Number of multigrain packets sold by store C, D and E = 72, 54 and 84 respectively.Number of whole wheat packets sold by store A and E = 44 and 72 respectively.So, number of multigrain packets sold by store F =&nbsp;$\dfrac{72+54+84}{3}=70$Number of whole wheat packets sold by store F =&nbsp;$\dfrac{44+72}{2}=\dfrac{116}{2}=58$So, total number of packets sold by store F = $70+58=128$Selling Price of each packet of store F =&nbsp;₹&nbsp;$240\times\ 1.2=288$So, total revenue collected by store F =&nbsp;₹&nbsp;$288\times\ 128$ = ₹ $36864$So, option B is the correct answer.

Total number of packets sold by store C = 80+72 = 152 Total number of packets sold by store E = 84+72 = 156 Hence, percentage less =&nbsp;$\dfrac{152-156}{156}\times\ 100\%=-\dfrac{4}{156}\times\ 100\%=-2.56\%$ Among the given options, option (B), $3\%$ less is the closest. So, option (B) is the correct answer.

Multigrain packets sold by store A and B = $92+64=156$ Whole-wheat packets sold by store C and D = $80+40=120$ So, required percentage increase =&nbsp;$\dfrac{156-120}{120}\times\ 100\%=\dfrac{36}{120}\times\ 100\%=30\%$ more. So, option (D) is the correct answer.

Each of the 4 people can have a birthday in any of the 12 calendar months. The total number of ways the birthdays can be distributed is =&nbsp;$12\times\ 12\times\ 12\times\ 12=12^4$ Now,&nbsp;for the birthdays to fall in exactly two calendar months, we must first choose which two months these will be. The number of ways of choosing =&nbsp;$^{12}C_2=\dfrac{12\times\ 11}{2}=66$ For each person, their birthday can be in one of the two chosen months. This gives $2^4=16$&nbsp;total possibilities. However, this includes two cases where all four birthdays fall in just one of the two months (either all in the first chosen month or all in the second). We must subtract these two cases to ensure the birthdays fall in exactly two months. So, number of ways =&nbsp;$2^4-2=16-2=14$ possibilities. Total number of favourable cases =&nbsp;$66\times\ 14$ So, required probability =&nbsp;$\dfrac{66\times\ 14}{12\times\ 12\times\ 12\times\ 12}=\dfrac{77}{1728}$ So, option A is the correct answer.

For the ninth one to be the last defective, among the first 8 watches, first 3 has to be defective. So, total number of ways of selecting =&nbsp;$^8C_3$ And total number of possible cases =&nbsp;$^{15}C_4$ (Out of the 15 watches, examined in order, any 4 can be defective) So, required probability =&nbsp;$\dfrac{^8C_3}{^{15}C_4}=\dfrac{\dfrac{8!}{3!\times\ 5!}}{\dfrac{15!}{4!\times\ 11!}}=\dfrac{8}{195}$ So, correct answer is option B.

1.) First and second in Mathematics: There are 30 students eligible for the first prize and 29 students remaining for the second prize. So, total number of ways of distributing the prize =&nbsp;$30\times\ 29$ 2.) First and second in Physics: There are 30 students eligible for the first prize and&nbsp;29 students remaining for the second prize. So, total number of ways of distributing the prize =&nbsp;$30\times\ 29$ 3.) First in Chemistry: There are 30 students eligible for the first prize. So, total number of ways of distributing the prize = 30 4.) First in English: There are 30 students eligible for the first prize. So, total number of ways of distributing the prize = 30 So, total number of ways of giving all the prizes =&nbsp;$\left(30\times\ 29\right)\times\ \left(30\times\ 29\right)\times\ 30\times\ 30=30^4\times\ 29^2$. So, correct answer is option D.

Evaluating A: $\dfrac{5}{9}\times\ \left(225.4-45.4\right)=\dfrac{5}{9}\times\ 180=100=\left(10\right)^2$ So, A= 10 Evaluating B: $\sqrt{1024}\times40+(20)^2+0.5\%$ of $9600+469=(?)^{3}$ =$32\ \times\ 40+400+\dfrac{0.5}{100}\times\ 9600+469=32\times\ 40+400+48+469=2197=13^3$ So, B=13 Evaluating C: $\dfrac{8^2}{16^2}\times\ 32^3\times\ \sqrt{\ 256}=\dfrac{1}{2^2}\times\ \left(2^5\right)^3\times\ 2^4=2^{17}$ So, C=17 Evaluating D: $18\dfrac{1}{3}+9\dfrac{2}{3}-10\dfrac{1}{3}=8+9\dfrac{2}{3}=1\dfrac{2}{3}+8+8=1\dfrac{2}{3}+16$ So, D = 16 So, the correct order of arrangement is $C&gt;D&gt;B&gt;A$

After the 2-year course gets completed,&nbsp;₹ 6 lakh loan will amount to:&nbsp;₹&nbsp;$6\left(1+\dfrac{8}{100}\right)^2$ Now, he has repayed half of the amount. So, amount outstanding after completion of course =&nbsp;$\dfrac{6}{2}\left(1+\dfrac{8}{100}\right)^2=3\left(1.08\right)^2=3.4992$ lakh Now after another 2 years, when rate of interest will be 10% p.a., the amount will become =&nbsp;$3.4992\left(1+\dfrac{10}{100}\right)^2=3.4992\times\ 1.1^2=4.234032$ lakh So, amount returned by Ravi =&nbsp;₹ $(3.4992+4.234032)=7.733232$ lakh. So, option A is the correct answer.

Let the efficiency of Bhushan be $x$ units/hr. So, efficiency of Amit be&nbsp;$1.5x$ units/hr. Given,&nbsp;Bhushan alone can do a piece of work in 30 hours. So, total work =&nbsp;$30x$ units Given,&nbsp;Bhushan starts working and had already worked for 12 hours. So, work done by Bhushan =&nbsp;$12x$ units So, remaining work =&nbsp;$30x-12x=18x$ units Now, this remaining work will be completed by both Amit and Bhushan. So, required number of hours =&nbsp;$\dfrac{18x}{x+1.5x}=\dfrac{18x}{2.5x}=\dfrac{180}{25}=7.2$ hours. So, correct answer is option (D).

cmat 2025 Complete Paper Solution | Slot 2

Instructions

Question 1.

Question 2.

Question 3.

Instructions

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

cmat 2025 Complete Paper Solution | Slot 2

Instructions

Question 1.

Question 2.

<div>The total number of packets sold by store 'C' is approximately what percent more or less than the total number of packets sold by store 'E' ?</div>

Question 3.

<div>Multigrain packets sold by stores 'A' and 'B' together are what percent more or less than whole-wheat packets sold by stores 'C' and 'D' together ?</div>

Instructions

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

Question 4.

<div>The probability that the birthdays of 4 different persons will fall in exactly two calendar months is :</div>

Question 5.

Question 6.

<div>A consignment of 15 wrist watches contains 4 defectives. The wrist watches are selected at random, one by one and examined . The ones examined are not replaced back. What is the probability that ninth one examined is the last defective?</div>

Question 7.

<div>In how many ways can the following prizes be given away to a class of 30 students, first and second in Mathematics, first and second in Physics, first in Chemistry and first in English?</div>

Question 8.

Question 9.

Question 10.

<div>Amit can do 50% more work than Bhushan in same time. Bhushan alone can do a piece of work in 30 hours. Bhushan starts working and had already worked for 12 hours when Amit joins him. How many hours should Bhushan and Amit work together to complete the remaining work?</div>