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### 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 n_{1} has a weighted average of A_{1}, and another sample n_{2} has a weighted average of A_{2}, then the weighted average, A,of the combined samples is determined by the alligation formula as: n_{1}(A - A_{1}) = n_{2}(A_{2} - A). Putting n_{1} = 80, A_{1} = 11, n_{2} = 160 and A_{2} = 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 n_{1} has an average price of A_{1}, and another sample n_{2} has an average price of A_{2}, then the price of the mixture, A, is determined by the alligation formula as: n_{1}(A - A_{1}) = n_{2}(A_{2} - A). Putting A_{2} = 24, A_{1} = 10, A = 20 we have n_{1} × (20 - 10) = n_{2} × (24 - 20), from where we get the required ratio as $n_1/n_2 = 2 : 5$.

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This Blog Post/Article "Alligations and Mixtures Quiz Set 001" by Parveen (Hoven) is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.

Updated on 2017-04-07.