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27 Aug 2024, 01:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
64% (02:22) correct
36%
(02:18)
wrong
based on 116
sessions
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Not Attempted Yet
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
(1) p = q
(2) P < q
(3) p > q
(4) p = q/2
(5) p = q/4
27 Aug 2024, 02:06
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To make things easier, look at what occurs to 100 when one increases 100 by a set amount and then solves for what value will bring one back to the initial amount:
Let the initial increase be 100%
\(100*2 = 200\)
Let the percent it decreases by be x
\(200*x = 100\)
\(x = \frac{1}{2}\) or \(0.5\) which is a \(50\)% decrease.
This means to start with a value, increase it by p% and then decrease it by q%, to get the same starting value p>q.
ANSWER C
Let the initial increase be 100%
\(100*2 = 200\)
Let the percent it decreases by be x
\(200*x = 100\)
\(x = \frac{1}{2}\) or \(0.5\) which is a \(50\)% decrease.
This means to start with a value, increase it by p% and then decrease it by q%, to get the same starting value p>q.
ANSWER C
27 Aug 2024, 02:42
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If we increase 100 by 20% => 120, and then decrease 120 by 20% => 96; we notice the final value decreased by 4%
From the above scenario we can derive an important property, i.e. whenever there is a successive increase (p%) and decrease (q%) for the original value to be retained after final operation p has to be greater than q; p>q
Alernatively,
You can also solve this problem as follows, It is basically derivation of the above property
In the first year, i.e. 1999, the final value at the end will be \(9(12) (1+\frac{p}{100}) (1-\frac{q}{100})\)
In the coming three years, \((1+\frac{p}{100})(1-\frac{q}{100})\) will repeat 3 more times.
Final value at the end of 2002 = \(9(12) (1+\frac{p}{100})^4 (1-\frac{q}{100})^4\)
We're given, \((1+\frac{p}{100})^4 (1-\frac{q}{100})^4 = 1\)
\(1-\frac{q}{100}+\frac{p}{100}-\frac{pq}{100} = 1\)
p - q = pq
p and q are both > 0, therefore pq > 0
p - q > 0 => p > q
Therefore, answer is c. p > q
From the above scenario we can derive an important property, i.e. whenever there is a successive increase (p%) and decrease (q%) for the original value to be retained after final operation p has to be greater than q; p>q
Alernatively,
You can also solve this problem as follows, It is basically derivation of the above property
In the first year, i.e. 1999, the final value at the end will be \(9(12) (1+\frac{p}{100}) (1-\frac{q}{100})\)
In the coming three years, \((1+\frac{p}{100})(1-\frac{q}{100})\) will repeat 3 more times.
Final value at the end of 2002 = \(9(12) (1+\frac{p}{100})^4 (1-\frac{q}{100})^4\)
We're given, \((1+\frac{p}{100})^4 (1-\frac{q}{100})^4 = 1\)
\(1-\frac{q}{100}+\frac{p}{100}-\frac{pq}{100} = 1\)
p - q = pq
p and q are both > 0, therefore pq > 0
p - q > 0 => p > q
Therefore, answer is c. p > q
