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DISHA PUBLICATION

ALL RIGHTS RESERVED © Copyright Publisher No part of this publication may be reproduced reproduced in any a ny form without prior permission of the publisher. The author and the publisher do not take any legal responsibility for any errors or misrepresent misrepresentations ations that might h ave crept in. We have tried and made our best efforts to provide accurate up-to-date information in this book. For further information about books from DISHA,































Preface About Te Book...

S

hortcuts in Quantitative Aptitude (with e-books) by Disha Publications is the most comprehensive book that focuses the fundamentals of quantitative aptitude with simple chapters, an array of examples and quick exercises. exercises . It is one of the best-selling best-sell ing reference guides that will help you get your elementary mathematics right. Te book has been designed designed to cover each each topic that is part of the Quantitative Ability syllabus. sy llabus. Shortcuts in Quantitative Aptitude is exceptional in a way that it conceives the short cut methods with which one can solve problems in no time with scientific yet student-friendly approach to discussing the topic. Hence, it not only enhances your efficiency but also helps you master the subject. Each chapter of Shortcuts in Quantitative Aptitude covers basic theory followed fol lowed by shortcut approaches and formulae and will help in learning various tips and tricks of Quantitative Q uantitative Aptitude. Aptitude. It It has been a bestseller in this segment for many years and is heralded as ‘a must read’ book on this topic. Te book is an attempt by Disha Publications to provide quality material to aspirants at a throwaway price.

Shortcuts in Quantitative Aptitude is also supported by ample practice material through e-books which cover: * Chapter-wise Solved Examples * Chapter-wise Practice Exercises with Hints and Solutions * Chapter-wise Cha pter-wise ests ests































We hope the book will prove to be an asset for all those aspiring for competitive examinations like UPSC (IAS Prelim), Banking, SSC, Insurance, Railway Recruitment Recr uitment Board Boa rd Examinations, Examination s, CDS, CBI, MBA, Sub-Inspectors of Police, CPO and various other competitive examinations.

Deepak Agarwal Agar wal































Index 1.

Number System & Simplification

1-21

2.

Algebraic Expression & Inequalities

22-28

3.

Average

29-32

4.

Percentage

33-39

5.

Ratio & Proportion

40-49

6.

Profit and Loss

50-55

7.

Simple & Compound Interest

56-60

8.

Time & Work

61-66

9.

Time, Speed and Distance

67-75

10.

Mensuration

76-87

11.

Clock and Calendar

88-92

12.

Permutation and Combination

93-99

13.

Probability

100-104

E-book 1: Solved Examples (Chapter 1 to 13) E-book 2: Practice Exercises with Hints and Solutions (Chapter 1 to 13) E-book 3: Tests (Chapter 1 to 13)































Disha’s MBA Books At A Glance

































Disha’s Banking Books At A Glance

































Disha’s General Knowledge At A Glance

































Chapter

   

 INTRODUCTION

Numbers

Real Numbers

Non-real or Imaginary Numbers

Set of all real numbers are represented by R.

a + ib, where i = –1, and a is any real number b is any non-zero-number

Rational Num b ers

Irrati on al Num be rs

Decimal Form q where p and q are integers and q ≠0 Examples: 5 , , 2 , 8 3 , because 3, 0 and

–4 3, 0, and 1 1 1 respectively. as

In this form, the number o f digits after decimal point is finite. Examples: 4.024, 5.008, 0.23, etc.

Integers Examples : 2, 3, .....

Examples :

Non-negative integer or Whole Numbers which includes includes zero and positive integers Examples : Examples : 0, 1, 2, 3, ...

(0) Examples :1, 2, 3, ...

q Form here q ≠ 0 and at least one of and q is not an integer Examples: 5 , 0.51 , 8.02 , 4.2 6 2.3

2, In this orm, number of digits after decimal point is infinite but just after decimal point or leaving some digits after decimal point a group of d igit (s) (one or more than one digits) repeats continuously. Examples: 5.4141 ....., 0.061245245...., etc. These irrati irrational onal numbers can be w ritte ritten n by puttin g a bar or recurring above the first group of digits(s) after decimal point which repeats further continuously. Examples: 5.4141... = 5.41, 0.2383838...

Zero

p

= correct 0.238, 0.061245245....

5,

3 , etc.; 2

because

2 and can be written as 2 and 5 1 1 espectively.

5

Non-terminating and Non-repeating (non-recurring) Decimal Form

In this this orm, number of digits after decimal point isinfinite but there is no group of digit(s) (one or more than one d igits) just after deci mal point or leaving some digits after decimal point which repeatss continuously repeat continuously.. Examples : 501.060060006..., 23.1424434444..., 0.009191191119..., –7.4010 02003... ., etc.

































