zisis wrote:
[quote2]\(X\) grams of water were added to 80 grams of a strong solution of acid. If as a result, the concentration of acid in the solution became \(\frac{1}{Y}\) times of the initial concentration, what was the concentration of acid in the original solution?
1. \(X = 80\)
2. \(Y = 2\)
(C) 2008 GMAT Club - m10#23
still doesnt get it...anyoen to help with a more thorough explanation?
While statement (2) might seem new and exciting, it actually provides identical information to (1).
To the explanation!
From (1), we know that we doubled the volume of the solution. There's a handy formula to remember for mixture questions:
(concentration 1)(volume 1) = (concentration 2)(volume 2)
or:
(C1)(V1) = (C2)(V2)
for short.
We know that the original volume is 80 grams (we're measuring in grams, so technically it's weight, not volume, but that doesn't matter) and that there's some acid in there as well. "Strong" has no specific meaning, so we don't know the quantity of acid.
So, based on (1), we know that:
(C1)(80) = (C2)(80 + 80)
(C1)(80) = (C2)(160)
(C1) = (C2)(160/80)
C1 = 2(C2)
so, from the original statement, we know that C1 is twice as strong as C2. Of course, this doesn't give us the actual value of C1, so it's insufficient.
From (2), we know that the new solution is 1/Y times as strong as the original. Since Y = 2, we know that the new solution is 1/2 times as strong. Also nothing about numbers, insufficient.
Now let's look at the statements together:
(1) C1 is twice as much as C2
(2) C2 is half as much as C1
When we stack them together like that, we quickly realize that each statement gives us identical information! Since neither statement was good enough by itself, they certainly won't be sufficient together: choose (E).