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# Must-solve CAT Questions on Ratio and Proportion PDF – Detailed Notes for revision

## Ratio and Proportion CAT Notes

### What are Ratio and Proportion?

**Ratio:Â **Ratio can be simply defined as the comparison of two different quantities. For example, there are two different quantities, “a” and “b”. The ratio of a and b can be **a:b** or **a/b**.

**Proportion:Â **Proportion is nothing but the comparison of two different ratios. If two sets of given quantities are increasing or decreasing in the same ratio, then the ratios are said to be directly proportional to each other.

### Properties of Ratio and Proportion in CAT

**Ratio Properties**

- A ratio need not be positive. However, if we are dealing with quantities of items, their ratios will be positive. In this concept, we will consider only positive ratios.
- A ratio remains the same if both antecedent and consequent are multiplied or divided by the same non-zero number, i.e.,
**a/b=pa/pb=qa/qb, p, qâ‰ 0**and**a/b=(a/p)/(b/p)=(a/q)/b/q),p, qâ‰ 0** - Two ratios in their fraction notation can be compared just as we compare

real numbers.

**a/b=p/q â‡” aq=bq****a/b>p/q â‡” aq>bp****a/b<p/q â‡” aq<bp**

- If antecedent > consequent, the ratio is said to be the ratio of greater inequality.
- If antecedent < consequent, the ratio is said to be the ratio of lesser inequality.
- If the antecedent = consequent, the ratio is said to be the ratio of equality.

**Proportion Properties**

A proportion is the equality of ratios. Hence a:b = c:d is a proportion. The first and last terms are called extremes, and the other two terms are called means. If four terms a, b, c, d are said to be proportional, then a:b = c:d. If three terms a, b, c are said to be proportional, then a:b = b:c.

If a:b = c:d is a proportion, then

- Product of extremes = product of means i.e., ad = bc
- Denominator addition/subtraction: a:a+b = c:c+d and a:a-b = c:c-d
- a, b, c, d,…. are in continued proportion means, a:b = b:c = c:d = ….
- a:b = b:c then b is called mean proportional and b2 = ac
- The third proportional of two numbers, a and b, is c, such that a:b = b:c
- ‘d’ is fourth proportional to numbers a, b, c if a:b = c:d

These are some of the important properties of both ratio and proportion. You can also check out the other important aspects of this topic, such as variation, mixtures and allegations, averages, etc. Download the **Ratio and Proportion Formula PDF**, which includes the important properties, formulas, tips, tricks and shortcuts to solve the questions involving ratio and proportion, mixtures and allegations, variation and averages.

## Top 20 Ratio and Proportion Questions for CAT Preparation

As mentioned before, these are the most important questions to practice and were taken from the previous CAT papers. After practising these questions, one can be confident to solve the ratio and proportion questions in the actual CAT exam. It is highly recommended to practice these 20 questions for the aspirants to ace the quant section in CAT. You can also download these questions in PDF, which consists of the video solutions for each and every question. The link to download the CAT ratio and proportion questions are given below.

Download Top 20 CAT Questions On Ratios & Proportions

Enrol to CAT 2023 Online Course

**Question 1:Â **A milkman mixes 20 litres of water with 80 litres of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he had sold. What is the current proportion of water to milk?

[CAT 2004]

a)Â 2 : 3

b)Â 1 : 2

c)Â 1 : 3

d)Â 3 : 4

**1)Â AnswerÂ (A)**

**Solution:**

After selling 1/4th of the mixture, the remaining quantity of water is 15 liters and milk is 60 liters. So the milkman would add 25 liters of water to the mixture. The total amount of water now is 40 liters and milk is 60 liters. Therefore, the required ratio is 2:3.

**Question 2:Â **A student took five papers in an examination, where the full marks were the same for each paper. His marks in these papers were in the proportion of 6 : 7 : 8 : 9 : 10. In all papers together, the candidate obtained 60% of the total marks. Then the number of papers in which he got more than 50% marks is

a)Â 2

b)Â 3

c)Â 4

d)Â 5

**2)Â AnswerÂ (C)**

**Solution:**

Let the marks in the five papers be 6k, 7k, 8k, 9k and 10k respectively.

