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A barrel contains a mixture of wine and water in the ratio 3 : 1. How much fraction of the mixture must be drawn off and substituted by water so that the ratio of wine and water in the resultant mixture in the barrel becomes 1 : 1 ?
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1 4 -
1 3 -
3 4 -
2 3
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Correct Option: B
Let the barrel contain 4 litres of mixture.
∴ Wine = 3 litres
Water = 1 litre
Let x litre mixture is taken out.
∴ Wine in (4 – x) litres mixture
| = | (4 − x) | |
| 4 |
On adding x litres water, water in mixture
| = (4 − x) × | + x | |
| 4 |
| = 1 − | + x | |
| 4 |
| = | = | ||
| 4 | 4 |
| ∴ | (4 − x) = | ||
| 4 | 4 |
| ⇒ 3 − | = 1 + | ||
| 4 | 4 |
| ⇒ 2 = | |
| 4 |
| ⇒ x = | = | ||
| 6 | 3 | ||
| ∴ Required answer = | = | ||
| 4 | 3 |
