Free Mixed Bag 3 Practice Test - CAT 

Question 1

N is a natural number. How many values of N exist, such that N2+24N+21 has exactly three factors?

A. 0
B. 1
C. 2
D. 3
E. >3


Solution : C

Solution:Option c

For N2+24N+21 to have exactly three factors, it must be square of a prime number.

Let N2+24N+21=a2 where a is a prime number.


(N+12+a)(N+12-a) = 123

Either N+12+a =123 and N+12-a = 1

Or N+12+a = 41 and N+12-a=3

In the first case N = 50 and a = 61.

In the second case N =10 and a =19

In either case N2+24N+21 is the square of a prime number. So two such values exist.

Question 2

In an arithmetic progression, the pth term is 1/q and the qth term is 1/p. Find the sum of the first ‘pq’ terms of the progression.

B. pq + 1
E. p*q


Solution : C

Solution: OPTION (c)

Using Assumption

Let p = 1 and q= 2

Then as stated 1st term is ½ and 2nd term is 1. So we have an AP with common difference ½.

So the AP is: ½, 1, 3/2, 2, 5/2….

Now sum of first pq terms : sum of first two terms = ½ + 1 = 3/2

Put p = 1 and q = 2 in the options:

Correct option is (c)

Question 3

Let A, B, C,... X,Y,Z be positive numbers such that AC=B,BD=C...XZ=Y.Given that A+B=1988 find max value of Y+Z?

A. 994
B. 1988
C. 1988
D. 3976
E. none of these


Solution : B


Let A = x and C =y

Then B = xy




G = x

H = xy

I = y

So we see that this is a sequence with a period six. So A = Y and B = Z

Given A+B = 1988, so max value of Y+Z = 1988.

Question 4

In a football team, 12 players are divided into three groups of 4 each and are given different jersey numbers numbering from 1 to 12 in such a way that the sum of the jersey numbers of each of the three groups is same. Find the sum of jersey numbers of each group.

A. 26
B. 34
C. 78
D. Can’t be determined


Solution : A

The question is about adding all the numbers from 1 to 12 and then dividing by 3. So, sum of numbers from 1 to 12: 12×132=78. So, sum of jersey numbers of each of the groups =783=26

Question 5

In the given figure three squares a, b and c of side 3, 5 and 8 units are placed adjacent each other. Find the area of the shaded region.    

A. 12.75
B. 13.75
C. 10
D. 15


Solution : B


Let x and y be the bases of smallest two triangles.

(y8)=(816); y=4.

(xy)=(x4)=(38); x=1.5.

Thus area of shaded region: 12 ((4*8)-(1.5*3))=13.75

Question 6

A Swedish watch company is assigning ID cards for its employees. The ID is made of 7 digits. How many ID numbers can exist such that at least one of the digits repeat?








None of these


Solution : C

Total number of combinations = No repeats + 1 number repeat + 2 number repeat +.....7 number repeat

Total number of combinations - No repeats = 1 number repeat + 2 number repeat +...... 7 number repeat

Total number of combinations - No repeats = Atleast one number repeat


Question 7

If f(x) = min (7x + 3, 8x – 6) for 0 < x < 4, then determine the maximum value of f(x).

A. 66
B. 26
C. 31
D. 28
E. 54


Solution : B


Equate the two terms to get the point of intersection


X=9, which is greater than the given constraint

Minimum/Maximum of two increasing function will also be an increasing function.

Here 7x + 3 and 8x – 6 are both increasing functions, so f(x) will also be an increasing function.

Based on the constraints given, max of f(x) will occur at x = 4 i.e. f(4) = min (31, 26) = 26

Hence option (b)

Question 8

Find the number of integral solutions to |x| + |y| + |z| = 15. 

A. 900
B. 901
C. 902
D. 903
E. none of these


Solution : C

option c

Case 1:When none of them are zero

Number of +ve integral solutions of x + y + z = 15 is 14C2 however, x y z can be -ve also total 8 ways for each set of x y z Total 14C2×8=728

Case 2:When one of them is zero

No of solutions = 3×4×14C1=168 Case 3: When two of them are zeroes

Number of solutions = 3×2=6

Total= 902, option (c)

Question 9

The integers P,Q,R and S shown on the number line below are all equally spaced. If R and S are equal to 712 and 713, respectively, what is the value of P?

A. 710
B. 711
C. 713
D. None of these


Solution : D

The distance between R and S is 713712=6×712

The distance between two consecutive points is constant.

Value of Q=7126×712=5×712

Value of P=5×7126×712=11×712

The correct answer is option (d).

