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Mixture Problem**

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by Ozlemg » Thu Jun 02, 2011 8:17 am
A portion of the 85% solution of chemicals was replaced with an equal amount of 20% solution of chemicals. As a result, 40% solution of chemicals resulted. What part of the original solution was replaced?

A. 5/13
B. 6/13
C. 9/13
D. 13/9
C: 13/6

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by Frankenstein » Thu Jun 02, 2011 8:27 am
Hi,
Concentrations of the two solutions are 0.85 and 0.2 and that of mixture is 0.4
So, using principle of allegations, liquid1/liquid2 = (0.4-0.2)/(0.85-0.4)=4/9
So, liquid2 is 9/13 by volume. So, the portion of liquid1 replaced = 9/13

Hence, C
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by SoCan » Thu Jun 02, 2011 8:38 am
Frankenstein wrote:Hi,
Concentrations of the two solutions are 0.85 and 0.2 and that of mixture is 0.4
So, using principle of allegations, liquid1/liquid2 = (0.4-0.2)/(0.85-0.4)=4/9
So, liquid2 is 9/13 by volume. So, the portion of liquid1 replaced = 9/13

Hence, C
I'm getting the hang of this way, I like it.

Algebraically:

.85(1-x) + 0.2(x) = 0.4
.85 - 0.85x + 0.2x = 0.4
0.65x = 0.45
x = 0.45/0.65
x = 9/13
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by cans » Thu Jun 02, 2011 8:56 am
let x part of original solution is replaced
thus 1-x part (remaining part) is 85% and x part is 20%
which is 40% =>
(1-x)*85 + x*20 = 40
45=65*x
=>x=9/13
IMO C
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