CMAT 2022 Question Paper

Given average age of three friends is 20 years

Sum of their ages = $$3\times\ 20$$ = 60 years

Given, their ages are in ratio 2:3:5. Let their ages be 2x, 3x and 5x

2x + 3x + 5x = 60

10x = 60

x = 6

Their ages are 12 years, 18 years and 30 years

Age of oldest friend = 30 years

Answer is option C.

Total number of students is given and percentage of students enrolled in different activity is given. Calculating number of students participating in each activity, we get

A) Boys enrolled in singing and handicraft = 140 + 365 = 505 (III)

B) Girls enrolled in swimming and drawing = 245 + 280 = 525 (II)

C) Girls enrolled in singing and drawing = 490 + 280 = 770 (IV)

D) Girls enrolled in dancing and handicraft = 350 + 385 = 735 (I)

Therefore, answer is option D.

AC is ladder, C is base of ladder, AB is wall

Given AC = 30 m and BC = $$\frac{1}{3}AC$$ = $$\frac{1}{3}\left(30\right)$$ = 10 m

Applying pythagoras theorem,

$$AB^2+\ BC^2=\ AC^2$$

$$AB^2$$ = $$30^2-10^2$$ = 800 

AB = $$20\sqrt{\ 2}$$

Answer is option B.

Total workforce = 60 lakhs

Male workforce in SA and MA sectors = 450 + 540 = 990

Total workforce is PT sector = 1140

% required = $$\frac{990}{1140}\times\ 100$$ = 86.8% $$\approx\ $$ 87%

Answer is option B.

Total workforce = 60 lakhs

Female workforce in SE and PR = 360 + 630 = 990

Male workforce in CM = 300

% required = $$\frac{990}{300}\times\ 100$$ = 330%

Answer is option A.

Total workforce = 60 lakhs

Female workforce in SA and MA = 2.7 + 7.2 = 9.9 lakhs

Female workforce in PT = 4.275 lakhs

Difference = 9.9 - 4.275 = 5.625 $$\approx\ $$ 5.6 lakhs

Answer is option C.

Let slower pipe fill the tank at the rate x units/ minute.

Faster pipe will fill tank at the rate of 3x units/ minute

Number of units filled by both the pipes in a minute = 4x

Total time taken to fill tank = 36 minutes

capacity of tank = 36$$\times\ $$4x = 144x

Time taken by slower pipe to fill the tank = $$\frac{144x}{x}$$ = 144 minutes

Answer is option D.

There are 6 red balls, 11 yellow balls and 5 pink balls

Total number of balls = 6 + 11 + 5 = 22

Probability of selecting red ball = $$\frac{6}{22}$$

After selecting one ball, there are 21 balls left

Probability of selecting yellow ball = $$\frac{11}{21}$$

Therefore, probability = $$\frac{6}{22}\times\ \frac{11}{21}$$ = $$\frac{1}{7}$$

Answer is option C.

Statement I:

Given arithmetic mean of (7,8,9,a,b,10,8,7) is 9 and mode is 8

7+8+9+a+b+10+8+7 = $$9\times\ 8$$

49+a+b = 72

a+b = 23

It is mentioned that mode is 8, 8 occurred twice and 7 also occurred twice. Therefore, a or b should be 8

If a = 8, b = 15

If b = 8, a = 15

ab = 120

Statement I is true

Statement II:

a + b + ab = 84

1 + a + b + ab = 85

(1+a)(1+b) = 85

85 = 1 $$\times\ $$ 85 or 5 $$\times\ $$ 17

It cannot be 1$$\times\ $$85, as it is mentioned both a and b are positive integers

If 1+a = 5, 1+b = 17

      a =4 and b = 16

If 1+a = 17, 1+b = 5

     a = 16 and b = 4

In both cases, a+b = 20

Statement II is true

Answer is option A.

Given,

x + y = 3 and xy = 2

$$x^3+y^3\ =\ \left(x+y\right)\left(x^2+y^2-xy\right)$$

x + y = 3

squaring on both sides

$$\left(x+y\right)^2$$ = 9

$$x^2+y^2+2xy$$ = 9

$$x^2+y^2$$ = 9-4 = 5

$$x^3+y^3$$ = 3(5-2) = 9

Answer is option D.