Expert: Socrates - 2/21/2009

Question
hallo, here's my question

Two  40 % and 60 % solutions of sulfuric acid were mixed and then were added 5 kg of pure water, as a result they got 20 % sulfuric solution. If they had added 5 kg of 80 % solution of sulfuric acid instead of 5 kg of pure water, they would have got 70 % solution. what was the amount of 1st and 2nd solutions?

thank you beforehand...

Answer
x = the amount of the first solution

y = the amount of the second solution

the total amount of sulfuric acid doesn't change after adding 5kg of water , so

.4x + .6y = .2 (x+y+5)


Adding 5 kg of 80% sulfuric acid adds (.8)(5) = 4 kg of sulfuric acid.
So .4x + .6y + 4 is the total amount of sufuric acid in the new solution if this is added.

This solution is 70 % sulfuric acid , so


.4x + .6y + 4 = .7(x+y+5)


Solve these two equations together

.4x + .6y = .2 (x+y+5)

.4x + .6y + 4 = .7(x+y+5)

Multiply through by 10 to get rid of the decimals

4x + 6y = 2(x+y+5)

4x + 6y + 40 = 7(x+y+5)

simplify

2x + 4y = 10

3x + y = 5

solve and get

x = 1  , y = 2