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Question 1188514: A scientist needs 7 litres of a 50% alcohol solution for an experiment. He has a 30% and an 80% solution on hand in the lab. How many litres of the 30% and how many litres of the 80% solutions should he mix to make the 50% alcohol solution?
Found 3 solutions by josgarithmetic, Shin123, greenestamps:
Answer by josgarithmetic(39617)   (Show Source):
Answer by Shin123(626)  (Show Source):
You can put this solution on YOUR website! Let x be the amount of 30% alcohol solution you need, and let y be the amount of 80% alcohol solution you need.
We have that  , since the scientist needs 7 liters of solution, and  , since  represents how much alcohol is in the mixture, and 3.5 represents how much alcohol you need.
| Solved by pluggable solver: SOLVE linear system by SUBSTITUTION |
Solve:
 We'll use substitution. After moving 1*y to the right, we get:
 , or  . Substitute that
into another equation:
 and simplify: So, we know that y=2.8. Since , x=4.2.
Answer:  .
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Therefore, the scientist needs 4.2 liters of 30%, and 2.8 liters of 80%.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Here is a quick and easy way to solve this kind of problem informally, if a formal algebraic solution is not required.
(1) Consider the three percentages 30, 50, and 80 on a number line and observe/calculate that 50 is 20/50 = 2/5 of the way from 30 to 80.
(2) That means 2/5 of the mixture needs to be the higher percentage ingredient.
ANSWER: 2/5 of 7 liters, or 2.8 liters, of the 80% solution; the other 4.2 liters of 30% solution.
CHECK:
0.80(2.8)+0.30(4.2)=2.24+1.26=3.50
0.50(7)=3.50
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