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mixture problem

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by blaster » Mon Jul 19, 2010 2:11 am
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

is there any specific way to solve this question? i mean specific method. i'm solving by logic and can't remember any other method from OG :) would be glad if someone post here their method how you are solving this kind of question .
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by kmittal82 » Mon Jul 19, 2010 2:19 am
Let x amount of alcohol be added.

Thus, new amount of solution = 100 +x
Amount of alcohol in the new solution = 20 + x
Percentage of alcohol in this new solution = 25% of the alcohol content of the solution

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

Solve for x to get 20/3 i.e. (C)
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by selango » Mon Jul 19, 2010 2:22 am
100 litres of solution = 20% percent alocohol=20 litres of alcohol,rem 80 liters

20+x/80=25/75

20+x/80=1/3

60+3x=80

3x=20

x=20/3
--Anand--
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by outreach » Mon Jul 19, 2010 2:35 am
(C) 20/3
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by gmatmachoman » Mon Jul 19, 2010 2:37 am
In 100 litres , we have 20 litres of alcohol,

Now X litres are added

20 + X= 1/4 ( 100+X)

Solve for X

X= 20/3
Pick C
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by pradeepkaushal9518 » Mon Jul 19, 2010 3:01 am
100 liter solution having 20 liter alcohol

suppose x liter of alcohol is to be added

now solution will be 100+x liters and alcohol will be 20+x and alcohol percentage to be 25 %

so 20+x/100+x= 25/100= 1/4

100+x = 80+4x
20=3x
x=20/3

so C
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by GMATGuruNY » Mon Jul 19, 2010 5:31 am
blaster 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

is there any specific way to solve this question? i mean specific method. i'm solving by logic and can't remember any other method from OG :) would be glad if someone post here their method how you are solving this kind of question .
If you're not so good at setting up equations, you could plug in the answer choices, which represent how much alcohol should be added.

Answer choice C:

20/3 + 20 = 80/3 alcohol.
20/3 + 100 = 320/3 total solution.
alchohol/solution = (80/3) / (320/3) = 80/320 = 1/4 = 25%.

The correct answer is C.
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by kanagasu » Mon Jul 19, 2010 7:51 am
Here the solution that is to be added should be Alcohol.

So 80lts of the remaining solution now, should make 75% of the new Solution.

i.e. If x lts is the new Solution

then 80 = (3/4)*X => X = (80*4)/3

Alochol (which should make 25% of the new Solution) = 1/4*X = 80/3

Therefore, the amount of Alcohol to be added would be 80/3 - 20 = 20/3.

This is my first response on this website, hope to be more active here.
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by harshavardhanc » Mon Jul 19, 2010 10:14 am
you had 80 liters of water initially.

Observe that it is becoming 3/4 th of the solution in the second case.

=> 80 = 3/4 X OR 320/3 is the total solution in the second case.

you had 100 before, so the quantity added 100 - 320/3 = 20/3 :)
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by gmatmachoman » Wed Jul 21, 2010 3:05 am
harshavardhanc wrote:you had 80 liters of water initially.

Observe that it is becoming 3/4 th of the solution in the second case.

=> 80 = 3/4 X OR 320/3 is the total solution in the second case.

you had 100 before, so the quantity added 100 - 320/3 = 20/3 :)
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