Actual CAT Problems 1998-2006
CAT 1998
1.
n
3is odd. Which of the following statement(s) is(are) true?
I. n is odd.
II. n
2is odd.
III. n
2is even.
a. I only
b. II only
c. I and II
d. I and III
2.
(BE)
2= MPB, where B, E, M and P are distinct integers. Then M =
a. 2
b. 3
c. 9
d. None of these
3.
Five-digit numbers are formed using only 0, 1, 2, 3, 4 exactly once. What is the difference between
the maximum and minimum number that can be formed?
a. 19800
b. 41976
c. 32976
d. None of these
4.
A certain number, when divided by 899, leaves a remainder 63. Find the remainder when the same
number is divided by 29.
a. 5
b. 4
c. 1
d. Cannot be determined
5.
A is the set of positive integers such that when divided by 2, 3, 4, 5, 6 leaves the remainders 1, 2, 3,
4, 5 respectively. How many integers between 0 and 100 belong to set A?
a. 0
b. 1
c. 2
d. None of these
6.
How many five-digit numbers can be formed using the digits 2, 3, 8, 7, 5 exactly once such that the
number is divisible by 125?
a. 0
b. 1
c. 4
d. 3
7.
What is the digit in the unit’s place of 2
51?
a. 2
b. 8
c. 1
d. 4
8.
A number is formed by writing first 54 natural numbers in front of each other as 12345678910111213
... Find the remainder when this number is divided by 8.
a. 1
b. 7
c. 2
d. 0
CAT 1999
9.
Let a, b, c be distinct digits. Consider a two-digit number ‘ab’ and a three-digit number ‘ccb’, both
defined under the usual decimal number system, if
(ab)
2=
ccb
>
300,
then the value of b is
a. 1
b. 0
c. 5
d. 6
10.
The remainder when
847 is divided by 342 is
a. 0
b. 1
c. 49
d. 341
11.
If
n
=
1
+
x
where x is the product of four consecutive positive integers, then which of the following
is/are true?
A. n is odd
B. n is prime
C. n is a perfect square
a. A and C only
b. A and B only
c. A only
d. None of these
12.
For two positive integers a and b define the function h(a,b) as the greatest common factor (G.C.F) of
a, b. Let A be a set of n positive integers. G(A), the G.C.F of the elements of set A is computed by
repeatedly using the function h. The minimum number of times h is required to be used to compute
G is
a.
2
1
n
b. (n – 1)
c. n
d. None of these
13.
If n
2= 12345678987654321, what is n?
a. 12344321
b. 1235789
c. 111111111
d. 11111111
Directions for questions 14 to 16: Answer the questions based on the following information.
There are 50 integers a
1, a
2… a
50, not all of them necessarily different. Let the greatest integer of these 50
integers be referred to as G, and the smallest integer be referred to as L. The integers a
1through a
24form
sequence S1, and the rest form sequence S2. Each member of S1 is less than or equal to each member
of S2.
14.
All values in S1 are changed in sign, while those in S2 remain unchanged. Which of the following
statements is true?
a. Every member of S1 is greater than or equal to every member of S2.
b. G is in S1.
c. If all numbers originally in S1 and S2 had the same sign, then after the change of sign, the
largest number of S1 and S2 is in S1.
d. None of the above
15.
Elements of S1 are in ascending order, and those of S2 are in descending order. a
24and a
25are
interchanged. Then which of the following statements is true?
a. S1 continues to be in ascending order.
b. S2 continues to be in descending order.
c. S1 continues to be in ascending order and S2 in descending order.
d. None of the above
16.
Every element of S1 is made greater than or equal to every element of S2 by adding to each element
of S1 an integer x. Then x cannot be less than
a. 2
10b. the smallest value of S2
c. the largest value of S2
d. (G – L)
CAT 2000
17.
Let D be a recurring decimal of the form D = 0. a
1a
2a
1a
2a
1a
2..., where digits a
1and a
2lie between
0 and 9. Further, at most one of them is zero. Which of the following numbers necessarily produces
an integer, when multiplied by D?
a. 18
b. 108
c. 198
d. 288
18.
Consider a sequence of seven consecutive integers. The average of the first five integers is n. The
average of all the seven integers is
a. n
b. n + 1
c. k × n, where k is a function of n
d.
n
2
7
+
19.
Let S be the set of integers x such that
I.
100
≤ ≤
x
200
,
II. x is odd and
III. x is divisible by 3 but not by 7.
How many elements does S contain?
a. 16
b. 12
c. 11
d. 13
20.
Let x, y and z be distinct integers, that are odd and positive. Which one of the following statements
cannot be true?
a. xyz
2is odd
b. (x – y)
2z is even
c. (x + y – z)
2(x + y) is even
d. (x – y)(y + z)(x + y – z) is odd
21.
Let S be the set of prime numbers greater than or equal to 2 and less than 100. Multiply all elements
of S. With how many consecutive zeros will the product end?
a. 1
b. 4
c. 5
d. 10
22.
What is the number of distinct triangles with integral valued sides and perimeter 14?
a. 6
b. 5
c. 4
d. 3
23.
Let N = 1421 × 1423 × 1425. What is the remainder when N is divided by 12?
a. 0
b. 9
c. 3
d. 6
24.
