Mixture/ratio problem 2

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guerrero: Master | Next Rank: 500 Posts; Posts: 227; Joined: Sun Apr 08, 2012 4:53 amFollowed by: 3 members; Thanked: 12 times

by guerrero » Sat Mar 02, 2013 7:32 am

10% of a 50% alcohol solution is replaced with water. From the resulting solution, again 10% is replaced with water. This step is repeated once more. What is the concentration of alcohol in the final solution obtained?

(A) 3%
(B) 20%
(C) 25%
(D) 36%
(E) 40%

any easy/quick approach?

thanks In advance

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Source: Beat The GMAT — Problem Solving | Sat Mar 02, 2013 7:32 am

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by Brent@GMATPrepNow » Sat Mar 02, 2013 8:13 am

guerrero wrote:10% of a 50% alcohol solution is replaced with water. From the resulting solution, again 10% is replaced with water. This step is repeated once more. What is the concentration of alcohol in the final solution obtained?

(A) 3%
(B) 20%
(C) 25%
(D) 36%
(E) 40%

any easy/quick approach?

thanks In advance

The quick answer is: [spoiler](50%)(0.9)(0.9)(0.9)[/spoiler]. Here's why:

To set this up, imagine that you have 100 ml of solution, and imagine that the alcohol and the water are separated. So, you have 50 ml of water and 50 ml of alcohol.

When you remove 10% of the solution, you're removing 10% of the water and 10% of the alcohol. So, you're removing 5 ml of water and 5 ml of alcohol, to get 45 ml of water and 45 ml of alcohol remaining. From here, you'll add 10ml of water.

IMPORTANT #1: Each time we remove 10 ml of solution and replace it with 10 ml of water, the resulting solution is 100 ml.

IMPORTANT #2: All we need to do is keep track of the alcohol each time. Also notice that removing 10% of the alcohol is the same as leaving 90%.

So, we begin with 50 ml of alcohol.

Step 1: Remove 10% (i.e., keep 90%)
This leaves us with (50)(0.9) ml of alcohol

Step 2: Remove 10%
This leaves us with (50)(0.9)(0.9) ml of alcohol

Step 3: Remove 10%
This leaves us with (50)(0.9)(0.9)(0.9) ml of alcohol
(50)(0.9)(0.9)(0.9) equals approximately 36 ml.

So, our final mixture has a volume of 100 ml of which approximately 36 ml are alcohol. So, the concentration of alcohol is approximately 36%

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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

by GMATGuruNY » Sat Mar 02, 2013 8:38 am

guerrero wrote:10% of a 50% alcohol solution is replaced with water. From the resulting solution, again 10% is replaced with water. This step is repeated once more. What is the APPROXIMATE concentration of alcohol in the final solution obtained?

(A) 3%
(B) 20%
(C) 25%
(D) 36%
(E) 40%

The question stem should be asking for an APPROXIMATION.
I've made the necessary change.

Let the total volume = 100 liters.
Amount of alcohol = 50 liters.
Each time 10% of the total volume is removed, the amount of alcohol also decreases by 10%.
The first time, the amount of alcohol removed = .1(50) = 5 liters.
The next two times, the amount of alcohol removed each time will be a bit LESS than 5 liters, since each time we'll be removing 10% of a SMALLER amount of alcohol.
Thus, at the end of the entire procedure, the TOTAL amount of alcohol removed must be a bit LESS than 3*5 = 15 liters.
Thus, the total amount of alcohol REMAINING must be a bit MORE than 50-15 = 35 liters.

The correct answer is D.

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PiyushKashyap: Newbie | Next Rank: 10 Posts; Posts: 4; Joined: Wed Feb 27, 2013 10:35 am; Thanked: 2 times

by PiyushKashyap » Sat Mar 02, 2013 10:41 am

Remaining qty
------------- = ( (initial qty - replaced qty ) / initial qty ) ^ number of times process repeated.
Initial qty

X/50ml = ((50ml - 5ml ) / 50ml)^3
X=36.45ml

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