Question

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

Answer 1

Answer:

Michelle's age = 16

Step-by-step 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 % = $\frac{4.5}{100} = 0.045$

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

$(1.045)^{0} = 1 \ngeq 2$

for t = 1

$(1.045)^{1} = 1.045 \ngeq 2$

for t = 2

$(1.045)^{2} = 1.092 \ngeq 2$

for t = 3

$(1.045)^{3} = 1.14 \ngeq 2$

for t = 4

$(1.045)^{4} = 1.19 \ngeq 2$

for t = 5

$(1.045)^{5} = 1.25 \ngeq 2$

for t = 6

$(1.045)^{6} = 1.30 \ngeq 2$

for t = 7

$(1.045)^{7} = 1.36 \ngeq 2$

for t = 8

$(1.045)^{8} = 1.42 \ngeq 2$

for t = 9

$(1.045)^{9} = 1.48 \ngeq 2$

for t = 10

$(1.045)^{10} = 1.55 \ngeq 2$

for t = 11

$(1.045)^{11} = 1.62 \ngeq 2$  

for t = 12

$(1.045)^{12} = 1.69 \ngeq 2$

for t = 13

$(1.045)^{13} = 1.77 \ngeq 2$

for t = 14

$(1.045)^{14} = 1.85 \ngeq 2$  

for t = 15

$(1.045)^{15} = 1.93 \ngeq 2$

for t = 16

$(1.045)^{16} = 2.02 > 2$

It satisfies at t= 16

∴ we get

Michelle's age = 16