Question

(b) If Joan places money into an account drawing 4% interest compounded monthly, then how many years would it take before Joan doubles her money?

Answer 1

Solution-

We know that the formula for final amount (A) if compounded n times a year is

....(1)

nt = A (1+)"

Now, If Joan places ( A) money into an account, at 4% compounded monthly then let time taken in doubling of his money (P = 2A) is t years.

So, we have

Initial amount =A

Final amount = 2A

Interest rate = 4 % = 0.04

Compounded monthly or n = 12

Time = t years

On putting these values in equation (1), we get

$2A=A\left ( 1+\frac{0.04}{12} \right )^{12t}$

$\Rightarrow A\left ( 1+\frac{0.04}{12} \right )^{12t}=2A$

$\Rightarrow \left ( 1+0.003333 \right )^{12t}=2$

$\Rightarrow \left ( 1.003333 \right )^{12t}=2$

Taking log both sides, to get

log(1.003333)^12t = log(2)

12t × log(1.003333) = log(2)

t = log(2)/[12log(1.003333)]

t = 0.3010/(12×0.00144524)

t = 17.3575

t ≈ 17.36 years

Hence, it will take 17.36 years to double Joan's Money.