2

ê ê ê ê ê ê ê

Number System System & Simplification Simplification

The ten symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 are called digits. 1 is neither prime nor composite. 1 is an odd integer inte ger.. 0 is neither positive nor negative. 0 is an even integer. 2 is prime & even both. All prime numbers (except 2) are odd.

Natural Numbers : These are the numbers (1, 2, 3, etc.) that ar e used for for counting. It is denoted by N. There are infinite i nfinite natural numbers numbers and the smallest smallest natural number number is one (1).

Even Even numbers : Natural numbers numbers which are divisible divisible by 2 are even even numbers. It is denoted by E. E = 2, 4, 6, 8,... Smallest Small est even number is 2. There is no largest even number.

Odd numbers : Natural numbers which are not divisible divisible by 2 are odd numbers. It is deno denote ted d by O. O = 1, 3, 5, 7, ... Smallest odd number is 1. There is no n o largest odd odd number.

Æ

Based on divisibili , there ther e could b

pes of

ural

bers :

































Number Number System System & Simplification Simplification

3

Compo site si te Numbers Numbers : It is a natural n atural numbe n umberr that has atleast one divisor divisor different from unity and itself. Every composite composite number can be factorised factorised into in to its prime factors. For Example : 24 = 2 × 2 × 2 × 3. Hence, 24 is a composite number. The smallest composite composite number is 4.

Twin-prime Numbers: Pairs Pair s of such prime pr ime numbers whose difference is 2. Example: 3 and 5, 11 and 13, 17 and an d 19. How to check whether whet her a given number is prime or not ? Steps : (i) Find approximate appr oximate square squar e root of the given number. (ii) Divide the given number by every prime number less than the approximate square root. (iii) If the given number is exactly divisible by atleast one of the prime numbers, the th e number is a composite composite number num ber otherwise a prime pri me number. Example : Is 401 a prime numb n umber? er? Sol. Approximate Approximate square squar e root of 401 is 20. Prime numbers less than 20 are ar e 2, 3, 5, 7, 11, 13, 17 and 19 401 is not divisible by 2, 3, 5, 7, 7 , 11, 13, 17 or 19. \ 401 is a prime number. (Hint : Next prime number after 19 and 23, which is greater than 20, so we need not check further.) furth er.) Co-prime Numbers : Co-prime numbers are those numbers which are prime to each other other i.e., they don't have any common common factor factor other other than 1. Since these numbers do not have any common factor, factor, their HCF is 1 and their LCM is equal to product of of the numbe nu mbers. rs. Note : Co-prime numbe n umbers rs can be prime or composite composite numbers. Any two two prime numbers numbers are alway alwayss co-prime co-prime numbers. numbers.



































4


Whole Numb Numbers ers : The natural natu ral numbers numbers along with zero (0), ( 0), form the th e system system of whole whole numb n umbers. ers. It is denoted by W. There is no largest whole whole number number and The Th e smallest whole whole number is 0.

Intege nt egers rs : The number system system consistin consisting g of natural natur al numbers, num bers, their negative and zero is called intege int egers. rs. It is denoted by Z or I. The smallest and the largest integers cannot be determined.

The Number Number L ine : The number line lin e is a straight line between between negative infinity on on the th e left to positive positive infinity on the right.

-4 -3 -2 -1 0 1 2 3 4 Rational numbers : Any number number that th at can be put in the form of

p

q where p and q are integers and q ¹ 0 , is called a rational numb n umber. er.. • It is denoted by Q.

,


































5

Irrational numbers : The numb n umbers ers which are not rational r ational or which cannot

be put in the form form of

p

, where p and q are integers in tegers and q ¹ 0, is

q

called irrational irrati onal number. It is denoted by Q ' or Qc . 2 , 3, 5, 2 + 3, 3 - 5 , 3 3 are irrational irra tional numbers. numbers. NOTE :

(i ) Every Every positiv positivee irrat irration ional al numb number er has has a negative negative irrati irrationa onal l number corresponding corresponding to it. (ii)

2+ 3¹ 5 5- 3

¹

2

3´ 2

=

3´ 2

6¸ 2

=

6 2

=

=

6

3

(ii i) Some times, times, product product of two irration irrational al numbers numbers is a rational rational number. For example :

2´ 2

=

2´2

=2

































6


FRACTIONS A fraction is a quantity which expresses a part of the whole. Fraction =

Numerator Denominator

TYPES OF FRACTIONS : 1.

