CAT Averages Ratio Proportion Questions & Solutions

A sample of real CAT Averages Ratio Proportion past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 89 Averages Ratio Proportion questions in all — sign up free to practise them timed.

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 = $25+30=55$
Q2.CAT 2024
When Rajesh's age was same as the present age of Garima, the ratio of their ages was 3 : 2. When Garima's age becomes the same as the present age of Rajesh, the ratio of the ages of Rajesh and Garima will become
- 3 : 2
- 4 : 3
- 5 : 4
- 2 : 1
Show solution
Let's take Rajesh and Garima's ages to be R and G, respectively
From the given ratio, we can see that Rajesh is older than Garima, so let's take R=G+x

When Rajesh was of age G, which was x years ago, Garima was of G-x years old
Giving the ratio as $\frac{G}{G-x}=\frac{3}{2}$
This gives us G as 3x, which in turn gives R as 4x

We are asked the ratio when Gramia becomes 4x years old.
By that time, Rajesh will be 5x years old.

Giinv their ratio as $\frac{5x}{4x}=5:4$

Therefore, Option C is the correct answer.
Q3.CAT 2024
Rajesh and Vimal own 20 hectares and 30 hectares of agricultural land, respectively, which are entirely covered by wheat and mustard crops. The cultivation area of wheat and mustard in the land owned by Vimal are in the ratio of 5 : 3. If the total cultivation area of wheat and mustard are in the ratio 11 : 9, then the ratio of cultivation area of wheat and mustard in the land owned by Rajesh is
- 4 : 3
- 7 : 9
- 3 : 7
- 1 : 1
Show solution
We are told that Rajesh manages 20 hectares and Vimal manages 30 hectares

For Vimal, we know the distribution of the land between Wheat and Mustard, 5:3
So, wheat area will be, $\frac{5}{8}\left(30\right)$
Mustard area will be, $\frac{3}{8}\left(30\right)$

Similarly, let us assume that the distribution of crops between Wheat and Mustard to be k:1
Wheat will be, $\frac{k}{k+1}\left(20\right)$
Mustard will be, $\frac{1}{k+1}\left(20\right)$

We are told that total area of Wheat and Mustard is in the ratio 11:9

Adding them up we get,
$\dfrac{\left(\frac{150}{8}+\frac{20k}{k+1}\right)}{\left(\frac{90}{8}+\frac{20}{k+1}\right)}=\dfrac{11}{9}$

$\dfrac{\left(\frac{15}{8}+\frac{2k}{k+1}\right)}{\left(\frac{9}{8}+\frac{2}{k+1}\right)}=\dfrac{11}{9}$

$\frac{135}{8}+\frac{18k}{k+1}=\frac{99}{8}+\frac{22}{k+1}$

$\frac{36}{8}=\frac{22-18k}{k+1}$

$44-36k=9k+9$

$45k=35$

$k=\frac{7}{9}$

Hence the ratio of distribution of area between Wheat and Mustard for Rajesh is $\dfrac{7}{9}$
Q4.CAT 2022
Pinky is standing in a queue at a ticket counter. Suppose the ratio of the number of persons standing ahead of Pinky to the number of persons standing behind her in the queue is 3 : 5. If the total number of persons in the queue is less than 300, then the maximum possible number of persons standing ahead of Pinky is
Answer: 111
Show solution
Let the number of persons standing ahead and behind of Pinky be 3a and 5a.

Total number of persons = 3a + 5a + 1(including pinky) = 8a + 1

8a + 1 < 300

8a < 299

a < 37.375

Maximum value a can take is 37.

The maximum possible number of persons standing ahead of Pinky = 3a = 3*37 = 111
Q5.CAT 2021
Onion is sold for 5 consecutive months at the rate of Rs 10, 20, 25, 25, and 50 per kg, respectively. A family spends a fixed amount of money on onion for each of the first three months, and then spends half that amount on onion for each of the next two months. The average expense for onion, in rupees per kg, for the family over these 5 months is closest to
- 26
- 18
- 16
- 20
Show solution
Let us assume the family spends Rs. 100 each month for the first 3 months and then spends Rs. 50 in each of the next two months.

Then amount of onions bought = 10, 5, 4, 2, 1, for months 1-5 respectively.

Total amount bought = 22kg.

Total amount spent = 100+100+100+50+50 = 400.

