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Director
Joined: 09 Aug 2006
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Some of 50%intensity red paint is replaced with 25%
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17 Nov 2007, 08:05
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66% (01:20) correct 34% (01:03) wrong based on 548 sessions
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Some of 50%intensity red paint is replaced with 25% solution of red paint such that the new paint intensity is 30%. What fraction of the original paint was replaced? A. 1/30 B. 1/5 C. 2/3 D. 3/4 E. 4/5
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Director
Joined: 01 Apr 2008
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Name: Ronak Amin
Schools: IIM Lucknow (IPMX)  Class of 2014

Re: PS : intensity red paint
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24 Mar 2009, 22:37
Guys, I just happened to visit this link for mixture problems, and it really makes life easy: http://www.onlinemathlearning.com/mixtureproblems.htmlThey explained 3 different scenarios with mixtures: add, replace and mix. If you set up a table it becomes very easy. For this problem Orig Removed Added Result Concent 0.5 0.5 0.25 0.30 Amt 1 x x 1 So, orig  rem + add = result (0.5)  0.5 x + 0.25x = 0.30 0.5  0.25x = 0.30 0.25x = 0.20 x=4/5 Hope it helps:)




VP
Joined: 08 Jun 2005
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See attachment
Intensity = concentration
since the new ratio is 5:20 = 1:4 then 4/5 was replaced (the old ratio was 1:0 > only 50% paint).
the answer is (E)
Attachments
MixtureI.JPG [ 6.53 KiB  Viewed 11949 times ]



Director
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Re: PS : intensity red paint
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17 Nov 2007, 10:01
Amit05 wrote: Some of 50%intensity red paint is replaced with 25% solution of red paint such that the new paint intensity is 30%. What fraction of the original paint was replaced?
* 1/30 * 1/5 * 2/3 * 3/4 * 4/5
Could anyone please explain me the wordings of this problem .. I have never heard the intensity of paint ..
forming an equation is always useful:
suppose, the total of 25% sol and 50% sol = 1
25% solution of red paint = x
50%intensity red paint = 1x
0.25 x + 0.5 (1x) = 0.3
x = 4/5



Director
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Re: PS : intensity red paint
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17 Nov 2007, 10:31
Amit05 wrote: Some of 50%intensity red paint is replaced with 25% solution of red paint such that the new paint intensity is 30%. What fraction of the original paint was replaced?
* 1/30 * 1/5 * 2/3 * 3/4 * 4/5
Could anyone please explain me the wordings of this problem .. I have never heard the intensity of paint ..
Let total paint = 1
Let amount replaced = x
50 (1x) + 25x = 30
x = 4/5



Director
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KillerSquirrel wrote: See attachment Intensity = concentration since the new ratio is 5:20 = 1:4 then 4/5 was replaced (the old ratio was 1:0 > only 50% paint). > how?? the answer is (E)
KS, I initially used the same diagram as you to solve the problem (I learned it a document attached to one of the posts). I determined that the new ratio is 1:4 but didn't know how to find out what fraction was replaced. I still don't understand how to figure that out after reading your post. Can you please explain? Thanks a bunch.



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GK_Gmat wrote: KillerSquirrel wrote: See attachment Intensity = concentration since the new ratio is 5:20 = 1:4 then 4/5 was replaced (the old ratio was 1:0 > only 50% paint). > how?? the answer is (E) KS, I initially used the same diagram as you to solve the problem (I learned it a document attached to one of the posts). I determined that the new ratio is 1:4 but didn't know how to find out what fraction was replaced. I still don't understand how to figure that out after reading your post. Can you please explain? Thanks a bunch.
Since you started only with the 50% paint (lets assume 10 liter) and you ended with 2 liter of that paint (2/10 = 1/5) then you replaced 8/10 = 4/5 of that paint. This takes some time to get used to  but once you get the hang of it you will solve it every time (or get your money back !).



