two quantities so ratio is independent of the concrete units employed in the quantities compared. (ii) Ratio exists only between two quantities; both the quantities must be in the same units. Composition of Ratio I. Compounded Ratio When two or more ratios are multiplied term wise, the ratio thus, obtained is called their compounded ratio. Compounded ratio of (a: (a: b, b, c: d and e: f ) II. Reciprocal Ratio For any ratio a : b, the reciprocal ratio will be

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The application of it is in calculating the ratio of the wages distributed among the workers, which is equal to the reciprocal ratio of number of days taken by them to complete the work. Proportion When two ratios are equal, the four terms involved, taken in order are called proportional, and they are said to be in proportion. 1. The ratio of a to b is equal to the ratio of c to to d i.e. i.e. if

, we write a : b : : c : d

2. If a : b : : c : d Then we have ad = bc i.e. the terms a and d are called Extremes and, the terms b and c are called the Means. The term a, b, c and d are known by the name 1 st, 2nd, 3rd and the 4th proportion respectively. Continued Proportion: Three quantities are said to be in continued proportion, if the ratio of the first to the second is same as the ratio of the second to the third. Thus a, b and c are in continued proportion if a : b = b : c (or) Mean Proportion: If a, b, c are in continued continued proportion then second quantity ‘b’ is called the mean proportional between ‘a’ and ‘c’ and a, b and c are known as 1 st, 2nd and 3rd proportion respectively. Then b2 = ac (or) b = Ex.1

What is the mean proportiona proportionall to

Sol.

Let the mean proportion of

Ex.2

What should should be added to each of the numbers 19, 26, 26, 37, 37, 50 so that that the resulting number should should be in proportion? proportion? (1) 2

Sol.

and

(2) 3

be x

(3) – 2

(4) – 5

Let the required number be x. According to question, 19 + x, 26 + x, 37 + x and 50 + x are in proportion 19 + x : 26 + x = 37 + x : 50 + x

(19 + x) (50 + x) = (37 + x) (26 + x)

x 2 + 69x + 950 = x 2 + 63x + 962

x2 – x2 + 69x – 69x – 63x = 962 – 962 – 950

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Hence the resulting number = 2. Basics of Proportion (i) Let us take four quantities a, b, c & d such that they are in proportion proportion i.e. a : b : : c : d then,

This operation is called componendo and Dividendo. Dividendo . (ii) If

then

Partnership & Share If there is profit in the business run by two partners A and B then,

Ex.3

Saman begins begins business business with a capital capital of Rs. 50,000 50,000 and after after 3 months takes Manu Manu into partnership with a capital of Rs. 75000. Three months later Amandeep joins the firm with a capital of Rs. 1, 25,000. At the end of the year the firm makes a profit of Rs. 99,495. How much of this sum should Amandeep receive?

Sol.

Money invested by Saman for 12 months = Rs. 50,000 Money invested by Manu for 9 months = Rs. 75000 Money invested by Amandeep for 6 months = Rs. 1,25,000 Share of Saman: Manu: Amandeep = 50,000 × 12 : 75,000 × 9 ; 1,25,000 × 6 = 6,00,000 : 6,75,000 : 7,50,000 = 600 : 675 : 750 = 8 : 9 : 10 Total profit = Rs. 99,495. Profit of Amandeep =

= Rs.36, 850

PROPORTIONS Direct Proportion If A is directly proportional to B, then as A increases B also increases proportionally or in other words the proportional change occurs in the same direction. In general when A is directly proportional to B, then

= Constant. Constant.

Examples of Direct proportion are Time & Distance problems; time taken to travel a distance is directly proportional to the distance traveled when the speed is constant. This means the distance-traveled doubles if the time taken doubles provided speed remains constant. Remember: If X is directly proportional to Y, Back to Table of Contents

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Direct Relation In these types of cases an increase in one causes increase in the other but the increase is not proportional as in the case of direct proportion. For example: With simple interest the amount increases with increase of the number of years but not proportionally while on the other hand interest doubles or triples after 2 nd and 3rd years. So the increment in amount is in direct relation while increment in interest is direct proportion. Remember: If X is in direct relation to Y then, X = k 1+yk2 Ex.4

If 6 men can lay 8 bricks in one day, day, then how many men are required required to lay 60 60 bricks bricks in the same time? (1) 45 men

Sol.

(2) 40 men

(3) 60 men

(4) 50 men

Since the time is same so to do more work we need more persons. Hence this is the problem of

direct

proportion, i.e. Ex.5

The cost of New Year party party organized organized in TCY is directly related to the number of persons persons attending that party. party. If 10 persons attend the party the cost per head is Rs 250 and if 15 people attend, the cost per head is Rs. 200. What will be the total cost of the party if 20 persons attend it?

Sol.

This is the problem of direct relation Let the total cost of party is Cost = K 1 + K2 N (where K 1 & K2 are fixed and variable costs and N is number of persons) 250 x 10 = K 1 + 10K2 ………(1) 200 x 15 = K 1 + 15K2 ………(2) Solving them we get K 1 = 1500 and K 2 = 100 So total cost for 20 persons = 1500 + 20 x 100 = Rs. 3500

Indirect/Inverse Proportion A is in indirect proportion to B if as A increases, B decreases proportionally i.e. the proportional change occurs in the opposite direction. In general if A is in indirect proportion to B, then AB = constant. constant . Examples of Indirect proportion are Price & Quantity (expenditure remaining same), Number of men required & rate of work done (amount of work remaining same), Time & Speed problems for same distance. Ex.6

If 6 men can can build build a wall in 9 days days then 60 men can build a similar wall wall in ______ days?

Sol.

Work = Men x Days


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Ex.7

Quants Funda

A can do a piece piece of work in 12 days, days, B is 60% more efficient than than A. Find Find the number number of days required required for B to do the same piece of work.

Sol.

Ratio of the efficiencies is A : B = 100 : 160 = 5 : 8. Since efficiency is inversely inversely proportional to the number of days, the ratio of days taken to complete the job is 8 : 5. So, the number of days taken by B =

Ex.8

Ten years ago, the ratio of ages of A and B is 3 : 4, now, it is 4 : 5. What is the present age of A?

Sol.

10 years ago, let their ages be 3k and 4k. So, present ages are 3k + 10 and 4k + 10. The ratio is given as 4 : 5. k = 10 Percentage of A = 3k + 10 = 40 years.

Ex.9

Three friends A, B, C earn Rs. 2000 together. If they want to distribute this money such that ‘A’ should get Rs. 300

more than B and Rs. 100 more than C, in what ratio, they have to distribute the money? Sol.

A = 300 + B = 100 + C So, A = 300 + B and C = 200 + B A + B + C = 2000 B = 500

Ex.10

A = 800 and C = 700 So, the required ratio = 8 : 5 : 7

Four friends A, B, C, D share Rs. 10,500 in the ratio

, how much more money A and C together get than B

and D together? Sol.

The given ratio is

,

Multiply with 24 = 8 : 6 : 4 : 3. Take A = 8k, B = 6k, C = 4k, D = 3k. Total = 21k = 10,500 k = 500 Required answer = (A + C) – C) – (B + D) = (8k + 4k) – 4k) – (6k + 3k) = 3k = Rs. 1500 Ex.11

xyz has to distribute Rs. Rs. 3500 such that, for for every 2 rupees x takes, takes, y will take take 3 rupees rupees and for every 4 rupees rupees y takes, z will take 5 rupees. How much money z will get?

Sol.

We can write from the given information that, x : y = 2 : 3 ………(1) and y : z = 4 : 5 ………(2) Equation (1) × 4

35

x : y = 8 : 12





Equation (2) × 3

Quants Funda

y : z = 12 : 15

x : y : z = 8 : 12 : 15 z

will get

Time, Speed & Distance/ Boats and streams Questions taken from students’ forum 1. The size of the bucket is N kb. The bucket fills at the rate of 0.1 kb per millisecond. A programmer sends a program to receiver. There it waits for 10 milliseconds. And response will be back to programmer in 20 milliseconds. How much time the program takes to get a response back to the programmer, after it is sent? ANS: 30MILISEC 30MILISECOND OND 2. A car is filled with four and half gallons of oil for full round trip. Fuel consumption is 1/4 gallons more in going than coming. What is the fuel consumed in coming up? 3. A person who decided to go weekend trip, should not exceed 8 hours driving in a day Average speed of forward journey is 40 mph. Due to traffic on Sundays, the return journey journey average speed is 30 30 mph. How far he can select a picnic spot? spot? 4. A person was fined by traffic police for exceeding the speed limit by 10 mph. Another person was also fined for exceeding the same speed limit by twice the same. If the second person was travelling at a speed of 35 mph. Find the speed limit. 5. A bus started from bus stand at 8.00a m and after 30 min staying at a destination, it returned back to the bus stand with 50% faster speed. The destination is 27 miles from the bus stand. At what time it retur4ns to the bus stand (11.00)? 6. Wind flows 160 miles in 330 min. Time required for 80 miles is _________? 7. Given that with 4/5 full tank the vehicle can travel 12 miles. Find the distance travelled with1/3 of full tank. 8. A storm will move with a velocity 150 150 km towards the center 2 hr. At the same rate how much far will it it move in 3 hrs? 9. Refer to the figure below, a ship started from P and moves at a speed of I miles per hour and another ship starts from L and moving with H miles per hour simultaneously. Where do the two ships meet? ||---g---||---h---||---i---||---j---| ||---g---||---h--||---i---||---j---||---k---||---l--|---k---||---l---|| || Where P, G, H, I, J, K, L are the various stops in between the distances denoted by || . The values g, h, i, j, k, l denote the distance between the ports. Ans: Between I and J, closer to J 10. Given that A is travelling at 72 km/hr on a highway and B is travelling at a speed of 25 meters per second on a highway. Find the difference in their speeds in m/sec. Back to Table of Contents

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11. A car travels 12 km with a 4/5th filled tank. How far will the car travel with 1/3 filled tank? 12. A family, planning a weekend trip, decides to spend not more than a total of 8 hours driving. By leaving early in the morning, they can average 40 miles per hour on the way to their destination. Due to the heavy Sunday traffic, they can average only 30 miles per hour on the return trip. What is the farthest distance from home they can plan to go? (a) 120 miles or less (b) Between 120and 140 miles (c) 140 miles (d) Between 140 and 160 miles (e) 160 m iles or more

Theory, Examples & Explanations Time, Speed & Distance 1. The fundamental relationship between Distance (s), time (t) and speed (v) is given by: s = v x t. 2. Let v 1 and v2 be the velocity of the two vehicles and let v 1 > v2 If both the vehicles are moving in the same direction then their Relative Velocity = R.V. = v 1 – v2 If both the vehicles are moving in the opposite direction then their Relative velocity = R.V. = v 1 + v2 3. Average velocity = If x1 & x2 are the distances covered at velocities v1 & v2 respectively then the average velocity over the entire distance (x1 + x2) is given by 4. A man travels first half of the distance at a velocity v 1, second half of the distance at a velocity v 2 then,

Average velocity

= 5. If the distance is covered in three equal parts with different speeds v 1, v2 and v3 then, Average velocity = 6. For the same distance, the time is inversely proportion to the speed of the object. These types of problems can be solved as the problems of percentage. 7. When time is constant the ratio of speeds of the object is equal to the r atio of the distance covered by them i.e.


37




Ex.1

Quants Funda

If I decrease my speed by 20% 20% of original speed, I reach office 7 minute late. late. What What is my usual time and new time of reaching office?

Sol.. Sol

Since speed speed is decreased by 20% 20% i.e. 1/5 of the original. New speed speed will become become 4/5 of of the original speed. speed. For For the same distance, the time will become 5/4 of original time. Therefore new time increase by 1/4 of the original. This is given equal to 7 minutes. So Usual time = 7 × 4 = 28 minutes and New time = 28 + 7 = 35 minutes.

Ex.2

A train leaves Calcutta Calcutta for Mumbai, a distance of 1600 1600 kms at the same time a train leaves Mumbai to Calcutta. Calcutta. The trains meet at Nagpur which is at a distance of 700 kms from Mumbai. What is the ratio of the speeds of the t rains?

Sol.

From the problem, it is clear that when the first train travels a distance of 1600 – 700 = 900 km, the second train travels a distance equal to 900. So, the ratio of their speeds is 9: 7.

Trains (i) When a train approaches a stationary object (a tree, a stationary man, a lamp-post; we assume the length of the object to be infinitely small; provided its length isn’t mentioned) Time taken by the train to cross Pole = (ii) However, when a train approaches a platform, the time taken by the train to cross the platform is same as the time taken by the train to cross a distance equal to its own length plus the length of the platform at its own velocity. Time taken to cross the platform = (iii) For two trains having lengths l 1 & l2 and traveling in the same direction with speeds v 1 & v2 respectively (v1 > v2). Time taken to cross each other completely = (iv) Similarly, for two trains traveling in the opposite opposite direction: Time taken to cross each other completely = (v) If two trains/Object start at the same time from two points X & Y and move towards each other and after crossing they take a & b hrs respectively r espectively to reach opposite points Y and X, then Ex.3

A train 110 m long travels at 60 kmph. How long does it take? (a) To pass a telegraph post by the side of the track? (b) To pass a man running at 6 kmph in the same direction as the train? (c) To pass a man running at 6 kmph in the opposite direction? (d) To pass a station platform 240 m long? (e) To pass another train 170 m long, running at 40 kmph in the same direction? (f) To pass another train 170 m long, running at 60 kmph in the opposite direction? Back to Table of Contents

38




Sol.

Quants Funda

(a) Speed of train = Time taken to cross the telegraph post = (b) Speed of man = Time taken to pass the man = Length of the train/ Relative velocity = (c) Time = Length of the train/ Relative velocity = (d) Time = (Length of the train + Length of platform) / Relative velocity = (e) Speed of the second train =

mps

Time = Sum of the length of the two trains/ Relative velocity = (f) Time = Sum of the length of the two trains/Relative velocity =

Boats & Streams Let Speed of boat in still water = b km/hr Speed of stream = w km/hr Speed of boat with stream (Down Stream), D = b + w Speed of boat against stream (Up stream), U = b Speed of boat boat in still water water,, b = (D + U) Speed of stream, w = (D – U) Ex.4

A man can row 4.5 km/hr in still water. It takes him twice as long to row upstream upstream as to row downstream. What is the rate of the current?

Sol.

Speed of boat in still water (b) = 4.5 km/hr. It is given upstream time is twice to that of down stream. Downstream speed is twice to that of upstream. So b + u = 2(b – 2(b – u) u = = 1.5 km/hr.


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Circular Motion Circular Motion with two people Ex.5

Sachin and Saurav, as as a warm-up exercise, exercise, are jogging on a circular circular track. track. Saurav Saurav is a better athlete athlete and jogs at at 18km/hr while Sachin jogs at 9 km/hr. The circumference of the track is 500 m (i.e. ½ k m). They start from the same point at the same time and in the same direction. When will they be t ogether again for the first time?

Sol.

Method 1: Since Saurav is faster than Sachin, he will take a lead and as they keep running, the gap between them will keep widening. Unlike on a straight track, they would meet again even if Saurav is faster than Sachin. The same problem could be rephrased as “In what time would Saurav take a lead of 500 m over Sachin”? Every second Saurav is taking a lead of

over Sachin in Therefore; they would meet for

the first time after 200 sec. In general, the first meeting if both are moving in the same direction and after both have started simultaneously occurs after Time of first meeting = Method 2: For every round that Sachin makes, Saurav would have made 2 rounds because the ratio of their speeds is 1 : 2. Hence, when Sachin has made 1 full round, Saurav would have taken a lead of 1 round. Therefore, they would meet after Ex.6

[Here,

is Sachin’s speed.+

Suppose in the earlier problem when when would the two meet for the first time if they are moving in the opposite opposite directions?

Sol.

If the two are moving in the opposite directions, then Relative speed = 2.5 + 5 = 7.5 m/s. [Hence, time for the first meeting = Circumference / Relative speed =

Ex.7

If the speeds speeds of Saurav and and Sachin Sachin were 8 km/hr and 5 km/hr, then after what time will the two meet for the first time at the starting point if they start simultaneously?

Sol.

Let us first calculate the time Saurav and Sachin take to make one full circle. Time taken by Saurav = Time take by Sachin = Hence, after every 225 sec, Saurav would be at the starting point and after every 360 sec Sachin would be at starting point. The time when they will be together again at the starting point simultaneously for the first time, would be the smallest multiple of both 225 and 360 which is the LCM of 225 and 360. Hence, they would both be together at the starting point for the first time after LCM (225, 360) = 1800 sec = 0.5 hr. Thus, every half an hour, they would meet at the starting point. Back to Table of Contents

40




Note:

Quants Funda

From the solution you could realise that it is immaterial immaterial whether they move in the same direction direction or in the opposite.

Circular motion with three people: If three persons A, B, and C are running along a circular track of length d meters with speeds V a ,Vb ,Vc respectively. To find the time when all the three will meet for the first time, we have to calculate the relative time of the meeting of any one (A or B or C) among the three with other two runners and then calculate the LCM of these two timings. This will be the time when all the three runners will meet for the first time.

And to calculate when they all meet for the first time at the starting point, we have to take the LCM of the timings taken by all the runners separately to cover one full circular motion.

Ex.8

Let us now discuss the cases cases of circular motion with with three people: Laxman joins Saurav Saurav and and Sachin, Sachin, and and all of them them run in the same direction from the same point simultaneously in a track of length 500 m. Laxman moves at 3 km/hr, Sachin at 5 km/hr and Saurav at 8 km/hr. When will all of them be together again? a. for the first time? b. for the first time at the starting point?

Sol.

(a) Break the problem into two separate cases. In the first case, Saurav moves at the relative speed of (8 – (8 – 5) = 3 km/hr with respect to Sachin. At a relative speed of 3 km/hr, he would meet Sachin a fter every In the second case, Saurav moves at the speed of (8 – (8 – 3) km/hr = 5 km/hr with respect to Laxman. At a relative speed of 5 km/hr, he would meet Laxman after every If all the three have to meet, they would meet after every [LCM (10, 6)] min = 30 min or ½ hour. Hence, they would all meet for the first time after 30 min. (b) If we need to find the time after which all of them would be at the starting point simultaneously for the first time, we shall use the same method as in the case involving two people. At a speed of 8 km/hr, Saurav takes 225 sec. to complete one circle. At a speed of 5 km/hr, Sachin takes 360 sec. to complete one circle. At a speed of 3 km/hr, Laxman would take 600 sec. to complete one circle. Back to Table of Contents

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Hence, they would meet for the first time at the starting point after LCM (225, 360, 600) sec. = 1800sec. Ex.9

A thief is spotted by a policeman from from a distance of of 200 200 m. When the policeman starts a chase, the thief starts running. Speed of thief is 10 Kmph and that of policeman is 12 kmph. After how many hours will the policeman catch the thief?

Sol. Ex.10

A man steals a car at 1 : 30 pm & drives at 40 kmph. At 2 pm the owner owner starts chasing chasing his car at 50 kmph. kmph. At what time does he catch the man?

Sol.

Distance covered by the thief in 1h = 40 km Distance covered in Now, time taken to catch the thief = Relative velocity = 50 – 50 – 40 = 10 kmph (  Both are moving in same direction)

Time = 4 p.m. (  2 p.m. + 2) Ex.11

A and B started moving simultaneously simultaneously from from P towards Q and their respective speeds are 36 kmph and 15 m/s respectively. What is the distance between them after moving for 2 minutes after starting from P?

Sol.

Speed of A = 36 kmph Speed of B = 15 m/s So in 1 sec, B covers 5 m extra than A and so, the distance between them will be 5 m. So in 2 mins = 120 sec, the distance between will be 120 × 5 m = 600 m.

Ex.12

Two trains are are moving in opposite directions directions with the respective speeds of of 36 kmph and 45 kmph. kmph. They will will cross each other in 20 seconds. If they are moving in the same direction at t he same speed, how much time will they take to cross each other?

Sol.

Speeds = 36 kmph, 45 kmph = 10 m/s, 12.5 m/s If they are moving in opposite directions, relative velocity = 10 + 12.5 = 22.5m/s. And they take 20 seconds to cross each other. If they move in the same direction, relative velocity = 12.5 – 12.5 – 10 = 2.5 Since the velocity velocity in the second case case is times the velocity in the first case. So, they will take 9 times more time to cross each other. i.e., 20 × 9 = 180 sec = 3 min.

Ex.13

Two trains are moving in the same direction. The speed of the faster train is twice the speed of the slower slower train. train. The faster train takes 60 sec to overtake the slower train. If they move in opposite directions, how much time will they take to cross each other?

42





Sol.

Quants Funda

Assume ‘v’ is the speed of the slower train, so 2v is the speed of the faster train. If they move in the same direction, relative speed = 2v – 2v – v = v. If they move in opposite directions, relative velocity = 2v + v = 3v. Since, the velocity is three times, the time required is i.e.

Ex.14

Train A starts from city P to to city Q with a velocity of 40 kmph. Train Train B starts at at city P towards towards city Q, 1 hour after after train A with a speed of 60 kmph. If both the trains reach station Q simultaneously, what is the distance between cities, P and Q?

Sol.

Assume, the distance as d. (Since train B takes 1 hour less than that of train A) Therefore d =120 km

Ex.15

Two friends A and B are are moving towards Q and and P respectively respectively from P and Q respectively. respectively. The distance distance between P and Q is 600 m and the speeds of the friends, A and B are 6 m/s and 8 m/s respectively. How much time after A starts for P, does B have to start for Q so that they meet at the exact midpoint of P and Q?

Sol.

