How many ounces of water must be added to 10 ounces of 3 %

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

How many ounces of water must be added to 10 ounces of 3 %

by Brent@GMATPrepNow » Wed Oct 03, 2018 9:48 am

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

Brent Hanneson - Creator of GMATPrepNow.com

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

How many ounces of water must be added to 10 ounces of 3 %

by GMATGuruNY » Wed Oct 03, 2018 11:25 am

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

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

Alcohol percentage in the original solution: 3%.
Alcohol percentage in the added water: 0%.
Alcohol percentage in the mixture: 1%.

Let S = the original solution and W = the added water.
The following approach is called ALLIGATION -- a very efficient way to handle MIXTURE PROBLEMS.

Step 1: Plot the 3 percentages on a number line, with the percentages for S and W (3% and 0%) on the ends and the percentage for the mixture (1%) in the middle.
S 3%----------1%-----------0% W

Step 2: Calculate the distances between the percentages.
S 3%----2-----1%----1-----0% W

Step 3: Determine the ratio in the mixture.
The ratio of original solution to added water is equal to the RECIPROCAL of the distances in red.
S:W = 1:2 = 10:20.

The resulting ratio indicates that the 10 ounces of original solution must be combined with 20 ounces of water.

The correct answer is D.

For two similar problems, check here:

https://www.beatthegmat.com/ratios-frac ... 15365.html

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8083; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Sun Apr 07, 2019 5:28 pm

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

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