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/
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