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vkb16: Master | Next Rank: 500 Posts; Posts: 122; Joined: Mon Dec 15, 2008 12:03 am; Thanked: 2 times

mixtures - very tough

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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Source: Beat The GMAT — Problem Solving | Mon Oct 19, 2009 3:31 am

happybruce: Newbie | Next Rank: 10 Posts; Posts: 9; Joined: Wed Oct 07, 2009 8:10 am; Thanked: 2 times

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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BuckeyeT: Senior | Next Rank: 100 Posts; Posts: 76; Joined: Sat Jan 24, 2009 3:00 pm; Thanked: 11 times; GMAT Score:730

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.