CMAT Quant Questions [Important PDF]

Download CMAT Quant questions with solutions PDF by Cracku. Practice CMAT solved Quant Questions paper tests, which are the practice question to have a firm grasp on the Quant topic in the CMAT exam. Top 20 very Important Matrix Arrangement Questions for CMAT based on the questions asked in previous exam papers. Click on the link below to download the Matrix Arrangement Questions for CMAT PDF with detailed solutions.

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Question 1: The product of ages of Harris and Sharma is 240 .If twice the age of Sharma is more than Harris age by 4 years ,what is Sharma age in years?

a) 12

b) 20

c) 10

d) 14

e) Data adequate

1) Answer (A)

Solution:

let age of sharma = x

let age of harris = y

xy = 240

2x = y+4

on solving these two equations

y = 20 or -24

age must be positive, so y = 20

x = 12

answer is option A

Question 2: The denominator of a fraction is 2 more than thrice it’s numerator.If the numerator as well as denominator are increased by one,the fraction becomes ⅓ .what was that the original fraction?

a) 2/8

b) 3/11

c) 4/14

d) 5/17

e) Cannot be determined

2) Answer (E)

Solution:

Let the numerator = $x$

=> Denominator = $(3x+2)$

Fraction = $\frac{x}{3x+2}$

If the numerator as well as denominator are increased by one

=> $\frac{x+1}{3x+2+1}=\frac{1}{3}$

=> $\frac{x+1}{3x+3}=\frac{1}{3}$

=> $3x+3=3x+3$

It cannot be solved, thus $x$ can take any value and the fraction can be 2/8 , 3/11 , 4/14 ,….

=> Ans – (E)

Question 3: A group of 20 girls has average age of 12 years. Average of first 12 from the same group is 13 years and what is the average age of other 8 girls in the group?

a) 10

b) 11

c) 11.5

d) Cannot be determined

e) None of these

3) Answer (E)

Solution:

Let the age of each of the girl in the group be $x_1,x_2,x_3,…..,x_{20}$ years

Average age of 20 girls = 12

=> $\frac{(x_1+x_2+x_3+…..+x_{20})}{20}=12$

=> $(x_1+x_2+x_3+…..+x_{20})=12 \times 20=240$ ————(i)

Average of first 12 girls = 13

=> $\frac{(x_1+x_2+x_2+….+x_{12})}{12}=13$

=> $(x_1+x_2+x_3+…..+x_{12})=13 \times 12=156$ ———–(ii)

Subtracting equation (ii) from (i)

=> $(x_1+x_2+x_3+…..+x_{20})$ $-$ $(x_1+x_2+x_3+…..+x_{12}) = (240-156)$

=> $(x_{13}+x_{14}+…..+x_{20})=84$

Dividing above equation by 8

=> $\frac{(x_{13}+x_{14}+……+x_{20})}{8}=\frac{84}{8} = 10.5$

=> Ans – (E)

Question 4: If the height of a triangle is decreased by 40% and it’s base is increased by 40% , what will be the effect on its area?

a) No change

b) 16 % increase

c) 8% decrease

d) 16% decrease

e) None of these

4) Answer (D)

Solution:

area of triangle = 1/2(base*height)

after decreasing height by 40%, new height = 0.6H

after increasing base by 40%, new base = 1.4B

now, area of triangle = 1/2(1.4B*0.6H) = 0.84(1/2*B*H)

Area decreased by 100-84%= 16%

so the answer is option D.

Question 5: 0ne-eighth of a number is 17.25. What will 73% of the number be ?

a) 100.74

b) 138.00

c) 96.42

d) 82.66

e) None of these

5) Answer (A)

Solution:

Let the number be $8x$

Acc to ques,

=> $\frac{1}{8} * 8x = 17.25$

=> $x = 17.25$

$\therefore$ 73 % of the number = $\frac{73}{100} * 8x$

= 0.73 * 8 * 17.25 = 100.74

Question 6: What will be the the compound interest acquired on sum of Rs 12,000/- for 3 years at the rate of 10 % per annum ?

a) 2,652

b) 3,972

c) 3,960

d) 3852

e) None of these

6) Answer (B)

Solution:

Principal amount = Rs. 12,000

Time period = 3 years and rate of interest = 10% under compound interest.