The integers 34041 and 32506, when divided by a three-digit integer n, leave the same remainder.
What is the value of n?
a. 289
b. 367
c. 453
d. 307
25.
Each of the numbers
x , x ,
1 2L
, x , n
n≥
4,
is equal to 1 or –1. Suppose
1 2 3 4 2 3 4 5 3 4 5 6 n 3 n 2 n 1 n n 2 n 1 n 1 n 1 n 1 2 n 1 2 3
x x x x
+
x x x x
+
x x x x
+
L
+
x
−x
−x
−x
+
x
−x
−x x
+
x
−x x x
+
x x x x
=
0,
then
a. n is even
b. n is odd
26.
Sam has forgotten his friend’s seven-digit telephone number. He remembers the following: the first
three digits are either 635 or 674, the number is odd, and the number 9 appears once. If Sam were
to use a trial and error process to reach his friend, what is the minimum number of trials he has to
make before he can be certain to succeed?
a. 10,000
b. 2,430
c. 3,402
d. 3,006
27.
Let N = 55
3+ 17
3– 72
3. N is divisible by
a. both 7 and 13
b. both 3 and 13
c. both 17 and 7
d. both 3 and 17
28.
Convert the number 1982 from base 10 to base 12. The result is
a. 1182
b. 1912
c. 1192
d. 1292
CAT 2001
29.
Let x, y and z be distinct integers. x and y are odd and positive, and z is even and positive. Which
one of the following statements cannot be true?
a. y(x – z)
2is even
b. y
2(x – z) is odd
c. y(x – z) is odd
d. z(x – y)
2is even
30.
In a four-digit number, the sum of the first 2 digits is equal to that of the last 2 digits. The sum of the
first and last digits is equal to the third digit. Finally, the sum of the second and fourth digits is twice
the sum of the other 2 digits. What is the third digit of the number?
a. 5
b. 8
c. 1
d. 4
31.
Anita had to do a multiplication. In stead of taking 35 as one of the multipliers, she took 53. As a
result, the product went up by 540. What is the new product?
a. 1050
b. 540
c. 1440
d. 1590
32.
x and y are real numbers satisfying the conditions 2 < x < 3 and – 8 < y < –7. Which of the following
expressions will have the least value?
a. x
2y
b. xy
2c. 5xy
d. None of these
33.
In a number system the product of 44 and 11 is 3414. The number 3111 of this system, when
converted to the decimal number system, becomes
a. 406
b. 1086
c. 213
d. 691
34.
All the page numbers from a book are added, beginning at page 1. However, one page number was
added twice by mistake. The sum obtained was 1000. Which page number was added twice?
a. 44
b. 45
c. 10
d. 12
35.
If a, b, c and d are four positive real numbers such that abcd = 1, what is the minimum value of
(1 + a)(1 + b)(1 + c)(1 + d)?
a. 4
b. 1
c. 16
d. 18
36.
A set of consecutive positive integers beginning with 1 is written on the blackboard. A student came
along and erased one number. The average of the remaining numbers is
17
7
35
. What was the
number erased?
37
Let b be a positive integer and a = b
2– b. If
b
≥
4
, then a
2– 2a is divisible by
a. 15
b. 20
c. 24
d. All of these
38.
In some code, letters a, b, c, d and e represent numbers 2, 4, 5, 6 and 10. We just do not know
which letter represents which number. Consider the following relationships:
I. a + c = e, II. b – d = d and III. e + a = b
Which of the following statements is true?
a. b = 4, d = 2
b. a = 4, e = 6
c. b = 6, e = 2
d. a = 4, c = 6
39.
Let n be the number of different five-digit numbers, divisible by 4 with the digits 1, 2, 3, 4, 5 and 6, no
digit being repeated in the numbers. What is the value of n?
a. 144
b. 168
c. 192
d. None of these
CAT 2002
40.
If there are 10 positive real numbers
n
1<
n
2<
n ...
3<
n
10, how many triplets of these numbers
(
n , n ,n
1 2 3) (
, n , n ,n
2 3 4)
, ... can be generated such that in each triplet the first number is always
less than the second number, and the second number is always less than the third number?
a. 45
b. 90
c. 120
d. 180
41.
Number S is obtained by squaring the sum of digits of a two-digit number D. If difference between
S and D is 27, then the two-digit number D is
a. 24
b. 54
c. 34
d. 45
42.
A rich merchant had collected many gold coins. He did not want anybody to know about him. One
day, his wife asked, " How many gold coins do we have?" After a brief pause, he replied, "Well! if I
divide the coins into two unequal numbers, then 48 times the difference between the two numbers
equals the difference between the squares of the two numbers." The wife looked puzzled. Can you
help the merchant's wife by finding out how many gold coins the merchant has?
a. 96
b. 53
c. 43
d. None of these
43.
A child was asked to add first few natural numbers (i.e. 1 + 2 + 3 + …) so long his patience
permitted. As he stopped, he gave the sum as 575. When the teacher declared the result wrong, the
child discovered he had missed one number in the sequence during addition. The number he missed
was
a. less than 10
b. 10
c. 15
d. more than 15
44.
