Free Mixed Bag 13 Practice Test - CAT 

Question 1

How many solutions are there for the equation A+B+3C = 10 if A,B,C are whole numbers?

A. 25
B. 24
C. 23
D. 26
E. 22


Solution : D

Option (d)

It is given that A,B and C can take values ≥0

Give values to C from 0

At C=0, the equation changes to A+B=10. Number of solutions will be based on

arrangement of 10 zeroes and 1 one = 11C1=11

At C=1, the equation changes to A+B=7. Number of solutions= 8C1=8

At C=2, the equation changes to A+B=4. Number of solutions= 5C1=5

At C=3, the equation changes to A+B=1. Number of solutions= 2C1=2

Total number of solutions = 11+8+5+2 = 26

Question 2

 A grocer uses a faulty balance that balances 10kg in one pan with 11 kg in the other. He sells his product after hiking the cost price by 10%. If he cheats the farmer while buying and his customer while selling, find his actual gain in the entire transaction.

A. 22.22%
B. 30%
C. 22%
D. 33%


Solution : D

When he cheats the farmer, on buying 10 kg, he is actually buying 11 kg at the price of 10. he always makes a profit of 10%


Similarly, while selling, when he sells 10 kg, if the cost price is 10 rupees, he is selling it for 11 rupees.

Again now,  SP=1.1×1.1×CP

He also professes to sell 11 kg, when he is actually selling 10 kg

SP=1.1×1.1×1.1×CP33.1% overall profit

Answer is option (d)  33%

Question 3

Find the number of degrees in arc AB of the figure below if segments PA and PB are secants; AEC = 115, BFD = 130, and P = 30



Solution :

Angle AEC=115. Suppose there is a point X in the major segment of chord AC. AECX will be a cyclic quadrilateral. Therefore , Angle AEC + Angle AXC =180

Angle AXC= 65

Angle AOC=130 (Measure of arc AC at the centre O)

Similarly Angle BOD = 100

Therefore, Angle AOB (or measure of arc AB) + Angle COD (or measure of arc CD)= 360 - (100+130) = 130 ---- (1)

Now we know the theorem that if there are two secants PA and PB and P lies outside the circle , then Angle P = 12 ( measure of arc COD - measure of arc AOB) =30 (given in the ques) -- (2) ( measure of arc CD - measure of arc AB) =60 -- (3)

You can see that equations (1) and (3) are simultaneous equations in measure of arc AB and measure of arc CD .So, finally measure of arc AB is (130+60)2 = 95.

Question 4

 The area enclosed by the graph |x-1| + |y-1| = 2 is

A. 2
B. 4
C. 4√2
D. 8
E. 16


Solution : D

Option (d)


The above is the graph of |x| + |y| = 2. Area of the region enclosed = (product of diagonals)/2 = (4*4)/2 = 8.

Now we get |x-1|+|y-1| =2 by shifting the above graph but the area enclosed will remain same as the size and shape is not changing. So enclosed area = 8

Question 5

 In an examination, 53 passed in Maths, 61 passed in Physics, 60 in Chemistry, 24 in Maths & Physics, 35 in Physics & Chemistry, 27 in Maths & Chemistry and 5 in none. Total number of students who had appeared in the examination was 100.
Then, the number of students who passed in all subjects is___


Solution :

Option (c)

S = I + 2II + 3III = 53 + 61 + 60 = 174 --------- (a)

Given that II +3III = 24 + 35 + 27 = 86--------------(b)

(a)-(b) gives us the I +II= 88

Out of 100 students, 5 have passed in none

Therefore, the number of students who have passed in all 3 = 95-88 = 7

Question 6

Find N, given that (AX)² = NPA, where A,X,N,P are distinct positive integers

A. 2
B. 3
C. 1
D. 4
E. cannot be determined


Solution : B


AX is two digit number between 11 and 31. Since squares cannot end it 2 and 3, A cannot be 2 or 3 hence A =1. The only number between 11-19 where the condition holds is 192= 361. answer is option (b)

Question 7

In a class of 120 students numbered 1 to 120, there are 3 extracurricular activities to choose from. All even numbered students opt for Dance, whose numbers are divisible by 5 opt for Music and those whose numbers are divisible by 7 opt for Debate. How many opt for none of the three subjects?

A. 19
B. 41
C. 20
D. 31
E. 52


Solution : B

option (b)

A is the set of those who opted for Dance = 1202=60 students.

B is the set of those who opted for Music = 1205=24.

C is the set of those who opted for Debate = 1207=17.

The 10th, 20th, 30th…… numbered students would have opted for both Dance and Music. 12010=12.

The 14th, 28th, 42nd…. Numbered students would have opted for Dance and Debate.12014=8.

The 35th, 70th …. Numbered students would have opted for Music and Debate = 12035=3.

And the 70th numbered student would have opted for all three subjects.

S=101, II+3III = 23

III= 1, S-X= 20+2

101-22 = X


Therefore, number of students who did not opt for any of the three activities = 41

Question 8

f(f(f(k))) = 27. What is the sum of the digits of k ?

A. 3
B. 6
C. 9
D. 8
E. cannot be determined


Solution : B

Option (b)

Since k is odd then f(f(f(k))) =27

=> f(f(k+3)) = 27------------(1) since k is odd , k+3 will be even.

Hence, expanding (1) we get, --------------(2)

Here can either be even or odd

Case 1: let is even

Then (2) reduces to here sum of digits is 6

Case 2: let is odd

Then (2) reduces to => k = 45

Substituting in it comes to be 24 which is not odd hence there is a contradiction hence

k = 105 here sum of digits is 6 is the only solution.

Hence, choice (b) is the correct answer.

Question 9

A box has a ball and everyday starting with today two balls are added to the box for every ball present in it. After 4 weeks the balls are distributed equally among five boxes. The remaining balls are thrown away. How many were thrown away?

A. 0
B. 1
C. 2
D. 3
E. 4


Solution : B

option (b)

On the first day there will be three balls, on second day there will be 9, on the third day 27.

Therefore on the 28th day =328balls.We need to find the remainder when divided by 5, 28= 4k. Therefore, 1 ball will be thrown away.

Question 10

 Solve for a ,

A. -2≤a≤3
B. -3≤a≤2
C. x≤-2, x≥3
D. x<6
E. x>5


Solution : A

Option (a)

Check for a=3, it is satisfied. Option (b) and (e) ruled out

Check for a=1, is satisfied. Option (c) is out

Check for a =4, , not satisfied. Option (d) is out