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Some of 50%-intensity red paint is replaced with 25%

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Some of 50%-intensity red paint is replaced with 25% [#permalink] New post 17 Nov 2007, 07:05
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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
[Reveal] Spoiler: OA
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 [#permalink] New post 17 Nov 2007, 07:33
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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)

:)
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MixtureI.JPG
MixtureI.JPG [ 6.53 KiB | Viewed 6291 times ]

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Re: PS : intensity red paint [#permalink] New post 17 Nov 2007, 09:01
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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 = 1-x

0.25 x + 0.5 (1-x) = 0.3
x = 4/5
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Re: PS : intensity red paint [#permalink] New post 17 Nov 2007, 09:31
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 (1-x) + 25x = 30
x = 4/5
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 [#permalink] New post 17 Nov 2007, 09:35
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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 [#permalink] New post 17 Nov 2007, 09:46
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 !).

:)
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 [#permalink] New post 17 Nov 2007, 23:59
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.
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 [#permalink] New post 18 Nov 2007, 02:58
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 ?
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 [#permalink] New post 18 Nov 2007, 09:12
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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 [#permalink] New post 24 Mar 2009, 14:32
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?
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Re: PS : intensity red paint [#permalink] New post 24 Mar 2009, 21:37
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Guys, I just happened to visit this link for mixture problems, and it really makes life easy:
http://www.onlinemathlearning.com/mixture-problems.html

They 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:)
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Re: PS : intensity red paint [#permalink] New post 17 Nov 2009, 21:35
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Let there be x litres of 50% solution initially
Suppose y litres of 50% solution is removed

Then the remaining solution is (x-y).50
Now, equal amount of solution that is replaced(y) is added, therefore => .25y

(x-y).50 + .25y = .30x
.20x = .25y
y=4/5

Hope this helps

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Re: PS : intensity red paint [#permalink] New post 16 Jul 2010, 01:18
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 [#permalink] New post 14 Apr 2011, 18:08
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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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Re: [#permalink] New post 24 Oct 2011, 13:17
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 [#permalink] New post 24 Oct 2011, 20:00
+1 for E

25x+50(1-x) = 30

x=4/5
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Re: Some of 50%-intensity red paint is replaced with 25% [#permalink] New post 18 Oct 2013, 19:12
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Re: PS : intensity red paint [#permalink] New post 18 Oct 2013, 20:28
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 (x-16)/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 (x-16)/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% [#permalink] New post 16 Dec 2013, 05:05
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 50-0.25x = 30
x = 80

So the ratio is 80/100 = 4/5
E is the correct answer

Hope it helps
Cheers!
J :)
Re: Some of 50%-intensity red paint is replaced with 25%   [#permalink] 16 Dec 2013, 05:05
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