SOLUTION: You need 300 mL of a 40% alcohol solution. On hand, you have a 35% alcohol mixture and a 65% alcohol mixture. How much of each mixture will you need to obtain the desired solution?

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: You need 300 mL of a 40% alcohol solution. On hand, you have a 35% alcohol mixture and a 65% alcohol mixture. How much of each mixture will you need to obtain the desired solution?      Log On

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Question 957736: You need 300 mL of a 40% alcohol solution. On hand, you have a 35% alcohol mixture and a 65% alcohol mixture. How much of each mixture will you need to obtain the desired solution?

Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
T=amount of 35%; S=amount of 65%
T+S=300ml
T=300ml-S
0.35T+0.+0.65S=0.40(300ml) Substitute for T.
0.35(300ml-S)+0.65S=120ml
105ml-0.35S+0.65S=120ml Subtract 105ml from each side.
0.30S=15ml Divide each side by 0.30
S=50ml ANSWER 1: You need 50ml 0f 65% mixture.
T=300ml-50ml=250ml ANSWER 2: You need 250ml of 35% mixture.
CHECK:
0.35T+0.+0.65S=0.40(300ml)
0.35(250ml)+0.65(50)=0.40(300ml)
87.5ml+32.5ml=120ml
120ml=120ml