Question

What equal amount Q must be deposited at the beginning of each year for the next 5 years in a savings account 10% interest in order to accumulate 1000 at the end of year 6. The value of Q is?

First cost for special equipment = $1000 per every 50 years of its life cycle. If n= infinity and I = 9% then it’s EUAC is most nearly:

You obtain 1000 loan from the bank. The amount to repay the sum at the end of 6 years with 10% simple interest is?

Answer 1

Q1

The beginning of year n is same as the end of the year (n-1), so when we say the beginning of yr 1, it also means the end of yr 0

Now 5 payments are made into the account at EOY0, EOY1, EOY2, EOY3, EOY4

and we need future value at EOY6

i = 10%

Future value at EOY 6 required = 1000

Let each amount invest be Q

Present value of annuity at EOY0 = Q + Q *(P/A, 10%, 4 )

= Q *(1+3.16986)

= Q * 4.16986

Present value of 1000 at EOY0 = 1000 * (P/F, 10%, 6)

= 1000 * 0.5644739 = 564.474

Now both present value should be equal, therefore,

Q * 4.16986 = 564.474

Q = 135.37

Q2

First cost = 1000, interest rate = 9%, life =50 years

EUAC = 1000 *(A/P, 9%,50)

= 1000 * 0.0912268

= 91.23

Q3

P = 1000, r = 10%, t = 6 yrs

Simple interest = P*R*T

= 1000 *0.1*6 = 600

Total amount to be paid at the end of 6 years = Principal + Interest = 1000 + 600 = 1600