Question

A chemist has one solution containing $50$% acid and a second one containing $25$% acid. How much of each should be mixed to make $10$ litres of a $40$% acid solution.

Answer 1

Let $X$ be the amount of first solution and Y be the amount of second to be taken.

The amount of acid in $x=0.5x$

But another $50\%$ has to be water.

The amount of water $=0.5x$.

The amount of acid in $y=0.25y$

The amount of water in $y=0.75y$

The total amount of acid required in final solution $=0.4\times 10=4$ litres and remaining $6$ lit would be water.

Therefore,

$0.5x+0.25y=4$ .............(1)

$Or,0.5x=4-0.25y$

Similarly for water,

$0.5x+0.75y=6$ .................(2)

Substituting the value of $0.5x$ from Eq. (1) into Eq. (2) we have,

$4-0.25y+0.75y=6$

$4+0.5y=6$

$y=4$

$x=6$