s2.pdf - University of Toronto Scarborough Department of Computer Mathematical Sciences MAT A32H Winter 2017 Solutions#2 1 Section 5.1 26 We now apply

# s2.pdf - University of Toronto Scarborough Department of...

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University of Toronto Scarborough Department of Computer & Mathematical Sciences MAT A32H Winter 2017 Solutions #2 1. Section 5.1 26. We now apply what we have learned about interest to the concept of inflation. Let P be the average price of some goods and let t be the number of years for the price to double. When the rate of inflation r is r = 7 . 25 % = 0 . 0725, we want to solve 2 P = ( P 1 + 0 . 0725 365 365 t or, equivalently, 2 = 1 + 0 . 0725 365 365 t for t . Taking ln of both sides we have ln 2 = ln 1 + 0 . 0725 365 365 t = 365 t ln 1 + 0 . 0725 365 . Hence t = ln 2 365 ln 1 + 0 . 0725 365 9 . 5616 . The average price will double in about 9.56 years. 28. We first note that a service charge is paid from the account not into the account; hence, r = 1 % = 0 . 01. Now the future value is S = 1000 ( 1 0 . 01 ) 20 = 817 . 9069. Only about \$ 817.91 will remain in the account after 20 years. 2. Section 5.2 20. (a) Since the net present value, NPV , is the sum of the present values of the cash flows less the original principal, we have NPV = 13000 (1 . 03) 10 + 14000 (1 . 03) 12 +15000 (1 . 03) 14 +16000 (1 . 03) 16 35000 4379 . 9773. Hence NPV is about \$ 4379.98.

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