Proper fraction : If numerator is less than its denominator, then it is a proper fraction.

For example : 2.

2

,

6

5 18

Improper fraction : If numerator is greater than or equal to its denominator, then it is i s a improper fracti fraction. on.

For example :

5 18 13 , , 2 7 13

NOTE : If in a fraction, its numerator and denominator are of equal value then fraction is equal to unity i.e. 1. 3.

Mixed fraction : It consists consists of an integer in teger and a proper fraction.

For example : 1

1

3

2

7

5


































7

NOTE : Value of fraction is not changed by multiplying or dividing divi ding both the numerator or denominator by the same number.

For example : (i)

(ii)

5.

2 5

=

36 16

2´5 5´5

=

10

=

25

36 ¸ 4 16 ¸ 4

=

9

So,

4

2 5

=

36 16

10 25

=

9 4

Like fractions: Fractions with same denomina denominators. tors.

For example : 6.

So,

2 3 9 11 11 , , , 7 7 7 7

Unlike fractions : Fractions with different denominat denominators. ors.

For example :

2 4 9 9 , , , 5 7 8 2

NOTE : Unlike fractions can be converted into like fractions.

For example :

3 5

and

4 7

































8 9.

Number System System & Simplification Simplification Decimal fraction : Denominator with the powers of 10.

10. 10.

2

9

= (0.09) 10 100 Compari parisson of Fract ractio ion ns Comparison Compar ison of two faction can be easily easil y understan under stand d by the following example:

For example :

= (0.2),

3

7

, multiply mult iply each each fraction by the 5 9 LCM (45) (45) of their denominators 5 and a nd 9.

To compare two fraction

3 5 7 9

and

´ 45 = 3 ´ 9 = 27 ´ 45 = 7 ´ 5 = 35

Since 27 < 35

\

3 5

<

7 9

SHORT CUT METHOD 3 5

7 9 27 < 35

[Write the each product on their numerator side]



































Number Number System System & Simplification Simplification =4+ = 4+

9

27 + 40 + 36 45 103 45

= 4+2

13 45

=6

13 45

Roundi ng off of f (Appr (Appr oximati on) of Decimals Decimals : There Ther e are some some decimals in which numbe n umbers rs are a re found upto large number n umber of decimal places. For For examp example le : 3.4578, 21.358940789. 21.358940789. But many times we require decimal numbers upto a certain number of decimal places. Therefore, If the digit of the decimal place is five or or more than five, then the th e digit in the t he preceding decimal place is increased by one and an d if the digit in the last place is less than t han five, five, then the digit in the precedence precedence place remains unchanged.

































10


where and an d Thus,

m = number number of recurring recurring digits in decimal part part n = number number of non-recurring digits digits in decimals decimals part part p

form of 64.20132 = q =

In short; 0.a

=

0.ab =

0.abcd =

a 9

, 0.ab =

ab - a 90

ab 99

6420132 - 6420 99900 6413712

, 0.abc =

, 0.abc =

abcd - ab 9900

99900

=

abc 999,

a bc - a 990

, ab. c de =

534476 8325

etc. and

, 0.abc

=

abcde - abc 990

abc - ab 900 , etc.

,


































11

Divisibili ty b y 4 : A number is divisible by by 4 if the last 2 digits are ar e divisible divisible by 4, or if the t he last two digits are 0’s.

Divisibili ty b y 5 : A number number is divisible divisible by 5 if its unit’s un it’s digit is 5 or 0.

Divisibili ty b y 6 : A number is divisible by 6 if it is i s simultaneously divisible divisible by 2 and 3.

Divisib ivi sib lity li ty by 7 : A number is divisible by 7 if unit’s place digit is multiplied by 2 and subtracted subtracted from the remaining r emaining digits and the number obtained obtained is divisible divisible by 7. For example,

1680 7 = 1680 – 7 × 2 = 1666

































12


Divisibil ivisi bil ity b y 12 12 : A number is divisible divisible by 12 if the number is divisible by both both 4 and 3.

Divisibil ivisi bil ity b y 13 13 : A number number is i s divisible divisible by 13 if its unit’s place digit is i s multiplied multipli ed by 4 and added to the remainin re maining g digits and an d the number num ber obtained obtained is divisible divisible by 13. For example,

® 219 + 7 ´ 4 = 247 219 7 ¾¾ Again 24 7 ¾¾ ® 24 + 7 ´ 4 = 52, divisible by 13. Hence 2197 is divisible by 13.

