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Some part of the 50% solution of acid was replaced with the
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18 Nov 2006, 19:56
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85% (00:47) correct 15% (00:58) wrong based on 359 sessions
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Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution f acid was obtained. what part of the original solution was replaced? A. 1/5 B. 1/4 C. 1/2 D. 3/4 E. 4/5 OPEN DISCUSSION OF THIS QUESTION IS HERE: somepartofa50solutionofacidwasreplacedwithan68805.html
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Re: Some part of the 50% solution of acid was replaced with the
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18 Nov 2006, 20:22
X be the amount of 50% solution
n be the amount taken out from 50% solution and the same amount of 30% solution is added later
In the resultant solution, acid amount is X/2 â€“ n/2 + 3n/10 = 2X/5
N = Â½



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Re: Some part of the 50% solution of acid was replaced with the
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18 Nov 2006, 20:28
Suppose originally there was x litres of solution. Out of which y litres was removed.
After removing the amount y, xy litres of solution is left.
Amount of acid in xy = xy/2
y litres of solution is replaced by y litres of 30 % solution. Amount of acid in y = 3y/10.
In the new mixture, total acid = (xy)/2 + 30y/100
(xy)/2 + 3y/10 = (40 % of x)
Solving the equation gives y= x/2
Is the answer 1/2?



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Re: Some part of the 50% solution of acid was replaced with the
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19 Nov 2006, 06:50
perhaps the fastest way is :
1/2*x + 3/10* x = 0.4
>x= 1/2



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Re: Some part of the 50% solution of acid was replaced with the
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20 Nov 2006, 07:18
cool, short way to solve it, tobiastt! Thanks everyone
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Re: Some part of the 50% solution of acid was replaced with the
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20 Jun 2009, 08:38
Does anyone know how to set this problem up using the box method? Orig Add Remove Result Concent. Amt
Thanks.



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Re: Some part of the 50% solution of acid was replaced with the
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22 Jun 2009, 07:20
Hi all.. I thought the answer was within the question itself. 50% of original solution was replace hence the answer is 1/2 or 50%



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Re: Some part of the 50% solution of acid was replaced with the
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22 Jun 2009, 08:43
parimal wrote: Hi all.. I thought the answer was within the question itself. 50% of original solution was replace hence the answer is 1/2 or 50% Wow  you registered in 2006 and this is your first post! Better late than never! Welcome back
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Re: Some part of the 50% solution of acid was replaced with the
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23 Jun 2009, 08:12
Thanks.. yes.. I have been hiding in the lurks but hoping to be an active participant now.. cheers!



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Re: Some part of the 50% solution of acid was replaced with the
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15 Aug 2009, 18:50
50..................40 .........30......... 10..................20 10/20=1/2
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Re: Some part of the 50% solution of acid was replaced with the
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12 Nov 2009, 13:08
Let's assume original amount = 1L
Equation is :
0,5  0,5x + 0,3x = 0,4 => x = 0,5 = 1/2



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Re: Some part of the 50% solution of acid was replaced with the
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12 Nov 2009, 14:04
I3igDmsu wrote: Does anyone know how to set this problem up using the box method? Orig Add Remove Result Concent. Amt
Thanks. O R A D VOLUME X Y Y X CONCENTRATION 50 50 30 40 50X50Y+30Y=40X, Y=X/2



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Re: Some part of the 50% solution of acid was replaced with the
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12 Nov 2009, 16:19
Hermione wrote: Does anyone have a quick and easy way of solving this problem? I had to pick from the answer choices but I want to know how to solve it outright.
Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution f acid was obtained. what part of the original solution was replaced?
1/5 1/4 1/2 3/4 4/5 Let X = Part of acid solution replaced, then 1X will be the parts not replaced. (1X)*0.5 + 0.3*x = 0.4 0.5  0.5X +0.3X = 0.4 0.2X = 0.1 X=1/2 So the answer is C



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Re: Some part of the 50% solution of acid was replaced with the
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02 Feb 2010, 04:19
pierrealexandre77 wrote: Let's assume original amount = 1L
Equation is :
0,5  0,5x + 0,3x = 0,4 => x = 0,5 = 1/2 Thanks! This explanation really worked for me



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Re: Some part of the 50% solution of acid was replaced with the
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09 Apr 2010, 15:04
Wow the box method makes these problems so easy. can you explain a bit more on the box method. ORAD?can we use this for most of the misture problems .any limitations?



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Re: Some part of the 50% solution of acid was replaced with the
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Updated on: 16 Apr 2010, 22:57
Initially 50 ml acid and 100 ml of total suppose x part of it is removed. thus we are left with 50(1x) of acid and 100(1x) of total if we add x part of 30% acid solution (30ml acid and 100ml total) acid will be 30x and total = 100x so total acid = 50(1x)+ 30x = 5020x total solution = 100(1x)+100x = 100 since if it equal to 40%acid solution(40ml acid and 100ml total) => (5020x)/100 = 40/100 => 5020x = 40 => 20x=10 => x=1/2 This looks lengthy but we can easily skip these steps while solving it.
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Originally posted by gurpreetsingh on 16 Apr 2010, 12:37.
Last edited by gurpreetsingh on 16 Apr 2010, 22:57, edited 1 time in total.



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Re: Some part of the 50% solution of acid was replaced with the
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16 Apr 2010, 13:31
mst wrote: Suppose originally there was x litres of solution. Out of which y litres was removed.
After removing the amount y, xy litres of solution is left. Amount of acid in xy = xy/2
y litres of solution is replaced by y litres of 30 % solution. Amount of acid in y = 3y/10.
In the new mixture, total acid = (xy)/2 + 30y/100
(xy)/2 + 3y/10 = (40 % of x) Solving the equation gives y= x/2
Is the answer 1/2? Great explanation! Thank you.



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Re: Some part of the 50% solution of acid was replaced with the
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17 Apr 2010, 06:04
tobiastt wrote: perhaps the fastest way is : 1/2*x + 3/10* x = 0.4 >x= 1/2 cool, it's nice way to answer quickly .



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Re: Some part of the 50% solution of acid was replaced with the
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06 Nov 2010, 23:47
I am using a chart to solve this problem. Can anyone point out the error I am making? See image attached below.



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Re: Some part of the 50% solution of acid was replaced with the
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07 Nov 2010, 05:59
greatchap wrote: I am using a chart to solve this problem. Can anyone point out the error I am making? See image attached below. x/2 + 3x/10 is the total stuff which is equal to 4 Solving x/2 + 3x/10 = 4 gives x = 5 Since x is the amount you assumed was replaced out of 10, you get 50%. Nothing wrong with what you did. Note: Simply put, this question says that some 50% solution was mixed with some 30% solution to get 40% solution. Using weighted averages, you can find out that they must have been mixed in equal quantities or in other words, 50% of the original solution must have been replaced.
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Re: Some part of the 50% solution of acid was replaced with the
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