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 $10$ litres of a $40\%$ acid solution?
✓
Answer
Let $x$ litres of $50 \%$ solution be mixed with $y$ litres of $25 \%$ solution.
Accroding to the given condition,
$50\% of x + 25\% of y = 40\% of 10$
$\Rightarrow\frac{50}{100}\text{x}+\frac{25}{100}\text{y}=\frac{40}{100}(10)$
$50x + 25y = 40(10)$
$2x + y = 16 ...(i)$
Since the amount of each solutions adds to $10$ litres,
$x + y = 10 ...(ii)$
Subtract $(ii)$ from $(i).$
$x = 6$
Substituting $x = 6$ in $(ii),$ we get
$y = 4.$
Hence, $6$ liters of $50\%$ solution is to be mixed with $4$ litres of $25\%$ solution.
Accroding to the given condition,
$50\% of x + 25\% of y = 40\% of 10$
$\Rightarrow\frac{50}{100}\text{x}+\frac{25}{100}\text{y}=\frac{40}{100}(10)$
$50x + 25y = 40(10)$
$2x + y = 16 ...(i)$
Since the amount of each solutions adds to $10$ litres,
$x + y = 10 ...(ii)$
Subtract $(ii)$ from $(i).$
$x = 6$
Substituting $x = 6$ in $(ii),$ we get
$y = 4.$
Hence, $6$ liters of $50\%$ solution is to be mixed with $4$ litres of $25\%$ solution.
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