23_years_CAT_Topic_wise_Solved_Papers.pdf

23 Years

TOPIC-WISE SOLVED PAPERS (1994-2016)                 

Deepak Agarwal Shipra Agarwal

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6 2 0 7 _ D B E

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Head Of f i c e : B-32, Shivali k Main Road, Malviy a Nagar, New New Delhi-110017

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6 2 0 7 _ D B E

UNIT - I : QUANTITATIVE QUANTITATIVE ABILITY A BILITY CHAPT HAPTER ER 1 :

NUMBE NU MBER R SYSTEM SYSTEM — Types Types of Numbers;

Surds and Indices; Arithmetic Calculations; Division and Divisibility Test; Simplification and Rationalization; HCF and LCM; Fractions; Comparison of Fractions; Numeration Systems; Conversion between Numeration Systems. — Sets and their Representations; CHAP CHAPTER TER 2 : SET SET THE THEOR ORY Y — Types Types of Sets; Sets; Subsets Subsets;; Venn Diagrams Diagrams;; Operatio Operations ns on Sets. Relations; Types of Relations; Relations; CHAP CHAPTER TER 3 : FUNCTI FUNCTION ONS S — Relations; Functions; Domain, Co-domain and Range; Inverse Trigonometric Functions; Real Valued Functions; Arithmetic Combinations of Functions; Different types of Functions; Composition of Functions. CHAPT CHAPTER ER 4 : AV AVERAGE, ERAGE, RATIO RATIO & PROPORT PROPORTION ION — Avera ge ; Mixtu Mi xture re Probl Pr obl ems ; Uni tar y Rat io; io ; Comparison of Two Two or More Ratios; Ratios ; Composition of Ratios; Ratios Applications; Direct and Inverse Variation Variation Alligation; Partnership Partnership.. CHAPT APTER 5 : ALGE ALGEBR BRAA-11 — Elementary Algebra; Algebraic Expressions; Basic Rules of Algebra; Equations Eq uations and Solutions; Equation involving Fractional Expressions; Polynomial Equations; Equations with Fractions or Absolute Values; Simplifying Radicals; Rational Exponents; Problems Based on Ages. CHAPT HAPTER ER 6 : ALGEB ALGEBRA RA-2 -2 — Sequence and Series; Sum of Numbers; Logarithm; Arithmetic Geometric and Harmonic Mean; Progressions. Value CHAP CHAPTE TER R 7 : PERCEN PERCENT TAGE, PROFI PROFIT T & LOSS — Value and Percentage; Comparing Percentages; Simple and Compound Interest; Profit and Loss. CHAP HAPTER 8 : GEOM GEOMET ETRY RY — Basic Concepts in Geometry; Classification of Triangles; Quadrilateral; Coordinate Geometry; Equations of Parallel and Perpendicular Lines; Bisectors of Angles between two lines; Concurrence of Straight Lines; General Equation of Circles; Position of point with respect to a Circle. CHAPT HAPTER ER 9 : MENS MENSUR URAT ATIO ION N — Polygon; Circle; Surface Areas Volumes and Areas of Solids; Cube and Cuboid; Cylinder; Sphere; Pyramid; Conversion of Solid from One Shape to Another. Time and CHAPTER CHAPTER 10 : TIME, DISTANCE DISTANCE & WORK — Time Distance; Time, Work and Wages; Pipes and Cisterns; Calendar Cale ndar and Clocks; Problems on Trains, Boats, Aeroplane, Streams and Races. CHAPT APTER 11

PERMUTATION, PERMUTATION, COMBINATION

UNIT UNIT - II : DATA INTERP INTERPRETA RETATION TION,, DATA DATA SUFFICIENCY & REASONING CHAPT HAPTER ER 12 : LINE & BAR CHART — Caselet Based

Problems on Line Chart; Bar Chart (Histogram); (Histogram); Mix of both. Based Problems on Pie CHAPT CHAPTER ER 13 : PIE CHART CHART — Caselet Based Chart. CHAPT HAPTER ER 14 : DATA DATA TABULATION TABULATION — Caselet Based

Problems Proble ms on Data Tables; Combination of Data Table Table with with other graph. — Questions based on CHAPTER CHAPTER 15 : DATA DATA SUFFICIENCY SUFFICIENCY — Mathematical data and Logical Logic al Reasoning. — Questions based CHAPT CHAPTER ER 16 16 : LOGICAL LOGICAL REASONING REASONING — Questions on Fact, Judgement and Inference; Syllogism; Sentence and Conclusion; Arguments. CHAPTER 17 : ANALYTICA ANALYTICAL L REASONIN REASONING G — Arrangement;

Conditional Analysis; Relationships and Ass oci ati ons; Categ Ca teg ori sati on; Optim Opt imis isati ati on; Mathematical Reasoning; Decision Making.

UNIT - III : VERBAL AND READING COMPREHENSION — Synonyms and Antonyms; CHAPT CHAPTER ER 18 : VOCABULAR VOCABULARY Y — Analogy; Analogy; Odd One One Out; Dictionary Dictionary definition definition and Usage; Contextual Use of Word; Fillers. CHAPT HAPTER ER 19 : GRAMMAR — Spot the Error; Sentence

Improvement; Best way of writing the sentence; Grammatical incorrect or inappropriate sentences; Fill in the Blanks. CHAP HAPTER TER 20 : PARAGRAPH

CONSTRUCTION

— Parajumbles; Coherent paragraph; Missing or deleted part of sentences; Sentence Completion. Completi on.

