Bunuel
\(x\) 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) \(y = 2\)
Given: x grams of water were added to 80 grams of a strong solution of acid. As a result, the concentration of acid in the solution became \(\frac{1}{y}\) times of the original concentration Let A = the grams of ACID in the
original strong solution
So, the ORIGINAL concentration of acid \(=\frac{A}{80}\)
After, adding x grams of water the NEW concentration of acid \(=\frac{A}{80+x}\)
We can now write the following equation: \(\frac{A}{80+x}=(\frac{1}{y})(\frac{A}{80})\)
Simplify to get: \(\frac{A}{80+x}=\frac{A}{80y}\)
Divide both sides of the equation by A to get: \(\frac{1}{80+x}=\frac{1}{80y}\)
Cross multiply to get:
80+x = 80yTarget question: What was the concentration of acid in the original solution?In other words,
what is the value of \(\frac{A}{80}\)? Statement 1: x = 80 Aside: At this point all we know is that
80+x = 80y. As you can see, knowing the value of x won't help us, since we must determine the value of A to answer the target question
To see what I mean let's take our equation
80+x = 80y and replace x with 80 to get:
80+80 = 80yWhen we solve this equation for y we get: y = 2, which doesn't help us determine the value of A
Since we can’t answer the
target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: y = 2We can apply the same logic here that we applied for statement 1 to see that statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined Statement 1 tells us that x = 80
Statement 2 tells us that y = 2
When we plug these values into the equation
80+x = 80y, we get:
80+80 = 80(2)Once again we have no way to determine the value of A, which means we can't answer the
target question.
The combined statements are NOT SUFFICIENT
Answer: E
Cheers,
Brent