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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: 09 Nov 2018, 06:03
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Question Stats:

79% (00:49) correct 21% (01:30) wrong based on 362 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. $$\frac{1}{5}$$

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

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

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

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

M25Q30

Originally posted by chrissy28 on 12 Aug 2008, 13:27.
Last edited by Bunuel on 09 Nov 2018, 06:03, edited 3 times 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, 22:07
2
3
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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16 Jun 2013, 15:55
3
2
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
##### 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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12 Aug 2008, 14:16
The reason that 1/2 is correct is this:

What would happen if you have 1 liter of 60% solution (NaCl with water) and add 1 liter of 50% solution? I'm not chemist but lets mix $$H_2O$$ (water) with Sodium Chloride ($$NaCl$$).

The result will be a solution of 55%. 60% will have .6 liters of NaCl and the 50% solution will have .5 liters of NaCl. Together that's 1.1 liters out of the total of 2 liters.

Here we see that 50% solution becomes 40% from adding 30%. 40% is just the average between the two, so we know that there must now be equal parts 50% solution and 30% solution. The only way to get equal parts is if you have 1/2 50% solution and 1/2 30% solution meaning 1/2 was replaced.

chrissy28 wrote:

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

(A) 1/5
(B) 1/4
(D) 3/4
(E) 4/5

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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, 14:40
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, 09:35
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, 02:00
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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17 Jun 2013, 20:57
Hi Bunuel, What is wrong in this equation? Kindly guide me.

Let total solution =100
Replacement = x (say)

equation:
(100-x)*50% + X*30% = 40%
Solving this equation i am getting x= 248

I looked every way possible but could not figure out the problem in the equation....
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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 Jun 2013, 22:14
atalpanditgmat wrote:
Hi Bunuel, What is wrong in this equation? Kindly guide me.

Let total solution =100
Replacement = x (say)

equation:
(100-x)*50% + X*30% = 40%
Solving this equation i am getting x= 248

I looked every way possible but could not figure out the problem in the equation....

(100-x)*0.5+0.3x=100*0.4 --> x=50 --> 50/100=1/2.

Hope it's clear.
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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 Jun 2013, 23:03
1
atalpanditgmat wrote:
Hi Bunuel, What is wrong in this equation? Kindly guide me.

Let total solution =100
Replacement = x (say)

equation:
(100-x)*50% + X*30% = 40% Solving this equation i am getting x= 248

I looked every way possible but could not figure out the problem in the equation....

Try using below equation.
50% of (100-x) + 30% of x = 40% of 100
x=50litres.
So 50 of 100 litres got replaced would amount to 1/2 in ratio.
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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, 01:09
3
Did using allegation method

Refer diagram below

Bunuel, kindly update the OA.

Attachments

all.png [ 3.61 KiB | Viewed 5718 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, 05:11
3
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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19 Aug 2015, 11:02
I cannot understand how they divided 10/20 to get 10? Where did 20 come from?
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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 Aug 2015, 11:22
tinnyshenoy wrote:
I cannot understand how they divided 10/20 to get 10? Where did 20 come from?

It is from the alligation method

As shown in the post above, per the alligation method, the distrbution will be:

50 10 (=40-30) parts

40

30 10 (=50-40) parts

Thus, you see that you need 10 parts out of 20 (=(10+10)) parts of 50% solution, giving you 10/20 = 1/2 or 50% as your answer.

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

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08 Apr 2016, 23:38
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

w1 is the weight 50% acid solution which has concentration C1 = 50%
w2 is the weight of 20% acid solution which has concentration C1 = 30%
Concentration of mixture (average) = 40%

using weihted average formula w1/w2 = (c2 - avg) / (Avg-c1)
w1/w2 = (30-40)/(40-50)
w1/w2 = 1/1

w1 is 1 part and w2 is 1 part in a total 2 part of solution.

question is What part of the original solution was replaced?
w2 is what was replaced, hence 1 out of 2 i.e. 1/2 part 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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24 Mar 2017, 18:25
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, 03:21
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, 03:30
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, 16:13
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  [#permalink]

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09 Nov 2018, 05:47
Say there were 100L of soln initially and X L was replaced.
Reducing the problem to eqn : 50- 0.5x + 0.3x = 40 ==> x=50
So 1/2 of the initial soln to be replaced. Ans C.
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Re: Some part of a 50% solution of acid was replaced with an equal amount &nbs [#permalink] 09 Nov 2018, 05:47
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