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Someone plans to invest $10,000 in an account paying 3%

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Someone plans to invest $10,000 in an account paying 3% [#permalink] New post 18 Aug 2006, 12:10
Someone plans to invest $10,000 in an account paying 3% annual interest and compounded semi-annually. How much must he invest in another account paying 5% annual interests and compounded quarterly so that his annual income from the 2 accounts in the first year are the same?

Anyone know a shorter way to solve this problem

10,000(1+(3/200))^2= A(1+(5/400))^4
I tried solving it wasted too much time
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Re: Compounding Problem solving [#permalink] New post 18 Aug 2006, 19:17
apollo168 wrote:
Someone plans to invest $10,000 in an account paying 3% annual interest and compounded semi-annually. How much must he invest in another account paying 5% annual interests and compounded quarterly so that his annual income from the 2 accounts in the first year are the same?

Anyone know a shorter way to solve this problem

10,000(1+(3/200))^2= A(1+(5/400))^4
I tried solving it wasted too much time


this is definitely a tough question. to solve this, we need either a scientific calculator or some more information such as PVIF/FVIF so that we can simplify some difficult calculations that are not possible even with a simple calculator. this problem involves some basic financial concepts.

your formula gives you the total value of the investments at the end of year 1. but the question is asking for the interest amount that is equal for both investments. therefore the equations should be as under:

interest at the end of first year for first investment = the interest at the end of first year for the second investment

(10,000)(1.015)^2 - 10,000 = x (1.0125)^4 - x
x = 302.25/0.050945337
x = 5932.83

i used excel for calculations. hope this helps.
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 [#permalink] New post 18 Aug 2006, 20:22
THanks professor,

I was really suprise to find it in the question bank. Thought there was some way to simplify it without to much effort :-D Thanks
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Re: Compounding Problem solving [#permalink] New post 18 Aug 2006, 21:32
apollo168 wrote:
Someone plans to invest $10,000 in an account paying 3% annual interest and compounded semi-annually. How much must he invest in another account paying 5% annual interests and compounded quarterly so that his annual income from the 2 accounts in the first year are the same?

Anyone know a shorter way to solve this problem

10,000(1+(3/200))^2= A(1+(5/400))^4
I tried solving it wasted too much time

There is a shorter way:

We know
(x+y)^4 = x^4 + 4x^3 y+ 4xy^3+6x^2 y^2 + y^4
(x+y)^2 = x^2+2xy+y^2
when y is too small as compared to x then these can be reduced to
(x+y)^4 = x^4 + 4x^3 y
(x+y)^2 = x^2+2xy

As professor said the correct formula is
(10,000)(1+3/200)^2 - 10,000 = x (1+5/400)^4 - x

Applying the above trick we get:

10000(1 + 6/200)-10000 = x(1+5/100)-x
300 = 5x/100
x = 6000 approx.
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 [#permalink] New post 20 Aug 2006, 00:56
apollo168,

Is this the correct answer???
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 [#permalink] New post 20 Aug 2006, 01:03
Hi PS_Dahiya,

I think both your answer and the professor's are correct. In the question bank it list 9 thousand something as the correct answer, but I tend to view the answers in it with a lot more suspicion after having found a lot of dubious answers. I think professor used excel to compute so that should settle things :-D
  [#permalink] 20 Aug 2006, 01:03
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