Question

Suppose you deposit ​$1,500 in a savings account that pays interest at an annual rate of 5​%. If no money is added or withdrawn from the​ account, answer the following questions. a. How much will be in the account after 3 ​years? b. How much will be in the account after 18 ​years? c. How many years will it take for the account to contain ​$2,000​? d. How many years will it take for the account to contain ​$2,500​?

Answer 1

Explanation:

`A=P(1+(R)/(n))^(nt)`

A = Amount after t time period

P = Principal amount

R = Interest rate

n = Number of times interest applied per time period

We have:

a) P = $1,500

R =5%=0.05

n = 1

t = 3 years

`A=\$1,500* (1+(0.05)/(1))^(1* 3)`

`A=\$1,736.43`

The amount after 3 years in the bank will be $1,736.43.

b) P = $1,500

R =5%=0.05

n = 1

t = 18 years

`A=\$1,500* (1+(0.05)/(1))^(1* 18)`

`A=\$3,609.92`

The amount after 3 years in the bank will be $3,609.92.

c) P = $1,500

R =5%=0.05

n = 1

t = ?

A= $2,000

`\$2,000=\$1,500* (1+(0.05)/(1))^(1* t)`

`(\$2,000)/(\$1,500)= (1+(0.05)/(1))^(1* t)`

`\log (\$2,000)/(\$1,500)=t*\log(1.05)`

`0.1249=t* 0.02119`

`t=(0.1249)/(0.02119)=5.89 years`

5.89 years will it take for the account to contain ​$2,000.

d) P = $1,500

R =5%=0.05

n = 1

t = ?

A= $2,500

`\$2,500=\$1,500* (1+(0.05)/(1))^(1* t)`

`(\$2,500)/(\$1,500)= (1+(0.05)/(1))^(1* t)`

`\log (\$2,500)/(\$1,500)=t*\log(1.05)`

`0.2218=t* 0.02119`

`t=(0.2218)/(0.02119)=10.47years`

10.47 years will it take for the account to contain ​$2,000.