# What rate is necessary to double an investment of $4000 in 10 years? ##### 2 Answers Sep 30, 2016 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 $4000 {\left(1 + \frac{p}{100}\right)}^{10} = 8000$. So, $1 + \frac{p}{100} = {2}^{\frac{1}{10}} = 1.07177346 \ldots$This gives p =7.17735%, nearly. Sep 30, 2016 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: $S I = \frac{P R T}{100}$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 = \frac{100 S I}{P T}$$R = \frac{100 \times 4000}{4000 \times 10} = 10 \text{ years}$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ If the interest is compounded annually , the formula is $A = P {\left(1 + r\right)}^{n} \text{ } \leftarrow A$represents the TOTAL amount $8000 = 4000 {\left(1 + r\right)}^{10} \text{ } \leftarrow$where $r$= rate as a decimal $\frac{8000}{4000} = {\left(1 + r\right)}^{10}$${\left(1 + r\right)}^{10} \text{ = 2 } \leftarrow$now find tenth root of each side ${\left({\left(1 + r\right)}^{10}\right)}^{\frac{1}{10}} = {2}^{\frac{1}{10}}$$1 + r = 1.071773$$r = 0.071773 = \frac{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.......