When
2
256is divided by 17, the remainder would be
a. 1
b. 16
c. 14
d. None of these
45.
At a bookstore, ‘MODERN BOOK STORE’ is flashed using neon lights. The words are individually
flashed at the intervals of
2
1
s, 4
1
s and 5
1
s
2
4
8
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
46.
Three pieces of cakes of weights
4
1
lb, 6
3
lb and 7
1
lb
2
4
5
respectively are to be divided into parts of
equal weight. Further, each part must be as heavy as possible. If one such part is served to each
guest, then what is the maximum number of guests that could be entertained?
a. 54
b. 72
c. 20
d. None of these
47.
After the division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4
respectively. What will be the remainder if 84 divides the same number?
a. 80
b. 75
c. 41
d. 53
48.
If u, v, w and m are natural numbers such that
u
m+
v
m=
w ,
mthen which one of the following is
true?
a. m
≥
min(u, v, w)
b. m
≥
max(u, v, w)
c. m < min(u, v, w)
d. None of these
49.
6n 6n7
– 6
, where n is an integer > 0, is divisible by
a. 13
b. 127
c. 559
d. All of these
50.
How many numbers greater than 0 and less than a million can be formed with the digits 0, 7
and 8?
a. 486
b. 1,084
c. 728
d. None of these
CAT 2003 (Leaked Paper)
51.
How many even integers n, where
100
≤ ≤
n
200
, are divisible neither by seven nor by nine?
a. 40
b. 37
c. 39
d. 38
52.
A positive whole number M less than 100 is represented in base 2 notation, base 3 notation, and
base 5 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
a. 31
b. 63
c. 75
d. 91
Directions for question 53: Each question is followed by two statements, A and B. Answer each
question using the following instructions.
Choose (a) if the question can be answered by one of the statements alone but not by the other.
Choose (b) if the question can be answered by using either statement alone.
Choose (c) if the question can be answered by using both the statements together, but cannot be answered
by using either statement alone.
Choose (d) if the question cannot be answered even by using both the statements together.
53.
Is a
44< b
11, given that a = 2 and b is an integer?
A. b is even
B. b is greater than 16
54.
How many three digit positive integers, with digits x, y and z in the hundred's, ten's and unit's place
respectively, exist such that x < y, z < y and
x
≠
0
?
55.
If the product of n positive real numbers is unity, then their sum is necessarily
a. a multiple of n
b. equal to
n
1
n
+
c. never less than n
d. a positive integer
56.
The number of positive integers n in the range
12
≤ ≤
n
40
such that the product (n 1)(n
−
−
2)
K
3.2.1
is not divisible by n is
a. 5
b. 7
c. 13
d. 14
CAT 2003 (Re test)
Directions for questions 57 to 59: Answer the questions on the basis of the information given below.
The seven basic symbols in a certain numeral system and their respective values are as follows:
I = 1, V = 5, X = 10, L = 50, C = 100, D = 500 and M = 1000
In general, the symbols in the numeral system are read from left to right, starting with the symbol
representing the largest value; the same symbol cannot occur continuously more than three times; the
value of the numeral is the sum of the values of the symbols. For example, XXVII = 10 + 10 + 5 + 1 + 1
= 27. An exception to the left-to-right reading occurs when a symbol is followed immediately by a
symbol of greater value; then the smaller value is subtracted from the larger.
For example, XLVI = (50 – 10) + 5 + 1 = 46.
57.
The value of the numeral MDCCLXXXVII is
a. 1687
b. 1787
c. 1887
d. 1987
58.
The value of the numeral MCMXCIX is
a. 1999
b. 1899
c. 1989
d. 1889
59.
Which of the following represent the numeral for 1995?
I. MCMLXXV
II. MCMXCV
III. MVD
IV. MVM
a. Only I and II
b. Only III and IV
c. Only II and IV
d. Only IV
60.
What is the sum of all two-digit numbers that give a remainder of 3 when they are divided by 7?
a. 666
b. 676
c. 683
d. 77
61.
An intelligence agency forms a code of two distinct digits selected from 0, 1, 2, …, 9 such that the
first digit of the code is non-zero. The code, handwritten on a slip, can however potentially create
confusion when read upside down — for example, the code 91 may appear as 16. How many codes
are there for which no such confusion can arise?
a. 80
b. 78
c. 71
d. 69
62.
What is the remainder when 4
96is divided by 6?
a. 0
b. 2
c. 3
d. 4
63.
Let n (>1) be a composite integer such that n is not an integer. Consider the following statements:
A: n has a perfect integer-valued divisor which is greater than 1 and less than n
B: n has a perfect integer-valued divisor which is greater than
n
but less than n
64.
If a, a + 2 and a + 4 are prime numbers, then the number of possible solutions for a is
a. one
b. two
c. three
d. more than three
65.
Let a, b, c, d and e be integers such that a = 6b = 12c, and 2b = 9d = 12 e. Then which of the
following pairs contains a number that is not an integer?
a.
a
,
b
27 e
b.
a
c
,
36 e
c.
a bd
,
12 18
d.
a c
,
6 d
CAT 2004
66.
