Bunuel wrote:

Official Solution:

If after 200 grams of water were added to the 24%-solution of alcohol, the strength of the solution decreased by one-third, how much of the 24%-solution was used?

A. 180 grams

B. 220 grams

C. 250 grams

D. 350 grams

E. 400 grams

Let the weight of 24% solution used be \(x\) grams, then the weight of alcohol in it would be \(0.24x\). Next, as in the final solution strength decreased by \(\frac{1}{3}\) thus the strength of the final mixture became \(24*\frac{2}{3}=16 \%\).

Set the equation: \(0.24x=0.16(x+200)\), the weight of 16% alcohol in \(x+200\) grams of new solution comes only from (equals to) 24% alcohol in \(x\) grams of strong (initial) solution, as there is 0 grams of alcohol in water (0% alcohol solution). Solving gives: \(x=400\).

Answer: E

Hi I use a different method and solve it really quickly :

After the water has been added the concentration decrease by \(\frac{1}{3}\).

It means that \(\frac{1}{3}\) of the final mixture contains no alcohol and therefore is water.

Hence, 200g of water = \(\frac{1}{3}\) of final mixture

Final Mixture = 600g wich 200g are water and 400 are the initial solution