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PS 1000: Section 6 # 19

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PS 1000: Section 6 # 19

by envyk10 » Thu May 10, 2007 10:51 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

I got the answer by plugging in the values. Can somebody please suggest a better method.

Thanks
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by bww » Thu May 10, 2007 7:34 pm
Hi...is the answer 20/3?
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Re: PS 1000: Section 6 # 19

by bww » Thu May 10, 2007 8:49 pm
envyk10 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 be the amount of liters of pure alcohol that must be added. The % of alcohol of this x liters is 100%, or 1. Thus, the total amount of liters that must be added is 1x.

We already have 100 liters of the 20% alcohol solution. Thus, the total amount of liters that we already have is .20(100) or 20.

The solution we want is 100+x (what we already have + what we need) that is 25% alcohol. Thus, the total amount of liters combined is .25(100+x).

Put the three together: 1x+20=.25(100+x) and solve for x.
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by envyk10 » Fri May 11, 2007 5:07 am
Yes, the answer is C. 20/3
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by Cybermusings » Fri May 11, 2007 12:19 pm
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

20% of 100-liter = 20 liter of pure alcohol
(20+x)/(x+100) * 100 = 25
2000 + 100x = 25x + 2500
500 = 75x
x = 500/75 = 20/3..
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