Advertisement

### Question

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 $41/{2 × 47}$ = ${41/94}$.