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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 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
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16 Jun 2013, 23:07
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. Answer: C. Alternately you can do the following, say \(x\) part of the original solution was replaced then: \((1x)*0.5+x*0.3=1*0.4\) > \(x=\frac{1}{2}\). Answer: C.
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Re: Some part of a 50% solution of acid was replaced with an equal amount
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19 Nov 2006, 07:50
perhaps the fastest way is :
1/2*x + 3/10* x = 0.4
>x= 1/2




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Re: Some part of a 50% solution of acid was replaced with an equal amount
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01 Dec 2008, 07:11
Look at it this way.
Total part is 1. Out of this, x part was replaced.
Hence, (1x)*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
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30 Dec 2009, 15:40
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% acid60% 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: mixtureproblem78345.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/mixtureproblems.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
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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: \(\frac{50}{100} * (TP) = \frac{TP}{2}\) 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: \(\frac{40}{100} * T = \frac{TP}{2} + \frac{30}{100}* P\) Solving for P, we get \(P = \frac{T}{2}\) which gives the part as \(\frac{1}{2}\) of Original Therefore answer is CHope this helps! Cheers! JT
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Re: Some part of a 50% solution of acid was replaced with an equal amount
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23 Jan 2010, 03:00
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
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17 Sep 2012, 05:14
Alligative Method is such a timesaving 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
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16 Jun 2013, 16:55
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
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10 Jul 2014, 02:09
Did using allegation method Refer diagram below Answer = 1/2 Bunuel, kindly update the OA. Thanks in advance
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Re: Some part of a 50% solution of acid was replaced with an equal amount
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20 Jul 2015, 06:11
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
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03 Sep 2018, 04:21
Hello chetan2u . I was unable to interpret the question and the above solutions . could you please explain Thank you.
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Re: Some part of a 50% solution of acid was replaced with an equal amount
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03 Sep 2018, 04:30
SonGoku wrote: Hello chetan2u . I was unable to interpret the question and the above solutions . could you please explain 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% = (5040)/(5030)=1/2 ....
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Re: Some part of a 50% solution of acid was replaced with an equal amount
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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. Answer: C
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Re: Some part of a 50% solution of acid was replaced with an equal amount
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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.
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Re: Some part of a 50% solution of acid was replaced with an equal amount
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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 M1M2 5030 40 1010 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
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04 Sep 2019, 10:50
We know this, w1/w2=A2Avg/AvgA1 ie w1/w2=5040/4030 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
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