So, the total marks in all the 5 papers put together is 40k. This is equal to 60% of the total maximum marks. So, the total maximum marks is 5/3 * 40k

So, the maximum marks in each paper is 5/3 * 40k / 5 = 40k/3 = 13.33k

50% of the maximum marks is 6.67k

So, the number of papers in which the student scored more than 50% is 4

**Question 3:Â **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?

a)Â 10 : 1

b)Â 5 : 1

c)Â 20 : 1

d)Â 1 : 1

**3)Â AnswerÂ (A)**

**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 .

**Question 4:Â **Two liquids A and B are in the ratio 5 : 1 in container 1 and 1 : 3 in container 2. In what ratio should the contents of the two containers be mixed so as to obtain a mixture of A and B in the ratio 1 : 1?

a)Â 2 : 3

b)Â 4 : 3

c)Â 3 : 2

d)Â 3 : 4

**4)Â AnswerÂ (D)**

**Solution:**

Fraction of A in contained 1 = $\frac{5}{6}$

Fraction of A in contained 2Â = $\frac{1}{4}$

Let the ratio of liquid required from containers 1 and 2 be x:1-x

x($\frac{5}{6}$) + (1-x)($\frac{1}{4}$) =Â $\frac{1}{2}$

$\frac{7x}{12}$ = $\frac{1}{4}$

=> x = $\frac{3}{7}$

=> Ratio = 3:4

**Question 5:Â **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?

a)Â 16 : 9

b)Â 9 : 4

c)Â 9 : 16

d)Â 4 : 9

**5)Â AnswerÂ (B)**

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

**Question 6:Â **There are two containers: the first contains 500 ml of alcohol, while the second contains 500 ml of water. Three cups of alcohol from the first container is taken out and is mixed well in the second container. Then three cups of this mixture is taken out and is mixed in the first container. Let A denote the proportion of water in the first container and B denote the proportion of alcohol in the second container. Then,

a)Â A > B

b)Â A < B

c)Â A = B

d)Â Cannot be determined

**6)Â AnswerÂ (C)**

**Solution:**

Let the volume of the cup be V.

Hence, after removing three cups of alcohol from the first container,

Volume of alcohol in the first container is 500-3V

Volume of water in the second container is 500 and volume of alcohol in the second container is 3V.

So, in each cup, the amount of water contained is $\frac{500}{500+3V}*V$

Hence, after adding back 3 cups of the mixture, amount of water in the first container is $0+\frac{1500V}{500+3V} $

Amount of alcohol contained in the second container is $3V – \frac{9V^2}{500+3V} = \frac{1500V}{500+3V}$

So, the required proportion of water in the first container and alcohol in the second container are equal.

**Instructions**

DIRECTIONS for the following two questions: The following table presents the sweetness of different items relative to sucrose, whose sweetness is taken to be 1.00.

**Question 7:Â **Approximately how many times sweeter than sucrose is a mixture consisting of glucose, sucrose and fructose in the ratio of 1: 2: 3?

a)Â 1.3

b)Â 1

c)Â 0.6

d)Â 2.3

**7)Â AnswerÂ (A)**

**Solution:**

The relative sweetness of the mixture is (1*0.74 + 2*1 + 3*1.7) / (1+2+3) = 7.84/6 = 1.30

Option a) is the correct answer.