Director
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KillerSquirrel wrote: GK_Gmat wrote: KillerSquirrel wrote: See attachment Intensity = concentration since the new ratio is 5:20 = 1:4 then 4/5 was replaced (the old ratio was 1:0 > only 50% paint). > how?? the answer is (E) KS, I initially used the same diagram as you to solve the problem (I learned it a document attached to one of the posts). I determined that the new ratio is 1:4 but didn't know how to find out what fraction was replaced. I still don't understand how to figure that out after reading your post. Can you please explain? Thanks a bunch. Since you started only with the 50% paint (lets assume 10 liter) and you ended with 2 liter of that paint (2/10 = 1/5) then you replaced 8/10 = 4/5 of that paint. This takes some time to get used to  but once you get the hang of it you will solve it every time (or get your money back !).
KS, I hate to be a pest but I couldn't get my head around this. How did you get 2 liter? Also, didn't we start with a ratio of 2:1 (50%:25%)? As you can tell, I'm totally confused. Thanks for any explanation in advance.



Director
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KillerSquirrel wrote: See attachment Intensity = concentration since the new ratio is 5:20 = 1:4 then 4/5 was replaced (the old ratio was 1:0 > only 50% paint). the answer is (E)
So if I reiterate the question, can I say that some part of 50% solution was replaced by 25% solution and the resultant was 30 % solution ?



VP
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GK_Gmat wrote: KillerSquirrel wrote: GK_Gmat wrote: KillerSquirrel wrote: See attachment Intensity = concentration since the new ratio is 5:20 = 1:4 then 4/5 was replaced (the old ratio was 1:0 > only 50% paint). > how?? the answer is (E) KS, I initially used the same diagram as you to solve the problem (I learned it a document attached to one of the posts). I determined that the new ratio is 1:4 but didn't know how to find out what fraction was replaced. I still don't understand how to figure that out after reading your post. Can you please explain? Thanks a bunch. Since you started only with the 50% paint (lets assume 10 liter) and you ended with 2 liter of that paint (2/10 = 1/5) then you replaced 8/10 = 4/5 of that paint. This takes some time to get used to  but once you get the hang of it you will solve it every time (or get your money back !). KS, I hate to be a pest but I couldn't get my head around this. How did you get 2 liter? Also, didn't we start with a ratio of 2:1 (50%:25%)? As you can tell, I'm totally confused. Thanks for any explanation in advance.
The question starts with a statement that you have a cerain amount of paint in 50% concentration (only 50% concentration and none from 25% concentration).
Using my diagram I found the new ratio of 50% to 25% to be 1:4.
Is this answer the question what fraction of the original paint was replaced ? Yes it does ! since you started with only 50% paint (i.e 5/5) and were left with 1/5 from this paint  meaning 4/5 was replaced.
And if you want to see it in numbers then assume we started with 10 liter of 50% paint or 20 liter for that matter, and the new ratio will give you 2:8 paint or 4:16 depending. So the amount that was replaced is 8/10 = 4/5 or 16/20 = 4/5



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Re: PS : intensity red paint
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24 Mar 2009, 15:32
the attached pdf is a really good document, but I think the explanations can be done a little bit better as I still had trouble wrapping my head around some of the methods.
Anyone care to give a look and try?



Manager
Joined: 17 Aug 2009
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Re: PS : intensity red paint
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17 Nov 2009, 22:35
Let there be x litres of 50% solution initially Suppose y litres of 50% solution is removed
Then the remaining solution is (xy).50 Now, equal amount of solution that is replaced(y) is added, therefore => .25y
(xy).50 + .25y = .30x .20x = .25y y=4/5
Hope this helps
Consider kudos for the post



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Re: PS : intensity red paint
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16 Jul 2010, 02:18
One more way we start with options:
Let the fraction be 3/4 and total capacity be 40 liter:
Fraction removed = 30 lts
originally intensity is 50% i.e. paint=20 lts and other =20 lts now from this 30lt is removed so remaining will be, paint=5 lts and other=5 lts now solution with 25% added (of 30lt) new solution will be, paint = 5 + 7.5 = 12.5 lts Other = 5 + 22.5 = 27.5 now the ratio will be 12.5 / 40 = 33.75 and not equal to 30%, wrong answer
Let the fraction be 4/5 and total capacity be 100 liter:
Fraction removed = 80 lts
originally intensity is 50% i.e. paint=50 lts and other =50 lts now from this 80lt is removed so remaining will be, paint=10 lts and other=10 lts now solution with 25% added (of 80lt) new solution will be, paint = 10 + 20 = 30 lts Other = 10 + 60 = 70 litre now the ratio will be 30/100 = 30%, so this means this is the right choice



