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Some part of a 50% solution of acid was replaced with an
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Updated on: 19 Aug 2015, 12:12
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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, 14:27.
Last edited by ENGRTOMBA2018 on 19 Aug 2015, 12:12, edited 2 times in total.
Added the OA.



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Re: Some part of a 50% solution of acid was replaced with an
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12 Aug 2008, 15:10
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 Original Acid : Total ratio = 50:100 . let us say you replaced x part of this. So you are left with (1x) of the original. So volume of acid left is (1x)50. in the new solution.. you added x of 30% solution. i.e 30x acid (1x)50 + 30x is the volume of acid. The total volume is still 100. And this concentration is 40% [(1x)50 + 30x] / 100 = 40/100 solve to get x=1/2



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Re: Some part of a 50% solution of acid was replaced with an
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12 Aug 2008, 15:16
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
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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
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Re: Some part of a 50% solution of acid was replaced with an
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17 Jun 2013, 21:57
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
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Re: Some part of a 50% solution of acid was replaced with an
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18 Jun 2013, 00:03
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
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18 Jun 2013, 00:33
Ahhhh!! I got it....Thank you both of you guys....I hope silly mistakes won't trouble me on the Real Test...
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Re: Some part of a 50% solution of acid was replaced with an
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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
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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
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19 Aug 2015, 11:41
Can you please explain to me how you got the answer to this problem using the allegation methos?
Thanks



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Re: Some part of a 50% solution of acid was replaced with an
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19 Aug 2015, 11:44
tinnyshenoy wrote: Can you please explain to me how you got the answer to this problem using the allegation methos?
Thanks Look at: somepartofa50solutionofacidwasreplacedwithan68805.html#p1381700



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Re: Some part of a 50% solution of acid was replaced with an
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19 Aug 2015, 12:02
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
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19 Aug 2015, 12:22
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
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09 Apr 2016, 00:38
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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