Difference between revisions of "1997 PMWC Problems/Problem I9"
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− | 7500 | + | == Problem == |
+ | 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) | ||
+ | |||
+ | == Solution == | ||
+ | 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. | ||
+ | |||
+ | <cmath>\frac{48}{100}x + \frac{80}{100}(8-x) = \frac{40}{100}(10)</cmath> | ||
+ | <cmath>\frac{8}{25}x = \frac{12}{5}</cmath> | ||
+ | <cmath>x = \frac{15}{2}</cmath> | ||
+ | |||
+ | There are <math>1000 \cdot \frac{15}{2} = 7500</math> millilitres of the acid. | ||
+ | |||
+ | == See Also == | ||
+ | {{PMWC box|year=1997|num-b=I8|num-a=I10}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] |
Latest revision as of 13:26, 20 April 2014
Problem
A chemist mixed an acid of concentration with the same acid of concentration, and then added litres of distilled water to the mixed acid. As a result, he got litres of the acid of concentration. How many millilitre of the acid of concentration that the chemist had used? ( litre = millilitres)
Solution
Let the quantity of the 48% acid, in liters, be . Then the acid of 80% concentration has a volume of liters.
There are 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 |