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Bunuel
How many ounces of water must be added to a 30-ounce solution that is 40 percent alcohol to dilute the solution to 25 percent alcohol?

(A) 9
(B) 10
(C) 15
(D) 16
(E) 18

Let x = number of ounces of water to be added

(concentration of A)*(Volume of A) + (concentration of B to be added)*(Volume of B to be added)= (resultant concentration of new mixture)*(total volume of new mixture = A + B)

Water has an alcohol concentration of 0

(.40)*(30) + (0)(x) = (.25) (30 + x)

12 + 0x = 7.5 + .25x

4.5 = .25x

\(\frac{4.5}{.25}\) = 18 = x. Number of ounces of water to be added is
Hope it helps
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Bunuel
How many ounces of water must be added to a 30-ounce solution that is 40 percent alcohol to dilute the solution to 25 percent alcohol?

(A) 9
(B) 10
(C) 15
(D) 16
(E) 18

let x=ounces of water to be added
.6*30+x=.75(30+x)
x=18 ounces
E
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\(\frac{Total Alcohol}{total solution + added water} = \frac{1}{4}\)

\(\frac{12}{30+x} = \frac{1}{4} =>30+X = 12*4 => X = 48-30=18\)

Option: E

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Heseraj
\(\frac{Total Alcohol}{total solution + added water} = \frac{1}{4}\)

\(\frac{12}{30+x} = \frac{1}{4} =>30+X = 12*4 => X = 48-30=18\)

Option: E

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Nice Solution! Additional Approach yet simple as above

Alcohol before = Alcohol After
40% of 30 = 25% of (30+x)
40*30/25 = 30+x
48-30 = x = 18 = E
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