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 Q50  V34
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Re: Someone plans to invest $10,000 in an account paying 3% [#permalink]
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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Re: Someone plans to invest $10,000 in an account paying 3% [#permalink]
apollo168,

Is this the correct answer???
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Re: Someone plans to invest $10,000 in an account paying 3% [#permalink]
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



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