CMAT 2024 Slot 1 Question Paper

Let C be Computer Science, M be Mechanical, and E be Electronics

Total number of students = C + M + E = 240 - (1)

It is also given that C + E = M - (2)

M =  Girls in M + Boys in M

Girls in M =  $$\dfrac{42.5}{100}M$$

Boys in M = $$\dfrac{57.5}{100}M$$

C = Girls in C + Boys in C

Girls in C =  $$\dfrac{1}{3}\times\ \dfrac{57.5}{100}M\ =\ \dfrac{57.5}{300}M$$

Boys in E -  Girls in C = Boys in C - Girls in E

Boys in E  +  Boys in C = Girls in C + Boys in C

E = C - (3)

Girls in E = 28.  

From (1), (2) and (3) we get

C + M + E = C + 2C + C = 240

4C = 240

C = E = 60

M = 120

Girls in M = $$\dfrac{42.5}{100}M$$ = $$\dfrac{42.5}{100}\ \times\ 120\ =\ 51$$

Girls in C = $$\dfrac{57.5}{300}\ \times\ 120\ =\ 23$$

Girls in E = 28

Total number of Girls = 51 + 23 + 28 = 102

The correct answer is option C.

Let C be Computer Science, M be Mechanical, and E be Electronics

Total number of students = C + M + E = 240 - (1)

It is also given that C + E = M - (2)

M = Girls in M + Boys in M

Girls in M = $$\dfrac{42.5}{100}M$$

Boys in M = $$\dfrac{57.5}{100}M$$

C = Girls in C + Boys in C

Girls in C = $$\dfrac{1}{3}\times\ \dfrac{57.5}{100}M\ =\ \dfrac{57.5}{300}M$$

Boys in E - Girls in C = Boys in C - Girls in E

Boys in E + Boys in C = Girls in C + Boys in C

E = C - (3)

Girls in E = 28.

From (1), (2) and (3) we get

C + M + E = C + 2C + C = 240

4C = 240

C = E = 60

M = 120

Girls in M = $$\dfrac{42.5}{100}M$$ = $$\dfrac{42.5}{100}\ \times\ 120\ =\ 51$$

Girls in C = $$\dfrac{57.5}{300}\ \times\ 120\ =\ 23$$

Girls in E = 28

The percentage number of girls who opted for Mechanical to the number of students who opted for Electronics is,

$$\dfrac{Girls\ in\ M}{E}\ \times\ 100\ =\ \dfrac{51}{60}\ \times\ 100\ =85\%\ $$

The correct answer is option C.

Let C be Computer Science, M be Mechanical, and E be Electronics

Total number of students = C + M + E = 240 - (1)

It is also given that C + E = M - (2)

M = Girls in M + Boys in M

Girls in M = $$\dfrac{42.5}{100}M$$

Boys in M = $$\dfrac{57.5}{100}M$$

C = Girls in C + Boys in C

Girls in C = $$\dfrac{1}{3}\times\ \dfrac{57.5}{100}M\ =\ \dfrac{57.5}{300}M$$

Boys in E - Girls in C = Boys in C - Girls in E

Boys in E + Boys in C = Girls in C + Boys in C

E = C - (3)

Girls in E = 28.

From (1), (2) and (3) we get

C + M + E = C + 2C + C = 240

4C = 240

C = E = 60

M = 120

Girls in M = $$\dfrac{42.5}{100}M$$ = $$\dfrac{42.5}{100}\ \times\ 120\ =\ 51$$

Girls in C = $$\dfrac{57.5}{300}\ \times\ 120\ =\ 23$$

Girls in E = 28

Boys in M = 120 - Girls in M = 120 - 51 = 69

E = 60

Ratio = $$\dfrac{69\ -\ 60}{60}\ \times\ 100\ =\ 15\%$$ 

The correct answer is option A.

Let the two-digit number be xy.

(2x)(y/2) = yx

So, 2x = y or y/2=x

This means the digit in the unit's place is twice the digit in ten's place.

Statement I :

Mixing the pulses in ratio 7 : 3, i.e. $$7x$$ kg and $$3x$$ kg

we get the total price of the mixture as $$7x\times15+3x\times20=165x$$

So, the price of 1 kg is $$\dfrac{165x}{10x}=16.5$$

Hence, Statement I is true.

Statement II :

If the quantity of rice sold at 16 % profit is 750 kg, then the quantity of rice sold at 8 % profit is 250 kg.

