SOLUTION: A chemist has three different acid solutions. The first acid solution contains 15 % acid, the second contains 35 % and the third contains 80%. He wants to use all three solutions t
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Question 1041870: A chemist has three different acid solutions. The first acid solution contains 15 % acid, the second contains 35 % and the third contains 80%. He wants to use all three solutions to obtain a mixture of 100 liters containing 30% acid, using 2 times as much of the 80 % solution as the 35 % solution. How many liters of each solution should be used? Found 2 solutions by stanbon, josmiceli:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! A chemist has three different acid solutions.
The first acid solution contains 15 % acid, the second contains 35 % and the third contains 80%.
He wants to use all three solutions to obtain a mixture of 100 liters containing 30% acid, using 2 times as much of the 80 % solution as the 35 % solution.
How many liters of each solution should be used?
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0.15f + 0.35t + 0.80e = 0.30*100
e = 2t
f + t + e = 100
f+3t = 100
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0.15f + 0.35t + 0.80(2t) = 0.30*100
0.15(100-3t) + 0.35t + 1.6t = 0.30*100
15(100-3t) + 35t + 160t = 30*100
1500 - 45t + 195t = 3000
150t = 1500
t = 10 liters (amt. of 35% needed)
e = 2t = 20 liters (amt. of 80% needed
f = 100 - 3t = 70 liters (amt of 15% needed)
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Cheers,
Stan H.
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You can put this solution on YOUR website! Let = liters of 15% solution needed
Let = liters of 35% solution needed
Let = liters of 80% solution needed
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(1)
(2)
(3)
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(3)
(3)
(3)
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Multiply both sides of (1) by and
subtract (1) from (3)
(3)
(1)
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Substitute (2) into this result
and
(2)
(2)
(2)
and
(1)
(1)
(1)
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70 liters of 15% solution is needed
10 liters of 35% solution is needed
20 liters of 80% solution is needed
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check:
(3)
(3)
(3)
(3)
(3)
OK
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