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Re: PS : intensity red paint
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14 Apr 2011, 19:08
Let original solution be 100 ml and x ml of 25% solution be added to it so 0.30(100) = 0.25x + 0.50(100  x) 30 = 0.25x + 50  0.50x => 0.25x = 20 => x = 80 So 80/100 = 4/5 Answer  E
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Manager
Joined: 23 Oct 2011
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KillerSquirrel wrote: See attachment Intensity = concentration since the new ratio is 5:20 = 1:4 then 4/5 was replaced (the old ratio was 1:0 > only 50% paint). the answer is (E) KillerSquirre, I would like to thank you for your great contribution (attachment with mixture problems) that makes mixture problems way easier to handle. I have a question though, what does the 1:0 ratio mean?



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Re: PS : intensity red paint
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24 Oct 2011, 21:00
+1 for E
25x+50(1x) = 30
x=4/5



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Re: PS : intensity red paint
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18 Oct 2013, 21:28
bigfernhead wrote: the attached pdf is a really good document, but I think the explanations can be done a little bit better as I still had trouble wrapping my head around some of the methods.
Anyone care to give a look and try? Will try to elaborate on the PDF file, Problem 3 on page 2 Assume we all understand how ratio of (x16)/6 is achieved through rule of allegation. I struggled in understanding how that equal 3/1. Below is the explaination that i could come up. Assume that initial quantity of solution = 10 litres Problems states that 1/4 of the sol is removed > implies that 2.5 litre of solution is removed; and replaced with solution of X % of sugar. So the remaining fraction of original is 7.5 litres and the volume of new sol with x% of sugar = 2.5 litres Hence (x16)/6 = 3/1 Solving it we X = 34. Last problem on page 4 We know that initial conc of acid was 50% and the final was 40%. Also by rule of allegation, each solution of acid was mixed in equal proportion. Hence in the final solution that we had, both 50% solution and 30% solution were in equal proportion. Assume that initially we had 1 litre of 50% solution, then we need to remove 500ml of 50% and replace it with 40% of 500 ml solution. Then the final will be 30%. Hence 1/2 of 50% solution was removed. Hope it helps
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Re: Some of 50%intensity red paint is replaced with 25%
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16 Dec 2013, 06:05
Amit05 wrote: Some of 50%intensity red paint is replaced with 25% solution of red paint such that the new paint intensity is 30%. What fraction of the original paint was replaced?
A. 1/30 B. 1/5 C. 2/3 D. 3/4 E. 4/5 Use smart numbers Give 100 for the total of the original paint So intensity paint is 50 Then you'll have 500.25x = 30 x = 80 So the ratio is 80/100 = 4/5 E is the correct answer Hope it helps Cheers! J



Director
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Re: Some of 50%intensity red paint is replaced with 25%
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08 Nov 2015, 06:05
Equation
1/2x+1/4y=3/10(x+y)
x/y=1/4
Differential approach
503025
20x=5y
x/y=1/4
x=1/5 and y=4/5 of final solution
E



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Re: Some of 50%intensity red paint is replaced with 25%
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09 Nov 2015, 03:15
The ratio for the 50% solution is 1:1, for the 25% is 1;3 and for the 30% mixture is 3;7 I just do the weighted average to find the weight of the 50% solution in the new 30% solution \(\frac{x}{2}\)+\(\frac{1(1x)}{4}\)=\(\frac{3}{10}\) x = \(\frac{1}{5}\) So \(\frac{4}{5}\)of the old 50% solution was replaced by 4/5 of the 25% solution we get 0.2*50 + 0.8*25 = 30
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Re: Some of 50%intensity red paint is replaced with 25% &nbs
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