CHAPTER CHAPTER 21 : CRITICAL CRITICAL REASONING REASONING — Questions based

on Information of Passage; Description of the unseasonable man in the passage; Essence of the text. CHAPTER CHAPTER 22 : READING COMPREHENSION COMPREHENSION Based on Passagess based based on SOCIAL SCIENCES — Passage

History, Geography, Psychology, Political Thoughts, Though ts, Sociolog Sociol ogy, y, Economy and Business Busine ss

































































































































QUANTITATIVE APTITUDE

1. 2.

3. 4.

5.

6. 7.

8.

9. 10. 11.

12.

What is is the smallest smallest numbe numberr which which when when increa increased sed by 5 is comple completely tely divisi divisible ble by 8, 11 and and 24? (1994) (a) 26 264 (b) 25 259 (c) 26 269 (d) No None of these Which is is the least least number number that that must must be subtract subtracted ed from 1856, 1856, so that that the remain remainder der when when divided divided by 7, 12 and 16 will will leave leave the (1994) same remainder 4 (a) 137 (b) 1361 (c) 140 (d) 172 Two positi positive ve intege integers rs differ differ by by 4 and and sum of their their recip reciprocal rocalss is 10/21. 10/21. Then Then one one of the the numb number er is (1995) (a) 3 (b) 1 (c) 5 (d) 21 Three bells bells chime chime at an interval interval of 18, 24 and and 32 minutes minutes respective respectively ly.. At a certain certain time time they they begin begin to chime chime together together.. What length length (1995) of time will elapse before they chime together again? (a) 2 hours 24 minutes (b) 4 hours 8 minutes (c) 1 hour 36 minutes (d) 5 hours For For the the prod produc uctt n(n n(n+1 +1)( )(2n 2n+1 +1), ), n  N, which one of the following is not necessarily true? (1995) (a) It is is alwa always ys even even.. (b) (b) Divi Divisi sibl blee by by 3. 3. (c) (c) Always Always divisible divisible by the sum of the square square of first n natural natural number numberss (d) (d) Neve Neverr divi divisi sible ble by 237 237.. (1995) The rema remaind inder er obtaine obtained d when when a prime prime number number great greater er than than 6 is divi divided ded by 6 is (a) 1 or 3 (b) 1 or 5 (c) 3 or 5 (d) 4 or 5 Cost of 72 hens hens is Rs .....96.7 .....96.7..... ..... Then, what what will will be be the cost of hen, hen, where where two two digits in place of “ ....... ....... ” are not visible visible or are written written in illegible hand-writing? (1995) (a) Rs. 3.23 (b) Rs. 5.11 (c) Rs. 5.51 (d) Rs. 7.22 Three consecuti consecutive ve positive positive even numbers numbers are such that that thrice the first first number number exceeds exceeds double double the third third number number by 2 then the third number is (1995) (a) 10 (b) 14 (c) 16 (d) 12 6 5 – 1 is divisible by (1995) (a) 13 (b) 31 (c) 5 (d) None of these (1996) If a number number 774958A96B 774958A96B is is to be divisible divisible by 8 and 9, the values values of A and B, respectiv respectively ely,, will be be (a) 7, 7, 8 (b) 8, 8, 0 (c) 5, 5, 8 (d) No None of these If n is any any odd odd num number ber greate greaterr than than 1, 1, then then n(n2 – 1) is (a) divisible by 48 al always (b) divisible by 24 al always Find Find the the valu valuee of of

1 1

1

2

3

1 3

1 2

(c)

divisible by by 6 always

(1996) (d) None of of th these (1996)

4

3

4

3

3



1 2

1 2

































































































































2

15. 16.

17.

18.

Topic-wise CAT Solved Papers

(1997) If n is an an integer, integer, how many many values of n will will give an integral integral value value of (16n (16n 2 + 7n + 6) / n? (a) 2 (b) 3 (c) 4 (d) None of these A student, student, instead instead of finding finding the value value of 7/8th of a number, number, found found the value value of 7/18th of the number number.. If his answer answer differed differed from (1997) the actual one by 770, find the number. (a) 1584 (b) 2520 (c) 1728 (d) 1656 If m and and n are integers integers divisible divisible by by 5, which which of the following following is is not necessarily necessarily true? (1997) 2 2 (a) m – n is divisible by 5 (b) m – n is divisible by 25 (c) m + n is divisible by 10 (d) None of these Which Which of the the follo followi wing ng is true true?? (1997) (a)

19.

20.

2

7 3  (73 ) 2

(b)

2

7 3  (7 3 ) 2

(c)

2

7 3  (7 3 ) 2

(d) Non None of of th these ese

P, Q and R are three consecutive odd odd numbers in ascending order. If the value of three times P is three less than two times R, find the value of R. (1997) (a) 5 (b) 7 (c) 9 (d) 11 A, B and and C are define defined d as follow followss : (1997) A  (2.000004)  [( 2.000004) 2  (4.000008)] B  (3.000003)  [(3.000003) 2  (9.000009)] C  ( 4.000002)  [(4.000002) 2  (8.000004)]

21. 22.

23.

24. 25.

26. 27.

28.

