**Ratio and Proportions questions for NMAT:**

Download Ratio and Proportions Questions for NMAT PDF. Top 10 very important Ratio and Proportions Questions for NMAT based on asked questions in previous exam papers.

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**Question 1:Â **A sum of money is split among Amal, Sunil and Mita so that the ratio of the sharesÂ of Amal and Sunil is 3:2, while the ratio of the shares of Sunil and Mita is 4:5. If theÂ difference between the largest and the smallest of these three shares is Rs.400,Â then Sunilâ€™s share, in rupees, is

**Question 2: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

**Question 3:Â **The respective ratio between the present age of Manisha and Deepali is 5 : X. Manisha is 9 years younger than Parineeta. Parineeta’s age after 9 years will be 33 years. The difference between Deepali’s and Manisha’s age is same as the present age of Parineeta. What will come in place of X?

a)Â 23

b)Â 39

c)Â 15

d)Â none of these

**Question 4: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

**Question 5: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

**Question 6:Â **Two alcohol solutions, A and B, are mixed in the proportion 1:3 by volume. The volumeÂ of the mixture is then doubled by adding solution A such that the resulting mixtureÂ has 72% alcohol. If solution A has 60% alcohol, then the percentage of alcohol inÂ solution B is

a)Â 90%

b)Â 94%

c)Â 92%

d)Â 89%

**Question 7:Â **A solution, of volume 40 litres, has dye and water in the proportion 2 : 3. Water is added to the solution to change this proportion to 2 : 5. If one fourths of this diluted solution is taken out, how many litres of dye must be added to the remaining solution to bring the proportion back to 2 : 3?

**Question 8: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

**Question 9: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

**Question 10: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

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**Answers & Solutions:**

**1)Â Answer:Â 800**

Let the amount of money with Amal and Sunil be 6x and 4x. Now the amount of money with Mita be 5x. Difference between the largest and smallest amount is â‚¹400 i.e. 6x-4x=400 or 2x=400 or x=200 . Amount of money with Sunil is 200(4)=â‚¹800

**2)Â AnswerÂ (B)**

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%

**3)Â AnswerÂ (D)**

Let the present age of Manisha be 5a

The present age of Deepali = Xa

The present age of Praneeta = 5a+9

Praneeta’s age after 9 years = 5a+18=33

5a=15

a=3

The difference between Deepali’s and Manisha’s age is the same as the present age of Parineeta

Xa-5a=5a+9

3X-15=15+9

3X=39 ==> X=13

D is the correct answer.

**4)Â AnswerÂ (B)**

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.

**5)Â AnswerÂ (C)**

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

**6)Â AnswerÂ (C)**

Initially let’s consider A and B as one component

The volume of the mixture is doubled byÂ adding A(60% alcohol) i.e they are mixed in 1:1 ratio and the resultant mixture has 72% alcohol.

Let the percentage of alcohol in component 1 be ‘x’.

Using allegations ,Â $\frac{\left(72-60\right)}{x-72}=\frac{1}{1}$ => x= 84

Percentage of alcohol inÂ A = 60% => Let’s percentage of alcohol in B = x%

The resultant mixture has 84% alcohol. ratio = 1:3

Using allegations ,Â $\frac{\left(x-84\right)}{84-60}=\frac{1}{3}$

=> x= 92%

**7)Â Answer:Â 8**

Initially the amount of Dye and Water are 16,24 respectively.

To make the ratio of Dye to Water to 2:5 the amount of water should be 40l for 16l of Dye=> 16l of water is added.

Now, the Dye and Water arr 16,40 respectively.

After removing 1/4th of solution the amount of Dye and Water will be 12,30l respectively.

To have Dye and Water in the ratio of 2:3, for 30l of water we need 20l of Dye => 8l of Dye should be added.

Hence , 8 is correct answer.

**8)Â AnswerÂ (B)**

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.

**9)Â AnswerÂ (C)**

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.

**10)Â AnswerÂ (C)**

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.

We hope this Ratio and Proportions Questions for NMAT pdf for NMAT exam will be highly useful for your Preparation.