In a mixture, unbroken and broken rice grains are in the ratio 3 : 2. How much fraction of the mixture must be drawn off and substituted with broken grains, so that the ratio of unbroken and broken grains becomes 1 : 1?

Let quantity of unbroken and broken rice grains be $$3$$ and $$2$$ units respectively.

Let $$x$$ units of mixture is taken out

=> Unbroken rice grains in $$(5-x)$$ units of mixture = $$\frac{3}{5}(5-x)$$

Similarly, after adding broken rice grains, => Broken rice grains in mixture = $$\frac{2}{5}(5-x)+x$$

According to ques,

=> $$\frac{\frac{3}{5}(5-x)}{\frac{2}{5}(5-x)+x}=\frac{1}{1}$$

=> $$3-\frac{3}{5}x=2-\frac{2}{5}x+x$$

=> $$\frac{3x}{5}-\frac{2x}{5}+x=3-2$$

=> $$\frac{6x}{5}=1$$

=> $$x=\frac{5}{6}$$

$$\therefore$$ Required fraction = $$\frac{\frac{5}{6}}{5}=\frac{1}{6}$$

=> Ans - (B)