Bunuel wrote:
Official Solution:
If \(P@ = \frac{P}{P-1}\), what is the value of \(P@@\)?
A. \(\frac {P}{P-1}\)
B. \(\frac{1}{P}\)
C. \(P\)
D. \(2 - P\)
E. \(P - 1\)
Solution 1. Pick numbers. If \(P = 2\). Then \(P@ = \frac{2}{2 - 1} = 2\). If \(P@ = P\), then \(P@@ = P\).
Solution 2. Algebra: \(P@ = \frac{P}{(P-1)}\). Take this value and substitute it for \(P\) to find \(P@@\). We will get \(\frac{\frac{P}{(P-1)}}{\frac{P}{(P-1)} - 1}\). Simplify to get the following expression: \(\frac{\frac{P}{(P-1)}}{\frac{1}{(P-1)}}\). Dividing eliminates \(P-1\) and we get \(\frac{P}{1}\) or P.
Answer: C
BunuelThis question took me a minute to understand. I do not mean to come out questioning your explanation, but I am not sure it was complete.
The picking number explanation might not be complete. It is my understanding that 2 is a unique number and should be avoided when using the testing cases approach. For example, if we set 4 = P the explanation P@=P, so P@@ = P does not work.
For example if P = 4: 4/(4-1) -> P@ = 4/3 this does not yield the result P@=P so you cannot then use the logic suggested above to find that P@@ = P. It is not until we substitute again that we find P@@=P.
If P@ = 4/3 then P@@ = (4/3)/((4/3)-1) -> (4/3)/((4/3)-(3/3)) -> (4/3)/(1/3) -> 4 which = P so therefore P@@=P
I think the above offers a more complete explanation of the substitution above. Bunuel, please correct me if my explanation is inefficient or incorrect.