The midpoint means exactly 300 m from both the sides. A can walk this 300 m at a speed 6 m/s speed in 50 sec. Whereas B can cover this distance in So B has to wait for (50 (50 – – 37.5) = 12.5 sec, if they want to reach the midpoint simultaneously

Ex.16

A river flows at a speed of 1.5 kmph and a boatman who can row his boat at a speed of 2.5 kmph in still water, takes

hours to go a certain distance distance up stream and and return to the starting starting point. point. What What is the total distance distance

covered by the boatman? Sol.

The speeds of the boatman upstream and downstream are 1 kmph and 4 kmph respectively. If the distance covered each way = x km, then total time taken to go upstream and downstream

The distance covered both ways is 12 km.


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Races & Games of Skill Races: Any contest of speed in running, riding, driving, sailing or rowing is called a race race.. The path on which the contests are held is called a race course. course. The point from which the race begins is called the starting point. point. The point where the race ends is called the winning post or the goal. goal. The person who first reaches the winning post is called the winner winner.. If all the contestants reach the goal at the same time then the race r ace is called a dead heat race. race . There are two types of races (a) Handicap Race: In this type of races 1st runner gives 2nd runner benefit of running for some time before 1st start the race. e.g. A can give B a handicap of 3 sec. mans they started from the same point but A start 3 sec after. (b) Headway/Start race: This means that runner A has started from the “K” distance behind runner B but at the same time. e.g. A gives B a start of 10 m means before starting A is on starting point and B is 10 meter forward from A and they start the race at the same time. In a contest with 2 participants, one is the winner and the other is the loser. (i) The winner can give/allow the loser a start of t seconds or x metres, i.e. start distance = x metres and start time = t seconds. (ii) The winner can beat the loser by t seconds or x metres, i.e. beat distance = x metres and beat time = t sec Interpretation from the statements given in the problems concerned with race (1) A beats B by X meter: This means that the winner of the race is A and B is X meters behind A when he crosses the finishing line. (2) A and B start together at P: This means both the runners have started from the same starting point. The ratio of the speed of the runners is the ratio of the distance covered by them in the time, in which winner reaches the finishing line. (3) A gives B a start of X metre: This means that A stands at starting point and B is X meter forward of the starting line at the starting of the ra ce. (4) A beats B by t seconds This states that after coming of A at the finishing point B will take more t seconds to cover the left distance. If we know that distance and time taken we can cover the speed of the looser? Winner's (A) time = Loser's (B) time – time – t A and B starts together at P, A finishes at Q, but t seconds before B finishes. Ex.17

A can run 330 metres in 41 seconds seconds and B in 44 seconds. By how many seconds will B win if he has 30 30 metres start? start?

Sol.

B runs 330 metres in 44 seconds.

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Quants Funda

B runs (330 – (330 – 30) metres in But A runs 330 metres in 41 seconds So, B wins by (41 – (41 – 40) seconds, i.e. 1 second. Ex.18

In one kilometer race, A beats B by 36 metres metres or 9 seconds. seconds. Find A's time over the course.

Sol.

Here A is the winner and B is the loser. The time taken by B to cover the distance of 36 metres = 9 seconds. Time taken by B to cover 1 kilometer = (1000 × 9)/36 = 250 sec. Therefore time taken by A to cover 1km = 250 – 250 – 9 = 241 seconds. Alternative method: Using formula:

Escalator Let us understand this concept with the help of an example. Ex.19

In an escalator Rajesh covers 4 steps with 3 steps of Suresh in the same time in a static escalator. escalator. When the escalator is moving Rajesh takes 24 steps while Suresh takes 21 steps to reach the top of the escalator. What are the total numbers of steps in the escalator?

Sol.

Assume that S be the number of steps of escalator that helped Rajesh and Suresh for the same time. Then,

For every every 4 steps of Rajesh, Suresh will take take 3 steps. steps.

So for 24 steps Rajesh take, Suresh will take 18 steps. But Suresh has taken taken 21 steps which are of the original original steps. steps. These extra steps taken are the result of extra time consumed by him and helped by the escalator. So, the equation will become Total number of steps in the escalator = 24 + S = 24 + 18 = 42 steps. Ex.20

In a 100m race, race, A can beat B by 10m, and C by 20m. If B and C run with the same velocity, velocity, By how many meters can B beat C in a 900 m race?

Sol.

From the first line of the question we can understand that when A runs 100m, B runs 90 m and C runs 80 m. So B can beat C by 10m in a 90 m race. So in a 900m race, B can beat C by 100m.

Ex.21

A can beat B by 44 metres in a 1760 1760 meter race, race, while in a 1320 metres race, B can beat C by 30 metres. By what what distance (in meters) will A beat C in i n a 880 meter race?

Sol.

45

When A runs 1760 metres, B runs 1716 metres.





Quants Funda

When B runs 1320 metres, C runs 1290 metres. When B runs 1716 metres, C runs 1290 (1719/1320) (1719/1320) = 1677 metres. A

B

C

1760

1716

1677

When A runs 880 metres, C runs 1677 (880/1760) (880/1760) = 838.5 metres. A beats C by (880 – (880 – 838.5) = 41.5 metres. Ex.22

Both A and B run a 2 km race. race. A gives B a start of 100 m and still beats him by 20 20 seconds. If A runs at a speed of 20 km per hour, find B's speed in kilometres per hour. (1) 17

Sol.

(2) 18

(3) 19

(4) 19.5

A covers a distance of 2 km. in (2/20) hour i.e. 360 sec. B covers the distance of (2000 – (2000 – 100) i.e. 1900 m in 360 + 20 i.e. 380 sec. B's speed = (1900/380) = 5 m/s =

Ex.23

J is

times as as fast fast as K. If If J gives K a start of 150 m, how how far far must must be the winning post so that the race race ends in a

dead heat? (1) 100 m Sol.

(2) 440 m

(3) 550 m

(4) 200 m

Race ends in a dead heat, i.e. times taken by J and K are the same.

Time & Work Questions taken from students’ forum 1. A can do do a piece piece of work in 20 days, where as B can do it in 12 days. In 9 days B does

of the work. How How many days will A

take to finish the remaining work? 2. A man, a woman, and a nd a child can do a piece of work in 6 days. And it is also known that a man alone can do it in 24 days and woman alone can do it in 16 days. Find the number of days taken by a child to complete the work (working along)? 3. A can copy 50 papers in 10 hours while both A & B can copy 70 papers in 10 hours. Then for how many hours required for B to copy 26 papers? Back to Table of Contents

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4. A is twice efficient than B. A and B can both work together together to complete a work in 7 days. How many many days does A alone take to complete the work? 5. A finish the work in 10 days. B is 60% efficient than A. How many days does B take to finish the work? 6. A work is done by two people in 24 min. One of them can do this work alone in 40 min. How much time required to do the same work by second person? 7. A finishes the work in 10 days and B in 8 da ys individually. If A works for 6 days then how many days should B work to complete remaining work? 8. Given 20 men take 15 days to complete a job. Find the number days required for 5 men to complete the work?

Theory, Examples & Explanations Time & Work To start with, let’s take an example. If 2 men take 10 days to build a wall, 1 man will take 20 days (not 5 days) to build the same wall. The question arises why? The reason is simple that man and time are inversely proportional to each other. When men will decrease the number of days required to complete the work will increase. Here men become half, so time will double and will complete the work in 20 days.

-------- (1), Where M & D are number of men and number of days

respectively. If we further break number of days to hours, then total hours = DH, where H are number of hours per da y. Now, our formula becomes To elaborate it further, let’s say M men take D days to build a room. Now if work is doubled (they have to build two rooms of same size) in same D no. of days, obviously they have to double their strength Or we can say that no. of men are directly proportional to the work to be finished. In mathematics, we can write it as M

W, -------------------------------------------------------------------------- (2)

Where W is the work to be finished. By combining (1) and (2), we get

(or)

, Where K is constant of proportionality (or)


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This is our general formula to solve time & work problems. It is also known as Work Equivalence Method. Majority of time & work questions can be divided into two types. Type I: Where efficiency of individual’s is not mentioned (as in the above example): Th ese are the cases where the man-days concept is utilized. Here the Rule of Fractions or the proportion (direct or indirect) concept can also be used. Type II: Where the efficiency is mentioned. Here the case of unit day’s or one day’s concept is utilized . Concept of unit time If A does a work in 10 days, then what is the one-day work done by A? The answer is one-tenth part of the total work. We can say that Unit time work = Work done in unit time is known as efficiency of the worker or we can say, that if a worker takes less number of days (than second worker) to finish a work, he will be more efficient. LCM Approach In this type of approach the total work is considered as the LCM of the time taken by the individuals and then unit time work is calculated. Here the work done is not 1(unity) but the LCM of the time taken by number of persons. Let us understand this with the help of an example. Ex.1

If Manni and Gopi Gopi finish a work in 10 and 15 15 days respectively, what will be number of days taken by both of them to complete the work when both work together?

Sol.

Let the total work is equal to the LCM of number of days of Manni & Gopi taken to do the work respectively. Work = LCM of 10 & 15 = 30 units One day work of Manni = One day work of Gopi = One day work of both = 5 units So number of days taken =

Simultaneous Working Problems It is always a good way if we try to solve such type of questions with the help of LCM approach. Let us understand the concept with the help of an example. Ex.2

A, B and and C can finish finish a work independently independently in 10, 12 12 and 15 days. days. C starts starts the work and after 1 day, B joins him. him. After 1 day of B, A also joins them but leave l eave 3 days before completion of the work, while B left two days before completion of work. What will be the total number of days taken to complete the work?

Sol.

Let total work is LCM (10, 12, 15) = 60 units Back to Table of Contents

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I day work: Let total days taken = N Days of A = (N – (N – 5), Days of B = (N – (N – 3), Days of C = N According to the question we have, 6(N – 6(N – 5) + 5(N – 5(N – 3) + 4N = 60 Solving we have N = 7 days Alternate Days concept In these, type of problems, we see the work done by two or more men on alternate days or hours. Ex.3

A can build a wall in 20 days days while B can build the same wall in 30 days. If they work on alternate alternate days in how many days, will the wall be completed if A start the job?

Sol.

Let total work = 60 units 1 day work of A = 3 units and 1 day work of B = 2 units 2 days work = 3 + 2 = 5 units To do 60 units, days required =

Negative Work Concept These types of problems revel about the problems of pipe and cistern in which one is inlet and other is outlet. Let us solve a puzzle “A monkey wants to climb a pole which is 24 m tall. In 1 minute, he can climb up by 3 m and in the next minute, he slips by 2 m. In how many minutes will the monkey reach the top of the pole? Ex.4

If A build the wall in 20 days and B can destroy destroy that wall in 30 days and work work on alternate days. What What will be the number of days required to build the wall for the first time?

Sol.

A can build 1/20 th of the wall in 1 day. Where as B will destroy 1/30 th of the wall in 1 day and since they are working on alternate days, So in 2 days, 1/20-1/30 = 1/60 th of the wall will be constructed. So, in 57 × 2 = 114 days, 57/60 of the work will be completed and on the 115th day, A will come and completes the remaining 3/60 = 1/20

th

th

work. So

it takes 115 days to construct the wall. Pipes and Cisterns Pipes and Cisterns problems are almost the same as those of Time and Work problems. Thus, if a pipe fills a tank in 6 hrs, then the pipe fills 1/6

th

of the tank in 1 hour. The only difference with Pipes and Cisterns problems is that there are

outlets as well as inlets. Thus, there are agents (the outlets) which perform negative work too. The rest of the process is almost similar. Inlet: A pipe connected with a tank (or a cistern or a reservoir) is called an inlet inlet,, if it fills the tank. The work done by this is taken as positive work.

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Outlet: A pipe connected with a tank is called an, outlet outlet,, if it empties the tank and the work done by this is taken as the negative work. Ex.5. Ex. 5.

Two pipes A and B can fill a tank in 36 hours and 45 hours respectively. If both the pipes are opened imultaneously, how much time will be taken to fill the tank?

Sol.

Part filled by A alone in 1 hour = 1/36 Part filled by B alone in 1 hour = 1/45 Part filled by (A + B) in 1 hour = 1/36+1/45 = 9/180 = 1/20 Hence, both the pipes together will fill the tank in 20 hours.

Ex.6. Ex. 6.

Pipe A can fill a tank in 36 hours and pipe B can empty em pty it in 45hours. If both the pipes are opened simultaneously, how much time will be taken to fill the t he tank?

Sol.

Total Volume of the tank = 180 units (LCM of 36 and 45) 1 hour work of A = 5 units 1 hour work of B = – = – 4 units 1 hour work of A + B = 1 units So, Total time taken = 180 hours

Ex.7. Ex. 7.

A can do a work in 30 days, and B can do in i n 40 days. If A and B work together for 10 days and A left, then C joined with B and completed the work in 10 days. How many days C alone can complete the work?

Sol.

A work only for 10 days, B work for 20 days and C work for 10 days. Let us assume C require X days to complete the th

work. So, in 1 day, A can do 1/30 of the work. B can do 1/40

th

of the work, and c can do 1/X of the work.

So, 10/30 + 20/40 + 10/X = 1. So X = 60 days. So, C can do the work in 60 days. Ex.8.

Four men can do work in 15 days. If a man left left the work after 5 days days and again joined joined after 5 more days, days, the remaining three work continuously till the end of work. How many more days than the estimated, it takes to complete the work?

Sol.

After 15 days, the work that is left is equal to the amount of work that can be done by 1 man in 5 days. This work can be done in 5/4 days by the 4 men. So it takes 5/4 more days to complete the work.

Ex.9. Ex. 9.

One man started a work on first day, the second day one more joined with him. The next day one m ore joined. Everyday one new person joined until the work gets completed. If the work is completed in 15 days, how many days it takes for 10 men to complete the same work, if they work regularly?

Sol.

Let us assume one man can do x amount of work in 1 day. So the amount of work, that can be completed

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On 1st day = x nd

2 day = 2x …………….. th

15 day = 15x Total = Σ15 x = 120 x But 10 men can do 10x work in 1 day. So, 10 men take 12 days to complete the work. Ex.10. Ex. 10.

B is twice as efficient as ‘A’ a nd ‘C’ is 50% more efficient than B. If B and C together can complete a work in 10 days,

how much time it takes for A and B to complete complete the work, if they work on alternate days starting with ‘A’? Sol.

If A can do x work in 1 day. B can do 2x C can do 3x B and C together takes 10 days

Total work = 50 x.

If A and B work on alternate days, In 2 days

x + 2x = 3x work will completed.

In 16 × 2 = 32 days 33rd day th

34 day

16 × 3x = 48x will be completed.

A will come and do x work. So, 48x + x = 49x will be completed. B will come and complete the work in 1/2 day.

So, answer = 33 1/2 days.

Heights & Distances Questions taken from students’ forum 1. A building with height D ft shadow up to G ft on a neighbor building. What is the height of the building with shadow up to C ft?


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Theory, Examples & Explanations Heights & Distances Angles and their relationship Angles are measured in many units’ viz. degrees, minutes, seconds, radians, gradients. Where 1 degree = 60 minutes, 1 minute = 60 seconds, π radians = 180° 180 ° = 200g 1 radian = 180°/π 180°/π and 1 degree = π/ 180 radians. The angle at the centre is of 1 radian

Basic Trigonometric Ratios In a right triangle ABC, if θ if θ be the angle between AC & BC

If θ If θ is one of the angle other then right angle, then the side opposite to the angle is perpendicular (P) and the sides containing the angle are taken as Base ( B) and the hypotenuse (H). In this type of triangles, we can have six types of ratios. These ratios are called trigonometric ratios.

Important Formulae For any angle θ: 1. sin2θ + cos2θ = 1 [Note sin2θ = (sin θ)2 and not (sin θ2)] 2

2

2. 1 + tan θ = sec θ 3. 1 + cot 2θ = cosec2θ Range of Values of Ratios If 0 ≤ θ ≤ 360o, then the values for different trigonometric ratios will be as follows.

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1. – 1. – 1 ≤ Sin θ ≤ 1 2. – 2. – 1 ≤ cos θ ≤ 1 3. – 3. – ∞ ≤ tan θ ≤ ∞ 4. – 4. – ∞ ≤ cot θ ≤ ∞ 5. – 5. – ∞ ≤ Sec θ ≤ -1 & 1 ≤ Sec θ ≤ ∞ 6. – 6. – ∞ ≤ Cosec θ ≤ – 1 & 1 ≤ Cosec θ ≤ ∞ Sign of Trigonometric ratios We divide the angle at a point (i.e. 360°) into 4 parts called quadrants. In the first quadrant all the trigonometric ratios are positive Values of trigonometric Ratio for some special angles: 0

0

30

0

45

0

60

0

90

Sin

0

1

C os

1

0

Tan

0

0

1

Properties of Triangle Sine rule In any triangle ABC if AB, BC, AC be represented by c, a, b respectively then we have R Circum – Radius = Cosine Rule In a triangle ABC of having sides of any size, we have the following rule;

Area of Triangle Area = Where S = semi Perimeter Back to Table of Contents

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Sets & Matrices Questions taken from students’ forum 1. The singularity matrix from a given set of matrices? (Hint det(A)=0) 2. In the class of 40 students, 30 speak Hindi and 20 speak English. What is the minimum possible number of students who speak both the languages? (a) 5

(b) 20

(c) 15

(d) 10

(e) 30

3. Three circles Venn diagram problems can be expected from this section. *I don’t remember the exact question, but it was about three languages Tamil, Hindi and Telugu+

Theory, Examples & Explanations Sets & Matrices Defining a Set "Set" is synonymous with the words "collection", "aggregate", "class" and is comprised of elements/objects/members. A set is a well-defined collection of of elements elements.. By well-defined elements it means that given a set and an element, it must be possible to decide whether or not the element belongs to the set. A set can be described in any one of the following ways. For example, the set of beautiful Actress of Bollywood or the set of Good Players of cricket in India are not sets as the world beautiful and Good are not well defined. What are the criteria of choosing an actress as beautiful and which player is said to be a good player of cricket. But on the other hand if we say Set of good player of cricket in India those has played at least 25 international games, will be a set as the word Good is now well defined. Representation of a Set A set can be described in two different ways. 1. Roaster Form: A set is described by listing elements, separated by commas, within brackets. For example, A = {a, e, i, o, u} is a set of vowels of English alphabets and a finite set and N = {2, 4, 6,…- is a set of even natural number and is a infinite set. Also, repetition of an element has no effect. For example {1, 2, 3, 2} is the same set as {1, 2, 3}. Back to Table of Contents

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2. Set Builder Form: A set can also be described by a characterizing property P(x) of its elements x. In such a case the set is described by {x | P (x) holds} or {x: P(x) holds}, which is read as 'the set of all x such that P(x) holds'. For example, A = {x | x is the root of x 2 + 5x – 5x – 6 = 0} Cardinal number of a set The number of elements contained in a set is known as the cardinal number of that set. On the basis of cardinal number a set can be an Empty Set or a Singleton Set. A set is said to be empty or void or null set if it has no elements and its cardinal number is 0. It is denoted by the symbol φ or { } such as Set of odd numbers divisible by 2. And a set is singleton set if the cardinal number is 1 i.e. it contains only one element, such as Set of lady Prime-minister of India. If a set A has 5 elements then its cardinal number is written as n(A) = 5, and read as “Cardinal number of set A is 5”. Subsets Let A and B be two sets. If every element of A is contained in B, then A is called the subset of B. If A is subset of B, we write A If A

B then a

A

B, which is read as "A is a subset of B" or "A is contained in in B". a

B. (The symbol

stands for "implies" and

stands for “belongs to”). Let P = a, b, c-, then

subsets of P are φ, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} i.e. it has 8 number of subsets and Empty set is the subset of every. Also every set is the subset of itself and it is improper subset of the set. Every set has one improper subset and other subsets as proper subsets. If A

B and also B

A, then we can can say every element of A is present in B and every element element of B is present in A.

These types of sets having same number and identical elements are known as Equal sets. Power Set The set of subsets of a set is the power set of the set. If A is a set then its Power set is denoted as P(A). If A = {a, b, c}, then P(A) = {φ { φ, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}} The cardinal number of any set A is 2n, where n = number of elements present in set A. Universal Set In any discussion in theory, there happens to be a set U that contains all sets under consideration. Such a set is called the universal set. Thus a set that contains all sets in a given context is called the universal set. For example, in plane geometry the set of all points in the plane is the universal set. Some Important Universal Sets: N = Set of all natural numbers = 1, 2, 3, 4,…Z or I Set of all integers = …– 3, 3, – – 2, 2, – – 1, 0, 1, 2, 3,..} +

Z = Set of all positive integers = 1, 2, 3,…- = N –

Z = Set of all negative integers = { – { – 1, 1, – – 2, 2, – – 3,..}

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W = Set of all whole numbers = 0, 1, 2, 3,…Z0 = The set of all non-zero integers = {± 1, ± 2, ± 3,..} Q = Q = The set of all rational numbers. = R = The set of all real numbers. R – –Q Q =The set of all irrational numbers. e.g

,

,

, ……. , e, log 2 etc… are are all irratio irrational nal numbers numbers..

