4th Ed. Guide 6 QC&DI: Pg 48 Formulas Example Question

[charles.kao15]

Question is as follows:

v& = 2v - 1

Quantity A:

(v&)&

Quantity B:

4v

As I follow the math, it makes sense to insert the defined value of "v&" into Quantity A, yielding:

(2v-1)&

this is where it loses me....The following is what I thought was the process:

Given the definition, we can replace the "v&" within the parenthesis to yield:

(2v - 1)&

Then given that we can multiply "&" into the parenthesis, yields:

2v& - &

Then, seeing that we have "v&" in the equation again we replace it with the definition "2v - 1":

2(2v - 1) - & => 4v - 2 - &

But the sample does not come to this conclusion. Somewhere along the way they dropped the &. The sample question runs

(2V-1)& => 2(2v-1) - 1 => 4v - 2 - 1 => 4v - 3

What did I miss?

[charles.kao15]

Nevermind, I figured it out. Completely forgot the methodology in which the GRE can use symbols.

v& can be interpreted as an f(x) type situation and (v&)& just is a form of f(f(x))....

[tommywallach]

Yep! Weird symbols are ALWAYS f(x) situations on the GRE. Also, this question has been asked many times in the forums before.

-t