Find all School-related info fast with the new School-Specific MBA Forum

It is currently 11 Jul 2014, 11:23

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

M03 #04

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
1 KUDOS received
Intern
Intern
avatar
Joined: 06 Jul 2009
Posts: 33
Followers: 0

Kudos [?]: 6 [1] , given: 0

M03 #04 [#permalink] New post 07 Jul 2009, 18:29
1
This post received
KUDOS
7
This post was
BOOKMARKED
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
Attachments

mimetex.cgi.gif
mimetex.cgi.gif [ 1.37 KiB | Viewed 8204 times ]

Kaplan GMAT Prep Discount CodesKnewton GMAT Discount CodesGMAT Pill GMAT Discount Codes
6 KUDOS received
Senior Manager
Senior Manager
avatar
Joined: 04 Jun 2008
Posts: 306
Followers: 5

Kudos [?]: 92 [6] , given: 15

GMAT Tests User
Re: M03 #04 [#permalink] New post 07 Jul 2009, 18:39
6
This post received
KUDOS
for such problems, when total quantity is not given, always assume it as 1 unit (litre, number, km, etc)

say 1 liter initial 50% solution was there,, ie concentration is 50%

say x liter is taken out (if you take total as 1, then x will be a fraction)

so 50%(1-x) alcohol remains in the original solution. To this x liter of 25% solution is added, ie, 25%(x) liter alcohol is added

this gives 30% of 1liter solution again

so 50(1-x) + 25x = 30*1

solve for x, which would be 4/5, or 80% of 1 liter
2 KUDOS received
Intern
Intern
avatar
Joined: 11 Jul 2009
Posts: 1
Followers: 0

Kudos [?]: 2 [2] , given: 1

Re: M03 #04 [#permalink] New post 26 Sep 2009, 12:22
2
This post received
KUDOS
Here is the how I solved the problem
Consider the quantity of solution as 1 litter.
It is given that 50% of the solution was alcohol.
Let X be the amount of alcohol removed from the solution.

The equation to remove X amount of alcohol from the 50% alcohol solution is
0.5 litters alcohol - 0.5 * X litters alcohol

The X litters removed from the solution is replaced by 25% alcohol solution.
0.5 litters alcohol - 0.5 * X litters alcohol + 0.25 * X litters alcohol

The resulting solution contains 30% alcohol. Therefore the final equation is
0.5 litters alcohol - 0.5 * X litters alcohol + 0.25 * X litters alcohol = 0.30 litters of alcohol.

If you solve for the above equation, you will get X as 0.8, which is 80%
Senior Manager
Senior Manager
avatar
Joined: 30 Aug 2009
Posts: 291
Location: India
Concentration: General Management
Followers: 2

Kudos [?]: 84 [0], given: 5

GMAT Tests User
Re: Percentage problem [#permalink] New post 11 Jan 2010, 10:17
amitanand wrote:
Hi Experts,

Can anyone help me in the below question:

Q: 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%

Thanks in Advance
-Amit


b) 20%
solution1 is 50% alcohol and solution2 is 25% alcohol. to get 30% alcohol solution we need solution1/ solution2 in ratio 1:4
Senior Manager
Senior Manager
User avatar
Joined: 21 Jul 2009
Posts: 367
Schools: LBS, INSEAD, IMD, ISB - Anything with just 1 yr program.
Followers: 14

Kudos [?]: 96 [0], given: 22

GMAT Tests User
Re: Percentage problem [#permalink] New post 11 Jan 2010, 13:05
Interesting question. My answer is E.

Let x be removed out of the first solution of 50% alcohol and x be added from the other solution of 25% alcohol to make a wholesome (a peg ??) 30% alcohol solution ( I don't take it with water though!!!)

Anyways, in equation form it is 50(1-x) + 25x = 30, which gives x = 4/5. so 80% quantity of the first solution is replaced with 80% quantity of second solution to get a wholesome 30% solution.
_________________

I am AWESOME and it's gonna be LEGENDARY!!!