**Question 8:Â **A stall sells popcorn and chips in packets of three sizes: large, super, and jumbo. The numbers of large, super, and jumbo packets in its stock are in the ratio 7 : 17 : 16 for popcorn and 6 : 15 : 14 for chips. If the total number of popcorn packets in its stock is the same as that of chips packets, then the numbers of jumbo popcorn packets and jumbo chips packets are in the ratio

a)Â 1 : 1

b)Â 8 : 7

c)Â 4 : 3

d)Â 6 : 5

**8)Â AnswerÂ (A)**

**Solution:**

The ratio of L, S, J for popcorn = 7 : 17 : 16

Let them be 7$x$, 17$x$ and 16$x$

The ratio of L, S, J for chips = 6 : 15 : 14

Let them 6$y$, 15$y$ and 14$y$

Given, 40$x$ = 35$y$, $x = \frac{7y}{8}$

Jumbo popcor = 16$x$ = 16 *Â $\frac{7y}{8}$= 14$y$

Hence, the ratio of jumbo popcorn and jumbo chips = 1 : 1

**Question 9:Â **Bottle 1 contains a mixture of milk and water in 7: 2 ratio and Bottle 2 contains a mixture of milk and water in 9: 4 ratio. In what ratio of volumes should the liquids in Bottle 1 and Bottle 2 be combined to obtain a mixture of milk and water in 3:1 ratio?

a)Â 27:14

b)Â 27:13

c)Â 27:16

d)Â 27:18

**9)Â AnswerÂ (B)**

**Solution:**

The ratio of milk and water in Bottle 1 is 7:2 and the ratio of milk and water in Bottle 2 is 9:4

Therefore, the proportion of milk in Bottle 1 is $\frac{7}{9}$ and the proportion of milk in Bottle 2 is $\frac{9}{13}$

Let the ratio in which they should be mixed be equal to X:1.

Hence, the total volume of milk is $\frac{7X}{9}+\frac{9}{13}$

The total volume of water is $\frac{2X}{9}+\frac{4}{13}$

They are in the ratio 3:1

Hence,Â $\frac{7X}{9}+\frac{9}{13} = 3*(\frac{2X}{9}+\frac{4}{13})$

Therefore, $91X+81=78X+108$

Therefore $X = \frac{27}{13}$

**Question 10:Â **If a, b, c are three positive integers such that a and b are in the ratio 3 : 4 while b and c are in the ratio 2:1, then which one of the following is a possible value of (a + b + c)?

a)Â 201

b)Â 205

c)Â 207

d)Â 210

**10)Â AnswerÂ (C)**

**Solution:**

a : b = 3:4 and b : c = 2:1 => a:b:c = 3:4:2

=> a = 3x, b = 4x , c = 2x

=> a + b + c = 9x

=> a + b + c is a multiple of 9.

From the given options only, option C is a multiple of 9

**Question 11:Â **Consider three mixtures â€” the first having water and liquid A in the ratio 1:2, the second having water and liquid B in the ratio 1:3, and the third having water and liquid C in the ratio 1:4. These three mixtures of A, B, and C, respectively, are further mixed in the proportion 4: 3: 2. Then the resulting mixture has

a)Â The same amount of water and liquid B

b)Â The same amount of liquids B and C

c)Â More water than liquid B

d)Â More water than liquid A

**11)Â AnswerÂ (C)**

**Solution:**

The proportion of water in the first mixture is $\frac{1}{3}$

The proportion of Liquid A in the first mixture is $\frac{2}{3}$

The proportion of water in the second mixture is $\frac{1}{4}$

The proportion of Liquid B in the second mixture is $\frac{3}{4}$

The proportion of water in the third mixture is $\frac{1}{5}$

The proportion of Liquid C in the third mixture is $\frac{4}{5}$

As they are mixed in the ratio 4:3:2, the final amount of water is $4 \times \frac{1}{3} + 3 \times \frac{1}{4} + 2 \times \frac{1}{5} = \frac{149}{60}$

The final amount of Liquid A in the mixture is $4\times\frac{2}{3} = \frac{8}{3}$

The final amount of Liquid B in the mixture is $3\times\frac{3}{4} = \frac{9}{4}$

The final amount of Liquid C in the mixture is $2\times\frac{4}{5} = \frac{8}{5}$

Hence, the ratio of Water : A : B : C in the final mixture is $\frac{149}{60}:\frac{8}{3}:\frac{9}{4}:\frac{8}{5} = 149:160:135:96$

From the given choices, only option C Â is correct.