Question 10

Considering all 2-digit natural numbers, how many values of "y" do not satisfy the equation |7x-5y|=3, given that "x" and "y" are positive integers.

A. 60
B. 65
C. 73
D. 48


Solution : B

Let us first look at the conventional approach.
7x - 5y = 3
7x = 5y+3 ----------------(1)
-7x + 5y = 3
7x+3 = 5y ----------------(2)
Solving equation (1), we get the first integral value for y, at which x is an integer at y = 5. Values of y will increase in steps of 7 (the coefficient of x). The next few values of "y" satisfying the equation will be 5,12,19.......96.
Number of terms = 13 (considering 2 digit numbers)
Solving equation (2), we get the first integral value for y at y = 2. Values of y will increase in steps of 7. Hence the second AP will be 2, 9, 16, 23...... 93.
Number of terms = 12 (considering 2 digit numbers)
Number of values of y which satisfy this equation = 25
Therefore, number of values which do not satisfy this equation = 90 - 25 = 65. The answer is option (b).

We know that there are 90 2-digit numbers.
The values which satisfy for "y" form an AP with a common difference = 7 (the coefficient of x).
Hence, the number of terms in that AP =907×2=24 or 25; since there are two APs.
The answer has to be either 90-24 = 66 or 90-25 = 65. Answer is option (b).

Question 11

If r,s are the roots of the equation x2-px+q=0, then what is the equation of the curve whose roots are ([1/r] + s) and ([1/s]+r)?

A. px2-q(1-p)x+(1+q2)=0
B. qx2+(p-q)x+(q2-1)=0
C. (p-q)x2+pqx-(p+1)(q+1)=0
D. qx2-p(q+1)x+(1+q)2=0
E. qx2-p(q+1)x+(1+q)=0


Solution : D

option d

put r=1 and s=1, then p=2 and q=1.

Roots of the curve’s equation = 2,2

Substitute values of p,q,r,s in options to check where 2,2 satisfy as roots. This happens only in option (d)

Question 12

Find the minimum value of the expression, P = 2x+1 + 1/ [4(x/2 + 3/2)] , given that x is real                                                                                 

A. ½
B. 1
C. ¼
D. 2
E. ¾


Solution : B

The expression can be rewritten as

2x+1 + 2 -2 (x/2 +3/2)

= 2x+1 + 2 -x-3

 Using the concept of AM ≥ GM

Therefore, the min value occurs when both the numbers are same. In that case 2(x+1) = 2-(x+3). X = -2. Thus the number = ½.so ½ +1/2 = 1.

Question 13

When 315! is divided by 1215x, remainder = 0. What is the maximum possible value for x?

A. 31
B. 15
C. 32
D. 33


Solution : A

The question is based on highest power of a number in a factorial.

Here since 1215 is composite number, prime factorize 1215 i.e 1215=35×5.

Required answer will be the highest power of 35 in 315!

(No need to find to find the highest power of 5 in 315! as that will always be more than that of 35)

To find out highest power of 35, we will first find the highest power of 3 and then divide it by 5.

Highest power of 3 in 315! = 155 (105 + 35 + 11 + 3 + 1)

Highest power of 35 in 315! = 31 (highest power of 5 we will get is 77)

Required answer is 31.

Question 14

A person has to move from A to B. He can only move upwards or towards the right. Find the probability that he passes through C?  

A. 12
B. 1021
C. 1324
D. 13
E. 1334


Solution : B

Option (b)

Totally there are 4 vertical distances and 5 horizontal distances VVVVHHHHH. Number of ways in

which a person can go from A to B via C= Arrangement of 9 things out of which 4 are identical and of one type and 5 are identical and of another type. = 9!4!×5!=126

Number of ways of reaching C from A= HHVV= 4C2=6. Number of ways of reaching B from C= HHHVV= 5C3=10

Required Probability = 10×6126=1021

Question 15

A house is at the center of the several flat paths which surround it: 4 straight paths that travel from the house to its circular moat, where they meet up with a perfectly circular path which borders the moat; that circular path circumscribes a square path which has its corners at the ends of the 4 straight paths—see the diagram to the right. If the total length of all of the pathways is q kilometers, then which expression represents the distance from the house to the circular moat?

A. q42+22+π km
B.  q2(2+22+π) km
C. q2+22+π km
D. 2q2+22+π km


Solution : B

Shortcut:- Using Assumption

Finding the total lengths of all the paths

Assuming radius of the circle =2

Straight paths= 2x4=8

Each side of the square = 22 x4=82

Circular moat perimeter= 4Π

q (Total) = 4( 2+22 + Π)

Distance from centre to the moat = 2

Look for 2 in the answer options. Answer is option (b)