Difference between revisions of "1997 PMWC Problems/Problem I9"
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| − | 7500 | + | == 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 <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]] | ||
Revision as of 15:26, 9 October 2007
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 
. Then the acid of 80% concentration has a volume of 
liters.
![\[\frac{48}{100}x + \frac{80}{100}(8-x) = \frac{40}{100}(10)\]](http://latex.artofproblemsolving.com/6/c/6/6c6f1ad4f907913a2032e8fdc1f38381ce2e16ee.png)
![\[\frac{8}{25}x = \frac{12}{5}\]](http://latex.artofproblemsolving.com/8/f/8/8f83d89ef1bc15c72e552bd3318c274a72118be4.png)
![\[x = \frac{15}{2}\]](http://latex.artofproblemsolving.com/5/f/a/5fac84d7f8b52a5fdbbcd6bb905de276b81b2d3f.png)
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 | ||