Divisibil ity b y 14 14 : A number is i s divisible by 14 if the th e number is divisib d ivisible le by both both 2 and a nd 7.


































13

Divisibil ivisi bil ity b y 19 19 : A number number is i s divisible divisible by 19 if its unit’s place digit is i s multiplied multipli ed by 2 and added to the remainin re maining g digits and an d the number num ber obtained obtained is divisible divisible by 19. For example,

® 4873 + 7 ´ 2 = 4887 4873 7 ¾¾ ® 488 + 7 ´ 2 = 502 488 7 ¾¾ divisible by 19. ® 50 + 2 ´ 2 = 54 not divisible 50 2 ¾¾ Hence 48737 is not n ot divisible by 19.

Properties of Divisi Divisibili bili ty (i) (ii) (iii)

The prod produc uctt of 3 co consecu nsecutiv tivee natura naturall numbers numbers is div divisib isible le by by 6. 6. The produc productt of 3 conse consecu cutiv tivee natural numb numbers ers,, the first first of of which which is even, is divisible di visible by 24. Diffe Difference rence betw between een any number number and the th e numbe number obtained btained by writing writing the digits in reverse order is divisible by 9.



































14


Complete Complete remainder remainder = [I divisor divisor × II remainder] + I remainder C.R. = d1r2

+ r1 C.R. = d1d 2 r3 + d1r2 + r1

Shortcut Approach Two different numbers n umbers x and y when when divided di vided by a certain cer tain divisor D leave remainder remain der r 1 and r 2 respectively. respectively. When the t he sum of them sis divided by by the same divisor, divisor, the remainder r emainder is r 3. Then,


































15

(ii) Division Metho Met hod d

To find the HCF of two two numbers by division division method, we divide the larger la rger number by the smaller smal ler number. Then we divide the smaller number n umber by the first remainder, remai nder, then first remainder r emainder by the second second remainder.. remain der.. and so on, till the t he remainder r emainder becomes becomes 0. The last diviso di visorr is i s the required requir ed HCF HCF..

Shortcut Approach To find the HCF of any number of given numbers, first find the

































16


Divide those numbers out of the given numbers n umbers by 2 which are a re divisible by 2 and write the t he quotient below below it. The given numbe n umber(s) r(s) that are not divisible divisible by 2 write writ e as it is below below it and an d repeat this t his step till t ill you do not find at least l east two numbers numbers that are a re not divisible divisible by any prime number. 2

5, 72, 19 196, 24 240

2

5, 36,

98, 120

2

5, 18,

49,

60

3

5,

49, 49

30 30

9,


































17

Prime Pri me factorisation of 9 × 10 = 3 × 3 × 2 × 5 Prime Prime factorisation of 15 = 3 × 5 Prime Pri me factorisation of 36 = 2 × 2 × 3 × 3 Here we see that both prime factors of 15 are present in the prime factorisation factorisation of 9 × 10 but one prime factor 2 of 36 is not prese pr esent nt in the th e LCM taken in step 1. So to find the LCM of 9, 10, 15 and 36; we multiply multipl y the LCM taken in i n step st ep 1 by by 2. Thus required LCM = 9 × 10 × 2 = 180

































18

2.


Su b t r ac t i o n s : Subtraction of two numbers numbers is same as the sum of a positive and a negative number. For Example : (+ 9) – (+ 2) = (+ 9) + (– 2) = 7 (– 3) – (– 5) = (– 3) + 5 = + 2.

NOTE : In subtraction subtraction of two two negative negative numbers, sign of second number

will change and become positive.



































Number Number System System & Simplification Simplification (iii) Curly Curly or midd middle le bracke rackets ts : { } (iv) Squa Square re or big big bracke rackets ts : [ ] The order for removal of brackets is (), {}, []

19

































20

(ii)

Number System System & Simplification Simplification am a

n

5

= am – n, if m > n

6

çç Example : 2 = 65- 2 = 63 ÷÷ 6

































Number Number System System & Simplification Simplification (ii) (a – b)3 = a3 – b3 – 3a2 b + 3ab2 Þ (a – b)3 = a3 – b3 – 3ab (a – b) (iii) a3 + b3 = (a + b) (a2 – ab + b2) (iv) a3 – b3 = (a – b) (a 2 + ab + b2)

21

































22

Chapter



Algebraic Algebraic Expression Expression & Inequalities

     

































Algebraic Algebraic Expression Expression & Inequalities

23

Hence Hence (3, –2) is a solution. The Th e following following table t able lists six possible values for x and the th e corresponding values val ues for for y, y, i.e. six solutions s olutions of the equation. x y

