Math Assignment

For problems 1a) and 1b), a loan company says that it will charge $0.49 per day for every $500 borrowed. George borrows $2000 for 30 days.

1a) What amount will George repay?

Answer:

We first compute the interest rate per day.

I = P x r x t , where I= $0.49, t=1, P=$500

r= I/(P x t) = .49/(500) = 0.00098 (note: this is daily interest rate)

Then we compute the interest for $2000

I= P x r x t, where P =2000, r= 0.00098, t=30

I= 2000 x 0.00098 x 30 = $58.80

Thus, George will pay $2058.80 in 30 days. (note: subject to rounding off errors)

1b) What annual interest rate is he paying to the company? Assume a 360 day year.

Answer:

From 1a, the interest per day is $0.49 for $500, and then we find the interest rate in 360 days by

r= I/ (P x t) = .49/(500 x 1/360) = .49/ (500 x 0.0028)= .35

Thus, the bank annual interest rate is 35%. (note: subject to rounding off errors)

2) If $3400 is deposited into an account earning 6.2% annual interest compounded quarterly, how much will be in the account in 5 years?

Answer:

P= C(1+r/n)t ,where C= initial deposit=$3400, r=annual interest rate=.062, n=# of times compounded yearly, t= numbers years =5, then

P= 3400(1+.062/4)5 = 3400(1 + 0.0155) 5= 3400 x 1.07994002824497346875= 3671.79609603290979375

Thus, after five years the account value will be $3671.80.