On January 1, 2004 two new societies s
1and s
2are formed, each n numbers. On the first day of
each subsequent month, s
1adds b members while s
2multiplies its current numbers by a constant
factor r. Both the societies have the same number of members on July 2, 2004. If b = 10.5n, what is
the value of r?
a. 2.0
b. 1.9
c. 1.8
d. 1.7
67.
Suppose n is an integer such that the sum of digits on n is 2, and 10
10< n 10
n. The number of
different values of n is
a. 11
b. 10
c. 9
d. 8
68.
Let
y
1
1
2
1
3
1
2
3 ...
=
+
+
+
+
What is the value of y?
a.
11 3
2
+
b.
11 3
2
−
c.
15
3
2
+
d.
15
3
2
−
69.
The reminder, when (15
23+ 23
23) is divided by 19, is
a. 4
b. 15
c. 0
d. 18
CAT 2005
70.
If
65 65 64 6430
– 29
R
30
29
=
+
, then
a. 0 < R
≤
0.1
b. 0 < R
≤
0.5
c. 0.5 < R
≤
0.1
d.R > 1.0
71.
If x = (16
3+ 17
3+ 18
3+ 19
3), then x divided by 70 leaves a remainder of
a. 0
b. 1
c. 69
d. 35
72.
let n! = 1 × 2 × 3 × … × n for integer n
≥
1
. If p = 1! + (2 × 2!) + (3 × 3!) + … + (10 × 10!), then p +
2 when divided by 11! Leaves a remainder of
73.
The digits of a three-digit number A are written in the reverse order to form another three-digit
number B. If B > A and B-A is perfectly divisible by 7, then which of the following is necessarily
true?
a. 100 < A < 299
b. 106 < A < 305
c. 112 < A < 311
d. 118< A < 317
74.
The rightmost non-zero digits of the number 30
2720is
a. 1
b. 3
c. 7
d. 9
75.
For a positive integer n, let p
ndenote the product of the digits of n and s
ndenote the sum of the
digits of n. The number of integers between 10 and 1000 for which p
n+ s
n= n is
a.81
b. 16
c. 18
d. 9
76.
Let S be a set of positive integers such that every element n of S satisfies the conditions
(a) 1000
≤
n
≤
1200
(b) every digit in n is odd
Then how many elements of S are divisible by 3?
a. 9
b. 10
c. 11
d. 12
CAT 2006
77.
If x = – 0.5, then which of the following has the smallest value?
1.
1x2
2.
1
x
3.
21
x
4. 2
X5.
−
1
x
78.
Which among
212
,
1 33
,
1 44
,
1 66
and
1 1212
is the largest?
1. 21/2
2. 31/3
3. 41/4
4. 61/6
5. 12/12
79.
A group of 630 children is arranged in rows for a group photograph session. Each row contains three
fewer children than the row in front of it. What number of rows is not possible?
1. 3
2. 4
3. 5
4. 6
5. 7
80.
The sum of four consecutive two-digit odd numbers, when divided by 10, becomes a perfect square.
Which of the following can possibly be one of these four numbers?
1. 21
2. 25
3. 41
4. 67
5. 73
81.
When you reverse the digits of the number 13, the number increases by 18. How many other
two-digit numbers increase by 18 when their two-digits are reversed?
A
nswers and Explanations
CAT 1998
1. c If n3 is odd, then n should also be odd. Hence, n2
should also be odd. And n2 will again be odd and not
even. So only I and II are true.
2. b Since MPB is a three-digit number, and also the square of a two-digit number, it can have a maximum value of 961 viz. 312. This means that the number BE should be
less than or equal to 31. So B can only take the values 0, 1, 2 and 3. Since the last digit of MPB is also B, it can only be 0 or 1 (as none of the squares end in 2 or 3). The only squares that end in 0 are 100, 400 and 900. But for this to occur the last digit of BE also has to be 0. Since E and B are distinct integers, both of them cannot be 0. Hence, B has to be 1. BE can be a number between 11 and 19 (as we have also ruled out 10), with its square also ending in 1. Hence, the number BE can only be 11 or 19. 112 = 121. This is not possible as
this will mean that M is also equal to 1. Hence, our actual numbers are 192 = 361. Hence, M = 3.
3. c The maximum and the minimum five-digit numbers that can be formed using only 0, 1, 2, 3, 4 exactly once are 43210 and 10234 respectively. The difference between them is 43210 – 10234 = 32976.
4. a The best way to solve this question is the method of simulation, i.e. take a number which when divided by 899 gives a remainder of 63. The smallest such number is (899 + 63) = 972. 972, when divided by 29 gives a remainder of 5. Hence, the answer is 5.
Students, please note that 899 itself is divisible by 29. Hence, the required remainder is the same as obtained by dividing 63 by 29, i.e. 5.
Shortcut:
Since 899 is divisible by 29, so you can directly divide the remainder of 63 by 29, so63
29 will give 5 as a remainder, option (a).
5. b Note that the difference between the divisors and the remainders is constant.
2 – 1 = 3 – 2 = 4 – 3 = 5 – 4 = 6 – 5 = 1
In such a case, the required number will always be [a multiple of LCM of (2, 3, 4, 5, 6) – (The constant difference)].