Average expense = $\frac{400}{22}=Rs.18.18\approx\ 18$
Q6.CAT 2021
A tea shop offers tea in cups of three different sizes. The product of the prices, in INR, of three different sizes is equal to 800. The prices of the smallest size and the medium size are in the ratio 2 : 5. If the shop owner decides to increase the prices of the smallest and the medium ones by INR 6 keeping the price of the largest size unchanged, the product then changes to 3200. The sum of the original prices of three different sizes, in INR, is
Answer: 34
Show solution
Let price of smallest cup be 2x and medium be 5x and large be y
Now by condition 1
we get $2x\ \times\ \ 5x\ \times\ y\ =800$
we get $x^2y\ =80$ (1)
Now as per second condition ;
$\left(2x+6\right)\times\ \left(5x+6\right)\ y\ =3200$ (2)
Now dividing (2) and (1)
we get $\frac{\left(\left(2x+6\right)\times\ \left(5x+6\right)\right)}{x^2}=40$
we get $10x^2+42x+36\ =\ 40x^2$
we get $\ 30x^2-42x-36=0$
$5x^2-7x-6=0$
we get x=2
So 2x=4 and 5x=10
Now substituting in (1) we get y =20
Now therefore sum = 4+10+20 =34
Q7.CAT 2019
In an examination, Rama's score was one-twelfth of the sum of the scores of Mohan and Anjali. After a review, the score of each of them increased by 6. The revised scores of Anjali, Mohan, and Rama were in the ratio 11:10:3. Then Anjali's score exceeded Rama's score by
- 26
- 32
- 35
- 24
Show solution
Let the scores of Rama, Anjali and Mohan be r, a, m.

It is given that Rama's score was one-twelfth of the sum of the scores of Mohan and Anjali

r= $\ \frac{\ m+a}{12}$ ———-(1)

The scores of Rama, Anjali and Mohan after review = r+6, a+6, m+6

a+6:m+6:r+6 = 11:10:3
Let a+6 = 11x => a= 11x-6
m+6=10x => m=10x-6
r+ 6 =3x => r = 3x-6
Substituting these values in equation (1), we get

3x-6= $\ \frac{\ 21x-12}{12}$

12(3x-6) = 21x-12

x=4

Anjali's score exceeds Rama's score by (a-r)=8x=32
Q8.CAT 2018
In an examination, the maximum possible score is N while the pass mark is 45% of N. A candidate obtains 36 marks, but falls short of the pass mark by 68%. Which one of the following is then correct?
- $N \leq 200$ .
- $243 \leq N \leq 252$ .
- $201 \leq N \leq 242$ .
- $N \geq 253$ .
Show solution
Total marks = N
Pass marks = 45% of N = 0.45N
Marks obtained = 36
It is given that, obtained marks is 68% less than that pass marks
=>the obtained marks is 32% of the pass marks.
So, 0.32 * 0.45N = 36
On solving, we get N = 250
Hence, option B is the correct answer.
Q9.CAT 2017
The average height of 22 toddlers increases by 2 inches when two of them leave this group. If the average height of these two toddlers is one-third the average height of the original 22, then the average height, in inches, of the remaining 20 toddlers is
- 30
- 28
- 32
- 26
Show solution
Let the average height of 22 toddlers be 3x.
Sum of the height of 22 toddlers = 66x
Hence average height of the two toddlers who left the group = x
Sum of the height of the remaining 20 toddlers = 66x - 2x = 64x
Average height of the remaining 20 toddlers = 64x/20 = 3.2x
Difference = 0.2x = 2 inches => x = 10 inches
Hence average height of the remaining 20 toddlers = 3.2x = 32 inches
Q10.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 .
Q11.CAT 1999
Total expenses of a boarding house are partly fixed and partly varying linearly with the number of boarders. The average expense per boarder is Rs. 700 when there are 25 boarders and Rs. 600 when there are 50 boarders. What is the average expense per boarder when there are 100 boarders?
- 550
- 580
- 540
- 560
Show solution
Let the fixed income be x and the number of boarders be y.

x + 25y = 17500

x + 50y = 30000

=> y = 500 and x = 5000

x + 100y = 5000 + 50000 = 55000

Average expense = $\frac{55000}{100}$ = Rs.550.
Q12.CAT 1997
The value of each of a set of coins varies as the square of its diameter, if its thickness remains constant, and it varies as the thickness, if the diameter remains constant. If the diameter of two coins are in the ratio 4 : 3, what should be the ratio of their thickness' be if the value of the first is four times that of the second?
- 16 : 9
- 9 : 4
- 9 : 16
- 4 : 9
Show solution
Value of coin = $k (2r)^2 t$ (where k is proportionality constant, 2r is diameter and t is thickness)
So (value of first coin) = 4 (value of second coin)