Operations of sets Union of Sets: The union of two sets A and B, is a set containing all the elements present in set A or in set B. The union of set A and B is represented as A U B. So, A U B = {x | x

A or x

B}

For example: A = {5, 7, 8, 9, 11} and B = {car, house, ball, sofa), then A U B = {5, 7, 8, 9, 11, car, house, ball, sofa} Intersection of Sets: The intersection of two sets A and B, is a set containing all the common elements present in set A or in set B. The B. So, So, A

intersection of set A and B is represented as A

B = {x | x

A and x

B}

For example A = {5, 7, 8, 9, 11} and B = {3, 4, 5, 6, 7, 8}, then A

B = {5, 7, 8}

Difference of sets: A – B is the difference of set A from set B and is defined as the set of elements present in set A but not in set B. So, A – B = {x | x

A and x

B}

Complement of a set: The complement of set A in U the set of those elements which are present in Universal set but not present in set A. Compliment of A is denoted by A' The shaded part represents A'. So, A‘= {x | x

U and x

A}

(a) Also the Compliment of set A is defined as the difference of Universal set U and set A. i.e. ⇒ A‘ = U – A Facts and Rules: i.

A

A=A

&

A

A=A

ii.

A

φ=A

&

A

φ=φ

iii.

A

U=U

&

A

U=A

A

B=B

&

A

B=B

iv.

56

A


A




A’ =

Quants Funda

v.

A

&

A

A’ = φ

vi.

n (A

B) = n (A) + n (B) (B) – – n (A

vii.

n (A

B

viii.

n (A (A – – B) = n (A) – (A) – n (A

Ex.1

In a certain city only two newspapers A & B are published. It It is known known that 25% of the city population population read read A & 20%

B)

C) = n (A) + n (B) + n (C) (C) – – n (A

B) – B) – n (B

C) – C) –n n (C

A) + n (A B C)

B)

read B while 8% read both A & B. It is also known that 30% of those who read A but not B, look into advertisements and 40% of those who read B but not A, look into advertisements while 50% of those who read both A & B look into advertisements. What % of the population reads an advertisement? Sol.

Let A & B denote sets of people who read paper A & paper B respectively and in all there are 100 people, then n (A) (A) = 25, n(B) = 20, n (A

B) = 8. 8.

Hence the people who read paper A only i.e. n (A – B) = n (A) – (A) – n (A B only i.e. n (B – (B – A) = n (B) – (B) – n (A

B) = 25 25 – – 8 = 17. And the people who read paper

B) = 20 20 – – 8 = 12. Now percentage of people reading an advertisement = [(30 % of

17) + (40% of + 12) + (50% of 8)] % = 13.9 %. Venn Diagrams Venn diagram is the pictorial representation of the set and also the operation involved in the sets. We often use circles to represent the sets and overlapping of the circles to represent the common elements in two or more sets. The universal set U is represented by interior of a rectangle and its subsets are represented by interior of circles within the rectangle. Maximum and Minimum elements in a set Let us consider the formula, n (A and asked to find the value of n (A

B) = n (A) (A) + n (B) (B) – – n (A

B) and imagine that the value of n (A

B) is not given

B), then we will get the range of the values. Let us understand the concept of maximum

and minimum values with the help of an example. Ex.2

If in a Survey, Survey, Organized Organized by any N.G.O on the cold drinks drinks after after effects effects found that 80% of of the total people people like Coca Cola and 70% like Limca. What can be the minimum and maximum number of people who drink both the drinks?

Sol. Limca

Coca Cola 80 - X

Here n (L

X

70 - X

C) ≤ 100%

Using the formula, n (L) + n (C) – n (L

C) = n (L

80 + 70 – 70 – X ≤ 100

57


C) ≤ 100% Back to Table of Contents



Quants Funda

Solving we get X ≥ 50%, Also X ≤ 70% Minimum value = 50% and Maximum value = 70%.

Functions Questions taken from students’ forum 1. If g (0) = g (1)=1 and g (n)= g (n-1) + g (n – –2). 2). Find g (6) 2. G(0)=-1, G(1)=1, G(N)=G(N-1) - G(N-2), G(5)= ? 3. X 0 Y 0.00001

10

100

1000

9999

1.02

1.72

3.00

4.72

Find the relation between X and Y? Ans:- Y= log10(X) 4. Find the equation of a curve that intersects x at -1 when y=0 a nd x=0 when y=3 goes upward? (Can Expect Choices)

Theory, Examples & Explanations Functions Definition Suppose, if we take any system, the output will be a function of input. That means a function is a relation between input and output. For Example, there is a system, which finds the square of the given input. That means the output is a square 2

2

of the given input. This can be represented by output= (input) or f(x) = x . Where x is input and f(x) is output. Here f is called the function of x, which is defined as f(x) = x 2. So, if f(x) = x 2 , f(1) = 1 2 = 1, and f(2) = 2 2 = 4. 2

2

In general, if f(x) = x , f(a) = a . Ex.1

(a) If f(x) = 2x 2 – 2x + 1, find f( – 1).

Sol.

(a) We substitute – substitute – 1 in place of x. f( – f( – 1) = 2( – 2( – 1)2 1)2 – – 2 ( – ( – 1) + 1 = 5 2

2

(b) If f(t) = 3t – 1, find f(a 2)

2

2

(b) We substitute a in place t. f(a ) = 3(a ) – 1 = 3a – 1. Odd and Even Functions Odd function: A function f is said to be odd if it changes sign when the sign of the variable is changed. i.e. If f( – f( – x) = – = – f (x). For example: f (x) = sin x ; 0 ≤ x ≤ 2π is a odd function.

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Even function: A function f is said to be an even function if it doesn’t change sign when the sign of the variable is changed. i.e. if f(– x) 4

2

= f (x).For example f (x) = x + x and g (x) = cos x are even functions. NOTE: There are many functions which are neither odd nor even i.e it is not necessary for a function to be either even of to be odd. E.g: g (x) = 3x 3 + 4x2 – 9 is a function in x which is neither even nor odd. Composite Functions A composite function is the function of another function. If f is a function from A in to B and g is a function from B in to C, then their composite function denoted by ( g o f) is a function from A in to C defined by (g o f) (x) = g [ f(x)] e.g. if f(x) = 2x, and g(x) = x + 2, Then (gof)(x) = g [ f(x)] = g (2x) = 2x + 2 (fog)(x) = f [g(x)] = f(x + 2) = 2(x + 2) = 2x + 4 FACT: This shows that it is not necessary that (fog)(x) = (g of)(x). Ex.2

Let a function f n+1 (x) = f n (x) + 3 . If f 2(2) = 4. Find the value of f 6(2).

Sol.

We have, f n+1 n+1(x) = f n(x) + 3 So, f 3(2) = f 2(2) + 3 = 7 f 4(2) = f 3(2) + 3 = 10 f 5(2) = f 4(2) + 3 = 13 f 6(2) = f 5(2) + 3 = 16

Alternate Method: Since the function is increasing with constant value. So, f 6(2) = f 2(2) + 3( 6 – 6 – 2) = 4 + 12 = 16 2

Ex.3

If f(x) = 2x – 3, then f(2) = ?

Sol.

f(2) = 2(2) 2 – 3 = 5

Ex.4

f(x) =

Sol.

f(2) + f(3) =

if f(2) + f(3) = f(5), then K = =

f(5) =

Ex.5

f (n (n + 1) = 2f (n) + 4 & f(1) = 0. Then f (5) =

Sol.

f(2) = 2f(1) + 4 = 0 + 4 = 4 f(3) = 2 × 4 + 4 = 12

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f(4) = 2 × 12 + 4 = 28 f(5) = 2 × 28 + 4 = 60 Ex.6

f(x) = x + 5, g(x) = x2 – 3. Then gofog(3) is :

Sol.

gofog(3) = gof(3 2 – 3) = gof(6) = g(6 + 5) 2

= g(11) = 11 – 3 = 118. Ex.7

f(x) = 3x 2 – 2x + 4 then f(1) + f(2) + - - - - + f(20) =

Sol.

= =

= 8270

Theory, Examples & Explanations Sequence & Series A set of numbers whose domain is a real number is called a SEQUENCE and sum of the sequence is called a SERIES. If a1, a2, a3, a4… an … is a sequence, then the expression a 1 + a2 + a3+ a4+ a5  … an  … is a series. Those sequences whose terms follow certain patterns are called progressions. For example 1, 4, 7, 10, 13 ……. 7, 4, 1, – 1, – 2, 2, – – 5……… 1, 2, 4, 8, 16……… 8, 4, 2, 1, ½….…… 2

2

2

2

Also if f (n) = n2 is a sequence, then f(1) = (1) = 1, f(2) = 2 = 4, f(3) = (3) = 9, f (10) = 10 = 100 and so on. The nth term of a sequence is usually denoted by T n Thus T1 = first term, T 2 = second term, T10 = tenth term and so on. There are three different progressions Arithmetic Progression (A.P) Geometric Progression (G.P) Harmonic Progression (H.P) Arithmetic Progression (A.P.) It is a series in which any two consecutive terms have common difference and next term can be derived by adding Back to Table of Contents

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that common difference in the previous term. Therefore T n+1 – Tn = constant constant and called common difference (d) for all n

N.

Examples: 1. 1, 4, 7, 10, ……. is an A. P. whose first term is 1 and the common difference is d = (4 – 1) = (7 – (7 – 4) = (10 – (10 – 7) = 3. 2. 11, 7, 3, – 3, – 1 …… is an A. P. whose first term is 11 and the common difference d = 7 – 11 = 3 – 3 – 7, = – = – 1 – 3 = – = – 4. If in an A. P. a = first term, d = common difference = Tn – Tn-1 Tn = nth term (Thus T 1 = first term, T 2 = second term, T10 tenth term and so on.) l = last term, Sn = Sum of the n terms. Then a, a + d, a + 2d, a + 3d... are in A.P. th

n term of an A.P. The nth term of an A.P is given by the formula Tn = a + (n – 1) d Note: If the last term of the A.P. consisting of n terms be l, then l = a + (n – 1) d Sum of n terms of an A.P The sum of first n terms of an AP is usually denoted by S n and is given by the following formula:

Where ‘l ’ is the last term of the series. th

st

Ex.1

Find the series whose n term is

Is it an AP series? If yes, find 101 term.

Sol.

Putting 1, 2, 3, 4….. We get T 1, T2, T3, T4….. = 1, 5/2, 4, 11/2…… d1 = 3/2, d2 = 3/2, d 3 = 3/2 As the common differences ae equal Therefore the series is an AP T101 = a+ 100d = 1+100*(3/2) = 151

Ex.2

Find 8th, 12th and 16 th terms of the series; -6, - 2, 2, 6, 10, 14, 18…..

Sol.

Here a = -6 and d = -2 -2 – –(-6) (-6) = 4 Therefore T8 = – 6 + 7 × 4 = 22

[T 8 = a + 7d]

T12 = a + 11d = – = – 6 + 11 × 4 = 38

[T 12 = a + 11d]

T16 = a + 15d = – = – 6 + 15 × 4 = 54

[T 16 = a + 15d]

Properties of an AP I. If each term of an AP is increased, decreased, multiplied or divided by the same non-zero number, then the resulting sequence is also an AP.

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For example: For A.P. 3, 5, 7, 9, 11… If you add constant let us say 1 in each

4, 6, 8, 10, 12......

term, you get

This is an A.P. with common difference 2

If you multiply by a constant let us say

6, 10, 14, 18, 22…..

2 each term, you get

Again this is an A.P. of common difference 4

II. In an AP, the sum of terms equidistant from the beginning and end is always same and equal to the sum of first and last terms. III. Three numbers in AP are taken as a – a – d, a, a + d. For 4 numbers in AP are taken as a – 3d, a – a – d, a + d, a + 3d For 5 numbers in AP are taken as a – 2d, a – a – d, a, a + d, a + 2d IV. Three numbers a, b, c are in A.P. if and a nd only if 2b = a + c Ex.3

The sum of three numbers in A.P. is – 3, and their product is 8. Find the numbers.

Sol.

Let the numbers be (a – (a – d), a, (a + d). Then, Sum = – = – 3

(a – (a – d) + a + (a + d) = – = – 3

3a = – 3

a = – = – 1

Product = 8 (a – (a – d) (a) (a + d) = 8 a (a2 – d2) = 8 2

( –1) –1) (1 – (1 – d ) = 8 d2 = 9 d=±3 If d = 3, the numbers are – are – 4, 4, – – 1, 2. If d = – = – 3, the numbers are 2, – 2, – 1, 1, – – 4. Thus, the numbers are – are – 4, 4, – – 1, 2 or 2, – 2, – 1, 1, – – 4. Ex.4

A student purchases a pen for Rs. 100. At the end of 8 years, it is valued at Rs. 20. Assuming Assuming that the yearly depreciation is constant. Find the annual depreciation.

Sol.

Original cost of pen = Rs. 100 Let D be the annual depreciation. Price after one year = 100 – 100 – D = T1 = a (say) Price after eight years = T 8 = a + 7 ( – D) = a – a – 7D = 100 – 100 – D – 7D = 100 – 100 – 8D By the given condition 100 – 100 – 8D = 20

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8D = 80 D = 10. Hence annual depreciation = Rs. 10. Geometric Progression A series in which each preceding term is formed by multiplying it by a constant factor is called a Geometric Progression G. P. The constant factor is called the common ratio and is formed by dividing any term by the term which precedes it. In other words, a sequence, a 1, a2, a3… an… is called a geometric progression progression If

= con const stant ant for all n N 2

The General form of a G. P. with n terms is a, ar, ar ,…ar

n – –1 1

Thus if a if a = the first term r = the common ratio, Tn = nth term and Sn = sum of n terms General term of GP = T n = ar

n – –1 1

Ex.5

Find the 9 th term and the general term of the progression.

Sol.

The given sequence is clearly a G P with first term a = 1 and common ratio = r =(-1/2) =( -1/2) Now T9 = ar8 = And Tn = arn-1 = (-1/2) n-1

Sum of n terms of a G.P: Where r>1 Where r<1 Sum of infinite G.P: If a G.P. has infinite terms and –1 < r < 1 or Ex.6

< 1 , then sum of infinite G.P is

The inventor of the chess board board suggested suggested a reward of one grain of of wheat wheat for the first square, 2 grains grains for the second, 4 grains for the third and so on, doubling the number of the grains for subsequent squares. How many grains would have to be given to inventor? (There are 64 squares in the chess board).

Sol.

Required number of grains 2

3

= 1 + 2 + 2 + 2 + ……. To 64 terms = 1

=

Recurring Decimals as Fractions If in the decimal representation a number occurs again and again, then we place a dot (.) on the number and read it as that the number is recurring. e.g., 0.5 (read as decimal 5 recurring).

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This mean 0. 5 = 0.55555……. 0.

= 0.477777……

These can be converted into fractions fra ctions as shown in the example given below Ex.7

Find the value in fractions which is same as of 0.4

Sol.

We have 0.4

= 0.4373737……….

= 0.4  0.037  0.00037  0.0000037  ……….. =

…………

= = Properties of G.P I. If each term of a GP is multiplied or divided by the same non-zero quantity, then the resulting sequence is also a GP. For example: For G.P. is 2, 4, 8, 16, 32… If you multiply each term by constant

4, 8, 16, 32, 64..

This is a G.P

1, 2, 4, 8, 16 ..

This is a G.P

let say 2,you get If you divide each term by constant let say 2,you get II. SELECTION OF TERMS IN G.P. Sometimes it is required to select a finite number of terms in G.P. It is always convenient if we select the terms in the following manner: No. of terms

Terms

Common ratio

3 4 5 2

3

If the product of the numbers is not given, then the numbers are taken as a, ar, ar , ar , …. III. Three non-zero numbers a, b, c are in G.P. if and only if b 2 = ac. b is called the geometric mean of a & c IV. In a GP, the product of terms equidistant from the beginning and end is always same and equal to the product of first and last terms as shown in the next example. Back to Table of Contents

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Harmonic Progression (H.P.) A series of quantities is said to be in a harmonic progression when their reciprocals are in arithmetic progression. e.g. 1/3, 1/5, 1/7, ….. and 1/a, 1/ (ad), 1/(a2d)….. are in HP as their reciprocals 3, 5, 7 ….. and a, ad, a2d……. are in AP nth term of HP th

Find the n term of the corresponding AP and then take its reciprocal. If the HP be 1/a, 1/ (ad), 1/(a2d)….. The corresponding AP is a, ad, a2d……. Tn of the AP is a + (n-1)d In order to solve a question on HP, one should form the corresponding AP A comparison between AP and GP Description

AP

GP

Principal Characteristic

Common Difference (d)

Common Ratio (r)

th

n Term Mean

Sum of First n Terms

Mean

Arithmetic - Geometric Progression 2

3

a + (a+d)r + (a+2d)r + (a+3d)r  ……. Is the form of Arithmetic Geometric Progression (A.G.P) One part of the series is in Arithmetic progression and other part is a Geometric progression. The sum of n terms series is

–

The infinite term series sum is Arithmetic Geometric Series can be solved as explained in the example below: Relation between AM, GM and HM: A =Arithmetic mean = G = Geometric mean = H = Harmonic mean = Multiplying A and H, we get AH =

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= 2

3

Ex.8

Find the sum of 1 +2x +3x + 4x + …..

Sol.

The given series is an Arithmetic Arithmetic – – Geometric series whose corresponding corresponding A.P and G.P are 1, 2, 3, 4… And 1, x, x2, x3… respectively. The common ratio of the of the G.P is x. 2

3

= 1 +2x+ 3x +4x ….. x

= x+2x2+3x3……

------------- (i)

-------------- (ii)

Subtracting (ii) from (i), we get = 1 +[x+x 2+x3…… ]

x

x) = 1 + = Ex.9

If the first item of an A>P is 12, and 6 th term is 27. What is the sum of first 10 terms?

Sol.

a = 12, t6 = a +5d = 27

d =3

Therefore S10 = (10/2)[2*12 +(10 -1)3] = 255 Ex.10

If the fourth and and sixth terms of an A.P are 6.5 and 9.5. 9.5. What What is the 9

Sol.

a+3d = 6.5 and a+5d = 9.5

th

term of that A.P?

a =2 and d =1.5 Therefore t9 = a+8d = 14 Ex.11

What is the arithmetic arithmetic mean of first 20 terms os an A.P. Whose Whose first term is 5 and 4 th term is 20?

Sol.

a = 5, t4 = a+3d = 20 d =5 th

th

A.M is the middle number = average of 10 and 11 number = 50+55/2 = 52.5 Ex.12

The first term of a G.P is half half of its 4 th term. What is the 12 th term of that G.P, if its sixth term is is 6

Sol.

t 1 = t4 3

a = ar

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r3 = 2 t6=ar5 = 6 11

5

6

t12 = ar = ar

r = 24

Ex.13

If the 1 st and 5th terms of a G.P are 2 and 162. What is the sum of these five t erms?

Sol.

a=2 ar4 = 162 r=3 S5 = 2(35 – 1)/(3-1) = 242 2

3

Ex.14

What is the value of r +3r + 5r + …….

Sol.

Assume S = r +3r + 5r  ……. ----------(1)

2

3

rS = r2 + 3r3 + 5r4 ……. -----------(2) (1) – (1) – (2) S(1 – S(1 –r) r) = r + 2r 2 + 2r3  ……. 2

3

= r + 2r + 2r  …….. S= Ex.15

The first term of a G.P. 2 and common common ratio is 3. If the sum of fist n terms of this G.P is greater than 243 then the

minimum value of ‘n’ is

Sol.

> 243 > 243 > 244 n >5 So, minimum possible value of n is 6.

Ex.16 Sol.

+

+

+ ……. +

is

= = = = n

Ex.17

an = 2 + 1 then (a 1 + a2 + …….. + a 20) – a21 is:

Sol.

a1 + a2  ……..  a 20 = 21 + 1 + 2 2 + 1 + 2 3  1 ……..  2 20 + 1

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= 2(220 – 1) + 20 = 2 21 + 18 Therefore Answer = 17

Statistics Questions taken from students’ forum 1. Which of the following set of numbers has the highest Standard deviation? a)

1

0

1

0

1

0

b)

-1

-1

-1

-1

-1

-1

c)

1

1

1

1

1

1

d)

1

1

0

-1

0

-1

2. Which of the following has the highest Standard deviation a)5, -5,5,-5, 5,-5

b) 5,5,5,5,5,5

c) -5,-5,-5,-5,-5,-5

d) -5,5,-5,5,-5,5

3. Which will give good standard deviation a) (7,0,-7,0,7)

b)(7,-7,7,-7,7)

c)(1,0,-1,0,1)

Theory, Examples & Explanations Statistics Frequency Distributions For some sets of measurements, it is more convenient and informative to display the measurements in a frequency distribution. For example, the following values could represent the number of dependent children in each of 25 families living on a particular street. 1, 2, 0, 4, 1, 3, 3, 1, 2, 0, 4, 5, 2, 3, 2, 3, 2, 4, 1, 2, 3, 0, 2, 3, 1 . These data can be grouped into a frequency distribution by listing ach different value (x) and the frequency of occurrence for each value.


X

F

0

3

1

5

68




2

7

3

6

4

3

5

1

Total

25

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Measures of Central Location Two common measures of central location, often called “average,” for a discrete set of numerical values or measurements are the arithmetic means and the median. The average ( arithmetic mean) of n values is defined as the sum of the n values divided by n. For example, the arithmetic mean of the values va lues 5, 8, 8, 14, 15, and 10 is 60 ÷ 6 = 10.

If we order the n values from least to greatest, the median is defined as the middle value if n is odd and the sum of the two middle values divided by 2 if n is even . In the example above, n = 6, which is even. Ordered from least to greatest, the values

are 5, 8, 8, 10, 14, and 15. Therefore, the median is (8+10)/2= 9. Note that for the same set of values, the arithmetic mean and the median need not be equal, although they could be. For example, the set of values 10, 20, 30, 40, and 50 has arithmetic mean = median = 30. Another measure of central location is called the mode, which is defined as the most frequently occurring value. For the six measurements above, the mode is 8. Ex.

