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

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

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: #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.......