Which of the following is true about the value of the above three expressions? (a) All of them lie between 0.18 and 0.20 (b) A is twice of C (c) C is the smallest (d) B is the smallest (1997) P and Q are two two integers integers such that that (PQ) = 64. Which Which of the the following following cannot cannot be the value of P + Q? (a) 20 (b) 65 (c) 16 (d) 35 Five digit numbers numbers are formed formed using only 0,1,2,3,4 exactly once. What is the difference between the maximum maximum and minimum (1998) number that can be formed? (a) 19800 (b) 41976 (c) 32976 (d) None of these 3 n is odd. Which of the following statements is/are true? (1998) 2 2 I. n is odd II. n is i s odd III. n is even (a) I only (b) II only (c) I and II only (d) I and III only 2 (BE) = MPB, where B, E, M and P are distinct integers, then M = ? (1998) (a) 2 (b) 3 (c) 9 (d) None of these Three wheels wheels can complete complete respectiv respectively ely 60,36,24 60,36,24 revolutions revolutions per minute. minute. There is a red spot on each wheel that that touches touches the ground at time zero. After how much time, all these spots will simultaneously touch touch the ground again? (1998) (a) 5/2 seconds (b) 5/3 seconds (c) 5 seconds (d) 7.5 seconds A certain number number when when divided divided by 899 leaves leaves the remainder remainder 63. Find Find the remainder remainder when when the same same number number is divided divided by 29. (1998) (a) 5 (b) 4 (c) 1 (d) Cannot be determined A is the set set of positive integers integers such that that when when divided divided by 2,3,4,5 and 6 leaves leaves the remainders remainders 1,2,3,4 1,2,3,4 and 5 respectively respectively.. How (1998) many integer(s) between 0 and 100 belongs to set A? (a) 0 (b) 1 (c) 2 (d) None of these Number of students who who have opted opted the subjects subjects A, A, B, C are 60, 84, 108 respectively respectively. The examination examination is to to be conducted conducted for these

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Chapter 1 : N UMBER SYSTEM

32. 32. 33. 34.

35.

36. 37.

38. 39. 40. 41.

42. 43. 43. 44.

45.

n 2  12345678987654321, then , n  ?

3

(1999)

(a) 1246789 (b) 12345321 (c) 1111111 (d) 11111111 84 When 7 is divided by 342, what is the remainder? (1999) (a) 0 (b) 1 (c) 49 (d) 341 A, B, C are are three distinct distinct digits. digits. AB AB is a two two digit numb number er and CCB CCB is a three digit numbe numberr such that that (AB) (AB) 2 = CCB where CCB > 320. What is the possible value of the digit B? (1999) (a) 1 (b) 0 (c) 3 (d) 9 For the given given pair pair (x, y) of of positive positive intege integers, rs, such such that 4x – 17y = 1 and and x  1000, how many integer integer values of y satisfy the given conditions (1999) (a) 55 (b) 56 (c) 57 (d) 58 (2000) Conv Conver ertt 1982 1982 in base base 10 to bas basee 12 (a) 1129 (b) 1292 (c) 1192 (d) 1832 Let D be a recurrin recurring g decima decimall of the form D = 0.a 1 a2 a1a2a1a2.............where a1 and a2 lie between 0 and 9. Further at most one of them (2000) is zero. Which of the following numbers numbers necessarily produces an integer when multiplied by D? (a) 18 (b) 198 (c) 100 (d) 288 (2000) P is the product of all the prime prime numbers numbers between 1 to 100. Then the number number of zeroes at the end of P are (a) 1 (b) 24 (c) 0 (d) none of these N = 1421 × 1423 1423 × 1425 1425 what what is the remainde remainderr when when N is divide divided d by 12? (2000) (a) 0 (b) 1 (c) 3 (d) 9 (2000) xn is either –1 or 1 & n  4; If x1x 2 x 3 x 4  x 2 x 3 x 4 x 5  x 3 x 4 x 5 x 6  .....x n x1x 2 x 3  0 then n can be (a) o d d (b) even (c) prime (d) can't be determined There are two integers 34041 and 32506, when divided by a three-digit integer integer n, leave the the same remainder. remainder. What is the value of n? (2000) (a) 298 (b) 307 (c) 461 (d) can’t be determined (2000) If x, y and and z are odd odd integers integers then which which of the following following is necess necessarily arily false? false? (a) xyz is odd (b) (x – y) z is even (c) (x – y) (z + y) x is even (d) (x – y – z) (x + z) is odd (2000) 55 3  17 3  72 3 is divisible by (a (a) both 3 and 13 (b) both 7 and 17 (c) both 3 and 17 d) both 7 and 13 Out of 128 boxes of oranges, oranges, each each box contains contains at least 120 120 and at most 144 oranges. oranges. The number number of boxes boxes containing containing the the same number of oranges is at least (2001) (a) 5 (b) 10 103 (c) 6 (d) Ca Cannot be determined In a 4 - digit number, number, the sum of the first two two digits is equal to to that of the last two digits. The sum of the first and last digits is equal

































































































































4

51. 52.


