# Alligations and Mixtures Quiz Set 001

### Question 1

The average weight of a class of 80 students is 11 Kg, and the average weight of a class of 160 students is 11 Kg. What is the average weight of the two combined classes?

A

11 Kg.

B

12 Kg.

C

10 Kg.

D

14 Kg.

Soln.
Ans: a

If a sample n1 has a weighted average of A1, and another sample n2 has a weighted average of A2, then the weighted average, A,of the combined samples is determined by the alligation formula as: n1(A - A1) = n2(A2 - A). Putting n1 = 80, A1 = 11, n2 = 160 and A2 = 11, we get A = 11 Kg.

### Question 2

In what ratio should water and wine be mixed so that after selling the mixture at cost price a profit of 10% is made?

A

${1/10}$.

B

$1{2/9}$.

C

$1{3/4}$.

D

$2{7/12}$.

Soln.
Ans: a

The water will be sold at the price of wine. So the profit will be earned by selling water as wine. Profit% = $\text"Profit"/\text"Cost"$ × 100. If profit percent is 10%, then 1 part of pure water is sold at the cost of 10 parts of wine. So the required mixing ratio should be 1 : 10.

### Question 3

In what ratio should water and wine be mixed so that after selling the mixture at cost price a profit of 50% is made?

A

${1/2}$.

B

$1{1/2}$.

C

$2{1/2}$.

D

$1{3/4}$.

Soln.
Ans: a

The water will be sold at the price of wine. So the profit will be earned by selling water as wine. Profit% = $\text"Profit"/\text"Cost"$ × 100. If profit percent is 50%, then 1 part of pure water is sold at the cost of 2 parts of wine. So the required mixing ratio should be 1 : 2.

### Question 4

A mixture of milk and water contains 47 parts of milk and 6 parts of water. How much fraction of the mixture should be removed and replaced by water so that ratio of water and milk becomes equal?

A

${41/94}$.

B

$1{14/31}$.

C

$2{37/96}$.

D

$3{35/96}$.

Soln.
Ans: a

Let the volume of the mixture be 47 + 6 = 53 liters. If x liters of the mixture is removed and replaced by water, the volume of water in the new mixture is $6 - {6x}/53 + x$. The volume of the milk in the new mixture would be $47 - {47x}/53.$ Equating the two volumes and solving for x we get x = ${53 × 41}/{2 × 47}$. The fraction that must be removed = $1/53$ × ${53 × 41}/{2 × 47}$, which gives $41/{2 × 47}$ = ${41/94}$.

### Question 5

In what ratio should a vendor mix two types of pulses costing Rs. 10/Kg and Rs. 24/Kg respectively so as to get a mixture of Rs. 20/Kg?

A

${2/5}$.

B

$1{3/4}$.

C

$1{5/7}$.

D

$2{3/7}$.

Soln.
Ans: a

We shall use the alligation formula. If a sample n1 has an average price of A1, and another sample n2 has an average price of A2, then the price of the mixture, A, is determined by the alligation formula as: n1(A - A1) = n2(A2 - A). Putting A2 = 24, A1 = 10, A = 20 we have n1 × (20 - 10) = n2 × (24 - 20), from where we get the required ratio as $n_1/n_2 = 2 : 5$.