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

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Math Expert
Joined: 02 Sep 2009
Posts: 50619
The number of livestock in a farm at the beginning of year 2000 was 10  [#permalink]

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19 Aug 2015, 00:13
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Difficulty:

55% (hard)

Question Stats:

67% (01:44) correct 33% (01:53) wrong based on 134 sessions

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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.

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Re: The number of livestock in a farm at the beginning of year 2000 was 10  [#permalink]

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19 Aug 2015, 03:02
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1
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.

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.
Manager
Joined: 05 Dec 2015
Posts: 115
Re: The number of livestock in a farm at the beginning of year 2000 was 10  [#permalink]

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13 Apr 2016, 22:57
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?
Math Expert
Joined: 02 Sep 2009
Posts: 50619
Re: The number of livestock in a farm at the beginning of year 2000 was 10  [#permalink]

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14 Apr 2016, 10:28
1
mdacosta wrote:
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?

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.
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The number of livestock in a farm at the beginning of year 2000 was 10  [#permalink]

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27 May 2017, 16:55
1
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/
Senior Manager
Joined: 02 Apr 2014
Posts: 471
GMAT 1: 700 Q50 V34
The number of livestock in a farm at the beginning of year 2000 was 10  [#permalink]

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22 Mar 2018, 12:24
1
(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)
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2203
Re: The number of livestock in a farm at the beginning of year 2000 was 10  [#permalink]

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26 Mar 2018, 09:55

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
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Re: The number of livestock in a farm at the beginning of year 2000 was 10 &nbs [#permalink] 26 Mar 2018, 09:55
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# The number of livestock in a farm at the beginning of year 2000 was 10

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