1997 PMWC Problems/Problem I9

Problem

A chemist mixed an acid of $48\%$ concentration with the same acid of $80\%$ concentration, and then added $2$ litres of distilled water to the mixed acid. As a result, he got $10$ litres of the acid of $40\%$ concentration. How many millilitre of the acid of $48\%$ concentration that the chemist had used? ($1$ litre = $1000$ millilitres)

Solution

Let the quantity of the 48% acid, in liters, be $x$. Then the acid of 80% concentration has a volume of $10 - 2 - x = 8 - x$ liters.

\[\frac{48}{100}x + \frac{80}{100}(8-x) = \frac{40}{100}(10)\] \[\frac{8}{25}x = \frac{12}{5}\] \[x = \frac{15}{2}\]

There are $1000 \cdot \frac{15}{2} = 7500$ millilitres of the acid.

See Also

1997 PMWC (Problems)
Preceded by
Problem I8
Followed by
Problem I10
I: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
T: 1 2 3 4 5 6 7 8 9 10
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