–2 8

–1 6

0 4

1 2

2 0

3 –2



































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































Time & Work

64

WORK AND WAGES

Wages are distributed distribu ted in proportion to the work done done and in i n indirect in direct proportion to the time taken by by the individual. PIPES AND CISTERNS

The same principle of Time and Work Work is employ employed to solve the problems on Pipes and Cisterns. Cistern s. The only difference difference is that in this case, the work done is in terms of filling filling or emptying emptying a cistern (tank) and an d the time taken t aken is the time taken taken by a pipe or a leak (crack) to fill or empty a cistern respec r espectively tively.. Inlet : A pipe connected connected with a tank ta nk (or a cistern or a reservoir) reservoir) is called an inlet, if it fills it. Outlet : A pipe connected connected with a tank ta nk is called an outlet, if it empties it.

Shortcut Short cut Approach 1

Æ

If a pipe can fill a tank in i n x ho h ours, then the t he part filled in 1 hour =

Æ

If a pipe can can empty a tank tan k in y hours, then th en the part of the full tank emptied in 1 hour =

Æ

1

x

.

If a pipe can fill a tank in i n x hours and another pipe can empty the full tank in y hours, hours, then the net part filled in 1 hour, hour, when both both the æ1 1ö = pipes are opened opened ç x – y ÷ . è ø \ Time taken to fill the tank, when both the pipes are opened =

xy – x

Æ

If a pipe can can fills or empties tank in x hours and another a nother can fill or empties the same tank in y hours, hours, then t hen time taken to t o fill fill or empty the tank =

Æ

.

xy

+x

, when both the pipes are opened

If a pipe fills a tank in x hours and another fills the same tank is y hours, but a third one empties the full tank in z hours, hour s, and all of them é1 1 1ù are opened together, together, then net part filled in 1 hr = ê + – ú ëx y zû































Time & Work

Æ

65

A pipe can fill a tank tan k in x hrs. Due to t o a leak leak in the bottom bottom it is filled in y hrs. If the th e tank is full, the time taken by the leak to empty the tank =

xy – x

hrs.

cistern has a leak which can empty it in X hours. A pipe pipe which Æ A cistern admits Y litres of water per hour into the cistern is turned on and now the cistern is emptied in Z hours. Then the capacity of the cistern is

X+Y+Z Z-X

litres.

whose diameters are ar e X cm., Y cm. Æ A cistern is filled by three pipes whose and Z cm. respectively respectively (where (where X < Y < Z). Three Th ree pipes are running runn ing together. If the largest pipe alone will fill it in P minutes and the amount of water water flowing flowing in by each each pipe pi pe is proportion proportional al to the square squar e of its its diameter di ameter,, then th en the time in i n which the th e cistern will be filled by the three pipes is

é ù PZ2 ê 2 2 2 ú minutes. êë X + Y + Z úû ti mes faster and takes X minutes less time Æ If one filling pipe A is n times than the other filling pipe B, then the time they will take to fill a

é nX ù ú minutes. cistern, if both both the th e pipes are opened opened together, is ê 2 êë (n - 1) úû æ X ö A will will fill the cistern in çè ÷ minutes and B will take to fill the n - 1ø æ nX ö cistern çè ÷ minutes. n - 1ø Here, A is the faster filling pipe and B is the t he slower one.

Æ Two filling pipes A and B opened together can fill a cistern in t minutes. If the first filling pipe A alo alone ne takes X minutes more or less than t and the second second fill pipe B along takes Y minutes minut es more or less than t minutes, min utes, then t is given by [ t =

xy ] minutes.































Time & Work

66

ebooks Reference

Page No.

Solved Examples

–

S-29-36

Exercises with Hints & Solutions

–

E-61-70

Chapter Test

–

15-16

Past Solved Papers































Chapter

    



TIME, SPEED AND DISTANCE Speed The rate ra te at which any moving body body cov covers ers a particular distance is i s called its speed. Speed =

Distance Time

; Time =

Distance Speed

;

Distance = Speed Speed × time

Unit : SI unit of speed is metre per second (mps). It is also al so measured measur ed in kilometers kilometers per hour (kph) or miles per hour h our (mph).

