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If money is invested at x percent interest, compounded annually, the

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New post Updated on: 09 Mar 2019, 02:28
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If money is invested at x percent interest, compounded annually, the amount of the investment will double in approximately 72/x years. If Jolene invested $10,000 in a long-term bond that pays 6 percent interest, compounded annually, what will be the approximate total amount of the investment 35 years later?

(A) $20,000
(B) $40,000
(C) $60,000
(D) $80,000
(E) $120,000

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Originally posted by Noshad on 09 Mar 2019, 00:44.
Last edited by Noshad on 09 Mar 2019, 02:28, edited 2 times in total.
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New post 09 Mar 2019, 02:12
Noshad wrote:
If money is invested at x percent interest, compounded annually, the amount of the investment will double in approximately 72/7 years. If Jolene invested $10,000 in a long-term bond that pays 6 percent interest, compounded annually, what will be the approximate total amount of the investment 35 years later?

(A) $20,000
(B) $40,000
(C) $60,000
(D) $80,000
(E) $120,000


I think - the problem should read "If money is invested at x percent interest, compounded annually,
the amount of the investment will double in approximately 72/x"

The solution will be as follows:

At x=6%, the amount will double in \(\frac{72}{6} = 12\) years.

In 35 years, there are three 12 year terms. The cost of an investment will double 3 times.

Therefore, the total amount of the long bond will be 2*2*2*$10000 = $80000(Option D) in 35 years.
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New post 09 Mar 2019, 02:30
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If money is invested at x percent interest, compounded annually, the  [#permalink]

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New post 09 Mar 2019, 05:44
Noshad wrote:
If money is invested at x percent interest, compounded annually, the amount of the investment will double in approximately 72/x years. If Jolene invested $10,000 in a long-term bond that pays 6 percent interest, compounded annually, what will be the approximate total amount of the investment 35 years later?

(A) $20,000
(B) $40,000
(C) $60,000
(D) $80,000
(E) $120,000


Investment doubles in approximately \(= \frac{72}{x}\) years --------- (\(x =\) rate of interest)

Invested amount \(= $10,000\)

Rate of interest \(= 6\)%

Time \(= 35\) years

Therefore; Investment doubles in approximately \(= \frac{72}{6} = 12\) years

\(12*3 = 36\) years.

Therefore investment doubles approximately \(3\) times in \(35\) years.

ie; \(2*2*2*10,000 = $ 80,000\)

Answer D
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If money is invested at x percent interest, compounded annually, the   [#permalink] 09 Mar 2019, 05:44
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