SOLUTION: How many liters of 70% methyl solution should be added to 200 liters of a 20% methyl solution to obtain a 35% methyl solution? A chemist has one solution that is 20% acid and a

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Question 903950: How many liters of 70% methyl solution should be added to 200 liters of a 20% methyl solution to obtain a 35% methyl solution?
A chemist has one solution that is 20% acid and a second that is 65% acid. How many gallons of each should be mixed together to get 120 gallons of a solution that is 50% acid?

Found 3 solutions by ewatrrr, stanbon, richwmiller:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
.70x + .20(200L) = .35(200L + x), x = .15(200L)/.35
.65x + .20(120G-x) = .50(120G), x = .30(120G)/.15

Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website!
How many liters of 70% methyl solution should be added to 200 liters of a 20% methyl solution to obtain a 35% methyl solution?
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Equation:
alcohol + alcohol = alcohol
0.70x + 0.20*200 = (0.35)(x+200)
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70x + 20*200 = 35x + 35*200
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35x = 15*200
x = (3/7)200
x = 85.71 liters (amt. of 70% solution needed)
200-x = 114.29 liters (amt. of 20% solution needed)
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Your 2nd problem is solved the same way as your 1st.
Cheers,
Stan H.
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Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website!
The first problem is set up
0.7*x+0.2*200=0.35(200+x)
The second problem is set up differently than the first because the info we have is different.
a+b=120,
0.2*a+0.65*b=0.5*120
a=120-b
0.2*(120-b)+0.65*b=60
24-0.2b+0.65*b=60
0.45*b=36
b=80
a=120-b
a=40 L at 20% b=80 L at 65%
check
0.2*40+0.65*80=0.5*120
8+52=60
60=60
ok
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