Difference between revisions of "1997 PMWC Problems/Problem I9"
(sol) |
(→Problem) |
||
(One intermediate revision by one other user not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == 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) | + | 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 == | == Solution == | ||
Line 11: | Line 11: | ||
There are <math>1000 \cdot \frac{15}{2} = 7500</math> millilitres of the acid. | There are <math>1000 \cdot \frac{15}{2} = 7500</math> millilitres of the acid. | ||
− | == See | + | == See Also == |
{{PMWC box|year=1997|num-b=I8|num-a=I10}} | {{PMWC box|year=1997|num-b=I8|num-a=I10}} | ||
[[Category:Introductory Algebra Problems]] | [[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 |