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mixture problem

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by ankurmit » Sun May 08, 2011 1:43 am
A mixture of 125 gallons of wine and water contains 20% water .How much water must be added to the mixture in order to increase the percentage of water to 25% of the new mixture.

a 10 gals
b 8.5 gals
c 8 gals
d 6.66 gals
e 8.33 gals
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by vkb001 » Sun May 08, 2011 2:36 am
Two ways to solve this:

Method 1:

125 gallons of solution has 25 gallons of water (20%). Let's add x gallons of water to the solution to bump the concentration to 25%.

So, the new concentration is: (25 + x)/(125+x) = 25/100. Solve for x. x = 25/3 or 8.33 gallons.

Method 2:

You are adding solution which is 100% concentrated water (pure water, of course). Resulting solution is 25%. If you are adding x gallons of 100% concentrated water,

(125*20 + x*100)/(125+x) = 25

=> 125*20 + x*100 = 125*25 + x*25

=> 75*x = 125*5

=> x = 25/3 or 8.33 gallons
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by GMATGuruNY » Sun May 08, 2011 3:17 am
ankurmit wrote:A mixture of 125 gallons of wine and water contains 20% water .How much water must be added to the mixture in order to increase the percentage of water to 25% of the new mixture.

a 10 gals
b 8.5 gals
c 8 gals
d 6.66 gals
e 8.33 gals
Wine = .8*125 = 100 gallons.
The amount of wine is not changing and must be 75% of the new mixture:
100 = .75x
x = 100/.75 = 133.33 gallons.
Since the volume of the new mixture is 133.33 gallons, the amount of water that must be added = 133.33-125 = 8.33 gallons.

The correct answer is E.
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by ankurmit » Sun May 08, 2011 3:20 am
vkb001 wrote:
Method 2:

You are adding solution which is 100% concentrated water (pure water, of course). Resulting solution is 25%. If you are adding x gallons of 100% concentrated water,

(125*20 + x*100)/(125+x) = 25

=> 125*20 + x*100 = 125*25 + x*25

=> 75*x = 125*5

=> x = 25/3 or 8.33 gallons
Thanks dear..i was looking out this method only.

Thank you Mitch
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by venmic » Sun May 08, 2011 6:41 am
Mitch great understandng of the underlying concept

Thanks

GMATGuruNY wrote:
ankurmit wrote:A mixture of 125 gallons of wine and water contains 20% water .How much water must be added to the mixture in order to increase the percentage of water to 25% of the new mixture.

a 10 gals
b 8.5 gals
c 8 gals
d 6.66 gals
e 8.33 gals
Wine = .8*125 = 100 gallons.
The amount of wine is not changing and must be 75% of the new mixture:
100 = .75x
x = 100/.75 = 133.33 gallons.
Since the volume of the new mixture is 133.33 gallons, the amount of water that must be added = 133.33-125 = 8.33 gallons.

The correct answer is E.
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