Question

A chemist has a $20\%$ and a $40\%$ acid solutions. What amount of each solution should be used in order to make $300ml$ of a $28\%$ acid solution?

Answer 1

Let $x$ and $y$ represent the amount of $20$% and $40$% solutions that the chemist adds, respectively.

$\Rightarrow x+y=300\Rightarrow x=300-y$

and $20$% of $\left(x\right)+40$% of $\left(y\right)=28$% of $\left(300\right)$

$\Rightarrow \frac{20}{100}\times (300-y)+\frac{40}{100}\times \left(y\right)=\frac{28}{100}\times 300$

$\Rightarrow 6000-20y+40y=8400$

$\Rightarrow 20y=2400\Rightarrow y=120$

$\Rightarrow x=300-120=180$

Therefore, he must use $180$ml of $20$% and $120$ml of $40$% acid solution.