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Percentge

by Nycgrl » Thu Jul 10, 2008 5:36 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


PLS EXPLAIN
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by lordpapi63 » Thu Jul 10, 2008 6:07 pm
here is what I did...

assuming pure alcohol is 100% proof, let's call the content to be added 1.00n

algebraic expression:

let n replace what is being added...

100%n + (20%)100liters = 25% (n+100liters)

1n + (0.2)100 = 0.25 (n+100)

1n + 20 = 0.25n + 25

1n - 0.25n = 25 - 20

0.75n = 5

3/4n = 5

n = 20/3

not too sure....but I like to think I got it right :-)
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n = 20/3

by lightbulb » Thu Jul 10, 2008 9:53 pm
n*100/100 + 100 * 20/100 = (100 + n) * 25/100

n + 20 = 25 + n/4
3n/4 = 5
n = 20/3
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by wawatan » Fri Jul 11, 2008 12:31 am
(20+x)/(100+x)= 1/4

x= 20/3

answer:C IMO
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