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mixtures - very tough

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by vkb16 » Mon Oct 19, 2009 3:31 am
Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?

1/5
1/4
1/2
3/4
4/5

Please explain what the qs is asking.. the wording is difficult to understand

OA is c
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by happybruce » Mon Oct 19, 2009 4:32 am
just looking at the numbers alone, i'd say the answer is 1/2(="c") b/c only when 1/2 of 50% solution is mixed with 1/2 of 30% solution, one will get a whole of 40% solution.
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by BuckeyeT » Mon Oct 19, 2009 5:13 am
vkb16-

I hope this explanation helps. Let's suppose our 50% acid solution is A and our 30% solution is B.

"Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid..." should translate to...
A - xA + xB (where x is the amount replaced)
Basically, we're replacing x amount of solution A with x amount of solution B (as the question states "equal amount").

We know that the new solution is 40%, so:
A - xA + xB = .4
Substitute back in our original values
.5 -.5x +.3x = .4
-.5x +.3x = -.1
-.2x = -.1
x = 1/2

C.
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by rahul.s » Sat Jan 23, 2010 1:56 am
is there a way of plugging in numbers and solving it. i tried, but couldn't get it. could someone please let me know?
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by thephoenix » Sat Jan 23, 2010 2:13 am
another method to solve such q is attached
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