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How many liters of pure alcohol must be added to a 100-liter

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How many liters of pure alcohol must be added to a 100-liter [#permalink]

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31 Mar 2011, 09:25
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How many liters of pure alcohol must be added to a 100-liter solution that is 20 percent alcohol in order to produce a solution that is 25 percent alcohol?

(A) 7/2
(B) 5
(C) 20/3
(D) 8
(E) 39/4
[Reveal] Spoiler: OA

Last edited by Bunuel on 10 Dec 2013, 03:11, edited 1 time in total.
RENAMED THE TOPIC.

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Re: PS 1000 Section6 Question 19 [#permalink]

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31 Mar 2011, 09:32
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Acer86 wrote:
How many liters of pure alcohol must be added to a 100-liter solution that is 20 percent alcohol in order to produce a solution that is 25 percent alcohol?

(A) 7/2
(B) 5
(C) 20/3
(D) 8
(E) 39/4

Hi guys just a bit lost in this question...would appreciate if someone could help me out

20% Alcohol solution means; in the 100 liter solution, 20 liters of solution is alcohol and 80 liters other solvents.

If we add "x" liters of alcohol to the solution, the solution becomes "100+x" liters and alcohol, which was 20 liters, becomes 20+x liters.

According to the statement;
20+x = 25% of (100+x)
OR
20+x=(100+x)/4
80+4x=100+x
3x=20
x=20/3

Ans: "C"
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Re: PS 1000 Section6 Question 19 [#permalink]

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31 Mar 2011, 12:28
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Acer86 wrote:
How many liters of pure alcohol must be added to a 100-liter solution that is 20 percent alcohol in order to produce a solution that is 25 percent alcohol?

(A) 7/2
(B) 5
(C) 20/3
(D) 8
(E) 39/4

Hi guys just a bit lost in this question...would appreciate if someone could help me out

Here is my crack at it:

Given:
• 20% of solution is alcohol
• 80% of solution is other
• Total volume = 100 liters

Derived:
• 20% * 100 liters = 20 liters

Now, if we add additional alcohol to the solution, we are altering the volume, so it's no longer going to be just 100 liters, it's going to be 100 plus some arbitrary amount. This gives us 100 + x liters of solution after adding more alcohol. In addition, when we added more alcohol, there is not just 20 liters of alcohol anymore (we just added some more), there is going to be 20 + x liters of alcohol.

What we're wanting to figure out is: how much alcohol must be added to result in there being 25% alcohol.

As mentioned before, we started out with:

20% * 100 = 20 liters

But, now we have:

25% * (100 + x) = (20 + x)

Now, let's just solve for x:

Convert the percentage and multiply:
$${ \frac{25}{100} * (100 + x) = (20 + x) }$$

$${ \frac{(2500 + 25x)}{100} = (20 + x) }$$

Simply the left side:
$${ \frac{25(100 + x)}{100} = (20 + x) }$$

Cancel out terms:
$${ \frac{(100 + x)}{4} = (20 + x) }$$

Multiple both sides by 4 to get rid of the 4 in the denominator on the left:
$${ 4 \bigg( \frac{(100 + x)}{4} \bigg) = 4(20 + x) }$$

$${ (100 + x) = (80 + 4x) }$$

Subtract the x from the left:
$${ (100) = (80 + 3x) }$$

Subtract the 80 from the right:
$${ 20 = 3x }$$

Divide....success!
$${ x = \frac{20}{3} }$$

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Re: PS 1000 Section6 Question 19 [#permalink]

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31 Mar 2011, 20:24
Equating total volume of alcohol :

(x + 100) * 25/100 = x + 20

=> x/4 + 25 = x + 20

=> 3x/4 = 5

=> x = 20/3

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Re: PS 1000 Section6 Question 19 [#permalink]

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02 Apr 2011, 13:07
x+20 = 25/100(100+x)

x= 20/3

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Re: PS 1000 Section6 Question 19 [#permalink]

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09 Dec 2013, 21:44
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Using Alligations:

We have 20% alcohol and 80% water. As per the question in the final mixture alcohol should be 25% and water should be 75%. Hence we can say that 80% water will be mixed with 0% water to yield a solution that will contain 75% water. So using alligations:

80%---------0%

75

75-----------5
75:5 = 15:1

Say we added x units of alcohol to the original mixture. These x units will have 0% water. And from above we see 80% water solution is mixed with 0% water solution in 15:1.

so, 15/1 = 100/x
x= 100/15

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Re: How many liters of pure alcohol must be added to a 100-liter [#permalink]

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02 Dec 2014, 22:53
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Alcohol .......... Non-Alcohol .................... Total

