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

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Intern
Joined: 27 Dec 2009
Posts: 15
Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

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30 Dec 2009, 15:09
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Question Stats:

81% (00:56) correct 19% (01:19) wrong based on 94 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
Intern
Joined: 18 Oct 2009
Posts: 8
Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

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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.

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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01 Jan 2010, 10:35
1
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:
$$\frac{50}{100} * (T-P) = \frac{T-P}{2}$$

$$\frac{30}{100} * P$$

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

Hope this helps!

Cheers!
JT
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Joined: 17 Jan 2010
Posts: 26
Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

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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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Joined: 07 Dec 2014
Posts: 1088
Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

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

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03 Sep 2018, 04:21
Hello chetan2u .

I was unable to interpret the question and the above solutions .
Thank you.
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Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

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03 Sep 2018, 04:30
1
SonGoku wrote:
Hello chetan2u .

I was unable to interpret the question and the above solutions .
Thank you.

Hi..

The question means that initially a 50% solution of acid is there. 50%solution of acid means the percentage of acid is 50%, so of the solution is 2 litre , the acid is 50% or 1 litre.
Later some of it is changed with a 30% solution that is a solution which has 30% as acid, so if you took 2 litre, acid was 2000*30/100=600ml
And finally it became 40% acid...

These questions are best done through weighted average method..
So two solutions 50% and 30% are mixed together to get 40%..
Since 40% is half way between 30 and 50, we can say that we took half of each..
Otherwise the weight of 30% = (50-40)/(50-30)=1/2 ....
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Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

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11 Sep 2018, 17:13
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 the original amount of the 50% solution of acid = 10 liters (so 5 liters are acid). Let x = the amount of solution that was replaced. So 0.5x acid was removed and replaced with 0.3x acid. We can create the equation:

(5 - 0.5x + 0.3x)/(10 - x + x) = 0.4

(5 - 0.2x)/10 = 4/10

5 - 0.2x = 4

1 = 0.2x

5 = x

Since 5/10 = 1/2, we see that 1/2 of the original solution was replaced.

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Re: Some part of a 50% solution of acid was replaced with an equal amount &nbs [#permalink] 11 Sep 2018, 17:13
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