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 .
mixture problem
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- kmittal82
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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)
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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(C) 20/3
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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
Now X litres are added
20 + X= 1/4 ( 100+X)
Solve for X
X= 20/3
Pick C
- pradeepkaushal9518
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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
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
- GMATGuruNY
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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.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 .
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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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.
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.
- harshavardhanc
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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
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
Regards,
Harsha
Harsha
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Harsha bhai..wowoharshavardhanc 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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