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Some part of the 50% solution of acid was replaced with the [#permalink]
18 Nov 2006, 19:56

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Difficulty:

15% (low)

Question Stats:

87% (01:43) correct
13% (00:56) wrong based on 113 sessions

Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution f acid was obtained. what part of the original solution was replaced?

Re: PS: Mixture Problem [#permalink]
12 Nov 2009, 16:19

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Hermione wrote:

Does anyone have a quick and easy way of solving this problem? I had to pick from the answer choices but I want to know how to solve it outright.

Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution f acid was obtained. what part of the original solution was replaced?

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

Let X = Part of acid solution replaced, then 1-X will be the parts not replaced.

X be the amount of 50% solution
n be the amount taken out from 50% solution and the same amount of 30% solution is added later
In the resultant solution, acid amount is X/2 â€“ n/2 + 3n/10 = 2X/5
N = Â½

Re: PS: Mixture Problem [#permalink]
09 Apr 2010, 15:04

Wow the box method makes these problems so easy. can you explain a bit more on the box method. ORAD?can we use this for most of the misture problems .any limitations?

Re: PS: Mixture Problem [#permalink]
07 Nov 2010, 05:59

Expert's post

greatchap wrote:

I am using a chart to solve this problem. Can anyone point out the error I am making?

See image attached below.

x/2 + 3x/10 is the total stuff which is equal to 4 Solving x/2 + 3x/10 = 4 gives x = 5

Since x is the amount you assumed was replaced out of 10, you get 50%. Nothing wrong with what you did.

Note: Simply put, this question says that some 50% solution was mixed with some 30% solution to get 40% solution. Using weighted averages, you can find out that they must have been mixed in equal quantities or in other words, 50% of the original solution must have been replaced. _________________