It is currently 15 Dec 2017, 10:08

# Decision(s) Day!:

CHAT Rooms | Olin (St. Louis) R1 | Tuck R1 | Ross R1 | Fuqua R1

### 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

Author Message
Intern
Joined: 06 Jul 2009
Posts: 33

Kudos [?]: 9 [2], given: 0

### Show Tags

07 Jul 2009, 18:29
2
KUDOS
9
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 [ 1.37 KiB | Viewed 9459 times ]

Kudos [?]: 9 [2], given: 0

Senior Manager
Joined: 04 Jun 2008
Posts: 286

Kudos [?]: 161 [6], given: 15

### Show Tags

07 Jul 2009, 18:39
6
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

Kudos [?]: 161 [6], given: 15

Intern
Joined: 11 Jul 2009
Posts: 1

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

### Show Tags

26 Sep 2009, 12:22
2
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%

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

Senior Manager
Joined: 30 Aug 2009
Posts: 281

Kudos [?]: 194 [0], given: 5

Location: India
Concentration: General Management

### Show Tags

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%

-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

Kudos [?]: 194 [0], given: 5

Senior Manager
Joined: 21 Jul 2009
Posts: 364

Kudos [?]: 204 [0], given: 22

Schools: LBS, INSEAD, IMD, ISB - Anything with just 1 yr program.

### Show Tags

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!!!

Kudos [?]: 204 [0], given: 22

Senior Manager
Joined: 30 Aug 2009
Posts: 281

Kudos [?]: 194 [0], given: 5

Location: India
Concentration: General Management

### Show Tags

11 Jan 2010, 19:52
amitanand wrote:
Hi,
Thanks for the reply. But the OA is E(80%).
-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%

Kudos [?]: 194 [0], given: 5

Intern
Joined: 29 Dec 2009
Posts: 41

Kudos [?]: 18 [0], given: 1

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)

### Show Tags

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

Kudos [?]: 18 [0], given: 1

Senior Manager
Joined: 01 Feb 2010
Posts: 251

Kudos [?]: 62 [0], given: 2

### Show Tags

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.

Kudos [?]: 62 [0], given: 2

Manager
Joined: 10 Feb 2010
Posts: 186

Kudos [?]: 152 [0], given: 6

### Show Tags

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

Kudos [?]: 152 [0], given: 6

Math Expert
Joined: 02 Sep 2009
Posts: 42617

Kudos [?]: 135746 [0], given: 12708

### Show Tags

17 Feb 2011, 05:32
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%.

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

Kudos [?]: 135746 [0], given: 12708

Current Student
Status: Up again.
Joined: 31 Oct 2010
Posts: 526

Kudos [?]: 560 [1], given: 75

Concentration: Strategy, Operations
GMAT 1: 710 Q48 V40
GMAT 2: 740 Q49 V42

### Show Tags

17 Feb 2011, 12:36
1
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: http://gmatclub.com/forum/from-620-to-710-my-gmat-journey-114437.html

Kudos [?]: 560 [1], given: 75

TOEFL Forum Moderator
Joined: 16 Nov 2010
Posts: 1589

Kudos [?]: 607 [0], given: 40

Location: United States (IN)
Concentration: Strategy, Technology

### Show Tags

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%

_________________

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

GMAT Club Premium Membership - big benefits and savings

Kudos [?]: 607 [0], given: 40

Manager
Joined: 10 Jan 2011
Posts: 232

Kudos [?]: 83 [0], given: 25

Location: India
GMAT Date: 07-16-2012
GPA: 3.4
WE: Consulting (Consulting)

### Show Tags

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-------

Kudos [?]: 83 [0], given: 25

Intern
Joined: 11 Jul 2011
Posts: 2

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

### Show Tags

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 .

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

Math Forum Moderator
Joined: 20 Dec 2010
Posts: 1949

Kudos [?]: 2140 [0], given: 376

### Show Tags

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
_________________

Kudos [?]: 2140 [0], given: 376

Intern
Joined: 11 Jul 2011
Posts: 2

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

### Show Tags

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 .

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

Math Forum Moderator
Joined: 20 Dec 2010
Posts: 1949

Kudos [?]: 2140 [0], given: 376

### Show Tags

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

(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.
_________________

Kudos [?]: 2140 [0], given: 376

Intern
Joined: 14 Aug 2011
Posts: 5

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

Location: United States (PA)
WE: Pharmaceuticals (Pharmaceuticals and Biotech)

### Show Tags

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.

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

Manager
Joined: 30 Sep 2009
Posts: 115

Kudos [?]: 34 [0], given: 183

### Show Tags

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 [ 7.84 KiB | Viewed 5120 times ]

Kudos [?]: 34 [0], given: 183

Math Expert
Joined: 02 Sep 2009
Posts: 42617

Kudos [?]: 135746 [0], given: 12708

### Show Tags

28 May 2012, 23:31
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.
_________________

Kudos [?]: 135746 [0], given: 12708

Re: M03 #04   [#permalink] 28 May 2012, 23:31

Go to page    1   2    Next  [ 29 posts ]

Display posts from previous: Sort by

# M03 #04

Moderator: Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne 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®.