CAT Percentages Questions & Solutions

A sample of real CAT Percentages past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 33 Percentages questions in all — sign up free to practise them timed.

1. Q1.CAT 2025

   In a class, there were more than 10 boys and a certain number of girls. After 40% of the girls and 60% of the boys left the class, the remaining number of girls was 8 more than the remaining number of boys. Then, the minimum possible number of students initially in the class was

   Answer: 55

   Show solution

   Let say number of girls be $g$ and number of boys be $b$ .

   If 40% of the girls left, remaining number of girls =  $0.6g$

   Also if 60% of the boys left, remaining number of boys =  $0.4b$

   or, $0.6g=0.4b+8$

   or, $6g=4b+80$

   or, $3g=2b+40$

   So, the possible values of (b,g) are: (13,22),(16,24),(19,26),(22,28),(25,30),.....

   Now, $0.6g$ and $0.4b$ has to be an integer.

   So, for this $g$ and $b$ has to be a multiple of 5

   So, $b=25$ and $g=30$

   So, minimum possible number of students = $25+30=55$

2. Q2.CAT 2025

   The monthly sales of a product from January to April were 120, 135, 150 and 165 units, respectively. The cost price of the product was Rs. 240 per unit, and a fixed marked price was used for the product in all the four months. Discounts of 20%, 10% and 5% were given on the marked price per unit in January, February and March, respectively, while no discounts were given in April. If the total profit from January to April was Rs. 138825, then the marked price per unit, in rupees, was

   • 520

   • 525

   • 510

   • 515

   Show solution

   The total number of products is $120+135+150+165= 570$ . Therefore, the total cost must have been $570\times 240 = 136800$

   Let the marked price for the products be $X$ . The January, February, March, and April, selling prices would respectively be $0.8X$ , $0.9X$ , $0.95X$ , and $X$ ; based on the discounts given on each of the months.

   The revenue earned from selling the products in each of the months would be: $120*0.8X + 135*0.9X + 150*0.95X + 165*X = 525X$ .

   The profit earned would be: Revenue -  Cost, and therefore, 

   $525X - 136800 = 138825$

   $525X = 275625$

   $X = \dfrac{275625}{525} = 525$

   The correct answer is option B, Rs. 525.

3. Q3.CAT 2024

   In a group of 250 students, the percentage of girls was at least 44% and at most 60%.The rest of the students were boys. Each student opted for either swimming or running or both. If 50% of the boys and 80% of the girls opted for swimming while 70%of the boys and 60% of the girls opted for running, then the minimum and maximum possible number of students who opted for both swimming and running, are

   • 72 and 88, respectively

   • 75 and 96, respectively

   • 72 and 80, respectively

   • 75 and 90, respectively

   Show solution

   Total number of students is 250, and we are told that, The percentage of girls was at least 44% and at most 60%.

   So the number of girls range from,  $0.44\left(250\right)\le Girls\le0.6\left(250\right)$
   $110\le Girls\le150$

   Statement 1: 
   If 50% of the boys and 80% of the girls opted for swimming, that means if the total number of Boys is B, Girls is G where B+G=250. 
   Swimming is: 0.5B+0.8G

   Statement 2: 
   If 70%of the boys and 60% of the girls opted for running, that means
   Running is 0.7B+0.6G

   Total number of enrolments for swimming and running together will be
   (0.7B+0.6G)+(0.5B+0.8G)=1.2B+1.4G

   Using the overlapping principle, where I represents people who have enrolled only for one activity and II represents number of people who have enrolled for two activities. 
   We know that,  $I+II=250=B+G$
   $I+2II=1.2B+1.4G$

   Subtracting the two equations, 
   $II=0.2B+0.4G$
   $II=0.2\left(B+2G\right)$
   Using B+G=250
   $II=0.2\left(250+G\right)$

   G can at-most be 150 and at least 110. 

   So maximum value of II will be  $0.2\left(250+150\right)=80$

   Minimum value of II will be  $0.2\left(250+110\right)=72$
   

4. Q4.CAT 2022

   In an election, there were four candidates and 80% of the registered voters casted their votes. One of the candidates received 30% of the casted votes while the other three candidates received the remaining casted votes in the proportion 1 : 2 : 3. If the winner of the election received 2512 votes more than the candidate with the second highest votes, then the number of registered voters was

   • 50240

   • 40192

   • 60288

   • 62800

   Show solution

   Let the number of registered votes be 100x

   The number of votes casted = 80x

   Votes received by one of the candidates =  $\frac{30}{100}\times80x$ = 24x

   Remaining votes = 80x - 24x = 56x

   Votes received by other three candidates is  $\frac{56x}{6},\frac{2\times56x}{6},\ \frac{\ 3\times56x}{6}$

   It is given,

   28x - 24x = 2512

   4x = 2512

   x = 628

   The number of registered votes = 100x = 62800

   The answer is option D.

