Free Mixed Bag 5 Practice Test - CAT 

Question 1

In a triangle r is the radius of the incircle, R is the radius of the circumcircle and A is the area of the triangle. If h1, h2 and h3 are the 3 altitudes of the triangle, then 1h1+1h2+1h3=?

A. rA
B. 1A
C. 1r


Solution : C

Option (c)

Let a = side, h = height,

Since no conditions are mentioned and only variables are used, we can assume an equilateral triangle.

Let r =1. Then R = 2., h = 3. In an equilateral triangle r : R : h = 1 : 2 : 3

Side a = 2 3; Area= 34 a2 = 33.

1h1=1h2=1h3=13. Sum = 1

Only option c gives 1.


Question 2

In how many ways can 1000 be written as sum of 2 or more consecutive natural numbers?

A. 4
B. 10
C. 3
D. 16


Solution : C

Shortcut - Number of ways of writing a number as a sum of 2 or more consecutive natural numbers = number of odd factors of that number-1

1000=23×53. Number of odd factors = 4

Answer = 4-1 = 3

Question 3

Two guys A and B are walking down an escalator in the direction of the motion of the escalator. A takes two steps on the same time when B takes one step. When A covers 60 steps he gets out of the escalator while B takes 40 steps to get out of the escalator. The number of steps in the escalator when it is stationary___


Solution :

Let the escalator moves x steps when A walks down 60 steps. Total number of steps on a stationary escalator = x + 60.

 When A takes 60 steps, B should have taken 30 steps and the escalator x steps. So when B takes 40 steps, the escalator should have taken 43×x steps.

 So, 43×x+40 = x + 60 = Total number of steps in the escalator when it is stationary. So x = 60. Hence, total number of steps = 120

Question 4

ABCDEF is a regular hexagon. PR= 13 AF, QS= 13 BC and PQ is parallel to RS. Also TE= 13 FE and UD= 13 CD and TU is parallel to ED. Find the ratio of shaded regions to the hexagon ABCDEF. 

A. 1:4
B. 8:27
C. 17:54
D. 1:3


Solution : B

Solution: Option (b)

A regular hexagon can be divided into 54 equal triangles as follows

Of which the shaded part constitutes 1654 part.

Hence, the answer is option (b)

Question 5

The sum of 3rd and 15th elements of an arithmetic progression is equal to the sum of 6th, 11th and 13th elements of the same progression. Then which element of the series should necessarily be equal to zero?

A. 12th
B. 11th
C. 9th
D. 8th
E. 7th


Solution : A

SolnLet the first term be a

Given that a+2d+a+14d=a+5d+a+10d+a+12d

=> a+11d=0. The 12th term is necessarily zero. Answer is option (a)

Question 6

A,B,C and D are contesting for the district elections. Of the total population, 50% like A, 25% like B, 30% like C. All of those who don’t like any of these like D. All those who like B, like both A and C. All those who like C, like A as well. Number of people who like C and not B is what percent of those who like D?

A. 0
B. 10
C. 15
D. 25
E. 20


Solution : B

option (b)From the Venn diagram given, it's clear that no. of people who like C and not B are 30 - 25 = 5. People who like D are 100 - 50 = 50. Therefore, 550×100=10.

Question 7

If x, y, z are distinct positive real numbers then the expression below would be

A. Greater than 4
B. Greater than 5
C. Greater than 6
D. Less than 6
E. Less than 2.


Solution : C

Alternatively:- Use assumption

At x=1, y=2, z=3 (these are the 3 minimum distinct positive integers)

Value of the expression is 48/6 = 8

 Try for one more value x=3, y=2 and z=4

Again the value >6. Answer is option (C)

Question 8

The value of   

A. 2√2
B. 3
C. 3√3
D. 9+4√5
E. 3+2√5


Solution : C

Question 9

Given that x,y,z are all distinct. Then find the value of

A. x2
B. x+5
C. (x+2)2
D. (x-6)2
E. (x-4)2


Solution : A

Option (a) Assume values. Take x=1,y=2,z=3 and a=1

Substitute in equation and the result you get is 1. look in the answer options for 1. answer a

Question 10

In a game of twist, there are 10 indistinguishable coins which need to be arranged in the boxes below such that no row is empty and each box contains no more than one coin. In how ways can the coins be arranged in the following layout

A. 64
B. 63
C. 61
D. 60
E. none of these


Solution : A

 Total number of possible selection of boxes for the 10 coins = 12C10=66

Cases in which rows will be empty = 2, when row 2, 3 and 4 and completely filled and when row 1,2, and 3 are completely filled. In all other cases, atleast one coin will be in each row.

 Answer = 66 - 2 = 64

Question 11

ABCD is a square. O is the point of intersection of the diagonals of the square. Given that RS= 14 BC and QC= 14 DC. Find ratios of areas of  Shaded region and unshaded region ?

A. 1:8
B. 2:7
C. 1:6
D. 1:7


Solution : D

Option (d)
We can graphically divide the square into 16 equal parts as follows


The shaded region forms 2 out of the 16 parts. Hence, ratio of areas = 2:14 = 1:7


Question 12

If a and b are the roots of the equation x2+px+1=0 and ; c and d are the roots of the equation x2+qx+1=0, then (a-c)(b-c)(a+d)(b+d)=?

A. p2-q2
B. a2+b2
C. q2-p2
D. q2+p2
E. p2


Solution : C

Shortcut: Assumption

Option (c)

product of roots=1 is the only constraint.

Assume values that satisfy the constraint and substitute, take a= 1 and b= 1, then p= -2

Assume c=1/2 and d= 2; q= -5/2

 Question is = (a-c)(b-c)(a+d)(b+d)=9/4

Only option (c) satisfies this => q2 –p2 = 25/4 – 4 =9/4

Question 13

Find the last two digits of the expression 84289?

A. 24
B. 44
C. 64
D. 84


Solution : C


The question changes to (21)289×4289

Using the last two digits technique for 1 -

For all numbers ending in 1, last digit is always 1.

Second last digit is the product of 10th digit of the base and the unit digit of the power.


In this question 21289=__81

Using the last two digits technique based on the pattern of 2, we can arrive at the answer very fast.


Since, (210)odd power always ends in 24.

Answer =__81×__24×__56=__64

Question 14

If the graph of the function is symmetric about the y axis then determine the possible value of ‘n.’

A. 2
B. 4
C. 3
D. 8
E. 6


Solution : C


f(y) = (3y – 1)/[(yn)(3y + 1)] = [1/(yn)][ (3y – 1)/(3y + 1)] = h(y) g(y)

Consider g(y) = (3y – 1)/(3y + 1)

g(-y)= (3-y – 1)/(3-y + 1)= (1 – 3y)/(1 + 3y) = -g(y)

So, g(y) is an odd function.

Now, as f(y) is symmetric about the y axis, f(y) is an even function.

h(y)*odd function = even function

→ h(y) should be an odd function

h(y) = 1/yn

for h(y) to be odd, n has to be odd.

Hence option (c)


Put y=1 and y=-1, since it is mentioned that it is symmetric about the y axis. The values of both expressions should be equal. This will happen only if n is odd. Hence the answer is option (c)

Question 15

What is the number of distinct terms in the expansion of (p+q+r+s)10?

A. 245
B. 105
C. 378
D. none of these


Solution : D

option (d)

The number of distinct terms = arrangement of 10 zeroes and 3 ones = 13C3=286