Difference between revisions of "1997 PMWC Problems/Problem I9"

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== Problem ==
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A chemist mixed an acid of <math>48\%</math> concentration with the same acid of <math>80\%</math> concentration, and then added <math>2</math> litres of distilled water to the mixed acid. As a result, he got <math>10</math> litres of the acid of <math>40\%</math> concentration. How many millilitre of the acid of <math>48\%</math> concentration that the chemist had used? (<math>1</math> litre = <math>1000</math> millilitres)
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== Solution ==
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Let the quantity of the 48% acid, in liters, be <math>x</math>. Then the acid of 80% concentration has a volume of <math>10 - 2 - x = 8 - x</math> liters.
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<cmath>\frac{48}{100}x + \frac{80}{100}(8-x) = \frac{40}{100}(10)</cmath>
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<cmath>\frac{8}{25}x = \frac{12}{5}</cmath>
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<cmath>x = \frac{15}{2}</cmath>
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There are <math>1000 \cdot \frac{15}{2} = 7500</math> millilitres of the acid.
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== See Also ==
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{{PMWC box|year=1997|num-b=I8|num-a=I10}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 14:26, 20 April 2014

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