SOLUTION: You need a 40% alcohol solution. On hand, you have a 650 mL of a 20% alcohol mixture. You also have 90% alcohol mixture. How much of the 90% mixture will you need to add to obtain

Question 522149: You need a 40% alcohol solution. On hand, you have a 650 mL of a 20% alcohol mixture. You also have 90% alcohol mixture. How much of the 90% mixture will you need to add to obtain the desired solution?
You will need
mL of the 90% solution
to obtain
mL of the desired 40% solution.
Found 2 solutions by mananth, lowtek:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
--------percent ---------------- quantity
Alcohol Type I 20 ---------------- 650 ml
Alcohol type II 90 ---------------- x ml
Mixture 40 ---------------- 650 + x ml

20*650+90x=40(650+x)
13000+90x=26000 +40x
90x-40x =26000-( -13000)
50x=13000
/50
x= 260
260 ml of Alcohol type II (90%)

Answer by lowtek(4) (Show Source): You can put this solution on YOUR website!
Let x = Amount Invested at 40%
Let y = Amount Invested at 90%

Amount Invested = $2.96219e-154
Total Income = $0

x + y = 2.96219e-154
x = 2.96219e-154 - y

0.4x +0.9y = 0
40x +90y = 0

40(2.96219e-154 - y) +90y = 0
1.18488e-152 - 40y +90y = 0
50y = -1.18488e-152

y = $-2.36975e-154 Invested at 90%

x = 2.96219e-154 - -2.36975e-154

x = $5.33194e-154 Invested at 40%
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