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

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

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Updated on: 22 Jun 2019, 02:13
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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. $$\frac{1}{5}$$

B. $$\frac{1}{4}$$

C. $$\frac{1}{2}$$

D. $$\frac{3}{4}$$

E. $$\frac{4}{5}$$

M25Q30

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Originally posted by Hermione on 18 Nov 2006, 20:56.
Last edited by Bunuel on 22 Jun 2019, 02:13, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

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16 Jun 2013, 23:07
2
6
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. $$\frac{1}{5}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$\frac{3}{4}$$
E. $$\frac{4}{5}$$

Since the concentration of acid in resulting solution (40%) is the average of 50% and 30% then exactly half of the original solution was replaced.

Alternately you can do the following, say $$x$$ part of the original solution was replaced then: $$(1-x)*0.5+x*0.3=1*0.4$$ --> $$x=\frac{1}{2}$$.

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

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19 Nov 2006, 07:50
6
1
perhaps the fastest way is :
1/2*x + 3/10* x = 0.4
->x= 1/2
##### General Discussion
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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 Dec 2008, 07:11
4
2
Look at it this way.

Total part is 1. Out of this, x part was replaced.

Hence, (1-x)*50 (percent) + x*30 = 1*40
or, 50 - 20x = 40
or, 10 = 20x
or, x = 1/2....or, 50% of original solution was replaced.
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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
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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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
2
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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17 Sep 2012, 05:14
3
Alligative Method is such a time-saving method.

Here we go:

30%-----40%-----50%
Distances 1:1

This means that both contribute an equal amount of the final solution.

Straight C: 1/2

Cheers,
Der alte Fritz.
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Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

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16 Jun 2013, 16:55
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If 50% solution becomes 40% solution, then 40% solution must be equally weighted between 50% solution and 30% solution - i.e. $$\frac{1}{2}$$ of 50% solution and $$\frac{1}{2}$$ of 30% solution. If the solution must be equally weighted, then 30% solution must replace $$\frac{1}{2}$$ of 50% solution (assuming original solution was 50% solution of acid).

Just another way of saying the same thing. Cheers
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Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

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10 Jul 2014, 02:09
3
Did using allegation method

Refer diagram below

Bunuel, kindly update the OA.

Attachments

all.png [ 3.61 KiB | Viewed 7284 times ]

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

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20 Jul 2015, 06:11
4
chrissy28 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. $$\frac{1}{5}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$\frac{3}{4}$$
E. $$\frac{4}{5}$$

M25Q30

Responding to a pm:

It is worded to look like a replacement question but essentially, it is a simple mixtures problem only.

You have some 50% solution of acid and you remove some of it. Now to that 50% solution, you are adding 30% solution to get 40% solution. The ratio in which you mix the 50% solution and the 30% solution is given by w1/w2 = (50 - 40)/(40 - 30) = 1/1
Basically, you mix equal parts of 50% solution and 30% solution.
So initially you must have had 2 parts of 50% solution out of which you removed 1 part and replaced with 1 part of 30% solution. So you replaced half of the original solution.

The first question of this post discusses exactly this concept: http://www.veritasprep.com/blog/2012/01 ... -mixtures/
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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
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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
1
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  [#permalink]

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12 Feb 2019, 00:07
my thought process was to use the process of elimination strategy, so I started with the middle answer of 1/2.

if half of the original solution was replaced that means I am left with:

50% of 50% acid solution + 50% of 30% solution = new acid solution (which happens to be 40% acid)

.5x.5 = .25
.5x.3 = .15

.25+.15 = .4. So, my first attempt was the answer.

Let's say the new acid solution was 35% instead of 40%, then after trying the 1/2 answer, I would realize that I need more of the 30% solution to bring down the amount of acid. From this point, I can eliminate A,B, and C.

I would then try 3/4 of the 30% solution.

.25*.5 + .75*.3 = new solution
.125+.225 = .35, which is the answer I was looking for.

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

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12 Feb 2019, 00:28
chrissy28 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. $$\frac{1}{5}$$

B. $$\frac{1}{4}$$

C. $$\frac{1}{2}$$

D. $$\frac{3}{4}$$

E. $$\frac{4}{5}$$

M25Q30

so we have a mixture in which
Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid.

40% solution of acid was obtained

By allegation method, we get the ratio
M1--------M2
50----------30
-------40-----
10-----------10

Mixture will be in the ratio of 1:1

what part of the original solution was replaced?

Total mixture = 2

1/2 was replaced

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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04 Sep 2019, 10:50
We know this,

w1/w2=A2-Avg/Avg-A1

ie

w1/w2=50-40/40-30

w1/w2=1

originally 50% alcohol was 1
now it is 1/2

so reduction is 1/2
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Re: Some part of a 50% solution of acid was replaced with an equal amount   [#permalink] 04 Sep 2019, 10:50
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