=> $C.I. = P [(1 + \frac{R}{100})^T – 1]$

= $12,000 [(1 + \frac{10}{100})^3 – 1]$

= $12,000 [(\frac{11}{10})^3 – 1] = 12,000 (\frac{1331 – 1000}{1000})$

= $12 \times 331 = Rs. 3,972$

Question 7: Sarita started a boutique investing an amount of Rs. 50,000. Six months later Neeta joined her with an amount of Rs. 80,000. At the end of one year they earned a profit of Rs. 18,000. What is Sarita’s share in the profit?

a) Rs. 9000

b) Rs. 8000

c) Rs. 12000

d) Rs. 10000

e) None of these

7) Answer (D)

Solution:

Amount invested by Sarita = Rs. 50,000 and amount invested by Neeta = Rs. 80,000

Ratio of amount invested by Sarita : Neeta = 5 : 8

Time period in which Sarita invested = 12 months and Neeta = 6 months

Ratio of time periods of Sarita : Neeta = 2 : 1

=> Ratio of profits earned by Sarita : Neeta = $(5 \times 2)$:$(8 \times 1)$

= 5 : 4

Total profit earned = Rs. 18,000

$\therefore$ Sarita’s share in the profit = $\frac{5}{(5+4)} \times 18,000$

= $5 \times 2,000 = Rs. 10,000$

=> Ans – (D)

Question 8: P. Q and R have a certain amount of money with themselves. Q has 25% more than what P has, and R has ${1 \over 5}$th of what Q has. If P. Q and R together have Rs. 150, then how much money does P alone have? (in Rs.)

a) 40

b) 70

c) 80

d) 60

e) 50

8) Answer (D)

Solution:

Let P has = $Rs. 100x$

=> Amount with Q = $100x + \frac{25}{100} \times 100x = Rs. 125x$

=> Amount with R = $\frac{1}{5} \times 125x = Rs. 25x$

Total amount together = $100x + 125x + 25x = 150$

=> $x = \frac{150}{250} = \frac{3}{5}$

=> $x = 0.6$

$\therefore$ Amount with P alone = $100 \times 0.6 = Rs. 60$

Question 9: 12 men can finish a project in 20 days. 18 women can finish the same project in 16 days and 24 children can finish it in 18 days. 8 women and 16 children worked for 9 days and then left. In how many days will 10 men complete the remaining project ?

a) $10\frac{1}{2}$

b) 10

c) 9

d) $11\frac{1}{2}$

e) $9\frac{1}{2}$

9) Answer (B)

Solution:

12 men can finish the project in 20 days.

=> 1 day work of 1 man = $\frac{1}{12 \times 20} = \frac{1}{240}$

Similarly, => 1 day work of 1 woman = $\frac{1}{18 \times 16} = \frac{1}{288}$

=> 1 day work of 1 children = $\frac{1}{24 \times 18} = \frac{1}{432}$

8 women and 16 children worked for 9 days

=> Work done in 9 days = $9 \times (8 \times \frac{1}{288}) + (16 \times \frac{1}{432})$

= $9 \times (\frac{1}{36} + \frac{1}{27}) = 9 \times \frac{7}{108}$

= $\frac{7}{12}$

=> Work left = $1 – \frac{7}{12} = \frac{5}{12}$

$\therefore$ Number of days taken by 10 men to complete the remaining work

= $\frac{\frac{10}{240}}{\frac{5}{12}} = \frac{1}{24} \times \frac{12}{5} = \frac{1}{10}$

Thus, 10 men will complete the remaining the work in 10 days.

Question 10: What will come in the place of question mark (?) in the following series ?
2  9  28  65 ?

a) 96

b) 106

c) 126

d) 130

e) None of these

10) Answer (C)

Solution:

Each number is of the form $(n^3+1)$ where $n$ is a natural number

$1^3+1$ = 2

$2^3+1$ = 9

$3^3+1$ = 28

$4^3+1$ = 65

$5^3+1$ = 126

=> Ans – (C)

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