Question

A new machine is expected to cost $6000 and have a life of 5 years . maintenance cost will be $1500 the first year , $ 1700 the second year , $ 1900 the third year , $2200 the forth year , and $2300 the fifth year . how much should be deposited in a fund that earns 9% per year , compounded monthly , in order to pay for this machine ?

Answer 1

Year 0:

$$\dfrac{\$6000}{\left(1 + \dfrac{0.09}{12}\right)^{0\times 12}} = \$6,000$$

Year 1:

$$\dfrac{\$1500}{\left(1 + \dfrac{0.09}{12}\right)^{1\times 12}} = \$1,371.34$$

Year 2:

$$\dfrac{\$1,700}{\left(1 + \dfrac{0.09}{12}\right)^{2\times 12}} = \$1,420.91$$

Year 3:

$$\dfrac{\$1900}{\left(1 + \dfrac{0.09}{12}\right)^{3\times 12}} = \$1,451.88$$

Year 4:

$$\dfrac{\$2200}{\left(1 + \dfrac{0.09}{12}\right)^{4\times 12}} = \$1,536.95$$

Year 5:

$$\dfrac{\$2300}{\left(1 + \dfrac{0.09}{12}\right)^{5\times 12}} = \$1,469.01$$

Therefore, the total amount that should be deposited in a fund that earns 9% per year , compounded monthly , in order to pay for this machine is:

$6,000 + $1,371.34 + $1,420.91 + $1,451.88 + $1,536.95 + $1,469.01 = $\boxed{\color{red}{\$13,250.09}}$