Guide 6, Page 48, "Formulas" section

[kathleen.t.kao]

There is a sample QC problem on this page, and I'm unable to figure out how the book got its answer. Here's the problem:

Common information: v& = 2v-1
Quantity A: v&(&)
Quantity B: 4v

The book goes on to solve the problem like this:

v& = 2v-1
Rewrite Quantity A as (2v-1)&. Evaluate the formula one more time.
(2v-1)& = 2(2v-1)-1 (Wha...?)
= 4v-2-1
= 4v-3
Quantity A: (4v-3)-4v = -3
Quantity B: 4v-4v = 0
Therefore, Quantity B is bigger.

Am I losing it or did the book not complete the distribution in the first step? I really want to understand where I went wrong on this problem. Can someone help me out?

[tommywallach]

Hey Kathleen,

It seems you've ignored the fact that this is a function question, not just a formula. The ampersand signals that you have to put the target into a function.

When you see v& = 2v-1, you recognize that you're dealing with a FUNCTION question. So whenever you see something before the & symbol, you put it through the function (i.e. replace the "v" in "2v-1" with whatever came before the & symbol

So looking at quantity A, the first thing we see before the ampersand symbol is v&. That means we need to put that into the function of 2v-1. This brings us to

2(v&)-1

Now, we have to do the function again. This time around, there's only a v before the ampersand, so we'll put that into the equation, where v simply replaces the v (the first time around, "2v" replaced the v):

2(2v-1)-1

Now we distribute, and we do get 4v-2

Hope that helps. If not, make sure you reread the chapter on functions and do all the check your skills questions.

-t