SOLUTION: another question, sorry again =] How much of a 20% acid solution should be mixed with 80 mL of 5% acid solution to obtain a 12% acid solution? well, i tried it, but no luck...

Algebra ->  Expressions-with-variables -> SOLUTION: another question, sorry again =] How much of a 20% acid solution should be mixed with 80 mL of 5% acid solution to obtain a 12% acid solution? well, i tried it, but no luck...       Log On


   

Question 44769: another question, sorry again =]
How much of a 20% acid solution should be mixed with 80 mL of 5% acid solution to obtain a 12% acid solution?
well, i tried it, but no luck...
0.20(x mL) + 0.05(80 mL) = 0.12(80 + x)
but when i solve it, the answer doesn't come out reasonable..maybe im solving it wrong, i don't know, but please help =]
Found 2 solutions by josmiceli, shalu.srk:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
(amount of acid / total solution) x 100 = percent acid
(.2x + .05*80) / (x + 80) = .12
.12x + 9.6 = .2x + 4
.08x = 9.6 - 4
x = 5.6 / .08
x = 70 ml of 20% solution
check
(.20*70 + .05*80) / (70 + 80) = .12
(14 + 4) / 150 = .12
18/150 = .12
OK

Answer by shalu.srk(1) About Me  (Show Source):
You can put this solution on YOUR website!
0.20x%2B0.05%2880+mL%29=0.12%2880%2Bx%29
distribute all the things that can be distributed =]
0.20x%2B4=9.6%2B0.12x
0.20x=5.6%2B0.12x
0.08x=5.6
5.6%2F0.08=x=70+mL