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An investment compounds annually at an interest rate of 34.1

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An investment compounds annually at an interest rate of 34.1  [#permalink]

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New post 22 Jun 2013, 21:45
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An investment compounds annually at an interest rate of 34.1% What is the smallest investment period by which time the investment will more than triple in value?

A 3
B 4
C 6
D 9
E 12
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Re: An investment compounds annually at an interest rate of 34.1  [#permalink]

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New post 22 Jun 2013, 21:53
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Hey, not 100% sure what the best way of doing this would be, but here was my approach:

34.1% is just over 1/3, so essentially we want to solve the following equation:

(4/3)^x = 3

4/3 * 4/3 = 16/9 ~ a little less than 1.8
(a little less than 1.8)^2 ~ a little more than 3.

So our answer was (4/3)^4; i.e, 4 years of compounding.
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Re: An investment compounds annually at an interest rate of 34.1  [#permalink]

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New post 12 Aug 2013, 16:57
mattce wrote:
Hey, not 100% sure what the best way of doing this would be, but here was my approach:

34.1% is just over 1/3, so essentially we want to solve the following equation:

(4/3)^x = 3

4/3 * 4/3 = 16/9 ~ a little less than 1.8
(a little less than 1.8)^2 ~ a little more than 3.

So our answer was (4/3)^4; i.e, 4 years of compounding.


Can you explain this a little more?
Where do you come up with (4/2)^x = 3?
Why are you multiplying 4/3*4/3?
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Re: An investment compounds annually at an interest rate of 34.1  [#permalink]

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New post 13 Aug 2013, 06:15
1
jjack0310 wrote:
mattce wrote:
Hey, not 100% sure what the best way of doing this would be, but here was my approach:

34.1% is just over 1/3, so essentially we want to solve the following equation:

(4/3)^x = 3

4/3 * 4/3 = 16/9 ~ a little less than 1.8
(a little less than 1.8)^2 ~ a little more than 3.

So our answer was (4/3)^4; i.e, 4 years of compounding.


Can you explain this a little more?
Where do you come up with (4/2)^x = 3?
Why are you multiplying 4/3*4/3?


Assume initial amount is x
Annual Interest is 34.1% so after 1 year the amount will become x * (100+34.1)/100 => x*4/3

now we need to find n for x * (4/3)^n = 3x

or in other words n = 4
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Re: An investment compounds annually at an interest rate of 34.1  [#permalink]

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New post 13 Aug 2013, 09:10
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fozzzy wrote:
An investment compounds annually at an interest rate of 34.1% What is the smallest investment period by which time the investment will more than triple in value?

A 3
B 4
C 6
D 9
E 12



Compound Interest Formula:

\(A = P (1+r/100)^n\)
where A = Final amount after the Prinicipal 'P' is compouned for 'n' years at the rate 'r' per year.
From the question we understand -

\(3P >= P(1+34.1/100)^n\)

we need to find the value of 'n'
==> \(3>=(1.341)^n\)
==> for the equation to satisfy, n can take a value of >= 4

Since the question asks for minnimum value, the answer is '4'.
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Re: An investment compounds annually at an interest rate of 34.1  [#permalink]

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New post 13 Aug 2013, 12:56
fozzzy wrote:
An investment compounds annually at an interest rate of 34.1% What is the smallest investment period by which time the investment will more than triple in value?

A 3
B 4
C 6
D 9
E 12

............
compounded amount = p(1+r/100)^t
or, I + p = p (1+r/100)^t
or, 2p + p = p (1+r/100)^t
or, 3 = (1.342)^t

t has to be more than 3, so t = 4 (Answer)

but this is not gmat type problem.......i think
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Re: An investment compounds annually at an interest rate of 34.1  [#permalink]

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New post 24 Oct 2015, 12:24
I believe there is an easier approach.
Think in terms of simple interest.
34.1 is added each year. clearly, if you multiply this by 10, it gives 341, which is way over 200$ that we need to add in order to triple our investment.
Just by thinking this way, we can eliminate D and E, since E is more than 10 and 9..well..if we multiply by 9, it is still way over 200.
Take 6 now, well, 6 x34.1 is little bit over 200. Now we're getting closer. Since 6 y with simple interest will triple the investment, then compounded interest will have a greater return, so C is out.
Take A. suppose we have 100 initial investment. After first year, we have 134.1. Now, 13.41 is 10%. Suppose we have 30% return. Now we have 39+ interest in the second year. That gets us to 134.1+39+ which is 173+. again, 17.3 is 10%, 30% is 51+, 173+51 gets us to 224+, slightly more than 224, but not enough to be 300. SO a is out, and B is the answer.
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Re: An investment compounds annually at an interest rate of 34.1  [#permalink]

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New post 15 Oct 2018, 20:32
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Re: An investment compounds annually at an interest rate of 34.1   [#permalink] 15 Oct 2018, 20:32
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