**Ratio and Proportion Questions for CAT Set-4 PDF**

Download important CAT Ratio and Proportion Set-4 Questions PDF based on previously asked questions in CAT exam. Practice Ratio and Proportion Set-4 Questions PDF for CAT exam.

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**Question 1: **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) 2 : 3

b) 4 : 3

c) 3 : 2

d) 3 : 4

**Question 2: **A student gets an aggregate of 60% marks in five subjects in the ratio 10 : 9 : 8 : 7 : 6. If the passing marks are 50% of the maximum marks and each subject has the same maximum marks, in how many subjects did he pass the examination?

a) 2

b) 3

c) 4

d) 5

**Question 3: **One bacterium splits into eight bacteria of the next generation. But due to environmental condition only 50% survives and remaining 50% dies after producing next generation. If the seventh generation number is 4,096 million, what is the number in first generation?

a) 1 million

b) 2 million

c) 4 million

d) 8 million

**Question 4: **I have one-rupee coins, 50-paisa coins and 25-paisa coins. The number of coins are in the ratio 2.5 : 3 : 4. If the total amount with me is Rs. 210, find the number of one-rupee coins.

a) 90

b) 85

c) 100

d) 105

**Question 5: **Company ABC starts an educational program in collaboration with Institute XYZ. As per the agreement, ABC and XYZ will share profit in 60 : 40 ratio. The initial investment of Rs.100,000 on infrastructure is borne entirely by ABC whereas the running cost of Rs. 400 per student is borne by XYZ. If each student pays Rs. 2000 for the program find the minimum number of students required to make the program profitable, assuming ABC wants to recover its investment in the very first year and the program has no seat limits.

a) 63

b) 84

c) 105

d) 157

e) 167

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**Question 6: **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) 2 : 3

b) 1 : 2

c) 1 : 3

d) 3 : 4

**Question 7: **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) 2

b) 3

c) 4

d) 5

**Question 8: **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) 2 kg

b) 2.4 kg

c) 2.5 kg

d) None of these

**Question 9: **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

**Question 10: **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) 3 : 10

b) 1 : 3

c) 1 : 4

d) 2 : 5

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**Answers & Solutions:**

**1) Answer (D)**

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

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

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

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

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

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

=> Ratio = 3:4

**2) Answer (C)**

Let’s say he scored marks as $10x,9x,8x,7x,6x$ or total of $40x$ which is 60% of total maximum marks(T).

$\frac{T \times 60}{100}=40x$

So T (total maximum marks)=$\frac{400x}{6}$

Or Individual max. marks = $\frac{T}{5}=\frac{80x}{6}$

Passing marks =50% of individual max. marks =$\frac{40x}{6}=6.66x$

Hence he scored more than passing marks in four subjects as $10x,9x,8x$ and $7x$

and failed in one subject as scoring $6x$ marks which is less than passing marks of $6.66x$

**3) Answer (A)**

let’s say x is the initial number of bacterias :

So in 2nd generation no. of bacterias = \frac{8x}{2} = 4x

In 3rd generation, it will be = 16x

4th gen. = 64x

5th gen. = 256x

6th gen. = 1024x

7th gen. = 4096x

Hence x = 1 million

**4) Answer (D)**

Let’s say number of coins are 2.5x , 3x and 4x

So total amount will be = 2.5x + 3x(0.5) + 4x(0.25) = 210

So x = 42

And number of 1 rs. coins = 2.5x = 105

**5) Answer (A)**

For program to be profitable both companies must recover costs before they can start making profits.

Since, ABC wants to recoup investment in the first year and there is no limit of number of students, profits can only be shared after both companies can reach a situation of minimum profits (zero profit), which would be:

Let x be minimum number of students required to reach a situation of minimum profits (in this case 0).

Comparing total costs and revenues :

=> $400x + 100000 = 2000x.$

=> $x = \frac{100000}{(2000 – 400)} = 62.5 = 63$

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**6) Answer (A)**

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.

**7) Answer (C)**

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

**8) Answer (C)**

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

**9) Answer: 50**

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) Answer (B)**

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

It is given that 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%.

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

$\Rightarrow$ $3a+2b+c = 180$ … (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 = $\dfrac{2b + 7c}{9}$ = $\dfrac{90 – 4c + 7c}{9}$ = $\dfrac{90 + 3c}{9}$

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

Hence, option B is the correct answer.

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