5. Q5.CAT 2021

   A box has 450 balls, each either white or black, there being as many metallic white balls as metallic black balls. If 40% of the white balls and 50% of the black balls are metallic, then the number of non-metallic balls in the box is

   Answer: 250

   Show solution

   Let the number of white balls be x and black balls be y 
   So we get x+y =450       (1)
   Now metallic black balls = 0.5y
   Metallic white balls = 0.4x
   From condition 0.4x=0.5y
   we get 4x-5y=0     (2)
   Solving (1) and (2) we get
   x=250 and y =200
   Now number of Non Metallic balls = 0.6x+0.5y = 150+100 = 250
   
   

6. Q6.CAT 2021

   The total of male and female populations in a city increased by 25% from 1970 to 1980. During the same period, the male population increased by 40% while the female population increased by 20%. From 1980 to 1990, the female population increased by 25%. In 1990, if the female population is twice the male population, then the percentage increase in the total of male and female populations in the city from 1970 to 1990 is

   • 68.25

   • 68.75

   • 68.50

   • 69.25

   Show solution

   Let us solve this question by assuming values(multiples of 100) and not variables(x).

   Since we know that the female population was twice the male population in 1990, let us assume their respective values as 200 and 100.

   Note that while assuming numbers, some of the population values might come out as a fraction(which is not possible, since the population needs to be a natural number). However, this would not affect our answer, since the calculations are in ratios and percentages and not real values of the population in any given year.

   

   Now, we know that the female population became 1.25 times itself in 1990 from what it was in 1980.
   

   Hence, the female population in 1980 = 200/1.25 = 160

   Also, the female population became 1.2 times itself in 1980 from what it was in 1970.

   Hence, the female population in 1970 = 160/1.2 = 1600/12 = 400/3

   

   Let the male population in 1970 be x. Hence, the male population in 1980 is 1.4x.

   

   Now, the total population in 1980 = 1.25 times the total population in 1970.

   Hence, 1.25 (x + 400/3) = 1.4x + 160

   Hence, x = 400/9.

   Population change = 300 - 400/9 - 400/3 = 300 - 1600/9 = 1100/9

   percentage change =  $\frac{\frac{1100}{9}}{\frac{1600}{9}}\times\ 100\ =\ \frac{1100}{16}\%=68.75\%$

7. Q7.CAT 2019

   The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. Each of three vessels A, B, C contains 500 ml of salt solution of strengths 10%, 22%, and 32%, respectively. Now, 100 ml of the solution in vessel A is transferred to vessel B. Then, 100 ml of the solution in vessel B is transferred to vessel C. Finally, 100 ml of the solution in vessel C is transferred to vessel A. The strength, in percentage, of the resulting solution in vessel A is

   • 15

   • 13

   • 12

   • 14

   Show solution

   Each of three vessels A, B, C contains 500 ml of salt solution of strengths 10%, 22%, and 32%, respectively.

   The amount of salt in vessels A, B, C = 50 ml, 110 ml, 160 ml respectively.

   The amount of water in vessels A, B, C = 450 ml, 390 ml, 340 ml respectively.

   In 100 ml solution in vessel A, there will be 10ml of salt and 90 ml of water
   

   Now, 100 ml of the solution in vessel A is transferred to vessel B. Then, 100 ml of the solution in vessel B is transferred to vessel C. Finally, 100 ml of the solution in vessel C is transferred to vessel A

   i.e after the first transfer, the amount of salt in vessels A, B, C = 40, 120, 160 ml respectively.

   after the second transfer, the amount of salt in vessels A, B, C =40, 100, 180 ml respectively.

   After the third transfer, the amount of salt in vessels A, B, C = 70, 100, 150 respectively.

   Each transfer can be captured through the following table.

