A milk vendor generally sells 3 Grades of milk. Grade I is pure milk with no water mixed in it, Grade II is a mixture of milk and water in the ratio 3:2 and Grade III is a mixture of milk and water in the ratio 2:3. On a particular day he has x liters of Grade I and 3 liters of Grade III milk and he got an order to supply 7 liters of Grade II milk. The minimum value ofx (in litres) required to prepare 7 Its of Grade H milk by mixing Grade I milk, Grade III milk and water, is

He needs to supply 7 ltr of Grade II milk : which contain milk:water in 3:2 ratio.

So, He must need $$\left(7\times\frac{3}{5}\right)=4.2$$ ltr of milk.

But ,if we consider only Grade III mixture ,then we have only $$\left(3\times\frac{2}{5}\right)=1.2$$ltr of milk available.

(as Grade III mixed in 2:3 ratio).

So, minimum amount of Grade I milk needed was (4.2-1.2)=3 ltr ,by which he can make

7 ltr of Grade II milk as well as Grade H milk mixtures by adding water to rest of the mixtures.

A is correct choice.