# CAT Ratio and Proportion Questions PDF [Most Important]

Ratio and Proportion is one of the key topics in the CAT Quant Section. The questions on Ratio and Proportion are usually easy, and hence students should not ignore this topic. It is essential that you know the basics of the CAT Ratio and Proportion well and practice the questions. Also, do check out all the Ratio and Proportion questions for CAT from the CAT Previous Papers with detailed video solutions. This article will look into some important Logs questions for the CAT Exam. If you want to practice these important Ratio and Proportion questions, you can download the PDF, which is completely Free.

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**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 **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?

a)Â 2 kg

b)Â 2.4 kg

c)Â 2.5 kg

d)Â None of these

**3)Â AnswerÂ (C)**

**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

**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)

Checkout: **CAT Free Practice Questions and Videos**

**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.