Find mean of all factors of 10.

Sol.

Factors of 10 = 1, 2, 5, 10

Ex.

The arithmetic mean of 6, 8, 5, 7, x and 4 is 7. Find the value of x.

Sol.

Ex.

The mean of 40 numbers is 160. It was detected detected on on checking that the value of 165 was wrongly copied as 125 while computing the mean. Find the correct mean.

Sol.

= 160, n = 40 = 160 × 40 = 6400

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But 165 was wrongly copied as 125. Correction = correct reading – reading – wrong reading = 165 – 165 – 125 = 40 Correct Correct

= 6400 + 40 = 6440 =

Geometric Mean The geometric mean is the nth root of the product of n items of a series. Thus if X 1, X2, X3, …….. X n are the given n observations, and then their G.M is given by G.M = Ex.

The GM of 4, 6, 9 is

Sol.

If there are n numbers then GM = Therefore G.M of 4, 6, 9 =

=6

Median The median is that value of the variable, which divides the group in two equal parts. One part comprising all the values greater and the other, all the values less than median. Then median of the distribution may defined as that value of the variable which exceeds and is exceeded by same numbers of observations, observations, i.e., it is the value v alue such that number of observations above it is equal to the number of observations below it. Hence, the median is only positional average its value depends on position occupied by a value in the frequency distribution Median of individual observations: In case of individual observations x 1, x2, ……. xn to find the median we use the following algorithm. Step 1: Arrange the observations x 1, x2, ……. xn in ascending or descending order of magnitude. Step 2: Determine the total number of observations, say, n Step 3: If n is od odd, d, th then en me medi dian an is is the the val value ue of and Ex.

obse ob serva rvati tion on.. If If n is is eve even, n, th then en me medi dian an is is the the AM of th the e valu values es of

observation

(i) The following are the makes makes of 9 students students in a class. Find the median 34, 32, 48, 38, 24, 30, 27, 21, 35 (ii) Find the median of the daily wages of ten workers. $ 20, 25, 17, 18, 8, 15, 22, 11, 9, 14

Sol.

(i) Arranging the data in ascending order of magnitude, we have 21, 24, 27, 30, 32, 34, 35, 38, 48. Since, there are 9, and odd number of items, therefore median is the value of

observation, i.e., 32. Back to Table of Contents

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(ii) Arranging the wages in ascending order of magnitude, we have 8, 9, 11, 14, 15, 17, 18, 20, 22, 25, Since, there are 10 observations, therefore median is the arithmetic mean of

and

observations. So median = (15 +

17)/2 = 16. Median of discrete frequency distribution: In case of a discrete frequency distribution xi/f; i = 1, 2… n; we calculate the median by using the following algorithm Step 1: Find the cumulative frequencies (c.f.) Step 2: Find N/2, where N = Step 3: See the cumulative frequency (c.f.) just greater than N/2 and determine the co-responding value of the variable. Step 4: The value obtained in step III is the median Ex.

Sol.

Obtain the median for the following frequency distribution: x:

1

2

3

4

5

6

7

8

9

f:

8

10

11

16

20

25

15

9

6

Calculation of Median

X

F

c.f

1

8

8

2

10

18

3

11

29

4

16

45

5

20

65

6

25

90

7

15

105

8

9

114

9

6

120

N =120 Here N = 120

N/2 = 60. We find that the cumulative cumulative frequency just just greater than N/2, is 65 and the value of of x corresponding corresponding

to 65 is 5. Therefore, median is 5. Mode Mode is the value which occurs most frequently in a set of observations and around which the other items of the set cluster densely OR the mode of a distribution is the value at the point around which the item tends to be most heavily concentrated Back to Table of Contents

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Measures of Dispersion Measures of dispersion, or spread, for a discrete set of numerical values or measurements take many forms in data analyses. The simplest measure of dispersion is called the range, which is defined as the greatest measurement minus the least measurement. Since the range is affected by only the two most extreme values in the set of measurements, other measures of dispersion have been developed that are affected by every measurement. The most commonly used of these other measures is called the standard deviation. The value of the standard deviation for a set of n measurements can be calculated by (1) first calculating the arithmetic mean, (2) finding the difference between that mean and each measurement, (3) squaring each of the differences, (4) summing the squared values, (5) dividing the sum by n, and finally (6) taking the nonnegative square root of the quotient. The standard deviation can be roughly interpreted as the average distance from the arithmetic mean for n measurements. The standard deviation cannot be negative, and when two sets of measurements are compared, the one with the larger dispersion will have the larger standard deviation. Mean Deviation Mean deviation of a series is the arithmetic average of the deviations of various items from a measure of central tendency (either mean, median or mode) If X1, X2, X3…….. Xn are n given observations then the mean deviation (M.D.) about an average A, is given by M.D. (about an average A)

Where d =

i.e. modulus value or absolute value of the deviation (After ignoring the

negative sign) Ex.

The S.D of 2, 4, 6, 8, 10, 12 is:

Sol. Variate x 2

-5

25

4

-3

9

6

-1

1

8

1

1

10

3

9

12

5

25


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Clocks & Calendar Questions taken from students’ forum 1. Angle between hands of the clock at a t 3:25? 2. At what time between 4 and 5 o’clock are the hands of the clock at right angle/together?

Theory, Examples & Explanations Clocks & Calendar The dial of a clock is a circle whose circumference is divided divided into 12 parts, called hour spaces. Each hour space is further divided into 5 parts, called minute’s spaces. This way, the whole circumference is divided into 12 × 5 = 60 minute spaces. If we consider clock as a circular track and the two hands of clock minute hand and hour hand are just like two players o

running in the same direction. Let total length of the track is 360 and minute hand completes one full round in 1 hour while hour hand cover full round in 12 hours.

Speed of hour hand =

=

Speed of minute hand = Since they are moving in the same direction, so the relative speed of both the hands with respect to each other =

Time taken by minute hand to overtake hour hand = There are 4 types of problems on clocks: 1. To calculate the angle between the two hands when time is given.

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2. To calculate the time when both the hands will be at some angle. 3. Concept of slow and fast clocks. 4. Overall gain/loss Calculating the angle The angle between the two hands is given by the following formula Formula for the angle between the hands = Where H ---- Hour Reading & M---Minute Reading Ex.1

Calculate the angle between the two hands of clock when the clock shows 5: 25 p.m.

Sol.

Given time = 5: 25 p.m. Hence H = 5 and M = 25 We can apply the following direct formula to find the angle between the hands = Required angle =

=

Alternative Method: Since at 5: 25 the minute hand will be at 5 and the angle between them will be same as the distance covered in degree by the hour hand in 25 minutes. Required angle = distance of hour hand = speed × time =

=

Calculating the time To calculate the time when both the hands will be at some angle In one minute the net gain of minute hand over hour hand = If the gain is

then the time is 1 min.

If the gain is

then the time is

If the gain is

then the time is

Note: If between H and (H1) o’clock, the t he two hands are together at an angle

then required time =

minutes, where H is reading of hour. Ex.2

At what time between 4 and 5 o’clock are the hands of the clock together?

Sol.

Method 1: At 4 o’clock, the hour hand is at 4 and the minute hand is at 12. It means that they are 20 min spaces apart. To be together, the minute hand must gain 20 minutes over the hour hand. Now, we know that 55 min. are gained in 60 min. 20 min are gained in

74

=





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Therefore, the hands will be together at Alternate method: Using the formula: Required time = Here

=

(Hands of clock are together) and H =4

Requi Req uire red d ti time me =

= 21

Therefore, There fore, the hands hands will be be together together at 21

min mi n min past past 4.

Another formula: Between H and (H1) o’clock, the two hands will be together at 5 In th this is ca case se;; 5

= 21

min past H.

min mi n pa past st 4

Ex.3

At what time between 4 and 5 o’clock will the hand of clock be at right angle?

Sol.

At 4o’clock there are 20 min. spaces between hour and minute hands. To be at right angle, they should be 15 min spaces apart. So, there are two cases: Case I: When the minute hand is 15 min spaces behind the hour hand To be in this position, the min hand should have to gain 20 20 – – 15 = 5 min spaces. Now, we know that 55 min spaces are gained in 60 min. Therefore 5 min spaces are gained in Therefore they are at right angle at

past 4

Case II: When the minute hand is 15 min spaces ahead of the hour hand To be in this position, the min hand should have to gain 20 + 15 = 35 min spaces. Now, we know that 55 min spaces are gained in 60 min Therefore 5 min spaces are gained in Therefore they are at right angle at

past 4

Alternate Method: As the hands of the clock are at right angle therefore Also time is between 4 and 5 o’clock, no of hours = 4 Required time = =

75

= 38

or 5

min





Therefore they are at right angle 38

Quants Funda

or 5

min past 4

Another Formula: Between x and (x  1) o’clock the two hands are right angle at (5X In thi thiss cas case; e; the theyy wil willl be at rig right ht ang angle le at (5

min pas pastt 4 = 38

min past X or 5

min pas pastt 4

Concept of slow and fast clocks Too Fast and Too Slow: If a watch indicates 9.20 when the correct time is 9.10, it is said to be 10 minutes too fast. And if it indicates 9.00 when the correct time is 9.10, it is said to be 10 minute too slow. Ex.4

Two clocks clocks are set at 1 p.m. Fast clock clock gains gains 1 min for every hour. hour. Find the time when the fast clock shows shows 6 p.m.

Sol.

For every 60 min of true clock, the fast clock will show 61 min. For 61 minutes of fast clock, true time = 60 minutes For 300 minutes (5hrs) of fast clock; true time =

=

Actual time in the true clock =5 Overall gain/loss Afte Af terr eve every ry 65

min mi n=

min the two hands will coincide. If the hands of a clock coincide every ‘X’ min, then Gain/Loss per

day by a watch, is given by

[If answer is (+) then there will be gain and if (-) then there will be loss.]

CALENDAR There are 12 months in 1 year. Jan – Jan –31 31 (days), Feb – Feb –28; 28; 29 (for a leap year), Mar – Mar –31, 31, Apr – Apr – 30, May – May –31, 31, June – June – 30; July – July – 31; Aug – Aug – 31; Sept – Sept –30; 30; Oct – Oct – 31; Nov – Nov –30; 30; Dec – Dec – 31

Total 365 (ordinary year); 366 days (for a leap year) To find the number of weeks in an ordinary year: = 52 weeks + 1 odd day (remainder) To find the number of weeks in a leap year: = 52 weeks + 2 odd day (remainder) Leap year: It is so called as it comes after a leap of 3 years from the previous leap year. Back to Table of Contents

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Method to find whether a given year is a leap year or an ordinary year Every year which is not a century year (i.e. which is not a multiple of 100) is a leap year if and only if it is completely divisible by 4. Every century year is a leap year if and only if it is completely divisible by 400 or is an integral multiple of 400 (i.e. the remainder ought to be 0). e.g. 2000 is a leap year. 1900 is not. 1996 is a leap year, 1998 is not Odd day’s concept

To find the number of odd days in a century A century, i.e. 100 year has 76 ordinary year and 24 leap year = [(76 × 52) weeks + 76 days] – [(24 × 52) weeks + 24 × 2 days] = 5200 weeks + 124 days = 5200 weeks + 17 weeks + 5 odd days = 5217 weeks + 5 odd days Therefore, 100 years contain 5 odd days. Now, (i) 200 years contain 5 × 2 = 10, i.e., 3 odd days. (ii) 300 years contain 5 × 3 = 15 i.e., 1odd day. (iii) 400 years contain 5 × 4 + 1 = 21, i.e., no odd day. Similarly, 800, 1200 years etc. contain no odd day. Note: (i) 5 × 2 = 10 days = 1 week + 3 days i.e., 3 odd days (ii) 5 × 3 = 15 days = 2 weeks + 1 day i.e. 1 odd day. (iii) 400th year is a leap year therefore one additional day is added. Odd days and their numeral values When we have to calculate the number of days on any particular Extra Days and their numeral Values 0 → Sunday 1 → Monday 2 → Tuesday 3 → Wednesday 4 → Thursday

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5 → Friday 6 → Saturday th

Ex.5

Father of Nation Mahatma Gandhi died on 30 January 1948. What What was the day on which he died?

Sol.

Up to 1600 AD we have 0 odd days; up to 1900 AD we have 1 odd day. Now for in 47 year s we have 11 leap years and 36 normal years. Odd days from 1901 to 1947 = 11 x 2 +36 x1 = 22 + 36 =58 odd days = 8 weeks + 2 odd days Total odd days up to 31st December 1947 = 1 + 2 = 3 odd days 30 days of January contain only 4 weeks + 2 odd days th

So 30 January 1948 has total 5 odd days Day on 30th January 1948 = Friday. Ex.6

How does the number of odd odd days help us in finding finding the day of a week? (Please take take care of this point) When a specific day is given: Suppose a question like: Jan 1, 1992 was Wednesday. What day of the week will it be on Jan 1, 1993? If you recall, 1992 being a leap year it has 2 odd days. So, the above said day will be two days beyond Wednesday, i.e., it will be Friday. When no specific day is given: Here, we count days according to number of odd days. Sunday for 0 odd day, Monday for 1 odd day and so on. (i.e. from 0 to 6; 6 being Saturday) th

Suppose someone asks you to find the day of the week on 12 January 1979. 12th Jan, 1979 means 1978 year + 12 days Now, 1600 years have 0 odd day. 300 years have 1 odd day 78 years have 59 ordinary year + 19 leap years = 6 odd days.

Total no. of odd days = 0  1  6  12 = 19 or 5 odd days. So, the day was “Friday”. Back to Table of Contents

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Points not to be ignored: 1. 400th year is a leap year or a century multiple of 400 is a leap year, rests are not. 2. 100 years has 5 odd days 200 years has 3 odd days 300 years has 1 odd day 400 years has 0 odd days. And so on

Geometry (2D & 3D)/ Coordinate Geometry Questions taken from students’ forum 1. Number of faces, vertices and edges of a cube.

[Ans: 6,8,12]

2. The sides are given. which triangle is not possible: (A) (2,3,6)

(B) (3,2,4)

(C) (3,4,5)

(D) (3,3,3)

(E) (5,3,5)

3. Sum of slopes of 2 perpendicular straight lines is given. Find the pair of lines from the given set of options which satisfy the above condition? Note: Questions can be asked related to the sides of rectangle

Theory, Examples & Explanations Geometry (2D & 3D)/ Coordinate Geometry Properties of lines Intersecting Lines and Angles: Angles : If two lines intersect at a point, then opposite angles are called vertical angles and have the same measure. Perpendicular Lines: An angle that has a measure of 90 o is a right angle. Parallel Lines: If two lines that are in the same plane do not intersect, the two lines are parallel. Parallel lines cut by a transverse: If two parallel lines L1 and L2 are cut by a third line called the transverse. Polygon A closed plane figure made up of several line segments that are joined together is called a polygon. The sides do not cross each other. Exactly two sides meet at every vertex.

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Types of Polygons: Regular: all angles are equal and all sides are the same length. Regular polygons are both equiangular and equilateral. Equiangular: all angles are equal. Equilateral: all sides are the same length. Triangles A triangle is a polygon of three sides. A triangle with three sides of different lengths is called a scalene triangle. triangle. An isosceles triangle has two equal sides. The third side is called the base. The angles that are opposite to the equal sides are also equal. An equilateral triangle has three equal sides. In this type of triangle, the angles are also equal, so it can a lso be called an equiangular triangle. Each angle of an equilateral triangle must measure 60o, since the sum of the interior angles of any triangle must equal to 180 o. Obtuse angled triangle: is a triangle in which one angle is always greater than 90

o

Acute angled triangle: In which all angles are less than 90 o Right Angled Triangle: A triangle whose one angle is 90 o is called a right (angled) Triangle. Properties of a Triangle 1. Sum of the three angles is 180 o. 2. An exterior angle is equal to the sum of the interior opposite angles. 3. The sum of the two sides is always greater than the third side. 4. The difference between any two sides is always less than the third side. 5. The side opposite to the greatest angle is the greatest side and the side opposite to the smallest angle will be the shortest side. Centroid: (a) The point of intersection of the medians of a triangle. (Median is the line joining the vertex to the mid-point of the opposite side. The medians will bisect the area of the triangle.) (b) The centroid divides each median from the vertex in the ratio 2 : 1. Orthocentre: The point of intersection of altitudes. (Altitude is a perpendicular drawn from a vertex of a triangle to the opposite side.) Circumcentre: The circumcentre of a triangle is the centre of the circle passing through the vertices of a triangle. It is also the point o f intersection of perpendicular bisectors of the sides of the triangle. If a, b, c, are the sides of the triangle, Δ is the area, then abc = 4R Δ where R is the radius of the circum circum-circle. -circle. Incentre: The point of intersection of the internal bisectors of the angles of a triangle

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Congruent triangles Two triangles ABC and DEF are said to the congruent, if they are equal in all respects (equal in shape and size). The tests for congruency (a) SAS Test: Two sides and the included angle of the first triangle are respectively equal to the two sides and included angle of the second triangle. (b) SSS Test: Three sides of one are respectively equal to the three sides of the other triangle. (c) AAS Test: Two angles and one side of one triangle are respectively equal to the two angles and one side of the other triangle. (d) RHS Test: The hypotenuse and one side of a right-angled triangle are respectively equal to the hypotenuse and one side of another right-angled triangle. Similar Triangles Two figures are said to be similar, if they have the same shape but not the same size. If two triangles are similar, the corresponding angles are equal and the corresponding sides are proportional. Test for similarity of triangles (a) AAA Similarity Test: Three angles of one triangle are respectively equal to the three angles of the other triangle. (b) SAS Similarity Test: Two sides of one are proportional to the two sides of the other and the included angles are equal. Properties of similar triangles: If two triangles are similar, the following properties are true: (a) The ratio of the medians is equal to the ratio ra tio of the corresponding sides. (b) The ratio of the altitudes is equal to the ratio of the corresponding sides. (c) The ratio of the circumradii is equal to the r atio of the corresponding sides. (d) The ratio of in radii is equal to the ratio of the corresponding sides. (e) The ratio of the internal bisectors is equal to the ratio of the corresponding sides. sides. (f) Ratio of areas is equal to the ratio of squares of corresponding sides. sides. Pythagoras Theorem The square of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the other two sides i.e. in a right 2

2

2

angled triangle ABC, right angled at B, AC = AB + BC

Pythagorean triplets are sets of 3 integers which can be three sides of a right-angled triangle. Examples of Pythagorean triplets are (3, 4, 5), (5, 12, 13), (7, 24, 25), (9, 40, 41) etc. Quadrilaterals A polygon with 4 sides, is a quadrilateral 1. In a quadrilateral, sum of the four angles is equal to 360°. 2. The area of the quadrilateral = ½ × one diagonal x sum of the perpendicular to it from vertices.

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Cyclic Quadrilatera Quadrilateral: l: If a quadrilateral is inscribed in a circle, it is said to be cyclic quadrilateral. 1. In a cyclic quadrilateral, opposite angles are supplementary. 2. In a cyclic quadrilateral, if any one side is extended, the exterior angle so formed is equal to the interior opposite angle. Circles If O is a fixed point in a given plane, the set of points in the plane which are at equal distances from O is a circle. Properties of the circle 1. Angles inscribed in the same arc of a circle are equal. 2. If two chords of a circle are equal, their corresponding arcs have equal measure. 3. A diameter perpendicular to a chord bisects the chord. 4. Equal chords of a circle are equidistant from the centre. 5. When two circles touch, their centres and their point of contact are collinear. 6. If the two circles touch externally, the distance between their centres is equal to sum of their radii. 7. If the two circles touch internally, the distance between the centres is equal to difference of their radii. 8. Angle at the centre made by an arc is equal to twice the angle made by the arc at any point on the remaining part of the circumference. 9. The angle inscribed in a semicircle is 90 o. 10. Angles in the alternate segments are equal. 11. If two chords AB and CD intersect externally at P, then PA × PB = PC × PD 12. If PAB is a secant and PT is a tangent, then PT 2 = PA × PB 13. If chords AB and CD intersect internally, then PA × PB = PC × PD 14. The length of the direct common tangent (PQ)

CO-ORDINATE GEOMETRY Rectangular axes (general) (i) Distance formula: If A (x 1, y1) and B (x2, y2) be two points, then

in particular, of a point

P(x, y) form O(0, 0) is (ii) Section formula: The point which divides the join of two distinct points A (x 1, y1) and B (x2, y2) in the ratio m1: m2 Internally, has the co-ordinates

m1 ≠ 0, m2 ≠ 0, m1 + m2 ≠ 0 and externally, is

In particular, the mid-point of the segment joining A (x1 y1) and B (x2, y2) has the coordinates Back to Table of Contents

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Straight Line Slope of a line Slope of a non-vertical line L is the tangent of the angle θ, which either of half ray of the line L makes with the positive direction of x-axis. In particular, (a) Slope of a line parallel of x-axis is zero. (b) Slope of a line parallel to y-axis is not defined. (c) Slope of a line equally inclined to both the axis is − 1 or 1. (d) Slope of a line making equal intercepts on the axis is − 1 or 1. (e) Slope of the line through the points A (x 1, y1) and B (x 2, y2) is (f) Slope of the line ax + by + c = 0, b ≠ 0, is – is – a / b. (g) Slopes of two parallel (non-vertical) lines are equal then m 1 = m2 (h) If m1 and m2 be the slopes of two perpendicular lines (which are oblique), then m 1m2 = – 1. Equation of a Line An equation of the form ax + by + c = 0 is called the general equation of a straight line, where x and y are variable and a, b, c are constants. Equation of a line parallel to X axis or Y - axis (i) Equation of any line parallel to x-axis is y = b, b being the directed distance of the line from the x-axis . In particular equation of x-axis is y = 0

(ii) Equation of any line parallel to y-axis is x = a, a being the directed distance of the line from the y-axis. In particular equation of y-axis is x = 0.

