Question

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?​

Answer 1

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.

∴(4x−p)×

4

3

:(4x−p)×

4

1

+p=1:1⇒3x−

4

3p

=x+

4

3p

⇒2x=

4

6p

⇒p=

6

2x×4

=

3

1

(4x)∴

3

1

part of mixture is drawn out.