SOLUTION: A chemist needs 130 mL of a 28% solution but has only 18% and 83% solutions available. Find how many mL of each solution should be mixed to get the desired solution.
The answer is
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The answer is
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Question 900786: A chemist needs 130 mL of a 28% solution but has only 18% and 83% solutions available. Find how many mL of each solution should be mixed to get the desired solution.
The answer is 110mL of 18%: 20mL of 83%
I don't know how to set it up. Found 2 solutions by richwmiller, MathTherapy:Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! a+b=130,
0.18*a+0.83*b=0.28*130
a=130-b
0.18*(130-b)+0.83*b=36.4
23.4-0.18b+0.83*b=36.4
0.65*b=13
b=20
a=130-b
a=110 mL of 18% b=20 mL of 83%
check
0.18*110+0.83*20=0.28*130
19.8+16.6=36.4
36.4=36.4
ok
You can put this solution on YOUR website! A chemist needs 130 mL of a 28% solution but has only 18% and 83% solutions available. Find how many mL of each solution should be mixed to get the desired solution.
The answer is 110mL of 18%: 20mL of 83%
I don't know how to set it up.
Let amount of 18% solution to be mixed, be E
Then amount of 83% solution to be mixed = 130 - E
Therefore, .18E + .83(130 – E) = .28(130)
.18E + 107.9 - .83E = 36.4
.18E - .83E = 36.4 – 107.9
- .65S = - 71.5
E, or amount of 18% solution = , or mL
Amount of 83% solution to be mixed: 130 – 110, or mL