$k (2r_1)^2 t_1 = 4 \times (k(2r_2)^2 t_2)$

or $\frac{t_1}{t_2} = \frac{9}{4}$ (As ratio of diameters 2r will be 9:4)

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CAT Averages Ratio Proportion Questions & Solutions

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 = $25+30=55$

Q2.CAT 2024

When Rajesh's age was same as the present age of Garima, the ratio of their ages was 3 : 2. When Garima's age becomes the same as the present age of Rajesh, the ratio of the ages of Rajesh and Garima will become

3 : 2
4 : 3
5 : 4
2 : 1

Show solution

Let's take Rajesh and Garima's ages to be R and G, respectively
From the given ratio, we can see that Rajesh is older than Garima, so let's take R=G+x

When Rajesh was of age G, which was x years ago, Garima was of G-x years old
Giving the ratio as $\frac{G}{G-x}=\frac{3}{2}$
This gives us G as 3x, which in turn gives R as 4x

We are asked the ratio when Gramia becomes 4x years old.
By that time, Rajesh will be 5x years old.

Giinv their ratio as $\frac{5x}{4x}=5:4$

Therefore, Option C is the correct answer.

Q3.CAT 2024

Rajesh and Vimal own 20 hectares and 30 hectares of agricultural land, respectively, which are entirely covered by wheat and mustard crops. The cultivation area of wheat and mustard in the land owned by Vimal are in the ratio of 5 : 3. If the total cultivation area of wheat and mustard are in the ratio 11 : 9, then the ratio of cultivation area of wheat and mustard in the land owned by Rajesh is

4 : 3
7 : 9
3 : 7
1 : 1

Show solution

We are told that Rajesh manages 20 hectares and Vimal manages 30 hectares

For Vimal, we know the distribution of the land between Wheat and Mustard, 5:3
So, wheat area will be, $\frac{5}{8}\left(30\right)$
Mustard area will be, $\frac{3}{8}\left(30\right)$

Similarly, let us assume that the distribution of crops between Wheat and Mustard to be k:1
Wheat will be, $\frac{k}{k+1}\left(20\right)$
Mustard will be, $\frac{1}{k+1}\left(20\right)$

We are told that total area of Wheat and Mustard is in the ratio 11:9

Adding them up we get,
$\dfrac{\left(\frac{150}{8}+\frac{20k}{k+1}\right)}{\left(\frac{90}{8}+\frac{20}{k+1}\right)}=\dfrac{11}{9}$

$\dfrac{\left(\frac{15}{8}+\frac{2k}{k+1}\right)}{\left(\frac{9}{8}+\frac{2}{k+1}\right)}=\dfrac{11}{9}$

$\frac{135}{8}+\frac{18k}{k+1}=\frac{99}{8}+\frac{22}{k+1}$

$\frac{36}{8}=\frac{22-18k}{k+1}$

$44-36k=9k+9$

$45k=35$

$k=\frac{7}{9}$

Hence the ratio of distribution of area between Wheat and Mustard for Rajesh is $\dfrac{7}{9}$

Q4.CAT 2022

Pinky is standing in a queue at a ticket counter. Suppose the ratio of the number of persons standing ahead of Pinky to the number of persons standing behind her in the queue is 3 : 5. If the total number of persons in the queue is less than 300, then the maximum possible number of persons standing ahead of Pinky is

Answer: 111

Show solution

Let the number of persons standing ahead and behind of Pinky be 3a and 5a.

Total number of persons = 3a + 5a + 1(including pinky) = 8a + 1

8a + 1 < 300

8a < 299

a < 37.375

Maximum value a can take is 37.

The maximum possible number of persons standing ahead of Pinky = 3a = 3*37 = 111

Q5.CAT 2021

Onion is sold for 5 consecutive months at the rate of Rs 10, 20, 25, 25, and 50 per kg, respectively. A family spends a fixed amount of money on onion for each of the first three months, and then spends half that amount on onion for each of the next two months. The average expense for onion, in rupees per kg, for the family over these 5 months is closest to

Show solution

Let us assume the family spends Rs. 100 each month for the first 3 months and then spends Rs. 50 in each of the next two months.

Then amount of onions bought = 10, 5, 4, 2, 1, for months 1-5 respectively.

Total amount bought = 22kg.

Total amount spent = 100+100+100+50+50 = 400.

