SOLUTION: The manager of a fish store has water that is 10% salt and the other water that is 25%. He needs to fill an aquarium with 10 gallons of water that is 20%. what is the answer?
Question 1096495: The manager of a fish store has water that is 10% salt and the other water that is 25%. He needs to fill an aquarium with 10 gallons of water that is 20%. what is the answer? Found 3 solutions by Alan3354, Fombitz, josmiceli:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! The manager of a fish store has water that is 10% salt and the other water that is 25%. He needs to fill an aquarium with 10 gallons of water that is 20%. what is the answer?
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Depends on the question.
Here's a similar problem.
Same problem, different numbers, actually.
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A scientist mixes water (containing no salt) with a solution that contains
40% salt. She wants to obtain 200 ounces of a mixture that is 5% salt. How many ounces of water and how many ounces of the 40% salt solution should she use
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W = Water
S = 40% salt (if available)
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W + S = 200
W*0 + S*40 = (W+S)*5
40S = 5W + 5S
5W = 35S
W = 200-S
5(200-S) = 35S
1000 - 5S = 35S
40S = 1000
S = 25
You can put this solution on YOUR website! Let A be the amount of 10% water and B the amount of 25% water.
1.
and
2.
From 1,
Substituting into 2,
Solve for B, then use either equation to solve for A.
You can put this solution on YOUR website! Let = gallons of 10% salt water needed
Let = gallons of 25% salt water needed
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(1)
(2)
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(2)
(2)
(2)
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Multiply both sides of (1) by
and subtract (1) from (2)
(2)
(1)
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and
(1)
(1)
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3.333 gallons of 10% salt water needed
6.667 gallons of 25% salt water needed
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check:
(2)
(2)
(2)
(2)
(2)
OK
