Let C be Computer Science, M be&nbsp;Mechanical, and E be&nbsp;Electronics Total number of students = C + M + E = 240 - (1) It is also given that C + E = M - (2) M =&nbsp; Girls in M + Boys in M Girls in M =&nbsp;&nbsp;$\dfrac{42.5}{100}M$ Boys in M =&nbsp;$\dfrac{57.5}{100}M$ C = Girls in C + Boys in C Girls in C =&nbsp;&nbsp;$\dfrac{1}{3}\times\ \dfrac{57.5}{100}M\ =\ \dfrac{57.5}{300}M$ Boys in E -&nbsp; Girls in C = Boys in C - Girls in E Boys in E&nbsp; +&nbsp; Boys in C = Girls in C + Boys in C E = C - (3) Girls in E = 28.&nbsp;&nbsp; From (1), (2) and (3) we get C + M + E = C +&nbsp;2C +&nbsp;C = 240 4C = 240 C = E = 60 M = 120 Girls in M =&nbsp;$\dfrac{42.5}{100}M$ =&nbsp;$\dfrac{42.5}{100}\ \times\ 120\ =\ 51$ Girls in C =&nbsp;$\dfrac{57.5}{300}\ \times\ 120\ =\ 23$ 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 ElectronicsTotal number of students = C + M + E = 240 - (1)It is also given that C + E = M - (2)M = Girls in M + Boys in MGirls in M = $\dfrac{42.5}{100}M$Boys in M = $\dfrac{57.5}{100}M$C = Girls in C + Boys in CGirls 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 EBoys in E + Boys in C = Girls in C + Boys in CE = C - (3)Girls in E = 28. From (1), (2) and (3) we getC + M + E = C + 2C + C = 2404C = 240C = E = 60M = 120Girls 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 = 28The 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 =&nbsp;$\dfrac{69\ -\ 60}{60}\ \times\ 100\ =\ 15\%$&nbsp; 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&nbsp;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&nbsp;$7x\times15+3x\times20=165x$ So, the price of 1 kg is&nbsp;$\dfrac{165x}{10x}=16.5$ Hence, Statement I is true. Statement II : If the quantity of rice sold at 16 % profit is 750 kg, then&nbsp;the quantity of rice sold at 8 % profit is 250 kg. The overall profit is&nbsp;$\dfrac{\left(16\times750+8\times250\right)}{1000}=\frac{14000}{1000}=14$ Hence, Statement II is also true

The Duplicate Ratio of 2 : 7 is the compound of it by itself i.e. 4 : 49The compound ratio of 2 : 7, 5 : 3 and 4 : 7 is 2*5*4 : 7*3*7 i.e. 40 : 147The ratio same as 2 : 7 is 4 : 14The ratio same as 5 : 6 is 25 : 30Hence, the answer is Option B.

A:&nbsp;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:&nbsp;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 =&nbsp;$\frac{k}{4}:\frac{k}{6}:\frac{k}{9}=\frac{9k}{36}:\frac{6k}{36}:\frac{4k}{36}=9:6:4$ Correct C:&nbsp;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 =&nbsp;$\frac{k}{2}:\frac{k}{3}:\frac{k}{4}=\frac{6k}{12}:\frac{4k}{12}:\frac{3k}{12}=6:4:3$ Incorrect D:&nbsp;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,&nbsp;$\frac{P}{Q}:\frac{Q}{R}:\frac{R}{P}=\frac{2k}{3k}:\frac{3k}{4k}:\frac{4k}{2k}=\frac{2}{3}:\frac{3}{4}:2=8:9:24$ 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 = $\frac{\text{no of green marbles}}{\text{Total no of mables}}$ =&nbsp;$\frac{3}{10}$ Probability of getting either a yellow or a green marble =&nbsp;$\frac{\text{no of yellow or green marbles}}{\text{Total no of mables}}$ = $\frac{4+3}{10}$ =&nbsp;$\frac{7}{10}$ 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. $\sqrt{\dfrac{9.5\times0.085}{0.0017\times0.19}}=\sqrt{50\times50}=50$

cmat 2024 Complete Paper Solution | Slot 1

Instructions

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cmat 2024 Complete Paper Solution | Slot 1

Instructions

Question 1.

<div>The total number of girls in the Engineering College is</div>

Question 2.

<div><p>The number of girls who opted Mechanical is how much percent of number of students who opted Electronics?</p></div>

Question 3.

<div>By what percent is the number of boys who opted for Mechanical is more than the total number of students who opted for Electronics?</div>

Instructions

<div>Instructions <br />For the following questions answer them individually</div>

Question 4.

<div>If the digit in the unit's place of a two-digit number is halved and the digit in ten's place is doubled, the number thus obtained is equal to the number obtained by interchanging the digits. Which of the following is definitely true?</div>

Question 5.

Question 6.

<div><p><img class="details-image" draggable="false" src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/12/image_HQww7rT.png" alt="">Choose the correct answer from the options given below:</p></div>

Question 7.

Question 8.

<div>A certain amount of money at compound interest grows to ₹66,550 in 3 years and ₹73,205 in 4 years. The rate percent per annum is :</div>

Question 9.

Question 10.

<div><p>Match List-I with List-II<br><img class="details-image" draggable="false" src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/12/image_biUElyz.png" alt="">Choose the correct answer from the options given below :</p></div>