The number of livestock in a farm at the beginning of year 2000 was 10

(1+p/100)(1-q/100) = 1

100^2 - 100q + 100p - pq = 100^2
100(p-q) = pq

we know, RHS is positive, so LHS must be positive as well, for that we need p > q => (A)

I picked A, but theoretically couldn't P=Q=0 and 100,000 becomes 100,000, or P>Q as in A.

Do I know it's A, because of the increased and then decreased that's specified?

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/

Solution:

Given:

• Livestock in the farm at the beginning of 2000 = 100,000

• Percent increase in livestock during 2000 = p%

• Percent decrease in livestock during 2001 = q%

• Livestock in the farm at the end of 2001 = 100,000

Working out:

We need to find the relation between p and q.

To solve this question, let us take an arbitrary value of p.

Let p = 10.

• Livestock in the farm at the beginning of 2000 = 100,000

• Percentage increase = 10%


o Value of livestock after the increase = 110*100,000/100 = 110,000

• Final value of livestock at the end of 2001 = 100,000


o Decrease in value wrt 2000: 110,000- 100,000= 10,000

o Percentage decrease = 10,000 / Final value


 Percent decrease = 10,000/ 110,000

 Or, 9.09 %

 And this is the value of q.

Thus, it is clear to us that \(q < p\)

Answer: Option A

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.