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16 Sep 2014, 01:52
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C
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Difficulty:
75%
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Question Stats:
54% (02:31) correct
46%
(02:34)
wrong
based on 71
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At the beginning of year 1, an investor puts \(p\) dollars into an investment whose value increases at a variable rate of \(x_n%\) per year, where \(n\) is an integer ranging from 1 to 3 indicating the year. If \(85 \lt x_n \lt 110\) for all \(n\) between 1 and 3, inclusive, then at the end of 3 years, the value of the investment must be between
A. $\(p\) and $\(2p\)
B. $\(2p\) and $\(5p\)
C. $\(5p\) and $\(10p\)
D. $\(10p\) and $\(25p\)
E. $\(25p\) and $\(75p\)
A. $\(p\) and $\(2p\)
B. $\(2p\) and $\(5p\)
C. $\(5p\) and $\(10p\)
D. $\(10p\) and $\(25p\)
E. $\(25p\) and $\(75p\)
16 Sep 2014, 01:52
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Official Solution:
At the beginning of year 1, an investor puts \(p\) dollars into an investment whose value increases at a variable rate of \(x_n%\) per year, where \(n\) is an integer ranging from 1 to 3 indicating the year. If \(85 \lt x_n \lt 110\) for all \(n\) between 1 and 3, inclusive, then at the end of 3 years, the value of the investment must be between
A. $\(p\) and $\(2p\)
B. $\(2p\) and $\(5p\)
C. $\(5p\) and $\(10p\)
D. $\(10p\) and $\(25p\)
E. $\(25p\) and $\(75p\)
The quick solution to this problem is to pick a convenient number in the allowed range of growth rates. If \(x_n\) is always between 85 and 110, then the most convenient growth rate to pick is 100. An annual growth rate of 100% is exactly equivalent to doubling one's money. Doubling one's money three times is equivalent to multiplying one's investment by a factor of 8 (\(= 2^3\)). The only range that includes $\(8p\) is the third range ($\(5p\) to $\(10p\)).
Computing the exact outer limits of the allowed range is much more cumbersome. We would have to cube 1.85 for the lower limit and 2.1 for the upper limit. The cube of 1.85 is 6.331625, and the cube of 2.1 is 9.261. These values fall between 5 and 10.
Answer: C
At the beginning of year 1, an investor puts \(p\) dollars into an investment whose value increases at a variable rate of \(x_n%\) per year, where \(n\) is an integer ranging from 1 to 3 indicating the year. If \(85 \lt x_n \lt 110\) for all \(n\) between 1 and 3, inclusive, then at the end of 3 years, the value of the investment must be between
A. $\(p\) and $\(2p\)
B. $\(2p\) and $\(5p\)
C. $\(5p\) and $\(10p\)
D. $\(10p\) and $\(25p\)
E. $\(25p\) and $\(75p\)
The quick solution to this problem is to pick a convenient number in the allowed range of growth rates. If \(x_n\) is always between 85 and 110, then the most convenient growth rate to pick is 100. An annual growth rate of 100% is exactly equivalent to doubling one's money. Doubling one's money three times is equivalent to multiplying one's investment by a factor of 8 (\(= 2^3\)). The only range that includes $\(8p\) is the third range ($\(5p\) to $\(10p\)).
Computing the exact outer limits of the allowed range is much more cumbersome. We would have to cube 1.85 for the lower limit and 2.1 for the upper limit. The cube of 1.85 is 6.331625, and the cube of 2.1 is 9.261. These values fall between 5 and 10.
Answer: C
04 Aug 2016, 06:30
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Hi There
I got little bit confused by the language . Growth rate of 85-100 ?
Is it normal to express growth rate as a mere number and not in percent/fraction ?
BR
Ashish
I got little bit confused by the language . Growth rate of 85-100 ?
Is it normal to express growth rate as a mere number and not in percent/fraction ?
BR
Ashish