   

   Percentage of salt in vessel A = $\ \frac{\ 70}{500}\times\ 100$

   =14%

8. Q8.CAT 2019

   The salaries of Ramesh, Ganesh and Rajesh were in the ratio 6:5:7 in 2010, and in the ratio 3:4:3 in 2015. If Ramesh’s salary increased by 25% during 2010-2015, then the percentage increase in Rajesh’s salary during this period is closest to

   • 10

   • 7

   • 9

   • 8

   Show solution

   Let the salaries of Ramesh, Ganesh and Rajesh in 2010 be 6x, 5x, 7x respectively

   Let the salaries of Ramesh, Ganesh and Rajesh in 2015 be 3y, 4y, 3y respectively

   It is given that Ramesh’s salary increased by 25% during 2010-2015,3y = 1.25*6x

   y=2.5x

   Percentage increase in Rajesh's salary = 7.5-7/7=0.07

   =7%

9. Q9.CAT 2018

   A 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1:3 by volume. This mixture is then mixed with an equal volume of 20% ethanol solution. If the resultant mixture is a 31.25% ethanol solution, then the unknown concentration of S is 
   

   • 30%

   • 40%

   • 50%

   • 60%

   Show solution

   Let the volume of the first and the second solution be 100 and 300.
   When they are mixed, quantity of ethanol in the mixture 
   = (20 + 300S)
   Let this solution be mixed with equal volume i.e. 400 of third solution in which the strength of ethanol is 20%.
   So, the quantity of ethanol in the final solution 
   = (20 + 300S + 80) = (300S + 100)
   It is given that, 31.25% of 800 = (300S + 100)
   or, 300S + 100 = 250
   or S = $\frac{1}{2}$ = 50%
   Hence, 50 is the correct answer.

10. Q10.CAT 2017

    The number of girls appearing for an admission test is twice the number of boys. If 30% of the girls and 45% of the boys get admission, the percentage of candidates who do not get admission is

    • 35

    • 50

    • 60

    • 65

    Show solution

    Let the number of girls be 2x and number of boys be x.

    Girls getting admission = 0.6x

    Boys getting admission = 0.45x

    Number of students not getting admission = 3x - 0.6x -0.45x = 1.95x

    Percentage = (1.95x/3x) * 100 = 65%

11. Q11.CAT 2002

    Flights A and B are scheduled from an airport within the next one hour. All the booked passengers of the two flights are waiting in the boarding hall after check-in. The hall has a seating capacity of 200, out of which 10% remained vacant. 40% of the waiting passengers are ladies. When boarding announcement came, passengers of flight A left the hall and boarded the flight. Seating capacity of each flight is two-third of the passengers who waited in the waiting hall for both the flights put together. Half the passengers who boarded flight A are women. After boarding for flight A, 60% of the waiting hall seats became empty. For every twenty of those who are still waiting in the hall for flight B, there is one air hostess in flight A. What is the ratio of empty seats in flight B to the number of air hostesses in flight A?

    • 10 : 1

    • 5 : 1

    • 20 : 1

    • 1 : 1

    Show solution

    Out of 200 of the seating capacity, 180 seats are filled out of which 108 are males and 72 are females. Remaining 20 seats are vacant. According to given condition seating capacity in both the planes is 120 . Considering flight A - we can find that 100 passenger in waiting hall will be taking fight A . So 80 people remain in in the waiting hall who will be taking flight B . Now for every 20 people taking flight B we have a air hostess in flight A . So in total there are 4 air hostess in flight A. Flight B having 120 as seating capacity, 40 remain vacant. So required ratio 40:4 = 10:1 .

12. Q12.CAT 2001

    Fresh grapes contain 90% water by weight while dried grapes contain 20% water by weight and the remaining proportion being pulp. What is the weight of dry grapes available from 20 kg of fresh grapes?

    • 2 kg

    • 2.4 kg

    • 2.5 kg

    • None of these

    Show solution

    Fresh grapes contain 90% water so water in 20kg of fresh pulp = (90/100)x20= 18kg. 
    

    In 20kg fresh grapes, the weight of water is 18kg and the weight of pulp is 2kg.

    The concept that we apply in this question is that the weight of pulp will remain the same in both dry and fresh grapes. If this grape is dried, the water content will change but pulp content will remain the same.

    Suppose the weight of the dry grapes be D.

    80% of the weight of dry grapes = weight of the pulp = 2 kg

    (80/100) x D =2 kg.

    D = 2.5 kg

CAT Percentages previous year questions with solutions — AthenaPrep