Bunuel wrote:

The number of livestock in a farm at the beginning of year 2000 was 100,000. During the year, the number increased by p%. During the next year 2001, there was a famine and the number decreased by q%. A census at the end of year 2001 revealed that the number of livestock in the farm was 100,000. Which of the following expressions is correct?

A. p > q

B. q > p

C. p = q

D. With the exception of 1 instance, p will be equal to q

E. There is no relation between p and q

Kudos for a correct solution.

Eliminate C, D, and E immediately.

1. "[W]hen you go up by a percent, then down by the same percent, you do not wind up where you started: that’s the trap."* p and q can never be equal when there is an original value involved. p ≠ q

Eliminate C and D.

2. E is nonsensical. The original 100,000 first increased; that population's return to original value was not random. In fact, not only are p and q related, when original value is the base, they are inversely proportional (see value-testing below). Eliminate E.

Now down to answers A and B. Let p% increase be 25 (i.e., a 25% increase).

3. To simplify, suppose 100 livestock instead of 100,000.

If that 100 increases by p% = 25, at the end of the good year, there are 100*1.25 = 125 livestock.

By what percentage q must that 125 decrease in order to return to 100?

By 1 - (fractional inverse of the increase).

The increase, expressed from decimal to fraction, is 1.25 = 1 \(\frac{25}{100}\) = 1\(\frac{1}{4}\) = \(\frac{5}{4}\)

Because percent increase and percent decrease are inversely proportional when original value is "constant," flip that fraction from \(\frac{5}{4}\)to \(\frac{4}{5}\) ------> 125 * \(\frac{4}{5}\)= 100.

So 1 - \(\frac{4}{5}\)= \(\frac{1}{5}\), or a 20% decrease.

Alternatively, \(\frac{4}{5}\)= .8 = 80%, which is a 20% decrease.

p% increase = 25, q% decrease = 20.

p > q

Answer A

*

https://magoosh.com/gmat/2012/understanding-percents-on-the-gmat/