Question

To m litres of a m% solution of acid, x litre of water is mixed to yield (m -10)% solution of acid. If m > 25, then x equals

• A. $\frac{10m}{m-10}$
• B. $\frac{5m}{m-10}$
• C. $\frac{2m}{m-10}$
• D. $\frac{m}{m-10}$
• E. None of these

Answer 1

The correct option is A $\frac{10m}{m-10}$

Here, quantity of acid =$m\times \frac{m}{100}=\frac{m^2}{100}$
After adding x litres of water.
New percentage $=\frac{m^2\times 100}{(m+x)\times 100}$
or $\frac{m^2}{m+}=m-10$
or $m^2-10m+mx-10x=m^2$
or x(m-10)=10m
or $x=\frac{10m}{m-10}$