Question

A mixture of 140 litre is prepared by mixing wine and rum having alcohol 42.5 % and 25% respectively in ratio 2 :5. To dilute the mixture by 2 % what amount of water is used
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Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

The amount of water used is 10 litres.

Step-by-step explanation:

Given:

A mixture of 140 litres is prepared by mixing wine and rum having alcohol 42.5 % and 25% respectively in ratio 2:5.

To dilute the mixture by 2 % what amount of water is used  

Solution:

A mixture of wine and rum = 2 : 5

Wine = $\frac{140 \times 2}{(2 + 5)}$= 40 litres

Rum = $\frac{140 \times 5}{(2 + 5)}$ = 100 litres

Alcohol in wine = 42.5 %

Alcohol in 40 litre wine = $40 \times 42.5 \times 100$ = 17 litres

Alcohol in rum = 25% = $\frac{25 \times 100}{100}$ = 5 litres

Total alcohol = 17 + 25 = 42 litres

Total mixture = 140 litres

Alcohol % = $\frac{42}{140} \times 100$ = 30%

Dilute the mixture by 2%, = 28%

Lets say W litre water is added,

28 = $\frac{100 \times 42}{(140 + W)}$

W= 10  litres.

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