Question

Suppose Lisa places $6000 in an account that pays 18% interest compounded each year. Assume that no withdrawals are made from the account.

Follow the instructions below. Do not do any rounding.

(a) Find the amount in the account at the end of 1 year.

(b) Find the amount in the account at the end of 2 years.

Answer 1

Explanation:

The formula for compounded interest is given by

`A = P\left(1 + (r)/(n)\right)^(nt)`

where

A = amount of money after a time t

P = principal or initial amount of money

r = interest rate in decimal form

n = number of times money bnb is compounded per unit time

t = unit time

a) For this problem, P = $6000, n = 1 since the money is compounded once a year, t = 1 year and r = 0.18 so after one year, Lisa's money amounts to

`A = (\$6000)\left(1 + (0.18)/(1)\right)= \$7,080`

b) After two years, the money is still compounded yearly so n = 1 but t = 2 years so Lisa's money amounts to

`A = (\$6000)\left(1 + (0.18)/(1)\right)^(2) = \$8354.40`

Answer 2

Explanation:

a)

Principal = $6000

Rate = 18%

Time = 1 year

A = P(1 + r/100)¹

A = 6000(1 + 18/100)

A = 6000(100 + 18/100)

A = 6000 × 118/100

A = 60 × 118

A = 7080

b)

Principal = $6000

Rate = 18%

Time = 2 years

A = P(1 + r/100)²

A = 6000(1 + 18/100)²

A = 6000(100 + 18/100)²

A = 6000(118/100)²

A = 6000 × 118/100 × 118/100

A = 6 × 118 × 118/10

A = 8354.4