Basic Conversions : (i)

• • • • • •

1 hour = 60 minutes = 60 × 60 seconds. 1 km = 1000 m 1 km = 0.6214 mile 1 mile = 1.609 km i.e. 8 km = 5 miles 1 yard = 3 feet 1 foot foot = 12 inches in ches 5

•

1 km/h =

•

1 m/sec =

•

1 miles/hr =

18

18 5

m/sec, km/h 22

ft/sec































Time, Speed and Distance Distan ce

68

Shortcut Short cut Approach

Æ Average speed =

Total Distan Distance ce Tota Totall time time

travelling a certain distance distance (d), if a man changes his h is speed speed in the Æ While travelling ratio ratio m : n, then then the ratio ratio of time taken becomes becomes n : m. B, is covered covered at ‘a’ Æ If a certain distance (d), say from A to B, km/hr and the same distance is covered again say from B to A in ‘b’ km/hr, then the average speed during the whole journey is given by :

æ 2ab ö Average speed = ç ÷ km/hr èa+bø Also, if t 1 and t 2 is time taken to travel from A to B and B to A respectively respectively,, the distance ‘d’ from A to B is given by :

æ ab ö d = (t 1 + t2) ç ÷ èa+bø æ ab ö d = (t1 – t2) ç è b - a ÷ø

æ t1t 2 ö ÷ d = (b – a) ç t t è 1 2ø covered at the rate r ate of v1 in time t 1 and Æ If first part of the distance is covered

the second part of the distance distan ce is covered covered at the th e rate of v2 in time

t2, then the average speed is

æ v1t1 + v2 t 2 ö çè t + t ÷ø 1

2

Relative Speed When two bodies are moving in same direction with speeds S 1 and S2 respectively, their relative speed is the difference of their speeds. i.e., Relative Speed = S1 – S2 , If S1 > S2 = S2 – S1, if S2 > S1 When two bodies bodies are ar e moving moving in opposite direction with speeds S 1 and S2


































69

Shortcut Approach If two persons persons (or vehicles or or trains) t rains) start at the same time in oppoopposite directions from two points A and B, and after crossing each other they take x and an d y hours respectively r espectively to complete complete the journey jour ney,, then

Speed of of first Speed of second

=

y x

Shortcut Short cut Approach Usual speed : If a man changes cha nges his speed to

a

of his usual speed, b reachss his destination late/earlier late/earli er by t minutes then,

Usual Usual time =

Change in time

æ b ö ç - 1÷ èa ø

Shortcut Approach A man covers a certain distance D. If he moves S 1 speed faster, he would would have ha ve taken taken t time less and if i f he moves moves S 2 speed slower, he would would have taken t time more. The Th e original speed is given by 2 ´ ( S1 ´ S2 ) S2 ~ S1

Shortcut Approach


































70

Shortcut Approach policemen sees a thief at a distance of d. He starts chasing the Æ A policemen thief who is running runni ng at a speed of of ‘a’ and policeman policeman is i s chasing with a speed of ‘b’ (b > a). a) . In this th is case, the th e distan dist ance ce covered covered by the thief

æ a ö when he is caught by the policeman, is given by d ç ÷. è b - a ø reach es the poin pointt B at t . Another Æ A man leaves a point A at t and reaches 1

2

man leaves the point B at t 3 and reaches the point A at t 4, then they will meet at t1 +

Æ

( t 2 – t1 ) ( t 4 – t1 ) ( t 2 – t1 ) + ( t 4 – t 3 )

Relation between between time ti me taken with two different different modes of transport : t2x + t2y = 2(t x + ty) where, tx = time when mode of transport x is used single way. ty = time when mode of of transport tr ansport y is used single way. way. t2x = time ti me when mode of tran transport sport x is i s used both both ways. t2y = time when mode of of transport tr ansport y is used both both ways.

TRAINS A train trai n is said to have crossed an object (stati

or moving) onl when


































71

Shortcut Approach tra in to cross a pole/a pole/a standing standi ng man Æ Time taken by a train

=

Length of train Speed of train

.

platform/bridge ridge etc. (i.e. a stationary Æ Time taken by a train to cross platform/b object object with some length) length )

=

length of train + length of platform/bridge etc. speed of train

.

train s with lengths L and L and with speeds S and Æ When two trains 1

2

1

S2 respectively, then (a) When When they are mo moving ving in the same same directi direction, on, time take taken n by by the faster train tr ain to cross the slower slower train trai n

=

L1 + L2 difference of their speeds

.

(b) (b) When When they are movi moving ng in the oppo opposit sitee directio direction, n, time take taken n by the trains to cross each each other

=

L1 + L 2 sum of their speeds

.

