GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Jan 2019, 23:33

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
• ### GMAT Club Tests are Free & Open for Martin Luther King Jr.'s Birthday!

January 21, 2019

January 21, 2019

10:00 PM PST

11:00 PM PST

Mark your calendars - All GMAT Club Tests are free and open January 21st for celebrate Martin Luther King Jr.'s Birthday.
• ### FREE Quant Workshop by e-GMAT!

January 20, 2019

January 20, 2019

07:00 AM PST

07:00 AM PST

Get personalized insights on how to achieve your Target Quant Score.

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

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 09 Aug 2006
Posts: 498
Some of 50%-intensity red paint is replaced with 25%  [#permalink]

### Show Tags

17 Nov 2007, 07:05
3
25
00:00

Difficulty:

45% (medium)

Question Stats:

68% (01:37) correct 32% (01:41) wrong based on 579 sessions

### HideShow timer Statistics

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
Director
Joined: 01 Apr 2008
Posts: 768
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014
Re: PS : intensity red paint  [#permalink]

### Show Tags

24 Mar 2009, 21:37
13
6
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

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:)
##### General Discussion
VP
Joined: 08 Jun 2005
Posts: 1133

### Show Tags

17 Nov 2007, 07:33
3
4
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).

Attachments

MixtureI.JPG [ 6.53 KiB | Viewed 12357 times ]

Director
Joined: 26 Feb 2006
Posts: 854
Re: PS : intensity red paint  [#permalink]

### Show Tags

17 Nov 2007, 09:01
2
4
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: 730
Re: PS : intensity red paint  [#permalink]

### Show Tags

17 Nov 2007, 09:31
3
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: 730

### Show Tags

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??

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: 1133

### Show Tags

17 Nov 2007, 09:46
1
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??

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: 730

### Show Tags

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??

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.
Senior Manager
Joined: 09 Aug 2006
Posts: 498

### Show Tags

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).

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: 1133

### Show Tags

18 Nov 2007, 09:12
3
2
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??

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

Attachments

Mixture Problems Made Easy.pdf [49.94 KiB]

Retired Moderator
Joined: 18 Jul 2008
Posts: 849
Re: PS : intensity red paint  [#permalink]

### Show Tags

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?
Manager
Joined: 17 Aug 2009
Posts: 175
Re: PS : intensity red paint  [#permalink]

### Show Tags

17 Nov 2009, 21:35
1
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: 104
Re: PS : intensity red paint  [#permalink]

### Show Tags

16 Jul 2010, 01:18

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
Retired Moderator
Joined: 16 Nov 2010
Posts: 1420
Location: United States (IN)
Concentration: Strategy, Technology
Re: PS : intensity red paint  [#permalink]

### Show Tags

14 Apr 2011, 18:08
1
1
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

_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings

Manager
Joined: 23 Oct 2011
Posts: 82

### Show Tags

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).

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?
Manager
Joined: 18 Jun 2010
Posts: 89
Re: PS : intensity red paint  [#permalink]

### Show Tags

24 Oct 2011, 20:00
+1 for E

25x+50(1-x) = 30

x=4/5
Manager
Joined: 23 May 2013
Posts: 104
Re: PS : intensity red paint  [#permalink]

### Show Tags

18 Oct 2013, 20:28
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.”

SVP
Joined: 06 Sep 2013
Posts: 1705
Concentration: Finance
Re: Some of 50%-intensity red paint is replaced with 25%  [#permalink]

### Show Tags

16 Dec 2013, 05:05
1
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

Hope it helps
Cheers!
J
Director
Joined: 23 Jan 2013
Posts: 559
Schools: Cambridge'16
Re: Some of 50%-intensity red paint is replaced with 25%  [#permalink]

### Show Tags

08 Nov 2015, 05:05
Equation

1/2x+1/4y=3/10(x+y)

x/y=1/4

Differential approach

50------30---25

20x=5y

x/y=1/4

x=1/5 and y=4/5 of final solution

E
Senior Manager
Joined: 23 Sep 2015
Posts: 378
Location: France
GMAT 1: 690 Q47 V38
GMAT 2: 700 Q48 V38
WE: Real Estate (Mutual Funds and Brokerage)
Re: Some of 50%-intensity red paint is replaced with 25%  [#permalink]

### Show Tags

09 Nov 2015, 02:15
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

$$\frac{x}{2}$$+$$\frac{1(1-x)}{4}$$=$$\frac{3}{10}$$

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
_________________

New Application Tracker : update your school profiles instantly!

Re: Some of 50%-intensity red paint is replaced with 25% &nbs [#permalink] 09 Nov 2015, 02:15

Go to page    1   2    Next  [ 23 posts ]

Display posts from previous: Sort by