**Question 12:Â **Two types of tea, A and B, are mixed and then sold at Rs. 40 per kg. The profit is 10% if A and B are mixed in the ratio 3 : 2, and 5% if this ratio is 2 : 3. The cost prices, per kg, of A and B are in the ratio

a)Â 17 : 25

b)Â 18 : 25

c)Â 19 : 24

d)Â 21 : 25

**12)Â AnswerÂ (C)**

**Solution:**

The selling price of the mixture is Rs.40/kg.

Let a be the price of 1 kg of tea A in the mixture and b be the price per kg of tea B.

It has been given that the profit is 10% if the 2 varieties are mixed in the ratio 3:2

Let the cost price of the mixture be x.

It has been given that 1.1x = 40

x = 40/1.1

Price per kg of the mixture in ratio 3:2 =Â $\frac{3a+2b}{5}Â $

$\frac{3a+2b}{5} = \frac{40}{1.1}$

$3.3a+2.2b=200$ ——–(1)

The profit is 5% if the 2 varieties are mixed in the ratio 2:3.

Price per kg of the mixture in ratio 2:3 =Â $\frac{2a+3b}{5}$

$\frac{2a+3b}{5} = \frac{40}{1.05}$

$2.1a+3.15b=200$Â ——(2)

Equating (1) and (2), we get,

$3.3a+2.2b = 2.1a+3.15b$

$1.2a=0.95b$

$\frac{a}{b} = \frac{0.95}{1.2}$

$\frac{a}{b} = \frac{19}{24}$

Therefore, option C is the right answer.

**Question 13:Â **Raju and Lalitha originally had marbles in the ratio 4:9. Then Lalitha gave some of her marbles to Raju. As a result, the ratio of the number of marbles with Raju to that with Lalitha became 5:6. What fraction of her original number of marbles was given by Lalitha to Raju?

a)Â $\frac{1}{5}$

b)Â $\frac{6}{19}$

c)Â $\frac{1}{4}$

d)Â $\frac{7}{33}$

**13)Â AnswerÂ (D)**

**Solution:**

Let the number of marbles with Raju and Lalitha initially be 4x and 9x.

Let the number of marbles that Lalitha gave to Raju be a.

It has been given that (4x+a)/(9x-a) = 5/6

24x +Â 6a = 45x – 5a

11a = 21x

a/x = 21/11

Fraction of original marbles given to Raju by Lalitha = a/9x (Since Lalitha had 9x marbles initially).

a/9x = 21/99

= 7/33.

Therefore, option D is the right answer.

**Question 14:Â **The scores of Amal and Bimal in an examination are in the ratio 11 : 14. After an appeal, their scores increase by the same amount and their new scores are in the ratio 47 : 56. The ratio of Bimalâ€™s new score to that of his original score is

a)Â 4 : 3

b)Â 8 : 5

c)Â 5 : 4

d)Â 3 : 2

**14)Â AnswerÂ (A)**

**Solution:**

Let the score of Amal and Bimal be 11k and 14k

Let the scores be increased by x

So, after increment, Amal’s score =Â 11k + x and Bimal’s score = 14k + x

According to the question,

$\dfrac{\text{11k + x}}{\text{14k + x}}$ = $\dfrac{47}{56}$

On solving, we get x = $\dfrac{42}{9}$k

Ratio of Bimal’s new score to his original score

=Â $\dfrac{\text{14k + x}}{\text{14k}}$

=$\dfrac{14k +\frac{42k}{9}}{14k}$

=$\dfrac{\text{168k}}{\text{14*9k}}$

=$\dfrac{4}{3}$

Hence, option A is the correct answer.

**Question 15:Â **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

a)Â 30%

b)Â 40%

c)Â 50%

d)Â 60%

**15)Â AnswerÂ (C)**

**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.

**Question 16:Â **There are two drums, each containing a mixture of paints A and B. In drum 1, A and B are in the ratio 18 : 7. The mixtures from drums 1 and 2 are mixed in the ratio 3 : 4 and in this final mixture, A and B are in the ratio 13 : 7. In drum 2, then A and B were in the ratio

a)Â 251 : 163

b)Â 239 : 161

c)Â 220 : 149

d)Â 229 : 141

**16)Â AnswerÂ (B)**

**Solution:**

It is given that in drum 1, A and B are in the ratio 18 : 7.