LCM of (2, 3, 4, 5, 6) = 60
Hence, the required number will be 60n – 1. Thus, we can see that the smallest such number is (60 × 1) – 1 = 59
The second smallest is (60 × 2) – 1 = 119
So between 1 and 100, there is only one such number, viz. 59.
6. c Let us find some of the smaller multiples of 125. They are 125, 250, 375, 500, 625, 750, 875, 1000 ... A five-digit number is divisible by 125, if the last three digits are divisible by 125. So the possibilities are 375 and 875, 5 should come in unit’s place, and 7 should come in ten’s place. Thousand’s place should contain 3 or 8. We can do it in 2! ways. Remaining first two digits, we can arrange in 2! ways. So we can have 2! × 2! = 4 such numbers.
There are: 23875, 32875, 28375, 82375. 7. b Since 2 has a cyclicity of 4,
i.e. 21 = 2, 22 = 4, 23 = 8, 24 = 16, 25 = 32, 26 = 64 ..., the
last digits (2, 4, 8, 6) are in four cycles.
∴ On dividing 51
4 , we get the remainder as 3.
∴ The last digit has to be 23 = 8
Shortcut:
Since cyclicity of the power of 2 is 4, so 251 can be
written in 24(12) + 3 or unit digit will be 23 = 8.
8. a The number formed by the last 3 digits of the main number is 354. The remainder is 2 if we divide 354 by 8. So the remainder of the main number is also 2 if we divide it by 8.
CAT 1999
9. a (ab)2 = ccb, the greatest possible value of ‘ab’ to be
31. Since 312 = 961 and since ccb > 300, 300 < ccb <
961, so 18 < ab < 31. So the possible value of ab which statisfies (ab)2 = ccb is 21. So 212 = 441,
∴a = 2, b = 1, c = 4.
10. b Note: 342 = 73 – 1. On further simplification we get,
(
)
28 3 28 28 342 1 (7 ) 343 342 342 342 + = = = = 342 N 1 1 342+ =342Hence, remainder = 1, optiion (b).
11. a Use the method of simulation, viz. take any sample values of x and verify that n is both odd as well as a perfect square.
12. b If there are n numbers, the function h has to be performed one time less.
13. c The square root is 11111111. 14. d None of the statements are true.
15. a S1 remains the same, but S2 changes.
16. d x must be equal to the greatest difference in the value of numbers of S1 and S2.
CAT 2000
17. b 99 × D = a1a2. Hence, D=a a1 2
99 . So D must be multiplied by 198 as 198 is a multiple of 99.
18. b Use any 7 consecutive numbers to check the answers.
(
1 2 3 4 5)
n 3 5 + + + + = = , average of 7 integers is(
1 2 3 4 5 6 7)
k 4 7 + + + + + + = = . So k = n + 1.Alternately, the average of the first 5 terms is the middle term which is third term, and the average of the first 7 terms is the middle term which is the fourth term. Hence, it is one more than the previous average. 19. d Numbers which are divisible by 3 (between 100 and 200) are 33. Numbers which are divisible by 21, i.e. LCM of 7 and 3 (between 100 and 200) are 5. Out of the 33 numbers divisible by 3, 17 are even and 16 are odd. Out of the 5 numbers divisible by 7, three are odd. Hence, the number of odd numbers divisible by 3 but not by 7 is (16 – 3) = 13.
20. d Take any three odd and positive numbers and check this out.
21. a There is only one 5 and one 2 in the set of prime numbers. Hence, there would be only one zero at the end of the resultant product.
22. c If the sides of the triangle are a, b and c, then a + b > c. Given a + b + c = 14.
Then the sides can be (4, 4, 6), (5, 5, 4), (6, 5, 3) and (6, 6, 2). Hence, four triangles are possible. 23. c N = 1421 × 1423 × 1425. When divided by 12, it shall
look like
(
1416 5) (
1416 7) (
1416 9)
. 12 + × + × +
Now the remainder will be governed by the term 5 × 7 × 9, which when divided by 12 leaves the remainder 3.
24. d Let r be the remainder. Then 34041 – r and 32506 – r are perfectly divisible by n. Hence, their difference should also be divisible by the same.
(34041 – r) – (32506 – r) = 1535, which is divisible by only 307.
25. a Each term has to be either 1 or –1.
Hence, if the sum of n such terms is 0, then n is even.
26. c There are two possible cases. The number 9 comes at the end, or it comes at position 4, 5, or 6.
For the first case, the number would look like: 9
635 ... ... 9. 674
In both these cases, the blanks can be occupied by any of the available 9 digits (0, 1, 2, ..., 8).
Thus, total possible numbers would be
2 × (9 × 9 × 9) = 1458. For the second case, the number 9 can occupy any of the given position 4, 5, or 6, and there shall be an odd number at position 7. Thus, the total number of ways shall be
2[3(9 × 9 × 4)] = 1944. Hence, answer is 3402. 27. d N can be written either (54 + 1)3 + (18 – 1)3 – 723 or
(51+ 4 )3 + 173 – (68 + 4)3 .
The first form is divisible by 3, and the second by 17. 28. c 1 2 19 82 1 2 16 5 2 1 2 13 9 1 – 1 – – The answer is 1192.
