SOLUTION: how many gallons of a 5% acid solution should be mixed with 30 gallons of a 10% acid solution to obtain a mixture that is 8% acid?
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Question 103174This question is from textbook Elementary and intermediate algebra
: how many gallons of a 5% acid solution should be mixed with 30 gallons of a 10% acid solution to obtain a mixture that is 8% acid?
This question is from textbook Elementary and intermediate algebra
Found 2 solutions by checkley75, stanbon:
Answer by checkley75(3666) (Show Source): You can put this solution on YOUR website!
.05x+.10*30=.08(30+x)
.05x+3=2.4+.08x
.05x-.08x=2.4-3
-.03x=-.6
x=-.6/-.03
x=20 gallons of 5% acid will be needed.
proof
.05*20+3=.08*50
1+3=4
4=4
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
how many gallons of a 5% acid solution should be mixed with 30 gallons of a 10% acid solution to obtain a mixture that is 8% acid?
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10% liquid DATA:
Amt = 30 gals ; amt of active ingredient = 0.10*30 = 3 gallons
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5% liquid DATA:
Amt = x gals ; amt of active = 0.05x gallons
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Mixture DATA:
Amt = 30+x gallons ; amt of active = 0.08(30+x) = 2.4 + 0.08x gallons
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EQUATION:
active + active = active
3 + 0.05x = 2.4 + 0.08x
0.6 = 0.03x
x = 0.6/0.03 = 20 gallons ( amt of 5% acid solution that must be added)
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Cheers,
Stan H.
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