(2002) When 2256 is divided by 17 the remainder would be (a) 1 (b) 16 (c) 14 (d) None of these At a book store, store, “MODERN BOOK BOOK STORE” STORE” is flashed using using neon neon lights. lights. The words are individu individually ally flashed flashed at interva intervals ls of 2

1 2

,4

1 4

,5

1 8

seconds respectively, and each word is put off after a second. The least time after which the full name of the

bookstore can be read again is (2002) (a) 49.5 seconds (b) 73.5 seconds (c) 1742.5 seconds (d) 855 seconds 53. After the division division of a number number successively by 3, 4 and 7, the remainders remainders obtained are 2, 1 and 4 respectively. respectively. What will will be the (2002) remainder if 84 divides the same number? (a) 80 (b) 76 (c) 41 (d) 53 m m m 54. If u, v, w and m are natura naturall numbers numbers such that u + v = w , then one of the following is true (2002) (a) 55. 56.

57. 58.

m  min( u , v, w )

(b)

m  max( u , v, w )

(c)

m  min( u , v, w )

(d) Non None of of th these ese

76n – 66n, where n is an integer > 0, is divisible by (2002) (a) 13 (b) 127 (c) 559 (d) All of these A positive positive whole whole number number M less than 100 is represented represented in base 2 notation, notation, base 3 notation, notation, and and base 5 notation. notation. It is found that in all three cases the last digit is 1, while in exactly two out of the three cases the leading digit is 1. Then M equals (2003C) (a) 31 (b) 63 (c) 75 (d) 91 How many many even even inte intege gers rs n, n, wher wheree 100  n (a) 40 (b) 37

 200, are divisible neither by seven nor by nine? (c)

39

(2003C)

(d) 38

The numb number er of positi positive ve integ integers ers n in the the range range 12  n  40 such that the product (n  1)( n  2)....3.2.1 is not divisible by n is (2003C) (a)

59.

5

(b) 7

What What is is the the rema remain inde derr whe when n 4 96 is divided by 6? (a) 0 (b) 2

(c)

13

(d) 14

(c)

3

(2003) (d) 4

Directions Directions for for Questions Questions 60 to 62 : Answer the questions on the basis of the information inf ormation given below. The seven basic symbols in a certain numeral system and their their respective values are as follow : I = 1 , V = 5 , X = 10, L = 50, C = 100, D = 500, and M = 1000 In general, the symbols in t he numeral system are read from left to right, starting with the s ymbol representing the largest value; the same symbol cannot occur continuously continuously more than three three times; the value of the numeral numeral is the sum of the values of the symbols. For example , XXVII = 10 + 10 + 5 + 1 + 1= 27. 27 . An exception exception to the left to right r ight reading occurs when a symbol is followed

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66.

Let a, b, c, d and e be integers such that a = 6b = 12c, and 2b = 9d = 12e. Then which of the following pairs contains conta ins a number (2004) that is not an integer? (a)

67. 68. 69. 70.

71.

72.

74.

 a , b     27 e 

(b)

If R 

30 65  29 65 30 64  29 64

77

 a , bd     12 18 

(d)

 a , c     6 d 

0  R  0.1

(b)

0.1  R  0.5

(c)

0.5  R  1.0

(d) R > 1.0

For For a posi positi tive ve int integ eger er n, n, let let pn denote the product of the digits of n, and s n denote the sum of the digits digits of n. The number of (2005) integers between 10 and 1000 for which p n + sn = n is (a) 81 (b) 16 (c) 18 (d) 9 If x = –0.5, –0.5, then which which of the the followin following g has the the smallest smallest value? value? (2006) (a)

76.

(c)

, then

1

75.

 a , c     36 e 

If a, a + 2 and a + 4 are prime numbers, then then the number of possible solutions for for a is (2004) (a) one (b) tw t wo (c) three (d) mo more than three 23 23 The rema remain inde der, r, when hen (15 (15 + 23 ) is divided by 19, is (2004 - 2 marks) (a) 4 (b) 15 (c) 0 (d) 18 3 3 3 3 (2005) If x = (16 + 17 + 18 + 19 ), then x divided by 70 leaves a remainder remainder of (a) 0 (b) 1 (c) 69 (d) 35 The digits of a three-digit three-digit numbe numberr A are are written written in the reverse reverse order to form another another three-digit three-digit numbe numberr B. If B > A and B–A is perfectly divisible by 7, then which which of the following is necessarily true? (2005 - 2 marks) (a) 100 < A < 299 (b) 106 < A < 305 (c) 112 < A < 311 (d) 118 < A < 317 2720 The righ rightm tmost ost non non-ze -zero ro digit digit of of the the numbe numberr 30 is (2005 - 2 marks) (a) 1 (b) 3 (c) 7 (d) 9

(a) 73.

5

2x

(b)

1

(c)

1

(d) 2x

(e)

1

x x x2 1/2 1/3 1/4 1/6 1/12 (2006) Whic Which h one one amon among g 2 , 3 , 4 , 6 and 12 is the largest? (a) 2 1/2 (b) 31/3 (c) 41/4 (d) 61/6 (e) 121/12 Consider four digit digit numbers for which which the first first two digits are are equal and the last two digits are also equal. How many such numbers numbers are perfect squares? (2007) (a) 1 (b) 3 (c) 2 (d) 4 (e) 0 How many many pairs pairs of of posit positive ive integ integers ers m, n satisfy

































































































































6

1. 2.

3.


