A barrel contains a mixture of wine and water in the ratio 3 : 1. How

1.	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 ? (a) \(\frac14\)(b) \(\frac13\) (c) \(\frac34\)(d) \(\frac23\)
Answer» (b) \(\frac13\) Let the barrel contains wine 3x litres and water x litres. Then, total mixture = 4x litres Let the part of mixture drawn out be p litres. ∴ (4\(x\) - p) x \(\frac34:(4x-p)\times\frac14+p=1:1\) ⇒ \(3x-\frac{3p}{4}:x-\frac{p}{4}+p=1:1\) ⇒ \(3x-\frac{3p}{4}=x+\frac{3p}{4}⇒2x=\frac{6p}{4}\) ⇒ p = \(\frac{2x\times4}{6}=\frac13(4x)\) ∴ \(\frac13\) part of mixture is drawn out.

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 ? (a) \(\frac14\)(b) \(\frac13\) (c) \(\frac34\)(d) \(\frac23\)

Answer»

(b) \(\frac13\)

Let the barrel contains wine 3x litres and water x litres.

Then, total mixture = 4x litres

Let the part of mixture drawn out be p litres.

∴ (4\(x\) - p) x \(\frac34:(4x-p)\times\frac14+p=1:1\)

⇒ \(3x-\frac{3p}{4}:x-\frac{p}{4}+p=1:1\)

⇒ \(3x-\frac{3p}{4}=x+\frac{3p}{4}⇒2x=\frac{6p}{4}\)

⇒ p = \(\frac{2x\times4}{6}=\frac13(4x)\)

∴ \(\frac13\) part of mixture is drawn out.