SOLUTION: How many gallons of 20% alcohol solution and 45% alcohol solution must be mixed to get 10 gallons of 25% alcohol solution?

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Question 1154014: How many gallons of 20% alcohol solution and 45% alcohol solution must be mixed to get 10 gallons of 25% alcohol solution?
Found 3 solutions by josmiceli, MathTherapy, greenestamps:
Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
Let = gallons of 20% solution needed
Let = gallons of 45% solution needed
---------------------------------------------------
(1)
(2)
-----------------------------------------
(2)
(2)
(2)
Multiply both sides of (1) by
and subtract (1) from (2)
(2)
(1)
-----------------------------


and
(1)
(1)
---------------------------
8 gallons of 20% solution are needed
2 gallons of 45% solution are needed
--------------------------------------------
check:
(2)
(2)
(2)
(2)
(2)
OK

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

How many gallons of 20% alcohol solution and 45% alcohol solution must be mixed to get 10 gallons of 25% alcohol solution?
I STRONGLY believe that these types of problems, when possible, need to be solved using ONLY 1 variable.
For one, they're less confusing, and 2) much, much less time-consuming.
Let the amount of 20% alcohol to mix be T
Then amount of 45% alcohol to mix = 10 - T
We then get: .2T + .45(10 - T) = .25(10)
.2T + 4.5 - .45T = 2.5
.2T - .45T = 2.5 - 4.5
- .25T = - 2
T, or
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


...and I prefer to solve this kind of problem using 0 variables....

Here is the quick and easy solution to this problem:

(1) The target 25% is 1/5 of the way from 20% to 45%. (20 to 45 is a difference of 25; 20 to 25 is a difference of 5; 5/25 = 1/5)
(2) That means 1/5 of the mixture has to be the 45% ingredient.

ANSWER: 1/5 of 10 gallons, or 2 gallons, of the 45% alcohol solution; 10-2=8 gallons of the 20% alcohol solution.


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