Senior Manager
Senior Manager
avatar
Joined: 30 Aug 2009
Posts: 291
Location: India
Concentration: General Management
Followers: 2

Kudos [?]: 84 [0], given: 5

GMAT Tests User
Re: Percentage problem [#permalink] New post 11 Jan 2010, 19:52
amitanand wrote:
Hi,
Thanks for the reply. But the OA is E(80%).
Could you please justify?
-Amit


ohk...read the q wrong...since we have 100% of original solution [50% alcohol] and after mixing it is only 20% so the 1st solution is replaced by 100 - 20 = 80%
Intern
Intern
User avatar
Joined: 29 Dec 2009
Posts: 42
Location: India
Concentration: Finance, Real Estate
Schools: Duke (Fuqua) - Class of 2014
GMAT 1: 770 Q50 V44
GPA: 3.5
WE: General Management (Real Estate)
Followers: 4

Kudos [?]: 10 [0], given: 1

Re: M03 #04 [#permalink] New post 12 Feb 2010, 06:27
I solved it using just a small variation of the methods explained above.

x = original quantity of solution
y = quantity of solution replaced

Therefore, the equation for the alcohol part of the solution is:

x(50%) - y(50%) + y(25%) = x(30%)
~ (x/2 )- (y/2) + (y/4) = 3x/10
Multiply throughout by 20,
10x - 10y + 5y = 6x
~4x = 5y
~y = 4x/5
~y = 80%x
Senior Manager
Senior Manager
avatar
Joined: 01 Feb 2010
Posts: 268
Followers: 1

Kudos [?]: 34 [0], given: 2

GMAT Tests User
Re: M03 #04 [#permalink] New post 12 Feb 2010, 07:38
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


Let x - amount of total mixture
y - amount of mixture replaced with 25% alcohol

Now,
0.5(x-y) + 0.25y = 0.3x
0.5(x-y) represents 50% alcohol quantity in new mixture
0.25y represents 25% alcohol quantity in new mixture
0.3x represents total alcohol quantity in new mixture

solving y = (4/5)x
hence y is 80% of x that means E.
Manager
Manager
avatar
Joined: 10 Feb 2010
Posts: 196
Followers: 2

Kudos [?]: 34 [0], given: 6

GMAT Tests User
Re: M03 #04 [#permalink] New post 12 Feb 2010, 18:45
1/2(x-y) + 1/4 y = 3/10 (x-y+y)
multiplying by 20
10 (x-y) + 5 y = 6x
4x=5y
y=4/5 x= 80% x
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18504
Followers: 3189

Kudos [?]: 21257 [0], given: 2543

Re: M03 #04 [#permalink] New post 17 Feb 2011, 05:32
Expert's post
This question was also posted in PS subforum. Below is my solution from there.

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%

Question can be solved algebraically or using allegation method.

Algebraic approach:

Initial solution is "half water/half alcohol mix" means it's 50% (0.5) alcohol solution.

Let the portion replaced be x and the volume of initial solution be 1 unit.

Then the amount of alcohol after removal of a portion will be 0.5(1-x) and the amount of alcohol added will be 0.25x, so total amount of alcohol will be 0.5(1-x)+0.25x. On the other hand as in the end 30% alcohol solution was obtained then the amount of alcohol in the end was 0.3*1.

So 0.5(1-x)+0.25x=0.3 --> x=0.8, or 80%.

Answer: E.

Other solutions: mixture-problem-can-someone-explain-this-100271.html
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

1 KUDOS received
Senior Manager
Senior Manager
avatar
Status: ready to boMBArd
Joined: 31 Oct 2010
Posts: 494
Location: India
Concentration: Entrepreneurship, Strategy
GMAT 1: 710 Q48 V40
WE: Project Management (Manufacturing)
Followers: 13

Kudos [?]: 28 [1] , given: 74

CAT Tests
Re: M03 #04 [#permalink] New post 17 Feb 2011, 12:36
1
This post received
KUDOS
E.

Those interested in learning the methods to solve such problems, they are very effectively explained here: http://www.onlinemathlearning.com/mixtu ... s.html#mix
_________________

My GMAT debrief: from-620-to-710-my-gmat-journey-114437.html

SVP
SVP
avatar
Joined: 16 Nov 2010
Posts: 1692
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 30

Kudos [?]: 272 [0], given: 36

GMAT Tests User Premium Member Reviews Badge
Re: M03 #04 [#permalink] New post 20 Feb 2011, 04:14
Let there be total 100 ml of mix

50 ml W + 50 ml A

Let x ml be the amount of new mix - 0.25x Alcohol

30 ml of Alcohol now in 100 ml

0.30 * 100 = 0.25x + 0.50(100-x)

30 = 0.25x + 50 - 0.50x

=> 0.25x = 20

=> x = 80

So 80 ml of original mix was replaced.