(b) (b) Requ Requir ired ed no. no. = LCM LCM of of ( 8, 11, 11, 24 24 ) –5 = 264 264 – 5 = 259 259 (d) Suppose least no. be x

(a)

1856  x  n ( LCM of 7,12,16)  4 or 1856 – x = n (336)+ 4 we should take n = 5 so that n(3 36) is nearest to 1856 and n (336)< 1856 1856 – x = 1680 + 4 = 1684 1 684 x = 1856 – 1684 = 172 Let Let tw two posi positi tive ve int integer egerss be x and y. .....(i)  x–y=4 1 and

4.

(b)

5.

(d) (d)

6.

(b)

7.

(c)

8.

(b)

x



1 y



xy

10 21

or

xy



10 21

.....(ii)

It is clear from second equation that x and y will be 3 and 7. L.C.M of 18, 24 and 32 = 288 Hence they would chime together after every 288 min. or 4 hrs. 8 min. It is clea clearr tha thatt for for n = 237 237 the the expr expres essi sion on n(n n(n + 1) (2n + 1) is divisible by 237. Hence option (d) is not necessarily true. It is clearly 1 or 5 Example : 7 divided by 6 leaves remainder 1 11 divided divide d by 6 leaves remainder 5 13 divided by 6 leaves remainder 1. Mult Multip iply ly each each opti option on by 72 and and fin find d out out the the res resul ultt which matches the visible digits. Clearly we see 72  5.51  396.72 Let x  2, x , x  2 be the 3 con secutive numbers then, 3 (x – 2) = 2 (x + 2) + 2 (according to the question) or 3x  6  2 x  6  x  12 Hence, the 3rd no. is 14.

1



1

1 3

3

 3

4 2

2 5

4 2 3 3

1

 1

1 20 3 12



3 12

3

11

1 33 1 33 4 33 45 15        12 21 3 21 7 21 21 7 1 1 16 4 For Qs. 13-14. Let quantity sold = q ; price = p and p', q' be the wrong price and quantity respectively and q' = q + 54 pq = sales = Rs 1148 = p'q'



 q  13.

14.

1148 p



1148 p'

 54

.....(i)

(b) (b) As q is an an inte intege gerr, both both p and and p' must must div divid idee 114 1148. 8. Now Now checking the options : (a) 1148 1148 is divisi divisible ble by by 82 and and 28 but does does not not satisfy (i) (b) (b) 1148 1148 is divisi divisible ble by by 41 and and also by 14. 14. It gives gives q = 28 and satisfies (i) (c) (c) 1148 1148 is divis divisibl iblee by 56 and and not not by 65. (d) (d) 1148 1148 is also also divisi divisible ble by 28 and 82 but does does not not satisfy (i). (a) (a) From From opti option on (b) of previo previous us ques questio tion, n, 28 is the the qua quanti ntity ty sold and the price is Rs 41. which is only possible in Alternatively : q' = q + 54, which case of 28 and 82 as given in options. 1148

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17.

18.

(c)

(b)

7

If m & n are are inte intege gers rs divi divisi sibl blee by by 5. 5. Then, (m + n) might be or might not be divisible by 10. For example: If m = 5 and a nd n = 10 then m + n = 15 which is not divisible by 10. But if m = 5, n = 25 then m + n = 30 which is divisible by 10. 7

32

 ( 7)

32

 7 while (73 )2  76  79 9

28.

2 Hence, 7 3  (73 ) 2

19.

20.

21.

22.

23. 24.

25

(c)

(d)

Let Let P, P, Q and and R be n, n, n + 2 and and n + 4 resp respeectiv ctivel ely y in in ascending order. According to the Question 3n = 2(n + 4) – 3 = 2n + 5  n =5 Thus, R = 5 + 4 = 9 A

2

 

1

; B

3



1

; C

4



So, number of rooms 

29.

1

18 6 8 4 16 4  B is the smallest. (d) Given Given PQ = 64 = 1 × 64 = 2 × 32 = 4 × 16 16 = 8 × 8. Corresponding values of P + Q are 65, 34, 20, 16. Therefore, P + Q cannot be equal to 35. (c) Maximum no. = 43210 Minimum no. = 10234 Hence, difference difference = 43210 – 10234 = 32976 3 (c) If n is odd then n and n 2 will also be odd. (b) (BE)2 = MPB If LHS has square, then according to question unit’s digit of RHS can be 0, 1, 4, 5, 6, 9 If B = 0, then (BE) 2 cannot be a three digit number If B  1 then LHS exceeds 3 digits d igits and is not compatible with RHS. So B = 1  E = 1 or 9 1 is rejected since B & E are distinct integers hence BE = 19  M = 3 (c) 1st 1st wh wheel eel ma makes kes 1 rev rev.. per per sec sec

LCM of 2, 3, 4, 5, 6 is 60. Hence the elements of A will be of the form 60n – 1, where n is any integer. Only for n is equal to 1 the number (60 – 1= 59) will be between 0 and 100. Hence, only one integer between 0 and 100 belongs to A. (d) (d) For Subje Subject ct A – 60 60 stu stude dent nts; s; Subj Subjec ectt B – 84 84 stude student nts; s; Subject C – 108 students HCF of 60, 84 and 108 is 12. So, each room co ntain 12 students at minimum. minimum. But each room contains students of only 1 subject

30.

31.

