2.2 Convert an interest rate of \( 10 \% \) pal compounded quarterly to an annual effective interest rate. A lawnower was traded in for 2.326 anter 10 years. The depreciation was caleulated at \( 9.2 \% \) p.a. on a straight-line basis. Determine the cost of the lawnmower 10 years ago.

A lawnmower was traded in for \( R 236 \) after 10 years. The depreciation was calculated at \( 9,2 \% \) p.a. on a straight-line basis. Determine the cost of the lawnmower 10 years ago.

2.2 Convert an interest rate of \( 10 \% \) pal compounded quarterly to an annual effective interest rate. A lawnower was traded in for 2.326 anter 10 years. The depreciation was caleulated at \( 9.2 \% \) p.a. on a straight-line basis. Determine the cost of the lawnmower 10 years ago.

Determine the number of months to complete paying for the balance.

₹ 6.050 is borrowed at \( 6.5 \% \) rate of indereat p.a. Find the inderest and the amount to be paid at the end of 3 years.

a. The future value of \( \$ 590 \) six years from now at 5 percent. b. The future value of \( \$ 1,000 \) saved each year for 10 years at 7 percent. c. The amount a person would have to deposit today (present value) at an interest rate of 9 percent to have \( \$ 600 \) five years from nov d. The amount a person would have to deposit today to be able to take out \( \$ 900 \) a year for 9 years from an account earning 7 percent.