CAT 2001
29. a Use the answer choices and the fact that: Odd × Odd = Odd
Odd × Even = Even Even × Even = Even
30. a Let the four-digit number be abcd. a + b = c + d ... (i)
b + d = 2(a + c) ... (ii) a + d = c ... (iii) From (i) and (iii), b = 2d From (i) and (ii), 3b = 4c + d
⇒3(2d) = 4c + d ⇒5d = 4c ⇒ d 4 5 c= Now d can be 4 or 8.
But if d = 8, then c = 10 not possible. So d = 4 which gives c = 5. 31. d Let the number be x.
Increase in product = 53x – 35x = 18x ⇒18x = 540 ⇒x = 30 Hence new product = 53 × 30 = 1590.
32. c The value of y would be negative and the value of x would be positive from the inequalities given in the question.
Therefore, from (a), y becomes positive. The value of xy2 would be positive and will not be the minimum.
On comparing (a) and (c), we find that x2 < 5x in 2 < x < 3.
2
x y 5xy
∴ > [Since y is negative.]
∴ 5xy would give the minimum value. 33. a The product of 44 and 11 is 484.
If base is x, then 3411 = 3x3+4x2+1x1+4×x0=484 480 x x 4 x 3 3+ 2+ = ⇒
This equation is satisfied only when x = 5. So base is 5.
In decimal system, the number 3111 can be written 406 5 1 5 1 5 1 5 3× 3+ × 2+ × 1+ × 0=
34. c Let the total number of pages in the book be n.
Let page number x be repeated. Then
∑
= = + n 1 i 1000 x i 1000 x 2 ) 1 n ( n = + + Thus, 1000 2 ) 1 n ( n ≤ + gives n = 44 Since 990 2 ) 1 n ( n = + (for n = 44). Hence x = 10. 35. c Take a = b = c = d = 1.
36. a Let the highest number be n and x be the number erased. Then n(n 1) – x 7 602 2 35 (n – 1) 17 17 + = = .
Hence, n = 69 and x = 7 satisfy the above conditions. 37. d a = b2 – b, b≥4
a2 – 2a = (b2 – b)2 – 2(b2 – b)
= (b2 – b)(b2 – b – 2)
Using different values to b≥4 and we find that it is divisible by 15, 20, 24.
Hence all of these is the right answer. 38. b From II, b = 2d
Hence, b = 10, d = 5 or b = 4, d = 2 From III, e + a = 10 or e + a = 4 From I, a + c = e or e – a = c
From III and I, we get 2e = 10 + c or 2e = 4 + c
⇒ e 5 c 2 = + ... (i) or e 2 c 2 = + ... (ii)
From (i), we can take c = 2, 4, 6, 10. For c = 2, e = 6
c = 6, c = 8 (Not possible)
c = 10, e = 10 (Not possible since both c and e cannot be 10)
From (ii), we have c = 2, 4, 6, 10. For c = 2, e = 3 (Not possible) c = 4, e = 4 (Not possible) c = 6, e = 5 (Possible) c = 10, e = 7 (Not possible)
Considering the possibility from B that c = 6 and e = 5 means e + a = 4
⇒a = –1 (Not possible)
Hence, only possibility is b = 10, d = 5, c = 2, e = 6. e + a = 10 ⇒a = 4
39. c The last two digits can be 12, 16, 24, 32, 36, 52, 56, and 64, i.e. 8 possibilites
Remaining digits can be chosen in 4P3=24ways. Hence, total number of such five-digit numbers = 24 × 8 = 192.
CAT 2002
40. c Total possible arrangements = 10 × 9 × 8
Now 3 numbers can be arranged among themselves in 3! ways = 6 ways
Given condition is satisfied by only 1 out of 6 ways. Hence, the required number of arrangements = 10 9 8 6 × × = 120 41. b Check choices Choice (b) 54 ⇒ S = (5 + 4)2 = 81
⇒ D – S = 81 – 54 = 27. Hence, the number = 54 42. d Let the number of gold coins = x + y
48(x – y) = X2 – Y2
48(x – y) = (x – y)(x + y) ⇒ x + y = 48
Hence the correct choice would be none of these.
43. d 2 n n 575 – x 2 + = 2 1150=n +n – 2x n(n+ ≥1) 1150 2 n + ≥n 1150
The smallest value for it is n = 34. For n = 34
40 = 2x ⇒x = 20
44. a
( )
24 64=(17 – 1)64=17n+ −( 1)64 = 17n + 1 Hence, remainder = 145. b Because each word is lit for a second,
5 17 41 7 21 49 LCM 1, 1, 1 LCM , , 2 4 8 2 4 8 + + + = ×
46. d HCF 9, 27, 36 HCF(9, 27, 36) 2 4 5 LCM (2, 4, 5) = 9 20 = lb
= Weight of each piece Total weight = 18.45 lb
Maximum number of guests = 18.45 20 41 9 × = 47. d 3(4(7x + 4) + 1) + 2 = 84x + 53 Therefore, remainder is 53. 48. d um+vm=wm 2 2 2 u +v =w
Taking Pythagorean triplet 3, 4 and 5, we see m < min (u, v, w)
Also 1' + 2' = 3' and hence m ≤ min (u, v, w) 49. d 76n– 66n
Put n = 1
6 6 3 3 3 3
7 – 6 =(7 – 6 )(7 +6 )
This is a multiple of 7 – 63 3=127and 73 + 63 = 559 and 7 + 6 = 13
50. c Number of ways for single digit = 2 2 digits = 2 × 3 = 6 3 digits = 2 × 3 × 3 = 18 4 digits = 2 × 3 × 3 × 3 = 54 5 digits = 2 × 3 × 3 × 3 × 3 = 162 6 digits = 2 × 3 × 3 × 3 × 3 × 3 = 486 Total = 728
CAT 2003 (leaked)
51. c There are 101 integers in all, of which 51 are even. From 100 to 200, there are 14 multiples of 7, of which 7 are even.