60



108



84

= 21 12 12 12 (b) The The dig digit it in the the uni unit’ t’ss pla place ce of 2 51 is equal to the remainder when 2 51 is divided by 10. 25 = 32 leaves the remainder 2 when when divided by 10. Then 250 = (25)10 leaves the remainder 2 10 = (25)2 which in turn leaves the remainder 2 2 = 4. Then 251 = 250× 2, when divided by 10, leaves the remainder 4 × 2 = 8. (c)

Given number  (1234...51525) 1000  354

Since 1000 = 8 × 125 So, remainder when 354 divided by 8 be 2 Required remainder = 2. (d) (d) Let Let the the fou fourr conse consecu cuti tive ve numb number er be be (a – 2), 2), (a – 1), a, (a + 1). Multiplying these, we get (a 4

 2a 3  a 2  2a ) , which will always be even.

By the problem we add 1. Thus the expression becomes (a 4

 2a 3  a 2  2a  1) ,

which is odd. This is also the perfect square of (a 2

 a  1) .

You can also take any four consecutive numbers and check for the validity.



































































































































8


Further also note that every 4th value of y (e.g. 3, 7, 11, ......) . .....) will give an integer value of x. So, number of values of y = 36.

(b) 12

1982 165 13 1

235 4

45.

(b)

2 9 1 1

38.

99D  a1a 2

D

a1a 2

99 Required number should be the multiple of 99. So we can get an integer when multiplied by D. Hence, 198 is the required number. (a) (a) There There are are only only 2 prim primee num numbe bers rs 5 & 2 betw betwee een n 1 & 100 100 which when multiplied multiplied will give zero in the end. Thus there will be only one zero at the end of the product of given number.

.....(1)

AD C

.....(2)

B  D  2( A  C )

.....(3)

 B  2D

Putting in (3), 3D  2(A  C)  2(A  A  D)

46.

100D  a1a 2 . a1a 2 100D = a 1a2 . D

ABCD

(1) – (2) gives B – D = D

D  0. a1a 2 Multiplied by 100 on both side



Let Let ABCD ABCD be the the 4-di 4-dig git num number ber According to the question, we have

= 58

Thus, 1982 (10) = 1192 (12) 37.

(a)

or 3D = 4A + 2D or D = 4A & B = 8A Putting these values in (2), C = A + 4A = 5A This can only be true for A = 1, hence C = 5. (d) Let Let the the oth other multi ultipl plie ierr be be x

 

47.

(a)

53x  35x  540

 x  30

New product = (53 × 30) = 1590 The The pro produ duct ct of 44 and and 11 11 is is 484 484.. But given product of 44 and 11 = 3414 (in number system) Here, 3x 3  4 x 2  1x1  4  x 0  484

 3x 3  4x 2  x  480 This equation is satisfied only when x = 5. In decimal system, the number 3111 can be written as

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52.

(c) (c)

9

Full Full name name of the the book booksto store re can be read read again again by taking taking LCM of the times



63.

5 17 41 , , 2 4 8

LCM of (5, 17, 17, 41 41) HCF of (2, 4, 8)



348 3485 2

13

So, S 

 1742.5 seconds

(d) (d) Acco Accord rdin ing g to to the the quest questio ion n the the requi require red d no. no. is is 3[4 (7x + 4) + 1] + 2 = 84x + 53 So the remainder is 5 53, 3, when the same number is divided by 84.

54.

(c)

We have u m  v m  w m where u, v, w, w, m are natural natur al numbers

64.

7 6n

65.

 6 6 n  7 6  6 6  (7 3 ) 2  (6 3 ) 2

 (7 3  63 )(7 3  63 )  (127)(7 3  63 )  (127)(559) 56.

This number is divisible by 127, 559 and 13. (d) (d) 63 and and 75 75 are are rul ruled ed out out as the their ir last last digit digit can can’t ’t be be 1. Converting to base 2, 3, and 5, we get

21  7

4 7 Put x = 2 and y = 6 and check for the options. By hit and trial all the 3 options can be proved wrong (d) Let n = 6 (b) 7 × 8 = 56

( c)

Therefore n  6  2.4 Now, Now, the the divisor of 6 are 1, 2, 3 If we take 2 as divisor then Statement I is true.

 66n , where n is integer > 0

Let n = 1, then 7 6 n

(d) x is prime say 7 y is not prime but com composite posite no. say 8, 9, 21 (a) 9 – 7 = 2

Take u = 2, v = 4, w = 6; then 2 m  4 m  6 m This will be true if m = 1 and 1 < min ( 2, 4 , 6) = 2 Hence, m < min (u, v, w) (d)

 (7n  3)  7 13  7  39  676 n 1

53.

55.

(b) Numb Number er is of the the form form = 7n 7n + 3; 3; n = 1 to 13

66.

n  2  1.

If we take 3 as divisor then 6  3  2.4, i.e. n  n Therefore statement II is true (d) Given a = 6b 6b = 12c = 27d 27d = 36 36e Multiplied and Divide by 108 in whole expression 108a 108 1



108b 18 1



108c 9 1

 1

108d 4 1



108e 3

































































































































10

71.

72.