Let us assume that in drum 2, A and B are in the ratio x : 1.

It is given thatÂ drums 1 and 2 are mixed in the ratio 3 : 4 and in this final mixture, A and B are in the ratio 13 : 7.

By equating concentration of A

$\Rightarrow$ $\dfrac{3*\dfrac{18}{18+7}+4*\dfrac{x}{x+1}}{3+4} = \dfrac{13}{13+7}$

$\Rightarrow$ $\dfrac{54}{25}+\dfrac{4x}{x+1} = \dfrac{91}{20}$

$\Rightarrow$ $\dfrac{4x}{x+1} = \dfrac{239}{100}$

$\Rightarrow$ $x = \dfrac{239}{161}$

Therefore, we can say that in drum 2,Â A and B are in the ratio $\dfrac{239}{161}$ : 1 or 239 : 161.

**Question 17:Â **The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. If three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%. If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%. A fourth solution, D, is produced by mixing B and C in the ratio 2 : 7. The ratio of the strength of D to that of A is

a)Â 3 : 10

b)Â 1 : 3

c)Â 1 : 4

d)Â 2 : 5

**17)Â AnswerÂ (B)**

**Solution:**

Let ‘a’, ‘b’ and ‘c’ be the concentration of salt in solutions A, B and C respectively.

It is given thatÂ three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%.

$\Rightarrow$ $\dfrac{a+2b+3c}{1+2+3} = 20$

$\Rightarrow$ $a+2b+3c = 120$ … (1)

If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%.

$\Rightarrow$ $\dfrac{3a+2b+c}{1+2+3} = 30$

$\Rightarrow$Â $3a+2b+c = 180$ … (2)

From equation (1) and (2), we can say that

$\Rightarrow$ $b+2c = 45$

$\Rightarrow$ $b = 45 – 2c$

Also, on subtracting (1) from (2), we get

$a – c = 30$

$\Rightarrow$ $a = 30 + c$

In solution D, B and C are mixed in the ratio 2 : 7

So, the concentration of salt in D = $\dfrac{2b + 7c}{9}$ =Â $\dfrac{90 – 4c + 7c}{9}$ =Â $\dfrac{90 + 3c}{9}$

Required ratio = $\dfrac{90 + 3c}{9a}$ =Â $\dfrac{90 + 3c}{9 (30 + c)}$ = $1 : 3$

Hence, option B is the correct answer.

**Question 18:Â **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 Rameshâ€™s salary increased by 25% during 2010-2015, then the percentage increase in Rajeshâ€™s salary during this period is closest to

a)Â 10

b)Â 7

c)Â 9

d)Â 8

**18)Â AnswerÂ (B)**

**Solution:**

LetÂ 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 thatÂ Rameshâ€™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%

**Question 19:Â **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

a)Â 26

b)Â 32

c)Â 35

d)Â 24

**19)Â AnswerÂ (B)**

**Solution:**

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

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

r=$\ \frac{\ m+a}{12}$ â€”â€”â€”-(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

**Question 20:Â **Amala, Bina, and Gouri invest money in the ratio 3 : 4 : 5 in fixed deposits having respective annual interest rates in the ratio 6 : 5 : 4. What is their total interest income (in Rs) after a year, if Bina’s interest income exceeds Amala’s by Rs 250?

a)Â 6350

b)Â 6000

c)Â 7000

d)Â 7250

**20)Â AnswerÂ (D)**

**Solution:**

Assuming the investment ofÂ Amala, Bina, and Gouri be 300x, 400x and 500x, hence the interest incomes will be 300x*6/100=18x, 400x*5/100=20x and 500x*4/100 = 20x

Given, Bina’s interest income exceeds Amala by 20x-18x=2x=250Â => x=125

Now, total interest income = 18x+20x+20x=58x = 58*125 = 7250

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