There are 11 multiples of 9, of which 6 are even. But there is one integer (i.e. 126) that is a multiple of both 7 and 9 and also even.
Hence the answer is (51 – 7 – 6 + 1) = 39
52. d Since the last digit in base 2, 3 and 5 is 1, the number should be such that on dividing by either 2, 3 or 5 we should get a remainder 1. The smallest such number is 31. The next set of numbers are 61, 91.
Among these only 31 and 91 are a part of the answer choices.
Among these, (31)10=(11111)2=(1011)3=(111)5 Thus, all three forms have leading digit 1.
Hence the answer is 91.
53. a Solution cannot be found by using only Statement A since b can take any even number 2, 4, 6…. But we can arrive at solution by using statement B alone.
If b > 16, say b = 17
54. c If y = 2 (it cannot be 0 or 1), then x can take 1 value and z can take 2 values.
Thus with y = 2, a total of 1 × 2 = 2 numbers can be formed. With y = 3, 2 × 3 = 6 numbers can be formed. Similarly checking for all values of y from 2 to 9 and adding up we get the answer as 240.
55. c The best way to do this is to take some value and verify.
E.g. 2, 1
2 and 1. Thus, n = 3 and the sum of the three numbers = 3.5.
Thus options 1, 2 and 4 get eliminated.
Alternative method:
Let the n positive numbers be a1, a2, a3 … an a1, a2, a3 … an = 1 We know that AM ≥ GM Hence 1 2 3 n 1 2 n1/ n 1 (a a a a ) (a a a ) n + + + …+ ≥ … or (a1 + a2 + a3 + … an) ≥ n
56. b From 12 to 40, there are 7 prime number, i.e. 13, 17, 19, 23, 29, 31, 37, which is not divisible by (n–1)!
CAT 2003 (Retest)
57. b MDCCLXXXVII = 1000 + 500 + 100 + 100 + 50 + 10 + 10 + 10 + 5 + 1 + 1 = 1787 58. a MCMXCIX = 1000 + (1000 – 100) + (100 – 10) + (10 – 1) = 1000 + 900 + 90 + 9 = 1999 59. c (I) MCMLXXV = 1000 + (1000 – 100) + 50 + 10 + 10 + 5 = 1975 (II) MCMXCV = 1000 + (1000 – 100) + (100 – 10) + 5 = 1995 (III) MVD = 1000 + (500 – 5) = 1495 (IV) MVM = 1000 + (1000 – 5) = 1995Therefore, the answer is (II) and (IV), i.e. option (c). 60. b Such numbers are 10, 17, …, 94.
These numbers are in AP. There are 13 numbers. 10 94 Sum 13 2 + ∴ = × = 52 × 13 = 676
61. c Total codes which can be formed = 9 × 9 = 81. (Distinct digit codes)
The digits which can confuse are 1, 6, 8, 9, from these digit we can form the codes = 4 × 3 = 12 Out of these 12 codes two numbers 69 and 96 will not create confusion.
Therefore, (12 – 2) = 10 codes will create a confusion. Therefore, total codes without confusion
62. d Remainder when 96 4
6
Let’s come down to basic property of dividing the power of 4 by 6, i.e. 1 4 4 6 = 2 4 4 6 = 3 4 4 6 = 4 4 4 6 =
Hence, any power of 4 when divided by 6 leaves a remainder of 4.
63. d Let n = 6
Therefore, n= 6≈2.4
Therefore, divisors of 6 are 1, 2, 3. If we take 2 as divisor, then n> >2 1. Statement A is true.
If we take 3 as divisor, then 6 > 3 > 2.4, i.e. n > 3 > n. Therefore, statement B is true.
64. a As any prime number greater than 3 can be expressed in the form 6n 1± , minimum difference between three consecutive prime numbers will be 2 and 4. The values that satisfy the given conditions are only 3, 5 and 7, i.e. only one set is possible.
65. d a = 6b = 12c and 2b = 9d = 12e.
Dividing the first equations by 12 and second by 36, we get a b c 12=2=1 and b d e 18=4=3 i.e. a b c 108=18=9 and b d e 18=4=3 a b c d e 108 18 9 4 3 ∴ = = = = ∴ a : b : c : d : e = 108 : 18 : 9 : 4 : 3. ∴ cd=94 is not an integer.
CAT 2004
66. a There will be an increase of 6 times. No. of members s1 will be in A.P.
