Question

Question #a2c15

Answer 1

After `9.327 (3dp)` years both amount will be same.

Explanation:

Let both amount will be same after `t` years.
Formula for continuously compounded interest is `A=P*e^(r/100*t)`
Wherte r = rate of interest , P=Principal , t is number of years.

At Bob's Bank `r=12% :' r/100=12/100=0.12 , P=4000`

At Charlie's Bank `r=6% :' r/100=6/100=0.06 , P=7000`

after `t` years `A_b=A_c :. 4000*e^(0.12t)=7000*e(0.06t)` or
`(e^(0.12t))/(e(0.06t)) = 7000/4000=7/4` Or

`e^((0.12-0.06)t) = 1.75 or e^(0.06t) =1.75` Taking `ln` on both sides we get, `0.06t *ln(e) =ln(1.75) or 0.06t = 0.5596; [ln(e)=1]` or

`t ~~ 0.5596/0.06 =9.327 (3dp)` years [Ans]