SOLUTION: 60 fluid ounces of a 15% muriatic acid solution is needed to kill algae in a pool. If the technician has a 5% solution and a pure solution on hand, how many ounces of each must be

Question 32058: 60 fluid ounces of a 15% muriatic acid solution is needed to kill algae in a pool. If the technician has a 5% solution and a pure solution on hand, how many ounces of each must be combined to create a 15% solution?
Found 2 solutions by stanbon, mukhopadhyay:
Answer by stanbon(48558) (Show Source):
You can put this solution on YOUR website!
Let amount of 5% solution be "x"
Then amount of 100% solution is "60-x"
Muriatic acid in 5% solution is 0.05x
Muriatic acid in 100% solution is 1(60-x)
EQUATION:
Acid + Acid = 0.15(60)=9 ml
0.05x+60-x=9 ml
-0.95x=-51
x=53.68 ml(amount of 5% solution)
60-x=6.32 ml (amount of 100% solution)
Cheers,
stan H.

Answer by mukhopadhyay(490) (Show Source):
You can put this solution on YOUR website!
Assume x ounces of 5% solution is to be mixed.
Total volume of the mix=60 ounces; so, 60-x ounces of pure (100%) solution is to be mixed to kill algae in the pool;
Thus, 5% of x + 100% of (60-x) = 15% of 60
=> .05x+(60-x) = .15*60
=>60-.95x=9
=>.95x=60-9
=>.95x=51
=>x=51/.95=53.68
So, about 54 ounces of 5% solution is to be mixed with 6 ounces of pure solution. For this type of questions, usually the numbers do not come so odd.