|
Author |
Message |
|
TAGS:
|
|
|
Director
Joined: 09 Aug 2006
Posts: 531
Followers: 2
Kudos [?]:
20
[0], given: 0
|
Some of 50%-intensity red paint is replaced with 25% [#permalink]
 17 Nov 2007, 08:05
Question Stats:
40% (02:35) correct
60% (01:47) wrong based on 0 sessions
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 ..
|
|
|
|
|
|
|
VP
Joined: 08 Jun 2005
Posts: 1172
Followers: 5
Kudos [?]:
78
[1] , given: 0
|
1
This post received
KUDOS
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 4748 times ]
|
|
|
|
|
|
Director
Joined: 26 Feb 2006
Posts: 919
Followers: 3
Kudos [?]:
28
[0], given: 0
|
Re: PS : intensity red paint [#permalink]
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 = 1-x
0.25 x + 0.5 (1-x) = 0.3
x = 4/5
|
|
|
|
|
|
Director
Joined: 09 Aug 2006
Posts: 776
Followers: 1
Kudos [?]:
18
[0], given: 0
|
Re: PS : intensity red paint [#permalink]
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 (1-x) + 25x = 30
x = 4/5
|
|
|
|
|
|
Director
Joined: 09 Aug 2006
Posts: 776
Followers: 1
Kudos [?]:
18
[0], given: 0
|
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.
|
|
|
|
|
|
VP
Joined: 08 Jun 2005
Posts: 1172
Followers: 5
Kudos [?]:
78
[0], given: 0
|
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
Joined: 09 Aug 2006
Posts: 776
Followers: 1
Kudos [?]:
18
[0], given: 0
|
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 !). Wow! Need to get my head around that. Thanks a lot!
|
|
|
|
|
|
Director
Joined: 09 Aug 2006
Posts: 776
Followers: 1
Kudos [?]:
18
[0], given: 0
|
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
Joined: 09 Aug 2006
Posts: 531
Followers: 2
Kudos [?]:
20
[0], given: 0
|
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
Joined: 08 Jun 2005
Posts: 1172
Followers: 5
Kudos [?]:
78
[0], given: 0
|
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
|
|
|
|
|
|
Director
Joined: 09 Aug 2006
Posts: 776
Followers: 1
Kudos [?]:
18
[0], given: 0
|
KillerSquirrel wrote: 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 Finally! I got it! Thanks a bunch for the explanation. Very helpful.
|
|
|
|
|
|
Retired Moderator
Joined: 18 Jul 2008
Posts: 1003
Followers: 7
Kudos [?]:
40
[0], given: 5
|
Re: PS : intensity red paint [#permalink]
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?
|
|
|
|
|
|
Director
Joined: 01 Apr 2008
Posts: 921
Schools: IIM Lucknow (IPMX) - Class of 2014
Followers: 8
Kudos [?]:
124
[4] , given: 18
|
Re: PS : intensity red paint [#permalink]
24 Mar 2009, 22:37
4
This post received
KUDOS
Guys, I just happened to visit this link for mixture problems, and it really makes life easy: http://www.onlinemathlearning.com/mixture-problems.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:)
|
|
|
|
|
|
Retired Moderator
Joined: 18 Jul 2008
Posts: 1003
Followers: 7
Kudos [?]:
40
[0], given: 5
|
Re: PS : intensity red paint [#permalink]
25 Mar 2009, 06:52
Thanks Economist! +1 Economist wrote: Guys, I just happened to visit this link for mixture problems, and it really makes life easy: http://www.onlinemathlearning.com/mixture-problems.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:) |
|
|
|
|
|
Director
Joined: 25 Oct 2008
Posts: 619
Location: Kolkata,India
Followers: 6
Kudos [?]:
93
[0], given: 100
|
Re: PS : intensity red paint [#permalink]
13 Aug 2009, 15:50
The only solution I understood was economist's!!
_________________
countdown-beginshas-ended-85483-40.html#p649902
|
|
|
|
|
|
Manager
Joined: 17 Aug 2009
Posts: 241
Followers: 2
Kudos [?]:
61
[0], given: 25
|
Re: PS : intensity red paint [#permalink]
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 (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
Consider kudos for the post
|
|
|
|
|
|
Manager
Joined: 12 Jun 2007
Posts: 129
Followers: 3
Kudos [?]:
32
[0], given: 2
|
Re: PS : intensity red paint [#permalink]
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
|
|
|
|
|
|
Manager
Joined: 25 Aug 2008
Posts: 238
Location: India
WE 1: 3.75 IT
WE 2: 1.0 IT
Followers: 2
Kudos [?]:
24
[0], given: 5
|
Re: PS : intensity red paint [#permalink]
16 Jul 2010, 02:44
KS.. Thanks a lot for explanation.
_________________
Cheers,
Varun
If you like my post, give me KUDOS!!
|
|
|
|
|
|
Intern
Joined: 06 Sep 2010
Posts: 47
Followers: 0
Kudos [?]:
1
[0], given: 0
|
Re: PS : intensity red paint [#permalink]
14 Apr 2011, 10:33
awesome explanation :D GK_Gmat wrote: 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 |
|
|
|
|
|
SVP
Joined: 16 Nov 2010
Posts: 1721
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 26
Kudos [?]:
228
[0], given: 34
|
Re: PS : intensity red paint [#permalink]
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
_________________
Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)
Find out what's new at GMAT Club - latest features and updates
|
|
|
|
|
|
|
Re: PS : intensity red paint
[#permalink]
14 Apr 2011, 19:08
|
|
|
|
|
|
|
|
|
|
|
close