Suppose two two trains tra ins or two bodies bodies are moving in the same direction Æ Suppose at u km/hr km/ hr and an d v km/hr respectively respectively such such that u > v, v, then their relative r elative speed speed = (u – v) km/hr. If their lengths length s be x km and an d y km respectively respectively,, then th en time taken by the faster faster train tr ain to cross cross the slower slower train tr ain (moving in the th e same direction) direction)

æx+yö =ç ÷ hrs. è u-vø

Æ


































72

(v – u) m/sec is called called the th e speed speed of of the train relative to man. Then 1 the time taken by the train to cross the man = seconds v-u run ning g at a speed of of u m/sec in a direction dir ection opposite opposite to Æ If a man is runnin that in which a train of length L meters is running with a speed v m/ sec, then (u + v) is called the speed of of the train relative to man. Then the th e time taken by the train tra in to cross the man 1 = seconds. v+u two trains start at the same time from from two points points A and and B towards towards Æ If two each other and after cr crossing, ossing, they take (a) and an d (b) hours in reachin r eaching g B and an d A respectively. respectively. Then, Th en, A’s speed : B’s speed =

(

)

b : a .

passes a platform of x m in t 1s, then time taken Æ If a train of length L m passes t2s by the same train to pass a platform of length y m is given as

æL+yö ÷ t1 èL+xø

t 2 = ç

tra ins start mov m oving ing tow t owards ards each other Æ From stations P and Q, two trains with the t he speeds a and b, respectively. respectively. When they t hey meet each other, it is found that one train covers distance d more than that of another train. In such cases, distance between stations P and Q is

æa+bö given as ç ÷ ´ d. èa bø

Æ The distance between P and Q is (d) km. A train with (a) km/h starts from station P towards Q and after a difference of of (t) hr another another train with (b) km/h starts start s from Q towards towards station P, P, then both both the trains will meet at a certain point after time T. T. Then,


































73

Shortcut Approach between two stations P and Q is d km. A train starts s tarts Æ The distance between from P towards towards Q and another an other train trai n starts from Q towards P at the same time and they meet at a certain point after t h. If train tr ain starting star ting from P travels with a speed of x km/h slower slower or faster than t han another a nother train, then (i)

æ d + tx ö Spee peed of of faste asterr tra train in = ç ÷ km/h è 2t ø

æ d - tx ö (ii) Spee Speed d of of slo slower train train = ç ÷ km/h è 2t ø tra in covers covers distance d sbetween sbetween two stations P and Q in t 1 h. If Æ A train the speed of train is reduced by (a) km/h, then the same distance will be covered covered in t2 h. (i)

Dista istanc ncee betw etween P and Q is

æ t1 t 2 ö ÷ km è t2 - t1 ø

d = aç

æ at 2 ö ÷ km/h (ii) Spe Speed of of the the train train = ç t t è 2 1ø

B OATS OATS AND STREAMS Stream : It implies that the water in the th e river is moving or flowing. flowing. Upstream Upstream : Going against aga inst the flow of the river. Going with the th e flow of the river. r iver.




































74

Shortcut Approach

Æ

Speed of boat boat with the stream (or downstream or D/S) = (X + Y) m/sec.

Æ

Speed of of boat boat against the stream (or upstream or U/S) = (X – Y) m/sec.

Æ

Speed of of boat boat in still water is X=

Æ

( X + Y ) + (X – Y )

=

Upstre tream + Downs ownsttream 2

2

Speed of the stream or current is Y =

=

(X + Y) – (X (X – Y) Y)

2 Downstre tream - Upstr stream eam 2

Shortcut Approach

Æ

A man man can row X km/h in still water. water. If in i n a stream str eam which which is flowing flowing of Y km/h, it takes him Z hours to row to a place and back, the distance between the two places is

Z( X 2 – Y 2 ) 2X


































75

Shortcut Approach

Æ

If speed speed of stream is a and a boat (swimmer) takes n times time s as long to row up as to row do down wn the th e river, then Speed of of boat boat (swimmer) (swimmer) in still water =

a (n + 1) (n - 1)

Note: This formula is applicable for equal distances.

Æ

If a man capable of rowing at the t he speed (u) m/sec in still water, rows the same distance up and down down a stream flowing flowing at a rate ra te of (v) m/ sec, then his average speed through the journey is Upstream ´ Downstre tream

=

Æ

Man 's rate in still water

=

(u - v) (u (u + v ) u

If boat's (swimmer's) speed in still water is a km/h and river is flowing with a speed of b km/h, then average speed in going to a certain place and an d coming coming back to starting point is given by by (a + b)(a - b) a

km/h.

































