It is currently 23 Jun 2017, 09:07

### 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 June
Open Detailed Calendar

# M03 #04

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Joined: 29 Mar 2012
Posts: 324
Location: India
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Re: M03 #04 [#permalink]

### Show Tags

14 Jun 2012, 01:59
ThrillRide wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
(A) 3%
(B) 20%
(C) 66%
(D) 75%
(E) 80%

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

OA explanation:
Find the difference between the original and final amounts of alcohol. Suppose we had 1 liter or 1 gallon of the initial solution, and has been replaced. We can build an equation using this information:

How do you solve this? I don't get how the OA came up with the equation below. thanks

Hi,

Using the alligation method:
we have A:B = 1:4 (refer attached diagram), so final solution must have 1 part of 50% solution & 4 parts of 25% solution,

thus, 4 parts of 50% solution should be removed and replaced by 25% solution
so, answer is 80% (E)

Regards,
Attachments

all.jpg [ 6.09 KiB | Viewed 1605 times ]

Senior Manager
Joined: 15 Sep 2009
Posts: 265
Re: M03 #04 [#permalink]

### Show Tags

21 Feb 2013, 07:21
abhi398 wrote:
can anyone solve this by allIgation method ??

i generally solve this type of problem using the alligation technique .. I am getting 20% ..

As mentioned in the other thread, you need to reverse the ratios. If you do this, you will the right answer, 80%. Reversing the ratios is the key step in the Alligation Method.

Please read the document in the link below for further information on how to apply the Alligation Method.

www.gmatclub.com/forum/demystifying-the ... 40276.html

Cheers,
Der alte Fritz.
_________________

+1 Kudos me - I'm half Irish, half Prussian.

Senior Manager
Joined: 15 Sep 2009
Posts: 265
Re: M03 #04 [#permalink]

### Show Tags

21 Feb 2013, 07:23
abhi398 wrote:
can anyone solve this by allIgation method ??

i generally solve this type of problem using the alligation technique .. I am getting 20% ..

Hi,

You need to reverse the ratios. That's the key step in the Alligation Method.

www.gmatclub.com/forum/demystifying-the ... 40276.html

Cheers,
Der alte Fritz
_________________

+1 Kudos me - I'm half Irish, half Prussian.

Intern
Joined: 22 Dec 2012
Posts: 16
GMAT 1: 720 Q49 V39
Re: M03 #04 [#permalink]

### Show Tags

21 Feb 2013, 13:50
lets consider we ve 1 litre of soln with us..
and we replaced X litre of the originl 50% with X lite 25% resulting in 30% solution (1 litre)..

so equating the water part

0.5 - 0.5X + 0.75X = 0.7
Solving for X we get .80 or 80 %
Intern
Joined: 11 Nov 2012
Posts: 7
Re: M03 #04 [#permalink]

### Show Tags

22 Feb 2013, 13:12
abhi398 wrote:
can anyone solve this by allIgation method ??

i generally solve this type of problem using the alligation technique .. I am getting 20% ..

Abhi .. by using the allegation technique in the diagram you will get the final ratio of alcohol : water in the mixture = 5:20 i.e. 1:4

that means that at the moment 5 parts of solution 1 part is alcohol, so the alcohol removed is 4 parts i.e. 4/5* 100 = 80%

other way let x be the part removed from the solution and let 1 L be the original solution so 50(1-x) + 25x = 30 (1) (Both sides are having percentages)

Solving for 25x = 20 4:5 or 80%
Manager
Joined: 12 Dec 2012
Posts: 230
GMAT 1: 540 Q36 V28
GMAT 2: 550 Q39 V27
GMAT 3: 620 Q42 V33
GPA: 2.82
WE: Human Resources (Health Care)
Re: M03 #04 [#permalink]

### Show Tags

23 Mar 2013, 08:24
solved this but still I feel I missed something ..0.5x+0.25y=0.3(x+y).....x/y =4/1 ie x=4/5=80%.But I do not know why I still feel I should have subtracted this 80% from 100% to be 20% ... anybody can help please?
_________________

My RC Recipe
http://gmatclub.com/forum/the-rc-recipe-149577.html

My Problem Takeaway Template
http://gmatclub.com/forum/the-simplest-problem-takeaway-template-150646.html

Math Expert
Joined: 02 Sep 2009
Posts: 39607
Re: M03 #04 [#permalink]

### Show Tags

23 Mar 2013, 08:53
TheNona wrote:
solved this but still I feel I missed something ..0.5x+0.25y=0.3(x+y).....x/y =4/1 ie x=4/5=80%.But I do not know why I still feel I should have subtracted this 80% from 100% to be 20% ... anybody can help please?

Check here: m03-80634.html#p873581

Hope it helps.
_________________
Intern
Joined: 01 Aug 2006
Posts: 35
Re: M03 #04 [#permalink]

### Show Tags

07 Feb 2014, 12:36
Let's say total of 10 liters; 5 water; 5 alcohol.
Total x removed; water = alcohol = 5 -(x/2)
25% of x added back = (0.25x).
Final volume of alcohol = 3.
So, [5-(x/2)] + 0.25x = 3
Solving x = 8 or 80%.
Manager
Joined: 20 Oct 2013
Posts: 76
Location: United States
Concentration: General Management, Real Estate
Re: M03 #04 [#permalink]

### Show Tags

21 Apr 2014, 06:34
1
KUDOS
The percentage replaced is x -> (1-x)*0.5+0.25x=0.3 -> 0.25x=0.2->x=80%
Re: M03 #04   [#permalink] 21 Apr 2014, 06:34

Go to page   Previous    1   2   [ 29 posts ]

Similar topics Replies Last post
Similar
Topics:
3 M03 Q 36 20 18 Sep 2012, 08:54
50 M03 #23 18 23 Jul 2013, 03:36
1 M03 #6 9 21 Sep 2011, 20:22
22 M03 #01 27 20 Apr 2014, 06:48
3 M03 #19 18 10 May 2014, 05:50
Display posts from previous: Sort by

# M03 #04

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderator: Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.