One point form Equation of a line (non-vertical) through the point (x 1, y1) and having slope m is y – y1 = m (x – x1). Two- points form Equation of a line (non-vertical) through the points (x 1, y1) and (x2, y2) is Slope-intercept form Equation of a line (non-vertical) with slope m and cutting off an intercept c from the y -axis is y = m x + c. c. Intercept form Equation of a line (non-vertical) with slope m and cutting off intercepts a and b from the x-axis and y-axis respectively is


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Logical Reasoning Questions taken from students’ forum Directions for Questions from 1 – 5: The office staff of the XYZ Corporation presently consists of three bookkeepers (A, B and C) and five secretaries (D, E, F, G and H). Management is planning to open a new office in another a nother city using three secretaries and two bookkeepers. To do so they plan to separate certain individuals who do not function well together. The following guidelines were established to set up the new office: I. Bookkeepers A and C are constantly finding fault with one another and should not be sent as a team to the new office. II. C and E function well alone but not as a team. They should be separated. III. D and G have not been on speaking terms for many months. They should not go together. IV. Since D and F have been competing for promotion; they should not be a team. 1. If A is to be moved as one of the bookkeepers, which of the following cannot be a possible working team? (a) ABDEH

(b) ABDGH

(c) ABEFH

(d) ABEGH

(e) ABFGH

2. If C and F are moved to the new office, how many combinations are possible? (a) 1

(b) 2

(c) 3

(d) 4

(e) 5

3. If C is sent to the new office, which member of the staff cannot go with C? (a) B

(b) D

(c) F

(d) G

(e) H

4. Under the guidelines developed, which of the following must go to the new office? (a) B

(b) D

(c) E

(d) G

(e) H

d. I and III only.

e. I, II and III.

5. If D goes to the new office which of the following is (are) true? I. C cannot go. II. A cannot go. III. H must also go. a. I only.

b. II only.

c. I and II only.

Directions for Questions from 5 – 8: An Airedale, a Boxer, a Collie, and a Doberman win the top four prizes in the kennel show. Their owners are Ms. Huntley, Mr. Grossman and Mr. Foster, Mr. Edwards's not necessarily in that order. Their dogs’ names are Jack, Kelly, Lad, and Max, not necessarily in that order.

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5. First prize is won by a) Mr. Edwards's dog b) Ms. Huntley's dog

c) MAX

d) JACK

e) LAD

c) Airedale

d) Wins second prize

e) Kelly

6. Mr. Grossman's dog a) Collie

b) Boxer

7. In which of the following statements are the dogs correctly listed in descending Order of their prizes A) KELLY; AIREDALE; Mr. EDWARD'S Dog B) BOXER ; Mr. GROSSMAN'S GROSSMAN'S Dog ; JACK C) Mr. EDWARD'S EDWARD'S Dog ; AIREDALE; AIREDALE; LAD a) A only

b) B only

c) C only

d) A and C only

e) B and C only

8. LAD is a) Owned by Mr. FOSTER

b) Owned by Mr. EDWARDS

c) Boxer

d) Collie

e) Wins third prize

Directions for Questions: There are 9 people I, J, K, L, M, N, O, P and Q living in a five storied building. The top floor has only one room and all other floors have two rooms. No rooms are vacant. No rooms carry two people. K and N live on the same floor. L is living in a floor lower than I. Q is living in an upper floor than N. Lives in the third floor. Note: Four questions were asked based on this paragraph about finding the floor and the specific person lives on. And also some conditions were given to state which one is valid.

Letter Series 1. How many V’s are there under the condition that, S should be followed by V and should not be followed by F? VSFTWELBVSLLKSMSVFLSDI 2. How many B's are there followed by G which are not followed by S in the following series: BBGMPQBGSKOBGASBBGDEFBGSTI 3. Find the number of Y followed by W but not followed by Z in the following series: YWRUDDYWZ

Number Series Complete the sequence a.

9, 10,11,13,15, __, 21, 28

b. 2, 7, 24, 77, __

[ANS: 238]

c.

[Ans: 15, 35]

3, 8, --, 24, --, 48, 63

d. 4, -5, 11, -14, 22, --e.

85

2, 5, 8 ...


[Ans: -27] Back to Table of Contents



f.

9, 10, 11, 13, 15,?, 21, 28

g.

4, -5, 11, -14, 22, ---

Quants Funda

Blood Relations Syllogism 1. All green are blue 2. All blue blue are white Conclusion: I) Some blue are green II) Some white are green III) Some green are not white IV) All white are blue

Odd man out Select the odd one out. 1. a. Java

b. Lisp

c. Smalltalk

d.Eiffel.

2. a. SMTP

b. WAP

c. SAP

d. ARP

3. a. Oracle

b. Linux

c. Ingress

d. DB2

4. a. WAP

b. HTTP

c. BAAN

d. ARP

5. a. LINUX

b. UNIX

c.SOLARIS

d. SQL SERVER

6. a. SQL

b. DB2

c.SYBASE

d. HTTP

Directions 1. ‘A’ moves 3 km east from his starting point. He then travels 5 km north. From that point he moves 8 km to the east. How far is ‘A’ from his starting point?

Coding – Decoding 1. In a certain format TUBUJPO is coded as STATION. The code of which string is FILTER 2. What is the code formed by reversing the First and second letters, the third and fourth letters and so on. of the word SIMULTANEOUSLY 3. In the word ORGANISATIONAL, if the first and second, third and forth, forth and fifth, fifth and sixth words are interchanged up to the last letter, what would be the tenth letter from right? 4. If TAFJHH is coded as RBEKGI, then RBDJK can be coded as ________ 5. If VXUPLVH is coded as SURMISE, then SHDVD can be coded as ________ If CSBWF is coded as BRAVE, then QMFBTF can be coded as ________ Back to Table of Contents

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Symbol based problems 1. Find the value of @@+25-++@1 Where @ denotes "square" + denotes "square root". 2. Find the value of M(373,5)+ M(373,5)+R(3.4)+T(7. R(3.4)+T(7.7)+R(5.8) 7)+R(5.8) Where M denotes modulus operation R denotes round-off T denotes truncation 3. Find the value of $%$6-%$% $%$6-%$%6 6 Where $ means Tripling % means change of sign

Theory, Examples & Explanations Logical Reasoning TYPES: 1. Prime number Series: Ex.1

2, 3, 5, 7, 11, 13, … (1) 15

(2) 17

(3) 18

(4) 19

Sol.

The given series is prime number series. The next prime number is 17. Answer: (2)

Ex.2

2, 5, 11, 17, 23, …, 41. (1) 29

Sol.

(2) 31

(3) 37

(4) 39

The prime numbers are written alternately. a lternately. Answer: (2)

2. Difference Series: Ex.1

2, 5, 8, 11, 14, 17,…, 23. (1) 19

(2) 21

(3) 20

(4) 18

Sol.

The difference between the numbers is 3. (17 + 3 = 20). Answer: (3)

Ex.2

45, 38, 31, 24, 17,…, 3. (1) 12

Sol.

(2) 14

(3) 10

The difference between the numbers is 7. (17 – 7 = 10) Answer: (3)

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(4) 9





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3. Multiplication Series: Ex.1

2, 6, 18, 54, 162,…, 1458. (1) 274 (2) 486 (3) 1236

Sol.

(4) 1032

The numbers are multiplied by 3 to get next number. (162 × 3 = 486) Answer: (2)

Ex.2

3, 12, 48, 192,…, 3072. (1) 768

Sol.

(2) 384

(3) 2376

(4) 1976

The numbers are multiplied by 4 to get the next number. (192 × 4 = 768) Answer: (1)

4. Division Series: Ex.1

720, 120, 24,…, 2, 1 (1) 12

(2) 18

(3) 20

Sol. Ex.2

Answer: (4) 32, 48, 72, 108,…, 243. (1) 130

Sol.

(4) 6

(2) 162

(3) 192

Number × 3/2 = ne next number. 32 32 × =48, 48

(4) 201 = 72 72, 72 72

= 108, 10 108

= 162

Answer (2) 5. n2 Series: Ex.1

1, 4, 9, 16, 25,…, 49 (1) 28

Sol.

(2) 30 2

2

2

(3) 32 2

(4) 36

2

The series is 1 , 2 , 3 , 4 , 5 …, The next number is 62 = 36. Answer: (4)

Ex.2

0, 4, 16, 36, 64,…, 144. (1) 100

Sol.

(2) 84

(3) 96

(4) 120

The series is 0 2, 22, 42, 62 etc. The next number is 102 = 100. ] Answer: (1)

6. n2 – 1 Series: Ex.1

0, 3, 8, 15, 24, 35, 48,… (1) 60

Sol.

(2) 62

(3) 63

The series is 1 2 – 1, 22 – 1, 32 – 1 etc. The next number is 8 2 – 1 = 63. Answer: (3)

88

(4) 64





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Another Logic: Difference between numbers is 3, 5, 7, 9, 11, 13 etc. The next number is (48 + 15 = 63). 2

7. n + 1 Series: Ex.1

2, 5, 10, 17, 26, 37,…, 65. (1) 50

Sol.

(2) 48

(3) 49

(4) 51

The series is 1 2 + 1, 22 + 1, 32 + 1 etc. The next number is 7 2 + 1 = 50. Answer: (1)

8. n2 + n Series (or) n2 – n Series: Ex.1

2, 6, 12, 20,…, 42. (1) 28

Sol.

(2) 30

(3) 32

(4) 36

The series is 1 2 + 1, 22 + 2, 32 + 3, 42 + 4 etc. The next number = 5 2 + 5 = 30. Answer: (2)

Another Logic: The series is 1 × 2, 2 × 3, 3 × 4, 4 × 5. The next number is 5 × 6 = 30. Another Logic: 2

2

2

2

2

The series is 2 – 2, 3 – 3, 4 – 4, 5 – 5. The next number is 6 – 6 = 30. 9. n3 Series: Ex.1

1, 8, 27, 64, 125, 216,… (1) 256

Sol.

(2) 343

(3) 365

(4) 400

The series is 1 3, 23, 33 etc. The missing number is 7 3 = 343. Answer: (2)

10. n3 + 1 Series: Ex.1

2, 9, 28, 65, 126, 217, 344,… (1) 513

Sol.

(2) 500 3

3

(3) 428 3

(4) 600 3

The series is 1 + 1, 2 + 1, 3 + 1 etc. The missing number is 8 + 1 = 513. Answer: (1)

LETTER SERIES Introduction: In these types of problems a series of the letters of alphabet will be given which follow a pattern or a sequence. The letter series mainly consists of skipping of the letters. To solve these types of problems, assign numbers 1 to 26 to the letters of the alphabet as shown below. In some cases it is useful to assign the numbers in the reverse order.

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1 2

3

4

5

6

7

8

9

10

11

12

13

A B

C

D

E

F

G

H

I

J

K

L

M

Z Y

X

W

V

U

T

S

R

Q

P

O

N

26 25

24

23

22

21

20

19

18

17

16

15

14

Here the table is showing both forward as well as reverse place value of any alphabet. A very important fa ct about the position of any alphabet is that both the sum of forward position and reverse position for any alphabet is always constant and equal to 27. Such as Sum of both positions of H is (8 + 19 = 27) or for W is (23 + 4 = 27). We can also remember the relative positions of these alphabets by just remembering the word EJOTY. Letters

E

J

O

T

Y

Position

5th

10th

15th

20th

25th

Just remember the word EJOTY and its values i.e. 5, 10, 15, 20, 25 e.g. If you are asked to complete the series F, K, P, U, __ Then from EJOTY, you know that values of F = 6, K = 11, P = 16, U = 21 i.e. difference is 5, so the answer should be 21 + 5 = 26 i.e. Z Various types of letter series are given below. TYPE – 1 One Letter Series: Ex.1

A, C, E, G, I, … (1) J

Sol.

(2) K

(3) L

(4) M

The series is (+ 2). i.e., A + 2 = C; C + 2 = E; E + 2 = G;

G + 2 = I.

The missing letter is I + 2 = K. Answer: (2) Another Logic: Skip one letter is I + 2 = K. After I skip J to get K; the missing letter is K. Note: "Skip" process saves time. Ex.2

A, B, D, G, K, ...…

(1) P Sol.

(2) N

(3) O

(4) L

The series is + 1, + 2, + 3 etc. The missing letter is (K + 5) = P. Answer: (1) Back to Table of Contents

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Skip Process: First no letter is skipped, then 1, 2, 3 etc. letters are skipped to get next letter. Skip 4 letters after 'K' to get P. Ex.3

B, E, H, K, N,… (1) P

Sol.

(2) O

(3) Q

(4) R

The series is + 3. The missing letter is N + 3 = Q. Answer: (3) Another Logic: Skip two letters to get the next letter. Skip Q, P after N to get Q. The missing letter is Q.

Ex.4

B, D, G, I, L, N,… (1) N

Sol.

(2) O

(3) P

(4) Q

The series is alternately + 2 and + 3. The missing letter is N + 3 = Q. Answer: (4) Another Logic: Skip one and two letters alternately to get the next letter. Skip two letters O, P a fter N to get Q.

Ex.5

B, C, E, G, K,… (1) M

Sol.

(2) N

(3) O

(4) P

If numbers are assigned, the series becomes prime number series. The next prime number is 13 and the corresponding letter is M. Answer: (1)

Ex.6

A, E, I, O,… (1) Q

Sol.

(2) R

(3) U

(4) S

(3) X

(4) Y

The series is a series of Vowels. Answer: (3)

Ex.7

A, D, I, P,… (1) U

Sol.

(2) V

If numbers are assigned, the series becomes square series. The next number is 5 2 = 25 and the corresponding letter is Y. Answer: (4)

Ex.8

D, F, H, I, J, L,… (1) M

Sol.

91

(2) N

(3) O

(4) P

If numbers are assigned, the series becomes composite number series.





Quants Funda

The next composite number is 14 and a nd the corresponding letter is N. Answer: (2) Ex.9

A, Z, B, Y, C, X, D,… (1) U

Sol.

(2) V

(3) W

(4) X

The sequence consists of two series A, B, C, D etc., and Z, Y, X, W etc. Answer: (3)

TYPE – 2 Two Letter Series: The first letters of the series follow one logic and the second letters follow another logic. Also, the first two letters, the next two letters and so on follow a logic. Ex.1

AM, BN, CO, DP, EQ,… (1) FG

Sol.

(2) FR

(3) GR

(4) ER

The first letters are A, B, C, D, E, F and the second letters are M, N, O, P, Q and R. Answer: (2)

Ex.2

AB, DE, GH, JK, JK, MN,… (1) OP

Sol.

(2) NO

(3) PQ

(4) RS

After every set of letters one letter is skipped. Skip O to get next two letters PQ. Answer: (3)

Ex.3

AA, CE, EI, GO,… (1) IU

Sol.

(2) IQ

(3) IR

(4) IT

The first letters follow a sequence of A, C, E, G, I. (+ 2 series) and the second letters are vowels. Answer: (1)

TYPE – 3 Three Letter Series: This sequence consists of 3 letters in each term. The first letters follow one logic, the second letters follow another logic and the third letters follow some other logic, (or the same logic in all the three cases) Ex.1

ABD, CDF, EFH, GHJ,… (1) IJK

Sol.

(2) IJL

(3) HIJ

(4) HIK

The first letters follow a sequence of A, C, E, G, I etc. The second letters follow a sequence of B, D, F, H, J etc. And the third letters form a sequence of D, F, H, J, L etc. Answer: (2)

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Ex.2

CKZ, DLY, EMX, FNW,… (1) GOV

Sol.

Quants Funda

(2) GOU

(3) GNU

(4) GNV

The first letters form a series of C, D, E, F, G etc. The second letters form a series of K, L, M, N, O etc, and the third letters letters form a series series of Z, Y, X, W, V etc. Answer: (1)

Ex.3

MAB, NEC, OIE, POG,… (1) QPH

Sol.

(2) QUH

(3) QUI

(4) QUK

The first letters letters form a series of M, N, O, P, P, Q etc. The second letters form Vowels; the third letters

form prime

number series (if numbers are assigned to letters). Answer: (4) Ex.4

ABC, CBA, DEF, FED, GHI, ..… (1) JKL

Sol.

(2) IHG

(3) DFE

(4) IJK

The second term is the reverse order of first term. In addition to the above types a number of other types can also be identified. Answer: (2)

CODING / DECODING Introduction: For conveying secret messages from one place to another, especially in Defence Services, coding is used. The codes are based on various principles/patterns such that the message can be easily be deciphered at the other end. Now-a-days, in certain competitive examinations, such questions are given to judge the candidates’ intelligence and mental ability. They are required to encode and decode words and sentences after observing the pattern and principles involved. These questions can be broadly classified into 5 main categories, as follows: (i) Coding with Letters of Alphabets (ii) Coding with Numerical Digits (Numbers) (iii) Mixed Coding (Both Alphabetical and Numerical) (iv) Coding with Arbitrary Signs / Symbols (v) Miscellaneous Type TYPE – 1 Coding with Letters of Alphabet: In these questions, the letters of the alphabets are exclusively used. These letters do not stand for themselves but are allotted some artificial values based on some logical patterns/analogies. By applying those principles or observing the pattern Back to Table of Contents

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involved, the candidates are required to decode a coded word or encode a word. These can be further classified into the following categories : Simple Analogical Letter Coding: These are also called arbitrary codes. There are 2 definite principles/pattern involved. Codes are based on the analogy of one example from which different codes are to be formed. Ex.1

If NETWORK is coded as O P C T R S Q, how is CROPS written in that code; is written in actual code?

Sol.

N=O

C=T

E=P

R=O

T=C

then

O=N

W=T

P=E

O=R

S=R

R=S K = Q Hence CROPS can be coded as TONER. Ex.2

Sol.

Ex.3

The code ‘TABLESTESF’ stands for the word ‘BELONGINGS’ how will you code the following : (1) LONG

(2) ON

(3) GIN

(4) SONG

(5) NO

(6) SING

(7) SINGS

(8) GONE

(9) IS

(10) GO

The coding is done as follows: (1) BLES

(2) LE

(3) STE

(4) FLES

(5) EL

(6) FTES

(7) FTESF

(8) SLEA

(9) TF

(10) SL

If INLAND is coded as BSTRSI, make codes of the following letters. (1) IN

Sol.

(2) TRSI

(6) LAID

(3) BSS

(4) RSI

(5) RS

(6) TRBI

(2) RATE

(3) SEAT

(4) NOT

(5) NOTE

(6) BEST

(2) HNTG

(3) FGNT

(4) OST

(5) OSTG

(6) WGFT

(5) SOM

(6) INMT

If PROMISED is coded as RMNIOSTD, decode the following codes. (1) RNST

Sol.

(5) AN

The coding is as follows: (1) HNTISO

Ex.5

(4) AND

If EWFGHONTISO stands for OBSERVATION, code the following letters. (1) RATION

Sol.

(3) INN

The coding is done as follows: (1) BS

Ex.4

(2) LAND

(2) MNIT

(3) DOI

(4) RMNST

The decoding is as follows: Back to Table of Contents

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(1) POSE Ex.6

Quants Funda

(2) ROME

(3) DIM

(4) PROSE

(5) SIR

(6) MORE

Column A contains certain words numbered numbered from (1) to (6). Column Column B goes with the codes for column A, but with different order. You have to match the words of column A with their respective coded word in column B. The pattern of coding used here is BLADES = CMBEFT.

Sol.

Column (A)

Column (B)

(1) BASE

(1) CBE

(2) BALE

(2) CBTF

(3) SALE

(3) CFE

(4) SAD

(4) CBMF

(5) BAD

(5) TBE

(6) BED

(6) TBMF

A (1) B (2),

A (2) B (4),

A (3) B (6),

A (4) B (5),

A (5) B (1),

A (6) B (3).

Letter Coding on Specific Pattern: In such questions, letters of alphabets are no doubt allotted artificial values but based on certain specific pattern/principles. The candidates are required first to observe the specific pattern involved and then proceed with coding or decoding; as the case may be. Ex.1

If POSTED is coded as DETSOP, how will be word SPEED be coded?

Sol.

A careful observation of the above example will reveal that letters of the first word have been

Ex.2

If GREET is coded as FQDDS, decode the following codes: (1) KDS

Sol.

(2) SNQD

(3) CNBI

(4) CDDO

(5) ONS

reversed

(6) ANRR

Here, each letter is allotted the value of its preceding preceding letter in the sequence; sequence; the pattern of coding used here is B = A, C = B. Based on this pattern, the answers to the above questions will be follows: (1) LET

Ex.3

(2) TORE

(3) DOCK

(4) DEEP

(5) POT

(6) BOSS

If A = E, how will you code the following words. (1) BLACK

(2) ACT

(3) BAT

(4) CADRE

(5) LOOT

(6) FOOL

Sol.

(1) FPEGO

(2) EGX

(3) FEX

(4) GEHVI

(5) PSSX

(6) JSSP

Ex.4

If “CAT” is coded as “DEBCUV”, how will you code “RACE”.

Sol.

The pattern of coding is such that each letter has been allotted value of 2 letters following the sequence, sequence, i.e. A = BC, B = CD, C = DE, etc. Hence, the word RACE will be coded as “STBCDEFG” Based on the above principle, try to code the following. Back to Table of Contents

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(1) FATHER Sol.

