SOLUTION: What quantity of 80% solution must be mixed with a 30% solution to produce 750 mL of a 50% solution?

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Question 163030: What quantity of 80% solution must be mixed with a 30% solution to produce 750 mL of a 50% solution?
Answer by jojo14344(1513) About Me  (Show Source):
You can put this solution on YOUR website!

Take note of the following:
80% solt'n="x"
30% solt'n="y"
So, x%2By=750ml -------------------------------> EQN 1
In EQN 1 we get, x=750-y --------------------> EQN 2
After the mixture,
0.80x%2B0.30y=0.50%28750ml%29 ---------------------> EQN 3
Subst. EQN 2 in EQN 3:
0.80%28750-y%29%2B0.30y=0.50%28750%29
600-0.80y=0.30y=375
600-375=0.80y-0.30y
225=0.50y -----> cross%28225%29450%2Fcross%280.50%29=cross%280.50%29y%2Fcross%280.50%29
highlight%28y=450ml%29, QUANTITY for the 30% solt'n.
Via EQN 2,
x=750-450
highlight%28x=300ml%29, QUANTITY for the 80% solt'n.
In doubt? Go back EQN 3,
0.80%28300%29%2B0.30%28450%29=0.50%28750%29
240%2B135=375ml
375ml=375ml
Thank you,
Jojo