=> Alcohol replaced in original mix = 80 * 50/100 = 40 ml

=> 40/50 * 100 = 80%

So answer is E.
_________________

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

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
Manager
avatar
Joined: 10 Jan 2011
Posts: 244
Location: India
GMAT Date: 07-16-2012
GPA: 3.4
WE: Consulting (Consulting)
Followers: 0

Kudos [?]: 18 [0], given: 25

Reviews Badge
Re: M03 #04 [#permalink] New post 20 Feb 2011, 10:26
Assume that there is 1 litter of mix

hence

0.5(1-X)+.25X = 0.3 where X is portion of alcohol
0.5-0.5X + 0.25X = 0.3
0.2 = .25X
X= 80%
_________________

-------Analyze why option A in SC wrong-------

Intern
Intern
avatar
Joined: 11 Jul 2011
Posts: 2
Followers: 0

Kudos [?]: 0 [0], given: 0

GMAT Tests User
Re: M03 #04 [#permalink] New post 24 Sep 2011, 05:46
if the question asked : what percentage of the [highlight]original mixture[/highlight] was replaced? you are correct

Q : what percentage of the [highlight]original alcohol[/highlight] was replaced? 80% would have been correct

In this case , 1 gal alcohol mix => original alcohol = 0.5 gal ; replaced alcohol = 0.8gal*0.25% = 0.2 gal
% of original alcohol replaced = (0.2gal)/(0.5gal) *100 = 40%

There might be mistake in this question .point out any mistake in my logic .
Math Forum Moderator
avatar
Joined: 20 Dec 2010
Posts: 2051
Followers: 125

Kudos [?]: 863 [0], given: 376

GMAT Tests User
Re: M03 #04 [#permalink] New post 24 Sep 2011, 06:17
ravijackson wrote:
if the question asked : what percentage of the [highlight]original mixture[/highlight] was replaced? you are correct

Q : what percentage of the [highlight]original alcohol[/highlight] was replaced? 80% would have been correct

In this case , 1 gal alcohol mix => original alcohol = 0.5 gal ; replaced alcohol = 0.8gal*0.25% = 0.2 gal
% of original alcohol replaced = (0.2gal)/(0.5gal) *100 = 40%

There might be mistake in this question .point out any mistake in my logic .


Both questions would give 80% because the ratio was 1:1,

So, replacing 80% of the solution means replacing 80% of the alcohol.

In the red-part, why did you further cut down the alcohol content to half? It should be:

0.8*0.5=0.4 g of alcohol replaced.

0.4 is 80% of 0.5
_________________

~fluke

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Intern
Intern
avatar
Joined: 11 Jul 2011
Posts: 2
Followers: 0

Kudos [?]: 0 [0], given: 0

GMAT Tests User
Re: M03 #04 [#permalink] New post 24 Sep 2011, 10:10
fluke wrote:
ravijackson wrote:
if the question asked : what percentage of the [highlight]original mixture[/highlight] was replaced? you are correct

Q : what percentage of the [highlight]original alcohol[/highlight] was replaced? 80% would have been correct

In this case , 1 gal alcohol mix => original alcohol = 0.5 gal ; replaced alcohol = 0.8gal*0.25% = 0.2 gal
% of original alcohol replaced = (0.2gal)/(0.5gal) *100 = 40%

There might be mistake in this question .point out any mistake in my logic .


Both questions would give 80% because the ratio was 1:1,

So, replacing 80% of the solution means replacing 80% of the alcohol.

In the red-part, why did you further cut down the alcohol content to half? It should be:

0.8*0.5=0.4 g of alcohol replaced.

0.4 is 80% of 0.5


0.8*0.5 gal of alcohol was removed from the original solution (0.8 gal of original mix was removed ) and replaced with 0.8*0.25 gal of alcohol ( 0.8 gal of new mix added whose alcohol conc is 25%) from the addition of the new mixture.

The question asks what % of original alcohol ( 0.4 gal ) was replaced ( 0.2 gal ) .

