## IIFT Ratio and Proportion Questions PDF

Download important IIFT Ratio and Proportion Questions PDF based on previously asked questions in IIFT and other MBA exams. Practice Ratio and Proportion questions and answers for IIFT 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

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

a) 1.3

b) 1

c) 0.6

d) 2.3

**Question 5: **Product M is produced by mixing chemical X and chemical Y in the ratio of 5 : 4. Chemical X is prepared by mixing two raw materials, A and B, in the ratio of 1 : 3. Chemical Y is prepared by mixing raw materials, B and C, in the ratio of 2 : 1. Then the final mixture is prepared by mixing 864 units of product M with water. If the concentration of the raw material B in the final mixture is 50%, how much water had been added to product M?

a) 328 units

b) 368 units

c) 392 units

d) 616 units

e) None of the above

**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 (A)**

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.

**5) Answer (B)**

Let the quantities of the chemicals X and Y, mixed to produce product M be $5c$ and $4c$ respectively.

X is prepared by mixing A and B in the ratio = 1 : 3

=> Quantity of B in X = $\frac{3}{4} \times 5c = \frac{15 c}{4}$

Y is prepared by mixing B and C in the ratio = 2 : 1

Quantity of B in Y = $\frac{2}{3} \times 4c = \frac{8 c}{3}$

Quantity of B in M = $\frac{15 c}{4} + \frac{8 c}{3} = \frac{77 c}{12}$

Now, 864 units of M was mixed with water to prepare the final mixture.

=> Total quantity of M = $9c = 864$ => $c = \frac{864}{9} = 96$

Concentration of raw material B in the final mixture is 50 %

=> Quantity of final mixture = $\frac{100}{50} \times \frac{77}{12} \times 96 = 1232$

$\therefore$ Quantity of water added to M = $1232 – 864 = 368$ units

**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)**

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. For example, 100 kg of fresh grapes will contain 90 kg of water and 10 kg pulp. If this grape is dried, the water content will change but pulp content will remain the same.

So, for 2kg of pulp,

80% of the weight of dry grapes = 2

20% water weight=2/4=0.5

Weight of dry grapes =water content+pulp= 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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