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Intern  Joined: 03 Dec 2010
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How many liters of pure alcohol must be added to a 100-liter  [#permalink]

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Question Stats: 80% (01:45) correct 20% (02:04) wrong based on 552 sessions

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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

Originally posted by Acer86 on 31 Mar 2011, 09:25.
Last edited by Bunuel on 10 Dec 2013, 03:11, edited 1 time in total.
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Re: How many liters of pure alcohol must be added to a 100-liter  [#permalink]

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6
4
Alcohol .......... Non-Alcohol .................... Total

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

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

Setting up the equation

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

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

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

##### General Discussion
Retired Moderator Joined: 20 Dec 2010
Posts: 1463
Re: PS 1000 Section6 Question 19  [#permalink]

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6
2
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"
Intern  Joined: 08 Dec 2010
Posts: 10
Re: PS 1000 Section6 Question 19  [#permalink]

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2
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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1
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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x+20 = 25/100(100+x)

x= 20/3

Intern  B
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Re: PS 1000 Section6 Question 19  [#permalink]

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1
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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How many liters of pure alcohol must be added to a 100-liter sol  [#permalink]

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1
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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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.

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

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

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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

Alcohol-------------------- Total solution--------------% of alcohol
20 -------------------------100 --------------------------20%
20+x-----------------------100+x ----------------------25%
(20+x)/(100+x) =25/100
x=20/3

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

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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

let x=liters of alcohol to be added
.2*100+x=.25(100+x)
x=20/3
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Re: How many liters of pure alcohol must be added to a 100-liter  [#permalink]

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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

$$\frac{Alcohol}{Solution} = \frac{20+x}{100+x} = \frac{1}{4}$$

$$\frac{Alcohol}{Solution} => 80 + 4x = 100 + x$$

$$\frac{Alcohol}{Solution} => 3x = 20$$

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

Hence, answer must be (C) $$\frac{20}{3}$$
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Re: How many liters of pure alcohol must be added to a 100-liter  [#permalink]

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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

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

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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

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

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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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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]

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I used a mixture chart to set-up the right equation:

----------------- OG -------- + CHANGE ------ = NEW
ALCOHOL..| 20 ---------| + x ----------- = (25/100 x 100 + x)
OTHER.....|
TOTAL.....| 100 ----------| + x ------------ = 100 + x

The original container contains 20 ML of alcohol.
The amount of alcohol added to get 25% is unknown, so we can add X under change.
We do know that the new mixture must contain 25% alcohol in the new container.

How can we express this algebraically?

We know the original container holds a 100 - which acquires an unknown amount (x).
So the new container holds a total of 100 + x.

Fill in the chart with this information and set-up the equation.

20 + x = 25/100(100+x)
x = 20/3
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Re: How many liters of pure alcohol must be added to a 100-liter  [#permalink]

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1
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.

Quote:
Why are we considering 100 lit. for w1. 100 lit. solution has 20% alcohol means w1=20 lit. Am I missing something ?

I thought (20/x) = (100-25)/(25-20)

I am not getting this part.

w1 and w2 are weights of solutions 1 and 2. They are not amount of alcohol in each solution. They are the weights of entire solutions.
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Re: How many liters of pure alcohol must be added to a 100-liter  [#permalink]

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what won't change is the remainder 80 litres of non-alcohol which will become 75% of the mix;
75/100(100+X) = 80; X = 20/3 Re: How many liters of pure alcohol must be added to a 100-liter   [#permalink] 29 Jan 2020, 03:14

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