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Mixture Problem

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Mixture Problem

by sme0928 » Fri Aug 14, 2009 3:42 pm
My question is on the explanation on the bottom, how does he arrive at (1-X) * 0,25?

Thanks,


800guy wrote:
OA coming when some people have answered..

A certain quantity of 40% solution is replaced with 25% solution such that the new concentration is 35%. What is the fraction of the solution that was replaced?
(A) 1/4
(B) 1/3
(C) 1/2
(D) 2/3
(E) ¾


Let X be the fraction of 40% solution kept
Then we have

X * 0,4 + (1-X) * 0,25 = 0,35
<=> 0,4*X + 0,25 - 0,25*X = 0,35
<=> 0,15*X = 0,10
<=> X = 2/3
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by ssmiles08 » Fri Aug 14, 2009 4:03 pm
here is how I did it.

let total solution = 100.

40% of original solution = 40(100) = 40.

amount x is taken out. = - .4(x)

same amount x is added with different % = + .25(x)

net % of total solution is .35(100) = 35.

so 40 -.4x + .25x = 35

x = 100/3

so 100/3 was the amount taken out and added in with different percentages. Total solution is 100.

so (100/3) / 100 = 1/3

IMO B.
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