As B > A  z > x ...(i) B – A = 99z – 99x = 99(z – x) As 99 is not divisible by 7 so (z – x) has to be divisible by 7. ...(ii) Using (i) & (ii), the only possible values of z and x are (8, 1) and (9, 2) So the minimum and maximum m aximum range of A are 108 and 299, which  106  A  305 (a) The number 302720 will have 2720, zero's. For the right most non-zero digit we have to check the power cycle of 3 and find when their multiplication again leads to a 3 as the right most digit. 31 = 3; 3 2 = 9; 33 = 27; 34 = 81; 35 = 243 Hence, 3 will appear after every fourth power of 3. Hence, 30 2720 = 32720 × 102720 = (34)680 × 102720 As the number 2720 is an exact multiple of 4, hence the last digit will be 1 similar to what we find in 3 4. (d) As xn – yn is divisible by x – y if n is odd. xn – yn = (x – y) (x n – 1y0 + xn – 2y1 +..........+ x 0yn – 1) Hence numerator becomes = (30 – 29) (30 64 +........+29 64) = 3064+........+ 29 64



R 

30 64  ......  29 64 64

64

30  29 Clearly the numerator is greater than the denominator.

76.

(a) (a)

77.

(a)

Since Since in in the the four four digit digitss numb number er firs firstt two two digit digitss are are equa equall and the last two digits are also equal, therefore we can suppose that the digit at the thousand and hundred place each each be x and the digit at the tenth and unit place each be y. Hence, the four digits number = 1000 x + 100 x + 10 y + y = 11(100 x + y) This number 11(100 x + y ) will be perfect square, if 100 x + y is of the form 11 n, where n is a perfect square Now 100 x + y = 11n  y = 11n – 100 x On checking, we get for the value n = 64 (a perfect square) only, y = 704 – 100 x, for which a single digit positive integral integral value 7 of x, the value of y = 4, which is the single digit positive integer. There is no single digit positive integral value of y for any other single positive integral value of x for the equation y = 704 – 100 x Hence, 7744 is the only for digits number. 1 m





4 n



1 12

12n + 48m – mn – 576 = – 576 m  12 

576

n  48 Since n is an odd, therefore, ( n – 48) is an odd.

...(i)

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Directions Directions for questions questions 1 to to 3 : Read the information given below and answer the questions that follow : Ghoshbabu is s taying at Ghosh Housing S ociety, ociety, Aghosh Colony, Colony, Dighoshpur, Calcutta. In Ghosh Housing Society 6 persons read daily Ganashakti and 4 read Anand Bazar Patrika; In his colony there is no pers on who reads both. Total number of persons who read these two newspapers in Aghosh Colony and Dighoshpur is 52 and 200 respectively. Number of persons who read Ganashakti in Aghosh Aghosh Colony and Dighoshpur is 33 and 121 respectively; while the persons who read Anand Anand Bazar Patrika in Aghosh Colony and Dighoshpur are 32 and 117 respectively. respectively. 1. 2. 3.

Numbe Numberr of perso persons ns in in Digho Dighoshp shpur ur who who read read only only Gana Ganasha shakt ktii is (a) 121 (b) 83 (c) 79 Number Number of perso persons ns in Aghosh Aghosh Colon Colony y who who read read both of these these new newspap spapers ers is is (a) 13 (b) 20 (c) 19 Number Number of pers persons ons in in Aghos Aghosh h Colon Colony y who who read read only only one one new newspap spaper er (a) 29 (b) 19 (c) 39

(1994) (d) 127 (1994) (d) 14 (1994) (d) 20

































































































































12

11.


(2000)

There There are are tw two disjo disjoin intt sets sets S1 and S2 where S1  {f (1), f (2), f (3).............} S2  {g (1), g (2), g (3).............} such that S 1  S2 forms the set of natural number. number. Also f (1)  f (2)  f (3)............. & g (1) < g (2) < g(3) and f (n) = g ((g (n)) + 1 then what is g (1)? (a)

12.

0

(b) 1

(c)

2

(d) can't be determined

Let Let ‘f’ be a funct functio ion n from from set A to set B for a set, XCB XCB defin definee f –1 (X) = {x  A : f ( x )  X} Then which of the following is (2000)

necessarily true for a subset U of X? (a) f ((f 1 ) U )  U

(b) f (f 1 ( U))  U

(c) f {f 1 ( U ))  U

(d) f (f 1 ( U ))  U

Directions Directions for questions questions 13 to 16 : Read the information given below and answer the questions ques tions that follow : A and B are two sets (e.g. A = mothers, B = women). The elements elements that could belong to both the sets (e.g. women who are mothers) is given by the set C = A.B. A.B. The elements which could belong to either A or B, or both, is indicated by the set D = A  B. A set that does not contain any elements is known as null null set, represented by is a mother, then C = A.B. is a null set , or C =

 (for example, if none of the women in the set B

 ).

Let ‘V’ signify the set of all vertebrates; ‘M’ the set all mammals; mammals; ‘D’ dogs, ‘F’ fish; ‘A’ ‘A’ alsatian and ‘P’, a dog named Pluto. 13.

Given that X = M.D is such such that X = D, which which of the the followin following g is true? true? (a) All dogs are mammals (b) Some dogs are mammals (c)

14.

X=



If Y = F.(D.V) F.(D.V),, is not not a null null set, set, it implies implies that that :

(2001)

(d) (d) All All mam mamma mals ls are are dog dogss (2001)

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Chapter 2 : S ET TEHORY

19. 20.

