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At the beginning of year 1, an investor puts p dollars into an investm

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At the beginning of year 1, an investor puts p dollars into an investm  [#permalink]

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New post 17 Nov 2014, 12:48
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Tough and Tricky questions: Percents and Interest Problems.



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 $\(75p\)


Kudos for a correct solution.

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Re: At the beginning of year 1, an investor puts p dollars into an investm  [#permalink]

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New post 17 Nov 2014, 23:43
1
for finding the range :

Max limit = p(1+110/100)^3=10p
Min limit = p(1+85/100)^3= 5p

so Answer is C
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Re: At the beginning of year 1, an investor puts p dollars into an investm  [#permalink]

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New post 17 Nov 2014, 23:52
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Bunuel wrote:

Tough and Tricky questions: Percents and Interest Problems.



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 $\(75p\)
Kudos for a correct solution.


Nice problem! I think the whole xn business is just a tricky way of saying this is a compound interest problem.

Since the min rate of interest is 86% the min value of investment will be (1+0.86)^3*p = (1.86^3)*p. Similarly max value will be (2.1)^3*p.

2.1^3 is 9.5 or you can guess little over 8 so 10p may be good estimate.
1.86^3 is trickier to estimate. I knew root(3) = 1.73. So 1.86^2 is close to 3 and 3*1.8 is 5.4. So 5p is a close lower bound.

So range is 5p to 10p.
Answer C.
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Re: At the beginning of year 1, an investor puts p dollars into an investm  [#permalink]

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New post 18 Nov 2014, 00:04
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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

The min. value for Xn for 1, 2 & 3 year can be 86%, 87% & 88% ( They can be as close to 85 as possible but for simplicity i am taking these rate and carefull about variable so cannot take all value same)
similarly Max value can be 109%, 108% & 107%
Now consider p 100
than solve by increasing it by saubsequent rate for both max and min for
for min rate it will be around 546 and for max it will be around 816 that is the range of final value and by puting p=100 in option u will get option C as the range.
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Re: At the beginning of year 1, an investor puts p dollars into an investm  [#permalink]

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New post 18 Nov 2014, 01:43
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Answer = C = $5p and $10p

Min limit \(= (1 + \frac{86}{100})^3 = (1 + 0.86)^3 = 1.86^3 =\) Below 8

Max limit \(= (1 + \frac{109}{100})^3 = (1 + 1.09)^3 = 2.09^3 =\) Above 8
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Re: At the beginning of year 1, an investor puts p dollars into an investm  [#permalink]

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New post 18 Nov 2014, 08:08
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1
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.
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Re: At the beginning of year 1, an investor puts p dollars into an investm  [#permalink]

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New post 07 Nov 2016, 23:49
Bunuel wrote:

Tough and Tricky questions: Percents and Interest Problems.



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 $\(75p\)
Kudos for a correct solution.


One question...How do we know that they are talking about compound interest and not simple interest?
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Re: At the beginning of year 1, an investor puts p dollars into an investm  [#permalink]

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New post 12 Nov 2016, 22:27
Even i have the same query as ShashankDave

Bunuel ,can you please clarify the below query

ShashankDave wrote:
One question...How do we know that they are talking about compound interest and not simple interest?

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Re: At the beginning of year 1, an investor puts p dollars into an investm  [#permalink]

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