On July 2nd , 2004, s1 will have n + 6 b members = n + 6 × 10.5 n
= 64n
No. of members in s2 will be in G.P
On July 2nd, 2004 Number of members in s2 = nr6
They are equal, Hence 64 n = nr6
67. a We have (1) 1010 < n < 1011
(2) Sum of the digits for 'n' = 2
Clearly-(n)min = 10000000001 (1 followed by 9 zeros and finally 1)
Obviously, we can form 10 such numbers by shifting '1' by one place from right to left again and again. Again, there is another possibility for 'n'
n = 20000000000
So finally : No. of different values for n = 10 + 1 = 11 ans. 68. d y 1 1 2 3 y = + + 3 y y 7 2y + ⇒ = + 2 2y 6y – 3 0 ⇒ + = –6 36 24 y 4 ± + ⇒ = –6 60 –3 15 4 2 ± ± = =
Since 'y' is a +ve number, therefore: 15 –3 y 2 = ans. 69. c 1523 =(19 – 4)23=19x+(–4)23 where x is a natural number. 23 23 23
23 =(19+4) =19y+(4) where y is a natural number.
(
)
23 23 23 23 15 +23 =19 x+y +4 +(–4) = 19 ( x +y)CAT 2005
70. d 65 65 64 64 30 – 29 1 30 +29 > as 3065 – 2965 > 3064 + 2964 3064 (30 – 1) > 2964 (29 + 1) 3064 × 29 > 2964 × 30 3063 > 2963 Hence option d. 71. a x = 163 + 173 + 183 + 193 is even number Therefore 2 divides x. a3 + b3 = (a + b) (a2 – ab – b2) ⇒ a + b always divides Therefore 163 + 193 is divisible by 35 183 + 173 is divisible by 3572. d If p = 1! = 1
Then p + 2 = 3 when divided by 2! remainder will be 1.
If p = 1! + 2 × 2! = 5
Then p + 2 = 7 when divided by 3! remainder is still 1.
Hence p = 1! + (2 × 2!) + (3 × 3!) + … + (10 × 10!) when divided by 11! leaves remainder 1
Alternative method:
P = 1 + 2.2! + 3.3!+ ….10.10!
= (2 –1)1! + ( 3 – 1)2! + (4 – 1)3! + ….(11 – 1)10! =2! – 1! + 3! – 2! + ….. 11! –10!
= 1 + 11!
Hence the remainder is 1. 73. b Let A = 100 x + 10y + z
⇒ B = 100z + 10y + x B - A = 99(z - x)
For B - A to be divided by 7, z - x has to be divisible by 7. Only possibility is z = 9, x = 2.
Biggest number A can be 299 Option b.
74. a
( )
( )
680 4 30Hence the right most non-zero digit is 1. 75. d 10 < n < 1000
Let n is two digit number. n = 10a + b ⇒ pn = ab, sn = a + b Then ab + a + b = 10a + b
⇒ ab = 9a ⇒ b = 9
There are 9 such numbers 19, 29, 33, … 99 Then Let n is three digit number
⇒ n = 100a + 10b + c ⇒ pn = abc, sn= a + b + c then abc + a + b + c = 100a + 10b + c
⇒ abc = 99a + 9b
⇒ bc = 99 + 9b a
But the maximum value for bc = 81
And RHS is more than 99. Hence no such number is possible.
Hence option d.
76. a The 100th and 1000th position value will be only 1 .
Now the possibility of unit and tens digits are (1, 3), (1, 9), (3, 1), (3, 7), (5, 5), (7, 3), (7, 9), (9, 1), (9, 7).
CAT 2006
77. 2 Go by option, put x=−1 2 (1) 2−2=1 4 (2) ⇒ = − − 1 1 2 x 1/ 2 (4) 2−1/ 2= 1 2 78. 2 LCM of 2, 3, 4, 6, 12 = 12 1226 1234 1243 1262 12121 ∴ 34 is greatestNote: n1/n is maximum when n = e (2.718). Among the options n = 3 is closest to the value of e.
79. 4 Let the no. of students in front row be x. So, the no. of students in next rows be x – 3, x - 6, x – 9…. so on
If n i.e. no. of rows be 3 then no. of students x + (x – 3) + (x – 6) = 630 3x = 639 x = 213 So possible similarly n = 4 x + (x – 3) + (x – 6) + (x - 9) = 630 4x – 18 = 630 =648= x 162 4 If n = 5 (4x – 18) + (x - 12) = 630 5x – 30 = 630 x = 120 Again possible. If n = 6 (5x - 30) + (x - 15) = 630 6x - 45 = 630 6x = 675 x ≠ Integer Hence n ≠ 6
80. 3 By options checking option (3), four consecutive odd numbers are 37, 39, 41 and 43. The sum of these 4 numbers is 160.
When divided by 10, we get 16 which is a perfect square.
∴ 41 is one of the odd numbers.
81. 2 Let the number be 10x + y so when number is reversed the number because 10y + x. So, the number increases by 18
Hence (10y + x) - (10x + y) = 9 (y - x) = 18 y - x = 2
So, the possible pairs of (x, y) is (3, 1) (4, 2) (5, 3) (6, 4), (7, 5) (8, 6) (9, 7)
But we want the number other than 13 so, there are 6 possible numbers are there i.e. 24, 35, 46, 57, 68, 79. So total possible numbers are 6.
