SOLUTION: You need 1040 mL of a 20% alcohol solution. On hand, you have a 15% alcohol mixture. You also have a 95% alcohol mixture. How much of each mixture will you need to add to obtain th

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Question 1197162: You need 1040 mL of a 20% alcohol solution. On hand, you have a 15% alcohol mixture. You also have a 95% alcohol mixture. How much of each mixture will you need to add to obtain the desired solution?
Found 3 solutions by josgarithmetic, greenestamps, math_tutor2020:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
v mL of the 95%
1040-v of the 15%

95v%2B15%281040-v%29=20%2A1040


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95v%2B15%2A1040-15v=20%2A1040
95v-15v%2B15%2A1040=20%2A1040
%2895-15%29v=20%2A1040-15%2A1040
v=%2820%2A1040-15%2A1040%29%2F%2895-15%29
highlight_green%28v=1040%28%2820-15%29%2F%2895-15%29%29%29 a convenient form from which to start computations.

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Here is a non-algebraic way to solve any 2-part mixture problem like this; this method is quick and easy if the numbers are "nice".

Think of it this way. You are starting with a 15% alcohol solution, and you are adding 95% alcohol solution, stopping when the mixture is 20% alcohol.

20% is 1/16 of the way from 15% to 95% (picture the three percentages on a number line, if it helps).

That means 1/16 of the mixture needs to be the 95% alcohol solution that you are adding.

1/16 of 1040 mL is 65mL, so you need 65 mL of the 95% alcohol solution and 1040-65 = 975 mL of the 15% alcohol solution.

ANSWER: 65 mL of 95%; 975 mL of 15%.

CHECK:
.95(65)+.15(975) = 208
.20(1040) = 208

Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

You require 1040 mL of a 20% alcohol solution.
That means you need 0.20*1040 = 208 mL of pure alcohol.

x = amount of the 15% solution (in mL)
15% of x = 0.15x = amount of pure alcohol from the 15% solution

1040-x = amount of the 95% solution
note that x and 1040-x combine to 1040

95% of (1040-x) = 0.95(1040-x) = amount of pure alcohol from the 95% solution

Combine the subtotal amounts of pure alcohol from the two batches.
The goal is to get 208 mL of pure alcohol.

0.15x + 0.95(1040-x) = 208
0.15x + 0.95(1040)+0.95(-x) = 208
0.15x + 988 - 0.95x = 208
-0.80x + 988 = 208
-0.80x = 208 - 988
-0.80x = -780
x = -780/(-0.80)
x = 975
You'll need 975 mL of the 15% solution

And you'll need 1040-x = 1040-975 = 65 mL of the 95% solution.