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Some part of a 50% solution of acid was replaced with an equal amount

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Joined: 27 Dec 2009
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Some part of a 50% solution of acid was replaced with an equal amount [#permalink]

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New post 30 Dec 2009, 15:09
1
2
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

83% (00:55) correct 17% (01:19) wrong based on 65 sessions

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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?

A. 1/5
B. 1/4
C. 1/2
D. 3/4
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Posts: 8
Re: Some part of a 50% solution of acid was replaced with an equal amount [#permalink]

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New post 30 Dec 2009, 15:40
1
C. 1/2

In the table below, this refers to the original solution, not the acid concentration. Sorry about the formatting...it should be a 3X4 table with one expression in each box.

Original Removed Added Result
Concentration 0.50 0.50 0.70 0.60
Amount 1 x x 1
Multiply 0.50 0.50x 0.70x 0.60

0.50 (Original amount) - 0.50x (Removed) + 0.70x (Added) = 0.60 (End Result)
0.20x = 0.10
x = 1/2

Explanation:
Because the question asks for the original solution, you can change the numbers to reflect that. If 50% of the solution is acid, then 50% of the solution is the original solution. Then you're removing 50% concentration of acid (which is also 50% original solution) and then adding the same amount back but with 30% concentration of acid (also 70% concentration of original solution). What I mean by that is if you remove 50% of the acid, then you are also removing 50% concentration of the original solution. If you are adding an amount with 30% solution of acid, then you are also adding 70% of the original solution. The end result should be 40% acid---60% original solution, which is the focus since the question asks for the original amount of the solution.

Looking at the table, the original amount is 50% of let's say, 1 unit. The original solution is 0.50. You remove 50% concentration of original solution of some amount x and then you add back the same amount x but with 70% of original solution concentration. The end result is 60% original solution. Multiply down, then remove and add terms and solve for x. The amount you replaced, or x, is 1/2.

Try this problem: mixture-problem-78345.html and see how you do.

I had trouble with mixture problems, too, but I came across the user Economist's link, http://www.onlinemathlearning.com/mixture-problems.html, and this is where I got the above strategy from. You should check it out.
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Re: Some part of a 50% solution of acid was replaced with an equal amount [#permalink]

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New post 01 Jan 2010, 10:35
Another way to look at this problem could be this.

Let the total solution be T and the part removed be P.

Now T has 50% acid
And P is removed from T.

Therefore left over acid (by quantity) would be:
\(\frac{50}{100} * (T-P) = \frac{T-P}{2}\)

to this P was added which had 30% acid in it. Hence added acid quanity is:
\(\frac{30}{100} * P\)

The resultant mixture had 40% acid in it. Hence the resultant acid quantity is:
\(\frac{40}{100} * T = \frac{T-P}{2} + \frac{30}{100}* P\)


Solving for P, we get \(P = \frac{T}{2}\) which gives the part as \(\frac{1}{2}\) of Original

Therefore answer is C

Hope this helps!

Cheers!
JT
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Re: Some part of a 50% solution of acid was replaced with an equal amount [#permalink]

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New post 23 Jan 2010, 03:00
1
since the replaced amount is equal, let x be the amount replaced.

so, A - xA + xB

A - xA + xB = 0.4

so,

0.5 - 0.5x + 0.3x = 0.4
-0.5x + 0.3x = -0.1
-0.2x = -0.1
\(x = \frac{1}{2}\)

IMO, it's C
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Re: Some part of a 50% solution of acid was replaced with an equal amount [#permalink]

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New post 24 Mar 2017, 19:25
samrand wrote:
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?

A. 1/5
B. 1/4
C. 1/2
D. 3/4


let x=part of original solution replaced
.5-.5x+.3x=.4
x=1/2
C
Re: Some part of a 50% solution of acid was replaced with an equal amount   [#permalink] 24 Mar 2017, 19:25
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