Question

Michelle's father deposited $1,000 into a savings account when she was born. The annual interest rate is

4.5%. Since opening the account, Michelle's father has never deposited or withdrawn money from it.

What age will Michelle be when her account reaches more than $2,000?

Answer 1

Michelle's age = 16

Explanation:

As we know,

Amount =
`P (1 + r)^(t)` where

P = Principal amount

r = rate of interest

t = time

Given ,

P = $1000

r = 4.5 % =
`(4.5)/(100) = 0.045`

Now,

Given that Amount reaches more than $2000

⇒
`P (1 + r)^(t)` > 2000

⇒
`1000(1 + 0.045)^(t)` > 2000

⇒
`(1.045)^(t)` > 2

Now,

we put the value of t from 0 , 1, 2, ..... untill the above equation does not satisfy

Now,

for t = 0

`(1.045)^(0) = 1 \\geq 2`

for t = 1

`(1.045)^(1) = 1.045 \\geq 2`

for t = 2

`(1.045)^(2) = 1.092 \\geq 2`

for t = 3

`(1.045)^(3) = 1.14 \\geq 2`

for t = 4

`(1.045)^(4) = 1.19 \\geq 2`

for t = 5

`(1.045)^(5) = 1.25 \\geq 2`

for t = 6

`(1.045)^(6) = 1.30 \\geq 2`

for t = 7

`(1.045)^(7) = 1.36 \\geq 2`

for t = 8

`(1.045)^(8) = 1.42 \\geq 2`

for t = 9

`(1.045)^(9) = 1.48 \\geq 2`

for t = 10

`(1.045)^(10) = 1.55 \\geq 2`

for t = 11

`(1.045)^(11) = 1.62 \\geq 2`

for t = 12

`(1.045)^(12) = 1.69 \\geq 2`

for t = 13

`(1.045)^(13) = 1.77 \\geq 2`

for t = 14

`(1.045)^(14) = 1.85 \\geq 2`

for t = 15

`(1.045)^(15) = 1.93 \\geq 2`

for t = 16

`(1.045)^(16) = 2.02 > 2`

It satisfies at t= 16

∴ we get

Michelle's age = 16