SOLUTION: You need 420 mL of a 35% alcohol solution. On hand, you have a 25% alcohol mixture and a 60% alcohol mixture. How much of each mixture will you need to obtain the desired solution?
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Question 1061149: You need 420 mL of a 35% alcohol solution. On hand, you have a 25% alcohol mixture and a 60% alcohol mixture. How much of each mixture will you need to obtain the desired solution?
You can put this solution on YOUR website! .
You need 420 mL of a 35% alcohol solution. On hand, you have a 25% alcohol mixture and a 60% alcohol mixture.
How much of each mixture will you need to obtain the desired solution?
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Let x be the amount (the volume in mL) of the 25% solution to be mixed.
Then the volume of the 60% solution is 420-x.
The "alcohol amount" equation is
0.25x + 0.6(420-x) = 0.35*420.
Simplify and solve for x.
0.25x + 252 - 0.6x = 147,
0.25x - 0.6x = 147 - 252,
0.35x = 105 ---> x = = 300.
Answer. 300 mL of the 25% solution and 420-300 = 120 mL of the 60% solution.
You can put this solution on YOUR website! Let = ml of 25% mixture needed
Let = ml of 60% mixture needed
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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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300 ml of 25% mixture is needed
120 ml of 60% mixture is needed
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check:
(2)
(2)
(2)
(2)
(2)
OK