Ex.5

(2) DATED

Quants Funda

(3) LATE

(4) FAKES

(5) MAIN

(1) G H B C U V F G S T

(2) E F B C U V F G E F

(3) M N B C U V F G

(4) G H B C L M F G TU

(5) N O B C J K O P

(6) Q R M N B C O P F G

(6) PLANE

Column (A) contains coded words and column (B) contains equivalent decoded words given in a different serial order. Match the words of the column (A) with column (B) and indicate the first and last letters of the coded word in column A from the answer choices. The pattern of coding is Q = P, S = R, U = T, etc. Column (1)

Column (B)

(1) USJN

(1) WORK

(2) CPOF

(2) SHORT

(3) MPPU

(3) FEET

(4) GFFU

(4) LOOT

(5) TIPSU

(5) BONE

(6) XPSL

(6) TRIM

Sol.

A (1) B (6),

Ex.6

If “EGHJKMKM” is the code for “FILL”, how will you code the following :

Sol.

A (2) B (5),

A (3) B (4),

A (4) B (3),

(1) QSDFRTSU

(2) SUDFKMKM

(3) EGDFDFKM

(4) CENPDFRT

(5) KMNPRTSU

(6) ACDFCERT

A (5) B (2),

A (6) B (1).

The pattern of coding is such that the sequence follows the letters in between, each pair of letters in the code. Pattern is AC = B, BD = C, CE = D, etc. (1) REST

(2) TELL

(3) FEEL

(4) DOES

(5) LOST

(6) BEDS

TYPE – 2 Coding with Numerical Digits: The pattern of coding with numerical digits is similar to that of coding with alphabets except the use of numerical digits with the assignment of some artificial va lues. The values are allotted based on some specific pattern which has to be discerned by the candidate in order to solve the problem in the quickest possible time. If TRAIN is coded as 23456, how will you code TIN and RAIN? The answer will be 256 for TIN and 3456 for RAIN. T = 2, R = 3, A = 4, I = 5, and N = 6. These values have been allotted arbitrary; based on logical relationship; the candidates will be able to solve the problem.


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Analogical Coding with Numerical Digits: Analogical coding with numerical digits involves the method of coding where the letters of alphabets are allotted numerical values and the pattern of coding is based on the analogy of the example given in the question. There are no set of principles or patterns involved. Candidates are required to study the examples given before getting started with the exercise. Ex.1

If SELDOM is coded as “1 2 4 3 6 5”, how will you code the following words? (A) DOES

(B) SOLE

(C) LED

(D) DOLE

(E) LODE

(F) ODE

Choices:

Sol.

(A)

(1) 3 6 2 1

(2) 6 2 3 1

(3) 1 6 3 2

(4) 6 2 1 3

(B)

(1) 1 4 6 2

(2) 1 6 4 2

(3) 1 4 2 6

(4) 1 6 2 4

(C)

(1) 4 3 2

(2) 3 2 4

(3) 4 2 3

(4) 4 2 6

(D)

(1) 3 6 4 1

(2) 3 6 4 2

(3) 3 6 2 4

(4) 3 6 4 1

(E)

(1) 4 6 2 3

(2) 4 6 3 2

(3) 6 3 2 4

(4) 4 3 6 2

(F)

(1) 6 2 3

(2) 2 3 6

(3) 6 3 2

(4) 6 3 4

If SELDOM stand for code 124365 which means S = 1, E = 2, L = 4, D = 3, O = 6, and M = 5. Based on this analogy, the correct answer will be, (A) (1)

Ex.2

(B) (2)

(C) (3)

(D) (2)

(E) (2)

(F) (3)

If “1 3 4 8 2 6 7 5 9” is the code for “O B S E R V A N T” how will you code the following words? (A) SERVANT

(B) SOBER

(C) BENT

(D) OVATE

(E) ORATE

(F) NOTES

Choices:

Sol.

(A)

(1) 4 8 2 6 7 5 9 (2) 4 8 2 6 7 6 0 (3) 4 8 2 6 7 5 0 (4) 4 2 8 6 7 5 9

(B)

(1) 4 1 3 8 2

(2) 4 1 3 8 1

(3) 4 3 1 8 2

(4) 4 1 3 2 8

(C)

(1) 3 8 9 5

(2) 3 8 5 9

(3) 3 5 8 9

(4) 9 8 3 5

(D)

(1) 1 7 6 9 8

(2) 1 7 6 8 9

(3) 1 6 7 9 8

(4) 9 8 7 6 1

(E)

(1) 1 7 2 9 8

(2) 1 2 7 9 8

(3) 1 2 7 8 9

(4) 8 9 2 7 1

(F)

(1) 9 1 5 8 4

(2) 5 1 9 8 4

(3) 5 9 1 8 4

(4) 8 4 9 1 5

In the example, you’ll observe O = 1, B = 3, S = 4, E = 8 R = 2, V = 6, A = 7, N = 5 and T = 9. Based on this analogy the correct answers will be : (A) (1)

Ex.3

Sol.

B) (1)

(C) (2)

(D) (3)

(E) (2)

(F) (2)

The code 6 7 4 5 3 2 7, stands for B E C A U S E. Decode the following codes (1) 4 5 3 2 7

(2) 6 5 2 7

(3) 4 5 2 7

(4) 6 7 7

(5) 4 7 5 2 7

(6) 3 2 7

(1) CAUSE

(2) BASE

(3) CASE

(4) BEE

(5) CEASE

(6) USE Back to Table of Contents

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Ex.4

Sol.

Quants Funda

If 4 0 6 5 4 2 5 7 is the code for S T A N D I N G, decode the following codes. (1) 4 0 6 7

(2) 6 5 4

(3) 4 0 6 2 5

(4) 42 5 2 5 7

(5) 4 6 5 4

(6) 4 0 2 5 7

(1) STAG

(2) AND

(3) STAIN

(4) DINING

(5) SAND

(6) STING

Coding with Specific Pattern: This is the pattern of coding which exhibits the natural correlation of Arabic numbers with alphabetic letters. For instance, alphabets A to Z are assigned the numeric codes from 1 to 26 where each letter gets the assignment in the pattern as follow A = 1, B = 2, C = 3, etc. The sequence is classified as follows : Forward sequence (e.g. A = 1, B = 2, etc.) Backward sequence (e.g. Z = 1, Y = 2, A = 26, etc.) Random Sequence (e.g. A = 2, B = 3 or A = 4, B = 6, C = 8 or any other pattern following a particular sequence). Forward Sequence: Ex.

If ‘PACE’ is code as 16-1-3-5, 16 -1-3-5, how will you code the following : (1) ACTED

Sol.

(2) BAIL

(1) 1-3-20-5-4

(3) RACE (2) 2-1-9-12

(4) FRAME

(5) GLAD

(3) 18-1-3-5

(6) GAIN

(4) 6-18-1-13-5 (5) 7-12-1-4

(6) 7-1-9-14 Backward Sequence: Ex.

If GREAT is coded as 20-9-22-26-7, how will you code the following words : (1) FATE

Sol.

(2) DATE

(1) 21-26-7-22 (2) 23-26-7-22

(3) MATE

(4) RATE

(5) GATE

(3) 14-26-7-22

(4) 9-26-7-22

(5) 20-26-7-22

Random sequence: The sequence will not follow a specific pattern of assignment as in other cases but will surely show a pattern at a strict analysis. The pattern can be established by various ways but in every case a set principle/pattern is involved which has to be discovered by careful examination of the example given in the question. Ex.1

If FRANCE is coded 9-21-4-6-8, code the following words after discerning the principle/pattern involved in this example : (1) INDIA

(2) CANADA

(3) GERMANY

(4) NEPAL

(5) PERU

(6) KENYA

Sol.

The pattern of assignment is read as given in the following table.

A B

C

D

E

F

G

H

I

J

K

L

M

N

O

P

4 5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

Q R

S

T

U

V

W

X

Y

Z

20 21

22

23

24

25

26

1

2

3

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Based on above pattern, the answers will be follows :

Ex.2

(1) 12-17-7-12-4

(2) 6-4-17-4-7-4

(5) 19-8-21-24

(6) 14-8-17-2-4

(3) 10-8-21-16-4-1 10-8-21-16-4-17-2 7-2

(4) 17-8-19-4-15

If BREAD is coded as “2-18-0-0“2 -18-0-0-4”, 4”, how will you code the following? (1) COOL

(2) COME

(3) BROOM

(4) GROOM

(5) SHEETAL

(6) CREAM

A strict analysis of the question reads that the vowels ‘E’ & ‘A’ are assigned the code ‘0’. The rest of the letters follow the regular sequence of numerical assignment, i.e. B = 2, C = 3, etc. Based on the answers are follows: Remaining letters of alphabet will follow the same order, Sol.

Ex.3

(1) 3-0-0-12

(2) 3-0-13-0

(3) 2-18-0-0-13

(4) 7-18-0-0-13

(5) 19-8-0-0-20-0 19-8-0-0-20-0-12 -12

(6) 3-18-0-0-13

above pattern, the

i.e. B = 2, C = 3, D = 4 etc.

If 6 – 6 – 12 12 – – 1 – 19 19 – – 8 = FLASH and 6 – 6 – 15 15 – – 15 15 – – 12 12 – – 9 – 19 19 – – 8 = FOOLISH, find the sum with all the letters put together. (A) LATE

(B) MAKE

(C) ICED

(D) ACT

(E) FACT

(F) LAND

Choices: (A)

(1) 38

(2) 59

(3) 56

(4) 58

(B)

(1) 32

(2) 30

(3) 34

(4) 36

(C)

(1) 22

(2) 23

(3) 21

(4) 24

(D)

(1) 23

(2) 22

(3) 20

(4) 24

(E)

(1) 41

(2) 51

(3) 21

(4) 30

(F)

(1) 42

(2) 29

(3) 31

(4) 30

Sol. (A) LATE = 12 + 1 + 20 + 5 = 38, hence the answer is (1). (B) MAKE = 13 + 1 + 11 + 5 = 30 hence the answer is (2). (C) ICED = 9 + 3 + 5 + 4 = 21 hence the answer is (3). (D) ACT = 1 + 3 + 20 = 24 hence the answer is (4). (E) FACT = 6 + 1 + 3 + 20 = 30 hence the answer is (4). (F) LAND = 12 + 1 + 14 + 4 = 31 hence the answer is (3). TYPE – 3 Mixed Coding (Letters + Digits): Mixed coding takes the pattern of coding with both the letters of alphabets and numerical assignment. The candidates are required to study the analogy given in question. Back to Table of Contents

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Ex.1

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If “A – 3 – T – 5 – D” stands for ACTED and “D1T5D” stands for “DATED”, how will you code the following (1) FADED

(2) LOCATE

(3) BAILED

(4) FAILED

(5) PRESS

(2) L15C1T5

(3) B 1 I 12E4

(4) F1I12E4

(5) P 18E 19S

(2) A 3 I 4

(3) 5A20I14G

(4) B 1 I 12

(5) K 9L 12

(2) ACID

(3) EATING

(4) BAIL

(6) DREAM Ans.

(1) F1D5D (6) D 18 E 1 M

Ex.2

Decode the following: (1) R 9 L 5 D (6) B 1 I 12E4

Ans.

(1) RILED

(5) KILL

(6) BAILED Ex.3

Ans.

Decode the following: (1) F 1 D 5 D

(2) A 9 D 9 N 7

(3) R 5 Q 21 I 18 E 4

(4) D 5 A 12 I 14 G

(5) O 2 S 5 R 22 E

(6) A 3 I 4

(1) F A D E D

(2) A I D I N G

(3) R E Q U I R E D

(4) D E A L I N G

(5) O B S E R V E

(6) A C I D

TYPE – 4 Miscellaneous Types Decoding the Rule Applied: This part of coding test required a careful examination of rules followed to code a certain word. Only after the analysis of the pattern applied in coding, you can decode them. Example: Study the five different ways of coding marked (1), (2), (3), (4) & (5). A specific rule has been applied to codify each of them. Can you find out the rule of coding applied in the question that follows; (1) N C E F R A (2) F A C R N E FRANCE

(3) E C N A R F (4) A C E F N R (5) F E R C A N

100

WORD

CODE

1. C A N A D A

CNDAAA

2. K E N Y A

KAEYN

3. N A T I O N S

SNOITAN

4. V A N D A N A

VNAAADN





Quants Funda

5. V A R D H M A N

NAMHDRAV

6. V A R I O U S

AIORUSV

7. C A R E E R

EERCAR

8. P O P U L A T I O N

PNOOPIUTLA

9. M E D I C I N E

MDCNEIIE

10. A P T I T U D E

ADEIPTTU

Answers: 1. (2)

2. (5)

3. (3)

4. (2)

5. (3)

6. (4)

7. (1)

8. (5)

9. (2)

10. (4)

Contrasting and Marking Comparisons: A set of words are given in column I and codes have been formed in column II. Here in such questions some alphabets/letter are underline in column I and a nd the corresponding codes in column II has been jumbled up thus making the question more difficult to correspond. To find the formula to decode these type of question some logical rule/principle is found by comparing or making contracts in all the questions. An example has been given below: Example In the following question the capital letters in column I are codified in small letters in column II. The small letters are not arranged in the same order on the capital letters. Study the column (I) and (II) together and determine the small letters for the corresponding underlined capital letter in column (I).

Keys:

Column (I)

Column (II)

1. D I G I T

wbzbm

2. T I G E R

mbzxk

3. F E V E R

xkyox

4. G I T A R

mtzbk

5. L I V E R

bexok

1. w

2. m

3. y

4. z

5. e

Explanation If we compare question (1) & (2) we find that there are 3 alphabets (T, I, G) common and there corresponding small letters will be (m, z, b) though not in the same order. This leaves us with (D and R ) with small alphabets (w and k). Therefore, we have now, Either ‘w or k’ is D’s code Now, if we taken (2) and (3), we find that ‘w’ is not present is column II of either (2) or (3) and D is not there in column II of either (2) & (3) the or conclude that D = w and therefore R = k. Now, carrying on with this finding, we see in question (3) and (5) there are two common elements in column I, V, E & R. Back to Table of Contents

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Quants Funda

Since E comes twice in (3), therefore code for E = x which leads to V = 0 and F = y in question is (1), I comes twice, this leads to I = b. So we are a re left with ‘T’ and ‘G’, which are either ‘z’ or ‘m’. Now, we cannot conclude anything more from these clues, but can fit in above observation to see what relation capital letters have with small letters.

Therefore, G = z and T = m Mathematical/Algebraic Operations

The code is always the sum of letters with the assignment of numbers put in the regular order. The order reads either in a forward sequence or a backward sequence. Consider the table given below. Ex.1

If DOLLY is 68, then how much will be SEEMA? (1) 65

Sol.

Ex.2

(3) 43

(4) 33

The coding is the sum of forward sequence of alphabets DOLLY

4 + 15 + 12 + 12 + 25 = 68

SEEMA

19 + 5 + 5 + 13 +1 = 43, hence the answer is (3).

If NEERAJ is 109, then how much will be SHEETAL? (1) 119

Sol.

(2) 86

(2) 98

(3) 125

(4) 100

The coding is the sum of backward sequence of alphabets: NEERAJ SHEETAL

13 + 22 + 22 + 9 + 26 + 15 = 109 8 + 19 + 22 22 + 22 + 7 + 26 + 15 15 = 119, hence the answer is (1).

BLOOD RELATIONS Introduction: In these tests the success of a candidate depends upon the knowledge of the blood relations, some of which are summarized below to help solve these tests. Mother's or father's son: Brother

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Mother's or father's daughter: daughter : Sister Mother's or father's brother: brother: Uncle Mother's or father's sister: sister : Aunt Mother's or father's father: Grandfather Mother's or father's mother: Grandmother Brother's son: Nephew Brother's daughter: Niece Uncle or aunt's son or daughter: Cousin Sister's husband: Brother – in – law Brother's wife: Sister in – law Son's wife: Daughter – in – law Daughter's husband: Son – in – law Husband's or wife's sister: Sister – in – law Husband's or wife's brother: Brother – in – law EXAMPLE: Directions for questions 1 – 3: Read the following information to answer the given questions. (i) A, B, C, D, E and F are six family members. m embers. (ii) There is one Doctor, one Lawyer, one Engineer, one Pilot, one Student and one Housewife. (iii) There are two married couples in the family. (iv) F, who is a lawyer, is father fa ther of A. (v) B is a pilot and a nd is mother of C. (vi) D is grandmother of C and is a Housewife (vii) E is father of F and is a Doctor. (viii) C is brother of A. 1. Which of the following statements is definitely true? (1) C is the brother of the Student

(2) F is the father of the Engineer

(3) A is the Engineer

(4) E is the father of the Pilot

2. How many female members are there in the family? (1) Two only

(2) Three only

(3) Three or Four

(4) Two or Three

3. How is A related to D? Back to Table of Contents

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(1) Grandson

(2) Granddaughter

Quants Funda

(3) Either granddaughter or grandson

(4) grandmother

Solutions 1 – 3: 1 Either A or C is Engineer, F is father fa ther of both A and C. Answer: (2) 2. Two or three; B and D (sex of A is not known). Answer: (4) 3. Either granddaughter or grandson. Answer: (3)

LOGICAL DIAGRAMS Introduction: There are three types of logical diagram tests generally asked by the examiner. These are explained below. TYPE – 1 In Type 1, questions are based on the concept of class. A class is a group or collection of objects, all having something in common. For example, a class of females will include all daughters and nieces in a group. There are three possible

relationships between any two different classes. 1. Class Containing Classes: All those females that fall into class of nieces are contained in class of females. The idea that one class may contain another is the most Fundamental logical principle underlying the logical diagram questions. Females

Nieces 2. Class Partially Containing Other Classes: Consider two classes, doctors and females. Since not all doctors are females, no class of doctors can entirely contain the class of females. The partial containment of one class by another class can be picturised in the following way: Doctors

Females

A

B

C

Now, in the above figure, the two-joined circles indicate that there are three classes.

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1. Those who are lady doctors. (B) 2. Those who are doctors, but not females. (A) 3. Those who are females, but not doctors. (C) 3. Classes Independent of Each Other The classes of all males and all females exclude each other, since no female can come into the class of males and also no male can be included into the class of females. In actual logical diagram tests, you will be working with three circles rather than two, with no new principles of relationships between classes. For example, let us take three different classes, Females, Doctors, Girls

Males

Females

Now, each of these classes will stand in one of the three types of relationships to the other, i.e. three different two-circle diagrams for females-girls, doctors-females, and doctors-girls. Instead of three different diagrams, you can represent all of these relationships by intersecting the three circles as in the figure below. Females

Doctors

Girls It will be helpful to familiarize yourself with the various patterns of three-circle relationships. relationships. Here are the seven most common patterns. Though more than seven patterns are possible, these are the patterns based on which questions are frequently asked. Diagram I 1. Insects, Butterflies, Mosquitoes 2. Males, Fathers, Boys

Diagram II 1. Females, Mothers, Lady Doctors 2. People, Players, Asians

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Diagram III 1. Males, Bachelors, Teachers 2. Females, Doctors, Girls

Diagram IV insects, Mosquitoes, Grapes 1. Mangoes, Fruits, Wheat 2. Men, Fathers, Girls

Diagram V 1. Hair, Nail, Teeth 2. Dogs, Cats, Birds

Diagram VI Entrances, Gates, Barriers 1. Aunts, Congressman, Fathers

Diagram VII Cousins, Sisters, Nieces 1. Teachers, Students, Professionals

Based on these principles, a variety of questions can be asked. TYPE – 2 In this type of questions, there are two sets of principles: 1. One to choose the figure that represents the logical relationship among the items (in the figure); 2. Another to find and measure the portion that represents a particular statement.


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Example: In the adjoining figure, and

represents citizens of Delhi,

represents males,

represents educated

represents unemployed. Find out which of the numbers denote the following:

1. Educated female outside Delhi citizens who are also employed. 2. Uneducated unemployed males who are citizens of Delhi. Answers: (1) 2 (2) 7 TYPE – 3 In this type of logical diagram question, you’ll be able to see through reasoning yourself and deduce the right answer. These are the “Logical Reasoning” questions falling under “Logical Diagrams” type questions; these reasoning questions become easier to understand if we solve them by making use of methods/ tips of logical diagram. Here you’ll be provided with a set of given statements which will be followed by another set of deductions / conclusions. The conclusions are supposed to follow from the question statements, and the statements (or assumptions) are to be taken as true even if they seem to be at a variance with commonly known facts or universal truths. The candidates are required to practice such questions and find out by themselves whether answering/solving such type of questions is easier by using the techniques of “Logical Reasoning” or “Logical Diagram” questions. Example: Statements: 1. Some Indians are Muslims. 2. Some Pakistanis are Muslims. Conclusions: (i) Every Muslim is either an Indian or a Pakistani. (ii) Some Muslims are Indians as well as Pakistanis. (iii) No Muslim is an Indian as well as a Pakistani. (iv) Some Muslims are neither Indians nor Pakistanis. Now, you have to choose your answer from one of the following alternatives (1) Only II follows

(2) Only III follows

(3) Either II or III follows (4) Only I, II and III follow

Answer: (3) Back to Table of Contents

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The following diagrams give a clear idea of what can be concluded from the above statements. Indians

Muslims

Pakistanis

Indians

Muslims

Fig (i)

Pakistanis

Fig (ii) Now, fig (i) infers that there are some Muslims who are Indians and are some who are Pakistanis. Also, there are

Muslims who are neither Indian nor Pakistani and fig (ii) denotes denotes – – there are some who are Pakistani, some who are both Indian and Pakistani; which can not be possible so this figure is not a proper explanation of the statements. So only figure (i) is true and statement 3 is true among all the statement. Answer: (3) MATHEMATICAL OPERATIONS These types of problems are common to be asked in good competitive examinations. In these some mathematical operations are inter – inter – changed among themselves such as if divide (÷) denotes multiplication (×), Greater to (>) denotes ()” etc are the type of statements given and on the basis of those statements we have to solve a given problem. Let us understand this in much wider concept with the help of an example. Ex.

