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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 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
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16 Jun 2013, 22:07




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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, 15:55
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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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: Please help me. I don't understand why (C) is the right answer...
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 (C) 1/2 *correct answer (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
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30 Dec 2009, 14:40
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, 09:35
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, 02:00
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 Jun 2013, 20:57
Hi Bunuel, What is wrong in this equation? Kindly guide me. Let total solution =100 Replacement = x (say) equation: (100x)*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
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17 Jun 2013, 22:14



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Re: Some part of a 50% solution of acid was replaced with an equal amount
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17 Jun 2013, 23:03
atalpanditgmat wrote: Hi Bunuel, What is wrong in this equation? Kindly guide me. Let total solution =100 Replacement = x (say) equation: (100x)*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 (100x) + 30% of x = 40% of 100x=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
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10 Jul 2014, 01:09
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, 05:11
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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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
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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 (=4030) parts 40 30 10 (=5040) 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
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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) / (Avgc1) w1/w2 = (3040)/(4050) 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
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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
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03 Sep 2018, 03:21
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, 03:30
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, 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. 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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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
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