20 ..................... 80 ................................. 100

20+x ................... 80 .............................. 100+x

Setting up the equation

$$\frac{75}{100} (100+x) = 80$$ OR

$$\frac{25}{100}(100+x) = 20+x$$

$$x = \frac{20}{3}$$

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How many liters of pure alcohol must be added to a 100-liter sol [#permalink]

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03 Jun 2015, 02:05
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We can solve such weighted average problem by understanding the underlying concept. You need to look at the difference between the alcohol level of the 20% solution and the pure alcohol.
_ The 20% solution has a alcohol level that is 5% lower than the level of alcohol of the final mixture, let's say -5 differential.
_ The pure (100%) alcohol has a alcohol level that is 75% higher than the level of alcohol of the final mixture, let's say +75 differential.
You need to make these differentials cancel out, so just multiply both differentials by respective volumes so that the positive will cancel out with the negative: 75x - 5.100 = 0 -> x = 500/75 = 20/3

Hope it's clear.

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Re: How many liters of pure alcohol must be added to a 100-liter [#permalink]

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03 Jun 2015, 23:02
Hi All,

Since this question is built around the concept of a "weighted average", there are several different ways that you can go about solving it. Since the question asks how much alcohol must be added, and the answer choices are NUMBERS, we can TEST THE ANSWERS and do some basic arithmetic to get to the solution.

We're told that a 100-liter solution is 20% alcohol. This means that 20 liters are alcohol and 80 liters are not alcohol. We're asked how many liters must be added to this 100-liter solution to get a solution that is 25% alcohol.

IF....we add 5 liters of alcohol to this existing mixture....
Total alcohol = 20+5 = 25 liters
Total liquid = 100+5 = 105 liters
25/105 is LESS than 25% alcohol (since 25/100 = 25%, 25/105 is LESS than 25%), thus Answer B CANNOT be the answer. We need MORE alcohol to raise the percentage.

Next, let's TEST ANSWER D: 8 liters
IF....we add 8 liters of alcohol to this existing mixture....
Total alcohol = 20+8 = 28 liters
Total liquid = 100+8 = 108 liters
28/108 is MORE than 25% alcohol (since 28/112 = 25%, 28/108 is MORE than 25%), thus Answer D CANNOT be the answer. We need LESS alcohol to raise the percentage.