The overall profit is $$\dfrac{\left(16\times750+8\times250\right)}{1000}=\frac{14000}{1000}=14$$

Hence, Statement II is also true

The Duplicate Ratio of 2 : 7 is the compound of it by itself i.e. 4 : 49

The compound ratio of 2 : 7, 5 : 3 and 4 : 7 is 2*5*4 : 7*3*7 i.e. 40 : 147

The ratio same as 2 : 7 is 4 : 14

The ratio same as 5 : 6 is 25 : 30

Hence, the answer is Option B.

A: If $$\frac{1}{x} : \frac{1}{y} : \frac{1}{z}$$ = 2 : 3 : 5, then x : y : z = 15 : 10 : 6

$$x:y:z\ =\ \frac{1}{2}:\frac{1}{3}:\frac{1}{5}=\frac{15}{30}:\frac{10}{30}:\frac{6}{30}$$

Hence, x : y : z = 15 : 10 : 6

Correct

B: If 4p = 6q = 9r, then p : q : r = 9 : 6 : 4

Let 4p = 6q = 9r = k

Therefore, p = k/4

q = k/6

r = k/9

Hence, p : q : r = $$\frac{k}{4}:\frac{k}{6}:\frac{k}{9}=\frac{9k}{36}:\frac{6k}{36}:\frac{4k}{36}=9:6:4$$

Correct

C: If 2A = 3B = 4C, then A : B : C = 3 : 4 : 6

Let 2A = 3B = 4C = k

then A = k/2

B = k/3

C = k/4

So, A:B:C = $$\frac{k}{2}:\frac{k}{3}:\frac{k}{4}=\frac{6k}{12}:\frac{4k}{12}:\frac{3k}{12}=6:4:3$$

Incorrect

D: If P : Q : R = 2 : 3 : 4, then $$\frac{P}{Q} : \frac{Q}{R} : \frac{R}{P}$$ = 9 : 8 : 24

Let P = 2k so Q and R will be 3k and 4k

SO, $$\frac{P}{Q}:\frac{Q}{R}:\frac{R}{P}=\frac{2k}{3k}:\frac{3k}{4k}:\frac{4k}{2k}=\frac{2}{3}:\frac{3}{4}:2=8:9:24$$

Incorrect

From 3 years to 4 years, the interest earned is 73205-66550 = ₹ 6,655

Assuming Rate of Interest = r%

Principal = 66,550

For 1 year from 3rd to 4th year, the amount becomes = ₹ 73,205

So, Interest = PRT/100

6655 = 66500*r*1/100

r = 10%

A glass jar contains 1 red, 3 green, 2 blue and 4 yellow marbles.

If a single marble is chosen at random from the jar, the probability of getting a yellow marble =$$\frac{\text{no of yellow marbles}}{\text{Total no of mables}}$$= $$\frac{4}{10}$$

Probability of getting a green marble = $$\frac{\text{no of green marbles}}{\text{Total no of mables}}$$ = $$\frac{3}{10}$$

Probability of getting either a yellow or a green marble = $$\frac{\text{no of yellow or green marbles}}{\text{Total no of mables}}$$ = $$\frac{4+3}{10}$$ = $$\frac{7}{10}$$

Probability of getting either a red or a yellow marble = $$\frac{\text{no of yellow or red marbles}}{\text{Total no of mables}}$$ = $$\frac{4+1}{10}$$ = $$\frac{5}{10}$$

So, only B, C are correct.

A. $$\sqrt{\dfrac{0.81\times\ 0.484}{0.064\times6.25}}=\sqrt{\dfrac{0.81\times\ 4.84}{0.64\times6.25}}=\sqrt{\dfrac{\left(0.9\right)^2\times\ \left(2.2\right)^2}{\left(0.8\right)^2\times\left(2.5\right)^2}}=\dfrac{0.9\times2.2}{0.8\times2.5}=0.99$$

B. $$\sqrt{\dfrac{0.204\times\ 42}{0.07\times3.4}}=\sqrt{\dfrac{20.4\times42}{3.4\times7}}=\sqrt{6\times6}=6$$

C. $$\sqrt{\dfrac{0.081\times0.324\times\ 4.624}{1.5625\times0.0289\times\ 72.9\times64}}=\sqrt{\dfrac{\left(0.9\right)^2\times\left(0.18\right)^2\times\left(6.8\right)^2}{\left(1.25\right)^2\times\left(0.17\right)^2\times\left(27\right)^2\times8^2}}=\dfrac{0.9\times0.18\times6.8}{1.25\times0.17\times27\times8}=0.024$$

D. $$\sqrt{\dfrac{9.5\times0.085}{0.0017\times0.19}}=\sqrt{50\times50}=50$$