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