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Some of 50%-intensity red paint is replaced with 25% [#permalink]
17 Nov 2007, 07:05

00:00

A

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E

Difficulty:

25% (low)

Question Stats:

64% (01:56) correct
35% (01:12) wrong based on 79 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?

Re: PS : intensity red paint [#permalink]
17 Nov 2007, 09:01

2

This post received KUDOS

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

Re: PS : intensity red paint [#permalink]
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 ..

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 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 !).

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.

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

Re: PS : intensity red paint [#permalink]
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.

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

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?

Re: Some of 50%-intensity red paint is replaced with 25% [#permalink]
18 Oct 2013, 19:12

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Re: PS : intensity red paint [#permalink]
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
_________________

“Confidence comes not from always being right but from not fearing to be wrong.”

Re: Some of 50%-intensity red paint is replaced with 25% [#permalink]
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

gmatclubot

Re: Some of 50%-intensity red paint is replaced with 25%
[#permalink]
16 Dec 2013, 05:05