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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