If ‘P’ denotes ‘divided by’; ‘Q’ denotes ‘added to’; ‘M’ denotes ‘subtracted from’; ‘B’ denotes ‘multiplied by’; then 1 8 B 1 2 P 4 M 8 Q 6 =? (1) 108

Sol.

(2) 46

(3) 17

(4) 52

(5) None of these

18B12P 4M8Q6 According to the given information put the signs assigned for each alphabet we get; = 18 × 12 ÷ 4 – 8 + 6 Now applying the concept of BODMAS to solve the above expression we get = 18 x 3 – 3 – 8 + 6 = 54 – 54 – 2 = 52 Answer: (4)

Ex:

Some symbols have been given different meaning. Read them correctly carefully and find out the correct one out of the four alternatives

SIGNS > Stands for ÷ v Stands for × < Stands for + ^ Stands for – for –

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+ Stands for = – Stands for > × Stands for < (1) 5 v 4 < 2 + 10 v 2 < 2

(2) 8 v 4 – 2 + 5 > 7 ^ 6

(3) 8 v 6 – 3 + 4 > 7 ^ 6

(4) 9 v 3 – 3 – 1 + 6 > 8 ^ 9 Sol.

5 v 4 < 2 + 10 v 2 < 2 gives 5 x 4 + 2 = 10 x 2 + 2 22 = 22 Answer: (1)

EXAMPLE: Directions for questions 1 – 5: In these questions, the symbols @, *, $, # and % are used with the following meanings as illustrated below: ‘P @ Q’ means ‘P is neither greater than nor equal to Q’. ‘P # Q’ means ‘P is not smaller than Q’. ‘P  Q’ means ‘P is not greater than Q’. ‘P % Q’ means ‘P is neither smaller than nor greater than Q’. ‘P $ Q’ means ‘P is neither smaller than nor equal to Q’. Now in each of the questions given below, assuming the given statements to be true, find which of the two conclusions I and II given below is/are definitely true? Give your answer as – as – (1) if only Conclusion I is true. (2) if only Conclusion II is true. (3) if only Conclusion I or II is true. (4) if neither Conclusion I nor II is true. (5) if both the Conclusion I and II are true. 1. Statements: T % B, M * B, J # B Conclusions: I. T & M

II. T % J

2. Statements: V # D, D * K, F $ K Conclusions: I. D @ F

II. V @ F

3. Statements: W # D, D % M, M * F Conclusions:

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I. D @ F

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II. F % D

4 Statements: H * R, R $ N, N @, K Conclusions: I. H @ K

II. K $ R

5 Statements: M * J, K $ J, K @ T Conclusions: I. T $ M

II. J @ T

Solutions 1 – 5: 1. T = B, M ≤ Q, J ≥ B ⇒ T ≤ J, M ≤ Q Q Answer: Answer: (4) 2. V ≥ D, D ≤ K, F > K ⇒

D ≥ K < F ⇒ F ⇒ D ≥ F ⇒ D @ F Answer: (1)

3 W ≥ D, D = M, M ≤ F ⇒

D ≤ F ⇒ D L F or D = F

⇒

D ≥ F or D = F

⇒

D @ F or D = F Answer: (3)

4. Answer: (4) 5 M ≤ J, K > J, K < T ⇒

M≤J
⇒

T>M⇒T≤M⇒T$M

Also J < T ⇒ J ≥ T ⇒ J @ T Answer: (5)

VISUAL REASONING Instructions: These are problems that are in the form of figures, drawings and designs. The problems may be in the form of series, analogies, classification, cube turning, turning, mirror image, paper folding, paper cutting, completion of incomplete pattern, figure perception, spotting the hidden designs or construction of square. Analogies: In these questions, there are two sets of figures viz. the problem figures and the answer figures. The problem figures are presented in two units. The first unit contains a pair of related figures and the second unit contains one figure and a question mark in place of the fourth figure. You have to establish a similar relationship between two figures and point out which one of the answer figures should be in place of the question mark.

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Problem Figures

?

Answer Figures

Consider the above problem figures. The second figure is related to the first figure in a certain way. That is the elements in the second figure are double the elements in the first figure. The first figure has one square and the second has two squares. The third and fourth figures should also have the same relationship as the first and second have. That means that the fourth figure should have two circles. Problem Figures

?

Answer Figures

Classification: In classification the problem figures themselves are the answer figures. Out of the five given figures tour are similar in a certain way. One figure is not like the other four. You have to identify the “odd man out”.

In the figures given below, of the five figures tour are straight lines whereas one is a circle. Thus the circle is the “odd Man out”. Series: The four figures given at the left are the problem figures. The next five are the answer figures. The problem figures make up a series. That means they change from left to right in a specific order. If the figures continue to change in the same Back to Table of Contents

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order what would the fifth figure be? In the example below, the line across the problem figures is falling down. Thus if the line continues to fall its fifth position would be lying flat i.e. it will be horizontal. Therefore the answer is (4).

Figure perception: In this type of problems, we have to count number of figures hidden in the given design

For example: 1.

The number of squares in the given figure is (1) 12

2.

(3) 14

(4) 15

The number of rectangles excluding squares in the above figure is (1) 12

3.

(2) 10

(2) 13

(3) 14

(4) 17

The number of triangles in the figure is (1) 54

(2) 48

(3) 69

(4) 70

Upon studying the figure one can easily state that the answer to the first question is (3), that to the second question is (4) and to the third question is (3). Cube turning: In this type of problems we have to deal with different positions of the same cube. For example: The drawing on the left in each of the following figures represents a cube. There is a different design on each of the six faces of the cube. Four other drawings of the cube are lettered (1), (2), (3) and (4). Point out which one of the four could be the cube on the left turned to a different position. The cube on the left may have been turned over or around or over and around.

After studying all the choices, one can infer that the answer is (3). Back to Table of Contents

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Problems on Dice: Sometimes we are given figures showing the same die in various positions. After looking at these figures, we have to find the number opposite a given number on the die. The procedure for finding such a number will be clear from the example given below. Two positions of a block are given below. When one is at the top, which number will be at the bottom?

(1) 3

(2) 6

(3) 2

(4) 1

(5) 4

In both the figures 2 is at the top. To get the position of second figure, we have to rotate the dice in the first f igure two times in clockwise direction. After rotating the dice two times in the same direction, 6 comes in the place of 1. So 6 is on the side opposite to the 1. ∴

Answer is (2).

Hidden Figure Test: Hidden figure test is one more type of problem that one may encounter in visual reasoning. A simple figure is given. One has to identify it in more complex figures. For Example:

Find the simple figures hidden in this complex figure.

By inspection one can say that the figure (4) is hidden in the above figure. Mirror Images: In this type of problems the reflection of a design is seen in mirrors placed in different manners. For Example: A plane mirror is kept horizontally below the figure and then one kept on its side. Choose the correct image in the second mirror.


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In the given problem the image of the question figure in the mirror kept horizontally below the figure is

The image of this in the mirror kept at the side of the first mirror is

∴

Answer is (1).

LOGICAL GAMES LOGICAL GAMES involve puzzles in which the relationships among the groups of objects, people, cities, activities etc. are given. These puzzles may deal with such things as making a group, seating arrangement, scheduling the activities etc. After reading and analyzing the statement, you’ll be asked to answer three to seven questions bout the relationships given, which require you to accurately interpret the information given as well as draw logical inferences about relationships. The analytical games can be categorized as follows: 1. Sequencing games 2. Grouping games 3. Matching games 4. Hybrid games 5. Mapping games SEQUENCING GAMES In these types of games you have to put the entities (persons, teaching, schedules etc.) in order. In a sequencing game, you may be asked to arrange/schedule the entities from north to south, left to right, top to bottom, or Monday through Friday etc. GROUPING GAMES In grouping games, you may be asked to organize the entities into groups or teams etc. It can be a selection or distribution problem e.g. selecting players or dividing the people into groups. In selection selection games you start with a large pool of entities and you have to select a smaller group from these. Tips for Sequencing and Grouping games Use short hand language to write the rules

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A and B both cannot be there

AB

A is done before B

A
Two things are done between A and B

A _ _ B or B _ _A

B is done two days/hours after A

A_B

“Picturizing a problem is more important than making the diagram of the problem. Short handing and diagram forming are only the tools to enhance your thinking and solve the question.”

1. Get the overview of the problem Establish the entities. Note the action 2. Picturize the problem mentally (understandin (understanding). g). Assemble the entities Use a simple diagram 3. Consider individual rules. Take time to understand the rules. Short hand the rules (brief and clear) 4. Combine rules Try deducing from the given set of rules 5. Answer: Read the question carefully and try to pre-phrase the answer. Use the elimination with the help of deductions you have made. Note: Don’t write the full name of cities, peoples etc., and the items sh ould be designated simply by their first letter. It’s unnecessary wastage of time. Try to start the diagram with definite or concrete relationship/condition relationship/condition Pay close attention to words like “could be”, “must be”, “may”, “not”, “except”, “necessarily”. Because Bec ause answer to question like “Which must be true” or “Which of the following may be true” will be different. Don’t get confused with the one-way one -way relationship. For example if A attends the seminar, then B also attends it. This means if A is present, B should also be present. Do not interpret it as if B is present A should also be present.

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Grouping games require us to answer the same basic questions: Who’s in and who’s out? Which group can include X, and who else can or cannot reside in a group with X? Tips to solve grouping game problems: 1. See what entities can, must or cannot be in what groups. 2. See what entities can, must or cannot be in the same group as other entities. 3. Notice whether the game asks you to put ALL of the entities into groups or asks you to select SOME of the entities for a smaller group. 4. Pay close attention to numbers: the number of entities in each group, the total number of entities available, the number of entities already chosen. 5. For ambiguous entity names or to differentiate group names from entity names, na mes, use upper case and lower case letters. MATCHING GAMES In this type of problem some persons with some pet names or professions or states or cities or names of their wives etc. are given but not in same order. You have to match the correct ones. Method to solve these types of problems: 1. Draw a table with name of the person vertically and quality or other parameter horizontally 2. Read the statement. Put the cross mark (×) if quality or parameter is not applicable. 3. Put the tick mark (√ ( √) if some quality or parameter is applicable. 4. If in a row or column, a tick mark (√ ( √) appears, then put cross marks (×) in all the remaining boxes in that row or column. 5. If in a row or column, all the boxes except one have cross marks (×), then put tick mark ( √) in that box. HYBRID GAMES It is a mixture of sequencing and grouping games. Mostly, these are considered to be the most difficult types of games. But not every game is a hybrid, and not all hybrid games are difficult. 1. Don’t panic. Organization is the key to hybrid games. 2. As in other games there is no one ‘correct diagram’ for hybrid games. 3. Try making as many deductions as possible. PROBLEMS SEQUENCING GAMES GAME 1 Back to Table of Contents

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A professor plans his teaching schedule to deliver eight topics B,C, H, P, Q, R, V, and W. of his subject. The topic must be taught one at a time in accordance with the following guidelines: H must be the fourth topic and W must be the sixth topic. Topic Q must be taught before topic H. Topic B and topic V cannot be taught consecutively. Topic C must immediately precede topic Q. Exactly two topics must be taught between topic P and topic Q.

1. Topic P must come immediately between which of the following pairs of topics? (1) Q and H

(2) C and W

(3) R and B

(4) H and W

2. What is the maximum number of topics that can be taught between the topics C and R? (1) Two (2) Three

(3) Four (4) Six

3. If Topic B is taught seventh, which of the following must be true? (1) Topic C is taught second.

(2) Topic V is taught third.

(3) Topic P is taught eighth.

(4) Topic R is taught third.

4. Which of the following pairs of courses cannot be taught consecutively? (1) Q and V

(2) W and V

(3) R and H

(4) B and R

Solutions: Since you’re asked to arrange eight topics in order, it makes senses to visualize the game by drawing eight slashes, and also number the slashes like this: –

–

–

–

–

–

–

–

1

2

3

4

5

6

7

8

Rule I is concrete rule .H is fourth and W is sixth. Build this right into your diagram Rule II: Q < H (As H is at 4th position, Q must be first, second or third) Rule III:

(not together)

Rule IV: CQ ( CQ ( C is immediately before Q. Rule I & II tell us Q must be first, second or third, so C must be first or second and Q can’t be at first position) Back to Table of Contents

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Rule V: Exactly two topic between P and Q (Sequence can be P _ C Q or Q or C Q _ _ P Now first not possible because Q must be second or third, so only C Q _ _ P Sequence is possible.) Q can only be taught second or third. If Q is taught second, C must be first and P must be fifth which is possible. But if Q is taught third, C must m ust be second and P must be sixth which is not possible as W must be taught sixth. rd

th

th

But that’s not all, Rule III said that B and V can’t taught consecutively .The only slot left are 3 , 7 , and 8 .Since B and V must be separated ,they can’t taught 7th,and 8th.Therefore ,either B or V must be taught third. By combining all the rules and deductions now the problem can be visualized as follow: B/V/R C

Q

B/V

H

P

W

_

_

1

2

3

4

5

6

7

8

Answer to the problems: 1. Answer: (4) 2. Answer: (4) C is definitely taught first and R can either be taught seventh or eighth. Since you are looking for most topics between two, so R will be taught eighth (C

Q

B/V

H

P

W

V/B

R)

3. If B is taught seventh then V must be taught third. Answer: (2) 4. All are possible except choice (3) i.e. R and H. Answer: (3) GAME 2 Hardy’s world ride is composed of six dragon bogies, numbered 1 to 6. Six children must be put into the six bogies, one child per bogie. The six children are Kailash, Mohair, Namarita, Onkar, Puneet, and Raman. Mohit must be in bogie 1 or 6. Onkar and Puneet must be in adjacent bogies. Kailash must be closer than Raman to the front of the dragon bogies. 1. Kailash CANNOT be in which one of the following bogies? (1) bogie 1

(2) bogie 2

(3) bogie 3

(4) bogie 6

2. If Onkar and Namarita are in adjacent bogies, and if Puneet is in bogie 6, Raman must be in which one of the following bogies? (1) bogie 1

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(2) bogie 2

(3) bogie 3


(4) bogie 4




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3. If Puneet is in bogie 1, which one of the following CANNOT be true? (1) Mohit is in bogie 6

(2) Namarita is in bogie 3

(3) Onkar is in bogie 2

(4) Kailash is in bogie 5

4. Mohit must be in bogie 6 if which of the following children is in bogie 2? (1) Namarita

(2) Onkar

(3) Puneet

(4) Raman

Solutions: Step –1: Establish the entities There are Kailash, Mohit, Namarita, Onkar, Puneet, and Raman. These can be casted as K, M, N, O, P, R. Step –2: Visualise the problem This is a sequencing game which can be visualized with six slots: 1

2

3

4

5

6

–

–

–

–

–

–

Step –3: Consider the individual rules Abbreviate wherever possible, express simple rules in visually direct shorthand. Rule- I: I: Mohit must be in bogie 1 or 6 can be abbreviated as 1

2

3

4

5

6

M

–

–

–

–

–

–

–

–

–

M

or –

Rule – II : Onkar and Puneet must be in adjacent bogies.

OP or PO

Rule – III: Kailash must be closer than Raman to the front of the dragon bogies K < R Step – 4: Answers to the questions 1. Kailash CANNOT be in which one of the following bogies? Sol. As K < R. R. R follows K. So K cannot occur last i.e. bogie no – 6. Answer: (4) 2. If Onkar and Namarita are in adjacent bogies, and if Puneet is in bogie 6, Raman must be in which one of the following bogies? Sol. As P is on 6 no. So O must be at 5 ( Rule no – II II)) and N must be at 4 (ON or NO given in the question) Back to Table of Contents

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1

2

3

4

5

6

M

–

–

N

O

P

Also K < R so K must be in bogie no. 2 and R in bogie No. 3 Answer: (3) 3. If Puneet is in bogie 1, which one of the following CANNOT be true? 1

2

3

4

5

6

P

O

–

–

–

M

Possible arrangements are P O K R N M or P O K N R M and P O N K R M Let’s check the options

(1) Mohit is in bogie 6

---- true

(2) Namarita is in bogie 3 ---- may be possible (3) Onkar is in bogie 2

---- true

(4) Kailash is in bogie 5

---- not possible (As K < R)

(5) Raman is in bogie 4

---- may be possible Answer: (4)

4. Mohit must be in bogie 6 if which of the t he following children is in bogie 2? If R will be in bogie no 2 then K must be in bogie no 1 ( Rule – III: K < R) R) 1

2

3

4

5

6

K

R

–

–

–

M

Therefore M must be in bogie no 6 (Rule – I) Answer: (4) GAME 3 The coach of the Sports Club must choose two two-person Badminton teams for an upcoming tournament. The players available are Chahail, Daman and Eshaan, who are experienced players; and Rajiv, Sahil, Tej, and Uday, who are novices. At least one experienced player must be in each team in the tournament. Daman and Sahil will be chosen only if the two are in different teams. If either Chahail or Tej is chosen, the other must also be chosen. Tej will not be chosen if Uday is chosen.

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Eshaan will not be chosen if Rajiv is chosen. 1. Which one of the following must be true? (1) Chahail and Rajiv cannot both be chosen.

(2) Daman and Tej cannot both be chosen.

(3) Uday and Chahail cannot both be chosen

(4) If Eshaan is chosen, Sahil cannot be chosen.

2. Which of the following is NOT an acceptable selection for the teams? (1) Team 1: Daman and Rajiv; Team 2: Chahail and Tej (2) Team 1: Chahail and Eshaan; Team 2: Tej and Rajiv (3) Team 1: Daman and Chahail; Team 2: Eshaan and Tej (4) Team 1: Eshaan and Sahil; Team 2: Daman and Uday 3. If Sahil is chosen and Tej is rejected for the tournament, which ones of the following must be the members of one of the teams? (1) Sahil and Daman

(2) Sahil and Chahail

(3) Daman and Rajiv

(4) Daman and Uday

4. If Uday is not chosen for the expedition, and Rajiv is chosen for team 1, which one of the following must be in team 2? (1) Eshaan

(2) Chahail

(3) Sahil

(4) Tej

Solutions: Step –1: Establish the entities Experienced players are Chahail, Daman, and Eshaan can be abbreviated as C, D & E Novices are Rajiv, Sahil, Tej, and Uday i.e. R, S, T, U Step –2: Visualise the problem This is a grouping game which can be visualized with two two person teams : GAME 3 The coach of the Sports Club must choose two two-person Badminton teams for an upcoming tournament. The players available are Chahail, Daman and Eshaan, who are experienced players; and Rajiv, Sahil, Tej, and Uday, who are novices. At least one experienced player must be in each team in the tournament. Daman and Sahil will be chosen only if the two are in different teams.

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If either Chahail or Tej is chosen, the other must also be chosen. Tej will not be chosen if Uday is chosen. Eshaan will not be chosen if Rajiv is chosen. T1 / T2 – – – – – – – – Step –3: Consider the individual rule Abbreviate where possible, express simple rules in visually direct shorthand Rule – I: At least one experienced players must be in each team in the tournament. Rule – II: Both Daman and Sahil will be chosen only if they are in different teams. teams . D | S Rule – III: If either Chahail or Tej is chosen, the other must also be chosen. CT or TC Rule – IV: Tej will not be chosen if Uday is chosen. Rule – V: Eshaan will not be chosen if Rajiv is chosen. Step – 4 Answers to the questions 1. Check the options: (1) Chahail and Rajiv cannot both be chosen. This is not true because we can choose C & R together. If we choose C then T will also be there. Fourth person will be D (Rule (Rule no – I & Rule no – V) T1 / T2 CT DR (2) Daman and Tej cannot both be chosen. This is also possible .SEE group made in option (1) (3) Uday and Chahail cannot both be chosen. This is not possible because if C is chosen, T should also be chosen ( Rule-III) Rule-III).. But U and T cannot be together (Rule ( Rule IV). IV). This option must be true (4) If Eshaan is chosen, Sahil cannot be chosen. This is not true if we can make T1 / T2 ES DU

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&

T1 / T2 ES CT





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(5) If Uday is chosen, Tej must also be chosen. Not possible (Rule (Rule no IV) IV) Answer: (3) 2. Which of the following is NOT an acceptable selection for the teams? All options except option (2 ( 2) are possible (Rule (Rule V: 3. If Sahil is chosen and Tej is rejected for the tournament, which one of the following must be the members of one of the teaming teams? If T is rejected then C can’t be chosen ( Rule V). But at least one-experienced players must be in each team in the tournament. So D & E must be chosen. If E is chosen then R can’t be chosen so the fourth person left is U. Possible teams are T1 / T2 SU DR Answer: (4) 4. If Uday is not chosen for the tournament, and Rajiv is chosen for team 1, which one of the following must be in team 2? If R is chosen, then experienced player with R will be C and D (E can’t Rule V) As C and T are a re always together. So T must be chosen. Possible teams are T1 / T2

&

RC DT

T1 / T2 RD CT

So T must be in team 2. Answer: (4) MAPPING GAMES Mapping games revolve around things like roads, messages relay thing with TO/FROM relationships GAME 4 A telecommunication company has six satellite towers in cities: New Delhi, Orissa, Panipat, Quilon, Rajkot R ajkot and Shimla Because of an antiquated technology, signals can be directly sent only from: New Delhi to Panipat Panipat to New Delhi New Delhi to Quilon

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Quilon to Panipat New Delhi to Shimla Rajkot to New Delhi Orissa to Rajkot Shimla to New Delhi A “relay” occurs when a tower receives a signal from another tower and sends it on to a third. A tower can relay a signal from one tower to another in any combination allowed by the above conditions. 1. Which tower cannot receive signals from any other tower? (1) New Delhi

(2) Orissa

(3) Panipat

(4) Quilon

2. Which of the following would require exactly one relay? (1) a signal sent from New Delhi to Shimla (3) a signal sent from Quilon to Orissa

(2) a signal sent from Orissa to Quilon (4) a signal sent from Rajkot to Quilon

3. A signal cannot possibly be sent from (1) Shimla to Quilon

(2) Rajkot to Panipat

(3) Panipat to Rajkot

(4) Shimla to Panipat

4. If the telecommunication system at Panipat fails, so that Panipat may send but not receive signals, which of the following would be IMPOSSIBLE? (1)Sending a signal from Orissa to New Delhi

(2) Sending a signal from Quilon to New Delhi

(3) Sending a signal from Orissa to Quilon

(4) Sending a signal from Shimla to Quilon

5. Quilon would be able to send signals to all other cities either directly or by relay if which of the following capabilities were added to the original list? (1) Sending signals from Orissa to Quilon

(2) Sending signals from Rajkot to Orissa

(3) Sending signals from Quilon to Shimla

(4) Sending signals from Panipat to Orissa

Solutions: Step – 1: Establish the entities There are six cities, which can be abbreviated as N, O, P, Q, R and S Step – 2: Visualise the problem This is a mapping game which can be visualized with a diagram.