What What is the small smallest est positi positive ve inte integer ger n such that g n = e? (a) 4 (b) 5 Upon Upon simp simpli lifi fica cati tion on,, f (a)

21.

(2003) (c)

2

(d) 3

f)}] equa equals ls  [f * { f  (f * f)}]

e

(b) f

(2003) (d) g

(d) h

Upon Upon simp simpli lifi fica cati tion on,, {a 10 * (f 10  g 9 )}  e 8 equals (a)

22.

13

e

(b) f

(2003) (d) g

(d) h

A survey survey on a sample sample of 25 new cars cars being sold sold at a local local auto dealer dealer was conduct conducted ed to see which which of the the three popular popular (2003) options — air-conditioning, radio and power windows - were already installed. The survey found 15 had air conditioning 2 had air conditioning and power windows windows but no radios 12 had radio

































































































































14

1.


(b) (b) Supp Suppos osee G and and A are are repr repres esen ente ted d by by Gan Ganas asha hakt ktii and and Anand Bazar Patrika respectively

Here,

n (A  L  B)  100  5  95 n (A  L)  n (L  B)  n( B  A)

Anand Bazar Ganashakti = 121 Patrika = 117

 3n( A  L  B)  20 n(ALB) = n(A) + n(L) + n(B) – n(A  L) – n(L  B)

121 – x

x

117 – x

– n(B  A) + n(A  L  B) 95  85  30  20  [n ( A  L)  n (L  B)

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Chapter 2 : S ET TEHORY

15

11.

BBC = 44

CNN = 30

(b) It is is giv giveen th that f(n) f(n) = g(g g(g(n (n)) )) + 1 Therefore, f (n )  g (g (n )) Also, g( g (1)  g( 2)  g(3) shows that the function g(x)

44 – x

30 – x x

is an increasing function. So for a natural number n, g(n )  n

 g (g( n ))  g ( n ) 160 – x DD = 160 According to the given conditions,

Thus, f (n )  g (n ) for every n or f (1)  g (1)  g (1) is the least number in S1  S 2 . Now, Now, S1  S2= set of natural numbers. Therefore, 1 in S 1  S2 is the smallest number. Thus, g(1) = 1.

































































































































16

19.

20.


(a)

g *g  h

(d)

 (g * g) * g  h * g  f  g * (g * g * g)  g * f  e  n = 4 f  [f * {f  (f * f )}]  f  [f *{f  h )}]

23.

 f   [f * e]  f   [f ] = h 21.

(a)

Clearly, a10 = a f 10  h * f * f 7  g * f 7  e * f 6  f * f 5  f * f  h 9

4

4

4

g  g * g * g  (e * g) * g  g * e  g e8  e

24.

Cars with no options = 25 – 23 = 2. (c) Empl Employ oyee eess who who does doesn’ n’tt hav havee VC VCD = 100 100 – 70 = 30% 30% Employees who who doesn’t have MWO = 100 – 75 = 25% Employees who doesn’t have AC = 100 – 80 = 20% Employees who doesn’t have WM = 100 – 85 = 15%  Total employees who doesn’t have atleast one of the four equipments equipments = 30 + 25 + 20 + 15 = 90%  Percentage of employees having all four gadgets = 100 – 90 = 10%. (b) (b) Putt Puttin ing g the the give given n infor informa matio tion n in the the form form of a venn venn diagram, we get

J 33

7

8

A

6 2 0 7 _ D B E

































































































































Directions Directions for questions 1 & 2 : Read the information given below and answer the questions that follow :

If md (x) = | x |, mn (x, y) = minimum of x and y and Ma (a, b, c, ....) = maximum of a, b, c, ....



































































































































18

10.


For what hat val value uess of a le le (a2 – 3a, a – 3) < 0 ? (a) 1 < a < 3

11

Lar

t val

(b) a < 0 and a < 3 of mi (2 +

2, 6 – 3x), when x > 0 is

(1995)

(c) a < 0 and a

 3

(d) a < 0 or a > 3 (1995)

6 2 0 7 _ D B E



































































































































Chapter 3 : F UNCTIONS

21.

19

If y = xx,, wh which ich of of the foll follow owing ing will will give give x2 as the final value ? (a) f (x, y) G (x, y)4

(b) G (f(x, y) f (x, y), F (x, y)/8

(c) (c) –F (x (x,, yy)) G (x (x,, yy)/l )/log og216

(d) –f (x, y) G (x, y) F (x, y)/F (3x, 3y)

(1999)



































































































































20


27.

There is a set of 'n' natural numbers. numbers. The function function 'H' is is such that it finds the highest common factor factor between any two two numbers. What is the minimum number of times that the function has to be invoked to find the H.C.F. of the given set of numbers? (1999) (a) 1/ 1/2 n (b) n – 1 (c) n (d) None of these

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Chapter 3 : F UNCTIONS

21

2



































































































































22

44. 44.


Suppose, for any real number x, [x] denotes the greatest integer less than or equal to x. Let L (x, y) = [x] + [y] + [x + y] and R(x,y)= [2x] + [2y]. Then it’s impossible impossible to find any two t wo positive real numbers x and y for which (a) L(x, y) = R (x, y)

(b) L(x, y)  R( R(x, y)

(c) L(x, y) < R (x, y)

(d) L(x, y) > R(x, y)

(2002)

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23_years_CAT_Topic_wise_Solved_Papers.pdf

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