Average expense = $\frac{400}{22}=Rs.18.18\approx\ 18$

Q6.CAT 2021

A tea shop offers tea in cups of three different sizes. The product of the prices, in INR, of three different sizes is equal to 800. The prices of the smallest size and the medium size are in the ratio 2 : 5. If the shop owner decides to increase the prices of the smallest and the medium ones by INR 6 keeping the price of the largest size unchanged, the product then changes to 3200. The sum of the original prices of three different sizes, in INR, is

Answer: 34

Show solution

Let price of smallest cup be 2x and medium be 5x and large be y
Now by condition 1
we get $2x\ \times\ \ 5x\ \times\ y\ =800$
we get $x^2y\ =80$ (1)
Now as per second condition ;
$\left(2x+6\right)\times\ \left(5x+6\right)\ y\ =3200$ (2)
Now dividing (2) and (1)
we get $\frac{\left(\left(2x+6\right)\times\ \left(5x+6\right)\right)}{x^2}=40$
we get $10x^2+42x+36\ =\ 40x^2$
we get $\ 30x^2-42x-36=0$
$5x^2-7x-6=0$
we get x=2
So 2x=4 and 5x=10
Now substituting in (1) we get y =20
Now therefore sum = 4+10+20 =34

Q7.CAT 2019

In an examination, Rama's score was one-twelfth of the sum of the scores of Mohan and Anjali. After a review, the score of each of them increased by 6. The revised scores of Anjali, Mohan, and Rama were in the ratio 11:10:3. Then Anjali's score exceeded Rama's score by

Show solution

Let the scores of Rama, Anjali and Mohan be r, a, m.

It is given that Rama's score was one-twelfth of the sum of the scores of Mohan and Anjali

r= $\ \frac{\ m+a}{12}$ ———-(1)

The scores of Rama, Anjali and Mohan after review = r+6, a+6, m+6

a+6:m+6:r+6 = 11:10:3

Let a+6 = 11x => a= 11x-6

m+6=10x => m=10x-6

r+ 6 =3x => r = 3x-6
Substituting these values in equation (1), we get

3x-6= $\ \frac{\ 21x-12}{12}$

12(3x-6) = 21x-12

x=4

Anjali's score exceeds Rama's score by (a-r)=8x=32

Q8.CAT 2018

In an examination, the maximum possible score is N while the pass mark is 45% of N. A candidate obtains 36 marks, but falls short of the pass mark by 68%. Which one of the following is then correct?

$N \leq 200$ .
$243 \leq N \leq 252$ .
$201 \leq N \leq 242$ .
$N \geq 253$ .

Show solution

Total marks = N
Pass marks = 45% of N = 0.45N
Marks obtained = 36
It is given that, obtained marks is 68% less than that pass marks
=>the obtained marks is 32% of the pass marks.
So, 0.32 * 0.45N = 36
On solving, we get N = 250
Hence, option B is the correct answer.

Q9.CAT 2017

The average height of 22 toddlers increases by 2 inches when two of them leave this group. If the average height of these two toddlers is one-third the average height of the original 22, then the average height, in inches, of the remaining 20 toddlers is

Show solution

Let the average height of 22 toddlers be 3x.
Sum of the height of 22 toddlers = 66x
Hence average height of the two toddlers who left the group = x
Sum of the height of the remaining 20 toddlers = 66x - 2x = 64x
Average height of the remaining 20 toddlers = 64x/20 = 3.2x
Difference = 0.2x = 2 inches => x = 10 inches
Hence average height of the remaining 20 toddlers = 3.2x = 32 inches

Q10.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 .

Q11.CAT 1999

Total expenses of a boarding house are partly fixed and partly varying linearly with the number of boarders. The average expense per boarder is Rs. 700 when there are 25 boarders and Rs. 600 when there are 50 boarders. What is the average expense per boarder when there are 100 boarders?

Show solution

Let the fixed income be x and the number of boarders be y.

x + 25y = 17500

x + 50y = 30000

=> y = 500 and x = 5000

x + 100y = 5000 + 50000 = 55000

Average expense = $\frac{55000}{100}$ = Rs.550.

Q12.CAT 1997

The value of each of a set of coins varies as the square of its diameter, if its thickness remains constant, and it varies as the thickness, if the diameter remains constant. If the diameter of two coins are in the ratio 4 : 3, what should be the ratio of their thickness' be if the value of the first is four times that of the second?

16 : 9
9 : 4
9 : 16
4 : 9

Show solution

Value of coin = $k (2r)^2 t$ (where k is proportionality constant, 2r is diameter and t is thickness)
So (value of first coin) = 4 (value of second coin)

$k (2r_1)^2 t_1 = 4 \times (k(2r_2)^2 t_2)$

or $\frac{t_1}{t_2} = \frac{9}{4}$ (As ratio of diameters 2r will be 9:4)

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