Mensuration

76

Chapter





MENSURATION Mensuration is the science of measurement of of the lenghts lenght s of lines, lines, areas ar eas of surfaces and volumes of solids.

Perimeter Perimeter is sum of all the sides. It I t is measured in cm, m m,, etc.

Area Ar ea The area are a of any figure is the th e amount amount of surface surface enclosed enclosed within its boundary boundary lines. This is measured in square unit like l ike cm2, m2, etc.

Volume If an object object is solid, then th en the th e space occupied occupied by such an a n object object is called its it s volume. volume. This is measured m easured in cubic unit like cm 3, m3, etc.

Basic Conversi Conversions ons : I

1

10 d

































Mensuration

77

TRIANGLE B

a h

C

c

b

A

Perimeter (P) = a + b + c Area (A) = s ( s – a ) ( s – b ) ( s – c ) where s =

Also, A =

a +b+c 2 1 2

and a, b and c are three thr ee sides of the trian tr iangle. gle.

´ bh ; where b ® base

h ® altitude

































Mensuration

78

Right igh t triangle

h

p

b A=

1 2

pb and h 2 = p2 + b2

(Pythagoras (Pythagoras triplet)

wh where p ® perpendicular b ® base h ® hypotenuse

RECTANGLE

b

Perimeter = 2 ( l + b)



































Mensuration

79

SQUARE

a

a

Perimeter = 4 × side = 4a Area = (side)2 = a2; where a ® side

PARALLELOGRAM b

































Mensuration

80

IRREGULAR QUADRILATERAL p

h1

q d

s

h2

r

Perimeter = p + q + r + s Area =

1 2

´ d ´ ( h1 + h 2 )

TRAPEZIUM a m

h

n

































Mensuration

81

Pathways athways out sid e

b

b + 2a

a + 2a

A = (l (l + + 2a) (b + 2a) – l b; b;

where l ® length

b ® breadth a ® width of the pathway

Pathways athways inside insi de – l –

2a

































Mensuration

82

CIRCLE Perimeter (Circumference) (Cir cumference) = 2pr = pd Area = pr 2; where here r ® radius d ® diameter and p =

22 7

r

or 3.14

Shortcut Short cut Approach The length and an d breadth of a rectangle are l and and b, then are of circle of maximum radius inscribed in that rectangle is

SEMICIRCLE Perimeter = pr + 2r

p b2 4

.

































Mensuration

83

Perimeter of segment = length of the arc ar c + length of segment segment AB =

pr q 180

+ 2r sin

q 2

RING

R 2 R 1

(

Area of ring = p R 22 – R 12

)



































84

Mensuration

Note: Increment in the value is taken as positive positiv e and decrement in in value is taken as negative. Positive result shows total increment and negative result shows total decrement.

CUBE A cube is a cuboid cuboid which has all its edges edges equal. Total surface area of a cube = 6a2 Volume of the th e cube = a 3

































Mensuration

85

h

r

PYRAMID A pyramid pyramid is a solid which can have any polygon polygon at its base and its edges

































Mensuration

86

l h

r

SPHERE

































Mensuration

87

Shortcut Short cut Approach If side of a cube or or radius (or diameter) of sphere sphere is increased by x%, 3 éæ ù x ö 1 00% then its volume increases by by êç1 + 100 ÷ - 1ú ´10 ø êëè úû

Shortcut Short cut Approach If in a cylinder or cone, height and radius both change by x%, then

































Clock and Calendar

88

Chapter



  

CLOCK Introduction



































Clock and Calendar

89

Relative elative pos iti on of the hands hands The position of the M.H. relative to the H.H. is said to be the same, whenever the M.H. is separated from the H.H. by the same number of minute divisio d ivisions ns and is on same side (cloc ( clockwise kwise or or anticlockwise) anticlockwise) of the

































Clock and Calendar

90

Shortcut Short cut Approach

Æ

Shortcut Approach for finding finding degrees degrees minutes and hours is

æ

ö

































Clock and Calendar

A century centur y (100 years) contains contain s = 24 leap years + 76 ordinary ordina ry years years = 24 × 2 + 76 = 124 odd days = 17 weeks + 5 odd days Similarly, 200 year yearss contains contain s 2 × 5 – 7 = 3 odd days 300 years years contains 3 × 5 – 14 = 1 odd day

91

































92

Clock and Calendar

NOTE : February in an ordinary ordinary year gives no odd days, but in a leap year gives one odd day day..

































Chapter



  















































































































































































































































































































































Shortcuts in Quantitative Aptitude Competitive Exams by Disha Experts (1)

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