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

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

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New post 19 Aug 2015, 01:13
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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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New post 19 Aug 2015, 04:02
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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.
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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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New post 13 Apr 2016, 23:57
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?
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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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New post 14 Apr 2016, 11:28
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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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New post 27 May 2017, 17:55
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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The number of livestock in a farm at the beginning of year 2000 was 10  [#permalink]

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New post 22 Mar 2018, 13:24
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(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)
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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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New post 26 Mar 2018, 10:55

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  [#permalink]

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Re: The number of livestock in a farm at the beginning of year 2000 was 10   [#permalink] 29 Mar 2019, 09:56
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