Question

A chemist has one solution containing 50% acid and a second one containing 25% acid. How much of each should be used to make 10 liters of a 40% acid solution? [ 4 MARKS]

Answer 1

Concept : 1 Mark
Application : 1 Mark
Calculation : 2 Marks

Let $x$ liters of $50\%$ solution be mixed with $y$ liters of $25\%$ solution.

According to the given condition,

$x+y=10\dots \dots \left(i\right)$

Also,

$\frac{50}{100}\times x+\frac{25}{100}\times y=\frac{40}{100}\times 10$

$\Rightarrow \frac{x}{2}+\frac{y}{4}=4$

$\Rightarrow 2x+y=16\dots \dots \left(ii\right)$

Solving (i) and (ii) we get,

$x=10-y$ [From (i)]

Substituting value of x in (ii)

$2x+y=16$

$2(10-y)+y=16$

$20-y=16\Rightarrow y=4$

Now, $x=10-y$

$x=10-4\Rightarrow x=6$

$\Rightarrow x=6,y=4$

$\therefore$ 6 liters of $50\%$ solution is to be mixed with 4 liters of $25\%$ solution.