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Step – 3: Consider the individual rule Rule No

8 S

1 N

P

5

2 6

O

3 R

4 Q

1. New Delhi to Panipat 2. Panipat to New Delhi 3. New Delhi to Quilon 4. Quilon to Panipat 5. New Delhi to Shimla 6. Rajkot to New Delhi 7. Orissa to Rajkot 8. Shimla to New Delhi Solutions: Step – 4 Answers to the questions 1. Which tower cannot receive signals from any other tower? (2) Orissa (see the diagram) 2. Which of the following would require exactly one relay? (1) a signal sent from New Delhi to Shimla ----- No relay (2) a signal sent from Orissa to Quilon ----- 2 relays (O – R – N – Q) (3) a signal sent from Quilon to Orissa ----- Not possible (4) a signal sent from Rajkot to Quilon ----- 1 relay (R – N – Q) Answer: (4)

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3. A signal cannot possibly be sent s ent from (1) Shimla to Quilon ---- Possible (S – (S – N – Q) (2) Rajkot to Panipat ---- Possible (R – (R – N – P) (3) Panipat to Rajkot ---- Not possible (S – (S – N – Q) (4) Shimla to Panipat ---- Possible (S – (S – N – P) Answer: (3) 4. If the telecommunication system at Panipat fails, so that Panipat may send but not receive signals, which of the t he following would be IMPOSSIBLE? Now diagram reduces to

S

O

N

R

P

Q

(1) Sending a signal from Orissa to New Delhi --- possible (O – R – –N) N) (2) Sending a signal from Quilon to New Delhi --- not possible (3) Sending a signal from Orissa to Quilon --- possible (O – R – –N N – –Q) Q) (4) Sending a signal from Shimla to Quilon --- possible (S – (S – N – –Q) Q) Answer: (2) 5. Quilon would be able to send signals to all other cities either directly or by relay if which of the following capabilities were added to the original list?

S

O

N

R

P

Q Back to Table of Contents

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Q can send signals to P, N, S but not to R and O. So Q would be able to send signals to all other cities only when O would be able to receive signals. (that is only in option (2) & option (4) ) But sending signals from R to O does not link O & Q. So Answer is (4). Another example of Mapping Game Five villages linked by roads. The roads run directly between: Village A and Village B Village B and Village C Village B and Village D Village D and Village C Village D and Village E There are no other roads that provide access to any of the villages. 1. How many different ways are there to travel by road from village A to village E without going through any village twice? (1) 1 (2) 2 (3) 3 (4) 4 Sol.

A

B

D

E

C We can see from the diagram that there are only two possible ways to travel from village A to village E that are: A – B – D – E and A – B – C – D – E Remember to: Start your map with an entity frequently mentioned in the rules. This will form a hub at the center of your map. Keep track of connections, not locations. Now --- draw a map! After drawing your map, think about its structure. Which entities are centrally positioned, forming nodes or hubs? Which are relatively cut off, forming dead ends? Back to Table of Contents

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GAME 5 Exactly four persons stand in a queue on the wait for their turn for collecting entrance ticket at ticket counter at a cinema theater, numbered 1 through 4 from first to last. Two of the persons are males and other two are females. Two out of four are doctors, one is lawyer and one is engineer. Exactly one of four is wearing a cap. The persons in a queue are standing according to the following conditions. The person wearing a cap is either at No. 1 or No. 4. Doctor is at No. 2 position. At least one male stands in line between the two females. One of the doctors is wearing a cap. 1. Which one of the following must be true for a person at No. 3 position? (1) She is a female. (2) He is a male. (3) Person is a lawyer. (4) Person does not wear a cap. 2. If person at No. 4 position is a male who wears a cap, then all of the following must be true EXCEPT. (1) No. 1 is a female. (2) No. 2 is a male. (3) Engineer is at No. 3 position. (4) Doctor is at No. 4 position. 3. If the two males stand in line immediately adjacent to each other, then which one of the following must be false? (1) A female wears a cap. (2) Engineer is a female. (3) Person at No. 3 is a male. (4) Both doctors are male. 4. If the two doctors stand in line immediately adjacent to each other, and if person at No 2 is a male, then which one of the following correctly describes person at N o. 1? (1) A female doctor wearing a cap. (2) A male doctor wearing a cap. (3) A female engineer without cap. (4) A male engineer without cap. Solutions: Step –1: Establish the entities M Stands for Male F Stands for Female D Stands for Doctor L Stands for Lawyer E Stands for Engineer

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Step – 2: Visualise the problem This is a hybrid game which can be visualized with a diagram 1234 – – – – – – – –

Step – 3: Consider the individual rule Two of the persons are males and other two are females.

Two out of four are doctors, one is lawyer, and one is engineer..

The person wearing a cap is either at No .1 or No. 4. Doctor is at No. 2 position. _ D _ _

Cap or Cap At least one male stands in line between the two females.

One of the doctors is wearing a cap. Step – 4 Answers to the questions 1. Which one of the following must be true for a person at No.3 position? (1) She is a female. ---- May be true (2) He is a male. ---- May be true (3) Person is a lawyer. ---- May be true (4) Person does not wear a cap. ---- Must be true (5) Person wears a cap. ---- False

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Answer: (4) 2. If person at No.4 position is a male who wears a cap, then all of the following must be true EXCEPT. F M FM – D – D

Cap (1) No. 1 is a female. ------ true (2) No. 2 is a male. ------ true (3) Engineer is at No. 3 position. ------ may be possible (4) Doctor is at No.4 position. ------ true (5) Exactly one person stands in the between the two doctors ------ true Answer: (3) 3. If the two male stand in line immediately adjacent to each other, then which one of the following must be false. FMMF D D – – – –

Cap

FMMF or

– D – D

Cap

(1) A female wears a cap. ---- May be true (2) Engineer is a female.. ---- May be true (3) Person at No.3 is a male. ---- Definite true (4) Both doctors are male. ---- False (5) Lawyer is a female. ---- May be true Answer: (4) 4. If the two doctors stand in line immediately adjacent to each other, and if person at No 2 is a male, then which one of the following correctly describes person at No.1?


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Answer: (1) (1) A female doctor wearing a cap IF – THEN (Conditional Statement) Here’s an If-Then RULE. This is going to be very helpful for the Grouping and Hybrid Games section. “ If A then B”: B ”: It means if given A, then B must be true. It also means that if we have not given B then A must not be true. So this conditional statement is equivalent to “If not B, then Not A ” but we can’t tell “if not A then ….. “and “if B then …….”. Be careful while applying this approach. For example: “If Amit attends the seminar then Ajay must attend it”.

We can deduce from it that if Ajay does not attend the seminar, then Amit must not attend it. But if Ajay attends the seminar then Amit may or may not attend it or if Amit does not attend the seminar, then whether Ajit attends the seminar or not that we can’t tell.

Data Interpretation Questions taken from students’ forum Questions based on Bar Diagram, Pie chart, etc… can be expected

There was a question based on Data table, containing company’s company ’s recruitment details and classified based on Number of persons joined and left who left every year. The following questions were asked based on the given details. 1. Average number of persons joined from 1995 - 1999 2. Which year has large differences in number of persons (Joined & Left) 3. If 10% of people leave the office in 1998 then, how many fresh candidates can be added in the next year?

Theory, Examples & Explanations Data Interpretation Data plays an important role in day to day life. If data is too large, it can be represented in precise form in a number of ways. Once data is represented in precise form, the user of that data has to understand it properly. The process of interpreting the data from its precise form is called Data Interpretation. Back to Table of Contents

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Now we will discuss different ways of representing data and we will see how we can extract the data from the given representations. Different ways of representing data: 1. Data Tables 2. Pie Charts 3. Two-Variable Graphs 4. Bar Charts 5. Venn Diagrams 6. Three-Variable Graphs 7. PERT Chart 8. Combination of 2 or more charts Now, we shall study these methods in detail. 1. Data Table: Here the entire data is represented in the form of a table. The data can be represented in a single table or in combination of tables. To understand it better, look at the following example.

From the above table, we can find the following: 1. Population of a particular city with respect to that in any other city for a given year. 2. Percentage change in the population of any city from one year to another. 3. The rate of growth of population of any city in any given year over the previous year. 4. The city, which has maximum percentage population growth in the given period. 5. For a given city, finding out the year in which the percentage increase in the population over the previous year was the highest. 6. Rate of growth of the population of all the cities together in a ny given year over the previous year. Back to Table of Contents

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2. Pie Chart In this, the total quantity is distributed over one complete circle. This circle is made into various parts for various elements. Each part represents share of the corresponding element as portion of the total quantity. These parts can be represented in terms of percentage or in terms of angle. Look at the following Pie-chart representing crude oil transported through different modes over a specific period of time.

The above pie chart can also be represented as below

We can find the following from the above pie chart. 1. The oil that has been transported through any mode if the total transported amount is known. 2 The proportion of oil transported through any mode with respect to any other mode. 3. The total oil transported, if the oil transported through any particular mode is known. 3. Two-Variable Graphs Here the data will be represented in the form of a graph. Generally it represents the change of one variable with respect to the other variable. Look at the following graph. Car sales in India in different years (in 000’s)


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From the above graph, we can calculate. 1. Percentage change in the sales of any brand in any year over the previous year. 2. Rate of growth of total sales of the cars (all the brands) in a given period. 3. Proportion of the sales of any brand with respect to those of any other brand in the given year. Bar Chart Bar Chart is also one of the ways to represent data. The data given in the above graph can also be represented in the form of bar chart as shown below.

Here also we can deduce all the parameters as we could do in the case of two-variable graph. 5. Venn Diagrams If the information comes under more than one category, we represent such data in the form of a Venn diagram. The following Venn diagram represents the number of people who speak different languages.

From the above Venn diagram, we can find 1. The number of people who can speak only English. 2. The number of people who can speak only Punjabi. 3. The number of people who can speak both Punjabi and Hindi. 4. The number of people who can speak all the three languages. 5. The number of people who can speak exactly one or two languages. Back to Table of Contents

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6. Three-Variable Graphs Look at the following example to understand the concept. The graph represents percentage of GRE, GMAT and CAT students in three institutes x, y, z.

The above diagram gives the percentage of students of each category (GRE, GMAT, CAT) in each of the institutes x, y, z.

Miscellaneous Questions taken from students’ forum 1. A plane moves from 9˚N 40˚E to 9˚N 40˚W. If the plane starts at 10:00 am and takes 8 hours to reach the destination. Find the local arrival time? 2. A Flight takes off at 2 A.M from Northeast direction and travels for 11 hours to reach the destination which is in North West direction. Given the latitude and longitude of source and destination. Find the local time of destination when the flight reaches there? 0

0

0

0

3. My flight takes off at 2 A.M from a place at 18 N 10 E and landed 10 hrs later at a place with coordinates 36 N 70 W. Find the local arrival time? a) 6:00 am

(b) 6:40am

(c) 7:40

(d) 7:00

(e) 8:00

[Hint: Every 1 deg longitude is equal to 4 minutes. If the direction is from west to east add time, else subtract time] Ans: (e) 8:00 4. Find the physical quantity represented by 5. Find the physical quantity represented by 6. In a two-dimensional array X [9, 7], with each element occupying 4 bytes of memory. If X [1, 1] is stored 3000. Find the address of X [8, 5].

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7. If the vertex [5, 7] is placed in the memory. First vertex X [1,1] address is 1245 and then address of X [5,7] is ---------8. Which of the following are orthogonal pairs? a) 3i+2j

b) i+j

c) 2i-3j

d) -7i+j

9. A, B and C are 8 bit numbers. A-11011011 B-01111 010 C-01101101 Find (A-B) u C 10. In a two-dimensional array A [9, 7], with each element occupying 2 bytes of memory. If A [1, 1] is stored in 3000. Find the memory of A [8, 5]? 11. You will will be be given given the the bit bit position position values for A, B and and C and using using the relati relation on (A (A

[Ans: 3106] B)

C you you have to constr construct uct the the truth truth

table. Then find the corresponding decimal number and choose the right option.


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Formula Booklet 1. Averages Simple averages = Weighted Average =

Geometric Mean = Harmonic Mean =

For two numbers, Harmonic Mean = 2. Percentage Change Change % =

%

Total Successive Change% =

%

3. Interest Simple Interest =

Compound Interest =

Population after n years P’ = 4. Growth Growth% =

SAGR or AAGR =

CAGR =

%

%

%

[Here, S. A. G. R. = Simple Annual Growth Rate, A. A. G. R . = Average Annual Growth Rate and C. A. G. R. = Compound Co mpound Annual Growth Rate]


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5. Profit and Loss Profit = SP − CP Loss = CP – CP – SP Percentage Profit =

Percentage Loss = 6. False Weights If an item is claimed to be sold at cost price, using false weights, then the overall percentage profit is given by Percentage Profit = 7. Discount Discount = Marked Price − Selling Price Discount Percentage = 8. Buy x and Get y Free If articles worth Rs. x are bought and articles worth Rs. y are obtained free along with x articles, then the discount is equal to y and discount percentage is given by Percentage discount = 9. Successive Discounts When a discount of a% is followed by another discount of b%, then Total discount = 10. Ratios If a: b = c: d , then a: b = c: d = (a + c): (b + d ) If 0
If a > b>0, then for a positive quantity x ,

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11. Proportions If a: b:: c:d or

, then

Alternendo Law

Invertendo Law

Componendo Law

Dividendo Law

Componendo and Dividendo Law If

, then

and p, q, r are real numbers, then 12. Successive Replacement

Where x is the original quantity, y is the quantity that is replaced and n is the number of times the replacement process is carried out. 13. Alligation Rule The ratio of the weights of the two items mixed will be inversely proportional to the deviation of attributes of these two it ems from the average attribute of the resultant mixture. x1

x2

x x2- x w1

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:

x – x1 w2





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14. Time, Speed and Distance Speed = Distance / Time Important Conversion Factors: 1 km/hr =

m/s and 1 m/s =

km/hr

15. Average Speed Average =

=

A man man travels travels first half of the the distanc distance e at a speed speed

, second second half half of of the the distanc distance e at a speed speed

then,, Average then Average Speed [Average speed is given

by harmonic mean of two speeds] S avg = If the time is constant, then average speed is gi ven by arithmetic mean of two speeds: S

avg

=

16. Relative Speed For Trains Time = For Boats and Streams Sdownstream = Sboat + Sstream Supstream = Sboat - Sstream Sboat =

Sstream = 17. Time and Work/Pipes and Cisterns Number of days to complete the work =

[This is our general formula to solve time & work problems. It is also known as Work Equivalence Method]


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18. Application of H.C.F. The greatest natural number that will divide x , y and z leaving remainders r 1, 1, r 2 and r 3, 3, respectively, is the H.C.F. of ( x − r 1), (y − r 2) and (z − r 3) 19. Application of L.C.M. The smallest natural number that is divisible by x , y and z leaving the same remainder r in each case is the L.C.M. of ( x , y and z) + r 20. H.C.F. and L.C.M. of Fractions H.C.F of fractions =

L.C.M of fractions = [Express all numbers as fractions in its simplest form] 21. Properties of Surds

22. Law of Indices If a and b are non – non – zero rational numbers and m and n are rational numbers, the


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If

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, then m = n

If

and m 0, then a = b if m is odd and a =

b if m is even

23. Laws of Logarithms

If If 24. Binomial Theorem If n is a natural number that is greater than or equal to 2, then according to the bin omial theorem: n

n

n 0

n

n-1 1

n

n-2 2

n

0 n

(x+a) = cox a + c1x a + c2x a ……….  cnx a Here,

25. Roots of Quadratic Equation 2

The two roots of the equation, ax +bx+c =0 are given by:

26. Algebraic Formulae

2

(a + b) (a − b) = a − b

142

2





2

2

2

2

2

2

(a + b) = a + 2ab + b

(a − b) = a − 2ab + b 2

2

2

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2

(a + b + c) = a + b + c + 2ab + 2bc + 2ca 3

3

2

2

3

3

3

2

2

3

(a + b) = a + 3a b + 3ab + b (a − b) = a − 3a b + 3ab − b 3

3

2

2

3

3

2

2

3

3

a + b = (a + b) (a − ab + b ) a − b = (a − b) (a + ab + b ) 3

2

2

2

a + b +c – 3abc = (a+ b + c) (a + b + c – ab ab – – bc bc – – ac) 27. Arithmetic Progression

28. Geometric Progression

29. Harmonic Progression

30. Sum of Important Series Sum of first n natural numbers


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Sum of the squares of the first n natural numbers

Sum of the cubes of the first n natural numbers

= 31. Factorial n! = 1 × 2 × 3 × … × ( n − 1) × n

n! = n × (n − 1)!

32. Permutations

33. Combinations

Important Properties:

34. Partition Rule Number of ways of distributing n identical things among r persons when each person may get an y number of things =

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35. Probability Probability of an event =

Odds in favour =

Odds against = 36. Pythagoras Theorem For right triangle ABC

2

2

2

AC = AB +BC

For acute triangle ABC

2

2

2

AC = AB + BC – 2 (BC) (BD)


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For obtuse triangle ABC

2

2

2

AC = AB + BC + 2*BC*BD 37. Area of Triangle When lengths of the sides are given

Area = Where, semi perimeter (s) = When lengths of the base and altitude are given


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Area = When lengths of two sides and the included angle are given

Area = For Equilateral Triangle

Area = For Isosceles Triangle


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Area = 38. Apollonius Theorem

If AD is the median, then: 2

2

2

2

AB + AC = 2(AD + BD ) 39. Angle Bisector Theorem

If AD is the angle bisector for a ngle A, then:

40. Area of Quadrilateral For Cyclic Quadrilateral Area = Where, semi perimeter (s) =


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If lengths of one diagonal and two offsets are given

Area = If lengths of two diagonals and included angle are given

Area = For Trapezium

Area =


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For Parallelogram

Area = bh For Rhombus

Area = For Rectangle

Area = lb


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For Square

Area = a

2

41. Polygon Number of Diagonals = The sum of all the interior angles = The sum of all exterior angles = 360 42. Area of Regular Hexagon

Area =


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43. Circle

Circumference C = Area (A) = Length of Arc (l) =

Area of Sector =

, where

, where

is in degrees.

is in degrees.

Perimeter of Sector = 44. Ellipse

If semi-major axis (OD) = a and semi-minor axis (OA) = b, Perimeter of the ellipse

Area of the ellipse


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45. Trigonometric Ratios For a right triangle, if P is the length of perpendicular, B is the length of base, H is the length of hypotenuse and

is

the angle between base and hypotenuse,

46. Distance between Points Distance between two points

and

is given by

AB =


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47. Right Prism

Lateral Surface Area (L.S.A.) = Perimeter of base × height Total Surface Area (T.S.A.) = L.S.A. + 2 × Area of base Volume ( V ) = Area of base × height 48. Cuboid

S.A. = 2(lh + bh) T.S.A. = 2(lh + bh + lb) Volume (V) = lbh Body Diagonal (d) =


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49. Cube

L.S.A = 4a

2

T.S.A = 6a

2

Volume (V) =a

3

Body Diagonal (d) = 50. Cylinder

Curved Surface Area (C.S.A.) = 2πrh 2π rh T.S.A. = 2πrh 2π rh  2πr 2πr

2

2

Volume (V) (V) = πr πr h


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51. Right Pyramid

L.S.A =

Perimeter of Base

Slant Height

T.S.A = L.S.A + Area of base Volume (V) =

Area of Base

Height

52. Cone

C.S.A. = πrl 2

T.S.A. = πrl π rl  πr πr Volume (V) =

Slant height (l) =

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53. Frustum of a cone

Volume (V) = 54. Sphere

C.S.A. = 4πr 4π r

2

2

T.S.A. = 4πr 4π r

Volume (V) =


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55. Hemisphere

C.S.A. = 2π 2πrr

2

2

T.S.A. = 3πr 3π r

Volume (V) =

56. Spherical shell

T.S.A = Volume (V) =


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TCS

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