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

Question

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.

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 .

ENGRTOMBA2018 · Answer

Number plugging in will be best strategy for this question.

Let p = 5%,

Per the question 100000(1.05)(1-q/100) = 100000 ---> q = 4.7%. Thus p > q and hence A is the correct answer.

mdacosta · Answer

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?

Bunuel · Answer

We are told that the number increased by p%, so p must be positive. The same way, we are told that the number decreased by q%, so q must be positive. We wouldn't have an increase or decrease at all if p or q were 0.

generis · Answer

Eliminate C, D, and E immediately.

1. " 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/

hellosanthosh2k2 · Answer

(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)

EgmatQuantExpert · Answer

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

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards