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At the end of the year 1998, Shepard bought nine dozen goats,

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At the end of the year 1998, Shepard bought nine dozen goats,  [#permalink]

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New post Updated on: 25 Jun 2017, 22:20
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At the end of the year 1998, Shepard bought nine dozen goats, Henceforth, every year he added p% of the goats at the beginning of the year and sold q% of the goats at the end of the year where p > 0 and q > O. If Shepard had nine dozen goats at the end of year 2002, after making the sales for that year, which of the following is true?

(1) p = q
(2) P < q
(3) p> q
(4) P = q/2
(5) P = q/4

Originally posted by theperfectgentleman on 25 Jun 2017, 08:22.
Last edited by Bunuel on 25 Jun 2017, 22:20, edited 1 time in total.
Renamed the topic and edited the question.
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Re: At the end of the year 1998, Shepard bought nine dozen goats,  [#permalink]

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New post 25 Jun 2017, 11:19
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When some value is increased and decreased by the same percentage, the result is always less than the original. For e.g., profit increased 10% in the first year and decreased 10% in the 2nd year would result in 100 -> 110 -> 99.

So, the increase percentage must be higher than the decrease percentage for the result to be the same. Hence, p > q.
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At the end of the year 1998, Shepard bought nine dozen goats,  [#permalink]

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New post 27 Jun 2017, 16:52
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theperfectgentleman wrote:
At the end of the year 1998, Shepard bought nine dozen goats, Henceforth, every year he added p% of the goats at the beginning of the year and sold q% of the goats at the end of the year where p > 0 and q > O. If Shepard had nine dozen goats at the end of year 2002, after making the sales for that year, which of the following is true?

(1) p = q
(2) P < q
(3) p> q
(4) P = q/2
(5) P = q/4

Good question.

1. When you increase by a percent, then decrease by that same percent, you do not end up where you began. That's a trap. Eliminate A.

2. In fact, if the original ends up at the same value, p% increase is inversely related to q% decrease. Forget 9 dozen. We just need to watch any quantity increase and decrease by percentages and return to its original value. Use 100.

3. If that 100 increases by p% = 25 at the beginning of the year, there are 100 * 1.25 = 125 goats that, um, Shepard :wink: , must herd. Or feed. Or whatever.

By what percentage q must that 125 decrease in order to return to 100? By 1 - (fractional inverse of the increase).

The increase, from decimal to fraction form, is 1.25 = 1\(\frac{25}{100}\) = 1\(\frac{1}{4}\) = \(\frac{5}{4}\)

Because percent increase and percent decrease are inversely proportional when the original quantity is the start and end value, flip the percent increase fraction from \(\frac{5}{4}\) to \(\frac{4}{5}\) to get the percent decrease multiplier. ------> 125 *\(\frac{4}{5}\) = 100

So 1 - \(\frac{4}{5}\)= \(\frac{1}{5}\) or a 20% decrease. Shepard sells 20% of the goats at the end of the year.

Alternatively, \(\frac{4}{5}\) = .8 = 80%, which is a 20% decrease.

4. p% increase = 25, q% decrease = 20. p > q

p > q. It doesn't matter that there are four years involved.

It wouldn't matter if there were 40 years involved.

Every year, the number of goats starts at the same value and returns to that same value;
the +25% and -20% just keep getting repeated.

p > q

Answer C
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Re: At the end of the year 1998, Shepard bought nine dozen goats,  [#permalink]

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New post 06 Jul 2018, 20:55
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Re: At the end of the year 1998, Shepard bought nine dozen goats, &nbs [#permalink] 06 Jul 2018, 20:55
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