AbhishekRyu wrote:How many ounces of water must be added to 10 ounces of 3 % alcohol and waer solution to reduce the solution to 1 % alcohol.

A)5
B)10
C)15
D)20
E)30

The original solution contains 10 ounces
3% of that is alcohol
3% of 10 = 0.3
So, the original solution contains 0.3 ounces of alcohol

Let x = the amount of water (in ounces) that must be added
So, the volume of the RESULTING solution = 10 + x ounces

IMPORTANT: Since we're adding water only, the volume of alcohol does not change.
In other words, there are 0.3 ounces of alcohol in the RESULTING solution

We want the RESULTING solution to be 1% alcohol
In other words: (volume of alcohol in solution)/(total volume of solution) = 1/100
Rewrite as: 0.3 /(10 + x) = 1/100
Multiply both sides by (10 + x) to get: 0.3 = (10 + x)/100
Multiply both sides by 100 to get: 30 = 10 + x
Solve: x = 20

Answer: D

Cheers,
Brent

AbhishekRyu wrote:How many ounces of water must be added to 10 ounces of 3 % alcohol and waer solution to reduce the solution to 1 % alcohol.

A)5
B)10
C)15
D)20
E)30

Initially, we have 0.03 x 10 = 0.3 ounces of alcohol and 9.7 ounces of water. To reduce the solution to 1% alcohol:

0.3/(10 + w) = 1/100

30 = 10 + w

20 = w

Alternate Solution:

To 10 ounces of 3% alcohol we will add x ounces of water (0% alcohol) to yield (x + 10) ounces of 1% alcohol. Changing the percents to decimals, we can express this in an equation as:

0.03(10) + 0.0(x) = 0.01(x + 10)

0.3 + 0 = 0.01x + 0.1

0.2 = 0.01x

20 = x

Answer: D