Question

What rate is necessary to double an investment of $4000 in 10 years?

Answer 1

Simple interest: 10%
Compound interest: 7.17735%, nearly.

Explanation:

The percentage interest p for simple interest is given by

4000(1+(10)(p/100))=8000 giving p = 10%.

For compound interest p, the equation to be solved is

`4000(1+p/100)^10=8000`.

So, `1+p/100=2^(1/10)=1.07177346...`

This gives p =7.17735%, nearly.

Answer 2

Simple interest rate `10% p.a.`
Coumpound interest `7.18 ~~ 7.2%` p.a.

Explanation:

The question does not indicate whether the interest is simple or compound interest.

Let's consider both..

The formula to calculate SIMPLE interest: `SI = (PRT)/100`

For the investment to double, the amount of interest has to be $4000.

($4000 invested + $4000 interest = $8000 in total.)

Making R the subject gives `R = (100 SI)/(PT)`

`R = (100 xx 4000)/(4000xx10) = 10 " years"`

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If the interest is compounded annually , the formula is

`A = P(1+r)^n" "larr A` represents the TOTAL amount

`8000 = 4000(1+r)^10" "larr` where `r` = rate as a decimal

`8000/4000 = (1+r)^10`

`(1+r)^10" = 2 "larr` now find tenth root of each side

`((1+r)^10)^(1/10) = 2^(1/10)`

`1+r = 1.071773`

`r = 0.071773 = 7.1773/100`

`r = 7.18%`
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In reality, neither of these is practical.
Investments do not earn simple interest.
Interest is usually compounded monthly, even daily, but rarely annually.
But the question leads to some nice math calculations.......