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Compound Interest

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Compound Interest

by engg.manik » Thu Oct 08, 2009 2:37 am
Q. A total of $1000 was invested at compounded annual interest rate. At the end of 12 years, the total value will be $4000. How many years are needed to reach a total of $8,000?
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by sanjana » Thu Oct 08, 2009 2:59 am
Is the OA : 11 Yrs?
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by papgust » Thu Oct 08, 2009 3:02 am
The key here is to find the rate of interest using the values 1000, 4000 in CI formula. From finding the rate of interest, as a second step, find the number of years requires using the values 1000, 8000 and r% in CI formula. This requires a lot of calculations.
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by rohan_vus » Thu Oct 08, 2009 3:09 am
I dont think you need to do lot of calculations

Consider this

A = P*( 1 + r/100)^n formula we all know for compound interest .

Substitue now

4000 = 1000*(1 + r/100)^12
Thus ( 1 + r/100) = 12th root of 4. -- eqn (1)

Now you got to find no of yrs for getting to 8000

So , 8000 = 1000*( 1 + r/100) ^n ---> 8 = (1+r/100)^n -- eqn (2)

Using eqn q and eqn 2 , you simpy get 8 = ( 12th root of 4) ^ n ==> 2 ^ 3 ==> n/6th root of 2 ==> 3 = n/6 and hence n = 18
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by ershovici » Thu Oct 08, 2009 3:14 am
First of all, it is wery interesting question.
Here is my explanation:
1- equotion for a compounded procent is S=D(1+%)^n, where n is teh number of time periods, D is the starting deposit, and S is the total sum.
2- we have 4000=1000(1+%)^12, from this we can find %
% = 2^1/6 - 1
3- now use the sasme reasoning to find solution, if
8000 = 4000(1+2^1/6 - 1)^x
2 = (2^1/6)^x
x = 6

My answer is 6+12(first timeperiod) = 18
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by tinmn8 » Thu Oct 08, 2009 7:46 am
I don't understand how you get the interest in the first part
4000= 1000(1+%)^12, I can simplify that to 4= (1+%)^12, but how do you solve from there?
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by srivas » Thu Oct 08, 2009 8:19 am
here

4000 = 1000*(1 + r/100)^12
Thus ( 1 + r/100) = 12th root of 4. this can be written as
(1+r/100) = 6ht root of 2 [2^2 = 4]--- I

8000 = 1000*(1 + r/100)^n then
( 1 + r/100) = nth root of 8 (8 = 2^3)

so 1+r/100 = 2^3/n ----- II

from I & II

2^1/6 = 2^3/n then 1/6 = 3/n and n = 18
Gmat710,, Hyd
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by okigbo » Wed Nov 18, 2009 12:58 am
Guys,

Do we have to do all those calculations? Please let me know if the reasoning below is incorrect.

Investment quadrupled in 12 years

Investment will double in 6 years

18 years is needed
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by palvarez » Wed Nov 18, 2009 10:13 am
okigbo wrote:Guys,

Do we have to do all those calculations? Please let me know if the reasoning below is incorrect.

Investment quadrupled in 12 years

Investment will double in 6 years

18 years is needed
In this problem, you can get away with such reasoning, because sqrt(4) = 4/2
What you are seeing a logarithmic proportion


Assume that investment triped in 6 years.
Does the investment double in 2 years or 3 years?
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