[Reveal] Spoiler:
C

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Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** Kudos [?]: 3780 [0], given: 173 Intern Joined: 22 Oct 2014 Posts: 27 Kudos [?]: 11 [0], given: 79 Re: How many liters of pure alcohol must be added to a 100-liter [#permalink] Show Tags 01 Aug 2015, 02:49 yenh wrote: We can solve such weighted average problem by understanding the underlying concept. You need to look at the difference between the alcohol level of the 20% solution and the pure alcohol. _ The 20% solution has a alcohol level that is 5% lower than the level of alcohol of the final mixture, let's say -5 differential. _ The pure (100%) alcohol has a alcohol level that is 75% higher than the level of alcohol of the final mixture, let's say +75 differential. You need to make these differentials cancel out, so just multiply both differentials by respective volumes so that the positive will cancel out with the negative:75x - 5.100 = 0 -> x = 500/75 = 20/3 Hope it's clear. Hi, could you please explain this concept in a simple form? or in graphical form? and Why the positive will cancel out with the negative? Kudos [?]: 11 [0], given: 79 Manager Joined: 24 May 2013 Posts: 85 Kudos [?]: 35 [0], given: 99 Re: How many liters of pure alcohol must be added to a 100-liter [#permalink] Show Tags 17 Mar 2016, 04:15 How many liters of pure alcohol must be added to a 100-liter solution that is 20 percent alcohol in order to produce a solution that is 25 percent alcohol? (A) 7/2 (B) 5 (C) 20/3 (D) 8 (E) 39/4 Alcohol-------------------- Total solution--------------% of alcohol 20 -------------------------100 --------------------------20% 20+x-----------------------100+x ----------------------25% (20+x)/(100+x) =25/100 x=20/3 Hence C Thanks Kudos [?]: 35 [0], given: 99 Director Joined: 07 Dec 2014 Posts: 884 Kudos [?]: 308 [0], given: 17 Re: How many liters of pure alcohol must be added to a 100-liter [#permalink] Show Tags 01 Apr 2017, 19:23 Acer86 wrote: How many liters of pure alcohol must be added to a 100-liter solution that is 20 percent alcohol in order to produce a solution that is 25 percent alcohol? (A) 7/2 (B) 5 (C) 20/3 (D) 8 (E) 39/4 let x=liters of alcohol to be added .2*100+x=.25(100+x) x=20/3 C Kudos [?]: 308 [0], given: 17 Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 3250 Kudos [?]: 1182 [0], given: 327 Location: India GPA: 3.5 WE: Business Development (Commercial Banking) Re: How many liters of pure alcohol must be added to a 100-liter [#permalink] Show Tags 01 Apr 2017, 22:35 1 This post was BOOKMARKED Acer86 wrote: How many liters of pure alcohol must be added to a 100-liter solution that is 20 percent alcohol in order to produce a solution that is 25 percent alcohol? (A) 7/2 (B) 5 (C) 20/3 (D) 8 (E) 39/4 $$\frac{Alcohol}{Solution} = \frac{20+x}{100+x} = \frac{1}{4}$$ $$\frac{Alcohol}{Solution} => 80 + 4x = 100 + x$$ $$\frac{Alcohol}{Solution} => 3x = 20$$ Or, $$x = \frac{20}{3}$$ Hence, answer must be (C) $$\frac{20}{3}$$ _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Kudos [?]: 1182 [0], given: 327 Manager Joined: 14 May 2017 Posts: 56 Kudos [?]: 3 [0], given: 112 Re: How many liters of pure alcohol must be added to a 100-liter [#permalink] Show Tags 03 Aug 2017, 15:52 This question can be solved in around 40s using x method. 20........100 25 75.........5 20 - is orginal alcohol% 100 - as new solution contains only alcohol 25 - new value of alcohol Final value is subatraction Now divide 15/5 to bring it to lower fraction. So the ratio will come 15:1 15x= 40, x= 20/3 1x=? 20/3. So that's our answer Posted from my mobile device Kudos [?]: 3 [0], given: 112 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7869 Kudos [?]: 18503 [0], given: 237 Location: Pune, India Re: How many liters of pure alcohol must be added to a 100-liter [#permalink] Show Tags 05 Dec 2017, 01:08 Acer86 wrote: How many liters of pure alcohol must be added to a 100-liter solution that is 20 percent alcohol in order to produce a solution that is 25 percent alcohol? (A) 7/2 (B) 5 (C) 20/3 (D) 8 (E) 39/4 Responding to a pm: Using scale method here: Working with the concentration of alcohol, w1/w2 = (A2 - Aavg)/(Aavg - A1) w1/w2 = (100 - 25)/(25 - 20) = 75/5 = 15/1 So for every 15 parts of the 20% solution, we must put 1 part of pure alcohol. For 100 liter solution, we need 100/15 = 20/3 liters of pure alcohol. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: How many liters of pure alcohol must be added to a 100-liter [#permalink]

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05 Dec 2017, 01:48
VeritasPrepKarishma wrote:
Acer86 wrote:
How many liters of pure alcohol must be added to a 100-liter solution that is 20 percent alcohol in order to produce a solution that is 25 percent alcohol?

(A) 7/2
(B) 5
(C) 20/3
(D) 8
(E) 39/4

Responding to a pm:

Using scale method here:

Working with the concentration of alcohol,

w1/w2 = (A2 - Aavg)/(Aavg - A1)

w1/w2 = (100 - 25)/(25 - 20) = 75/5 = 15/1

So for every 15 parts of the 20% solution, we must put 1 part of pure alcohol.

For 100 liter solution, we need 100/15 = 20/3 liters of pure alcohol.

Mam, why did we use A2 as 100. It is in litres and Aavg and A1 as percentage. I was doing A2 = 25, Aavg = 20, A1 = 0 as we are using pure alchohol.
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Re: How many liters of pure alcohol must be added to a 100-liter [#permalink]

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05 Dec 2017, 03:49
QZ wrote:
VeritasPrepKarishma wrote:
Acer86 wrote:
How many liters of pure alcohol must be added to a 100-liter solution that is 20 percent alcohol in order to produce a solution that is 25 percent alcohol?

(A) 7/2
(B) 5
(C) 20/3
(D) 8
(E) 39/4

Responding to a pm:

Using scale method here:

Working with the concentration of alcohol,

w1/w2 = (A2 - Aavg)/(Aavg - A1)

w1/w2 = (100 - 25)/(25 - 20) = 75/5 = 15/1

So for every 15 parts of the 20% solution, we must put 1 part of pure alcohol.

For 100 liter solution, we need 100/15 = 20/3 liters of pure alcohol.

Mam, why did we use A2 as 100. It is in litres and Aavg and A1 as percentage. I was doing A2 = 25, Aavg = 20, A1 = 0 as we are using pure alchohol.

We are working with the percentages of alcohol. 20% alcohol solution. 25% alcohol solution. Pure alcohol has 100% alcohol.

You would use 0 if you were working with the percentages of water.
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Re: How many liters of pure alcohol must be added to a 100-liter   [#permalink] 05 Dec 2017, 03:49
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