Even if you would consider the question to mean what % of original alcohol ( 0.4 gal ) was replaced with water ( 0.2 gal ) ; even then the ans would be 40%.

Again , open to discussion .
Math Forum Moderator
avatar
Joined: 20 Dec 2010
Posts: 2051
Followers: 125

Kudos [?]: 863 [0], given: 376

GMAT Tests User
Re: M03 #04 [#permalink] New post 24 Sep 2011, 10:45
ravijackson wrote:
0.8*0.5 gal of alcohol was removed from the original solution (0.8 gal of original mix was removed ) and replaced with 0.8*0.25 gal of alcohol ( 0.8 gal of new mix added whose alcohol conc is 25%) from the addition of the new mixture.

The question asks what % of original alcohol ( 0.4 gal ) was replaced ( 0.2 gal ) .

Even if you would consider the question to mean what % of original alcohol ( 0.4 gal ) was replaced with water ( 0.2 gal ) ; even then the ans would be 40%.

Again , open to discussion .


I'm going to do this backwards:

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?

1 G mix
0.5 g Original Alcohol
0.5 g Original Water
***********

Answer: 80% of original alcohol was replaced.

0.4g of alcohol from 0.5g of original alcohol was replaced, (80% of original alcohol replaced- removed AND something else was added), with 0.1g Alcohol+0.3g Water.

Also,
0.4g of water from 0.5g of original water was replaced, (80% of original water replaced- removed AND something else was added), again with
0.1g Alcohol+0.3g Water

So,
A+W
0.5+0.5

Removed:
0.4A+0.4W

Remaining
0.1A+0.1W

Added
(Placed back for alcohol)+(Placed back for water)
(0.1A+0.3W)+(0.1A+ 0.3W)

Total Content:
0.1A+0.1W+0.1A+0.3W+ 0.1A+ 0.3W
0.3A+0.7W

I don't see any problem with the wordings.
_________________

~fluke

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Intern
Intern
avatar
Joined: 14 Aug 2011
Posts: 5
Location: United States (PA)
WE: Pharmaceuticals (Pharmaceuticals and Biotech)
Followers: 0

Kudos [?]: 5 [0], given: 5

Re: Percentage problem [#permalink] New post 24 May 2012, 15:24
ravijackson wrote:
Q : what percentage of the [highlight]original alcohol[/highlight] was replaced? 80% would have been correct if the question asked : what percentage of the [highlight]original mixture[/highlight] was replaced?

In this case , 1 gal alcohol mix => original alcohol = 0.5 gal ; replaced alcohol = 0.8gal*0.25% = 0.2 gal
% of original alcohol replaced = (0.2gal)/(0.5gal) *100 = 40%

There might be mistake in this question .point out any mistakes in my logic .


I also arrived at the same answer. I think the question should as "what % of original mixture was replaced" or 40% should be one of the choices.
Manager
Manager
avatar
Joined: 30 Sep 2009
Posts: 110
Followers: 0

Kudos [?]: 13 [0], given: 78

GMAT ToolKit User
Re: M03 #04 [#permalink] New post 28 May 2012, 23:23
can anyone solve this by allIgation method ??

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

alli.png
alli.png [ 7.84 KiB | Viewed 3867 times ]

Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18504
Followers: 3189

Kudos [?]: 21257 [0], given: 2543

Re: M03 #04 [#permalink] New post 28 May 2012, 23:31
Expert's post
abhi398 wrote:
can anyone solve this by allIgation method ??

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


Check this: mixture-problem-can-someone-explain-this-100271.html

Hope it helps.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Re: M03 #04   [#permalink] 28 May 2012, 23:31
    Similar topics Author Replies Last post
Similar
Topics:
33 Experts publish their posts in the topic M03 #23 redbeanaddict 18 04 Oct 2008, 02:55
1 M03 #6 dczuchta 9 20 Sep 2008, 16:08
20 M03 #01 dczuchta 43 19 Sep 2008, 16:06
3 Experts publish their posts in the topic M03 #19 jjhko 18 15 Sep 2008, 11:36
1 m03 - #7 Liquid 13 28 Feb 2008, 21:12
Display posts from previous: Sort by

M03 #04

  Question banks Downloads My Bookmarks Reviews Important topics  

Go to page    1   2    Next  [ 29 posts ] 

Moderators: WoundedTiger, Bunuel



GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

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®.