Question

1. Suppose that an account earns 4.8% interest per annum. How much would need to be deposited now in order to have $10000 in 5 years if there interest is compounded (a) Monthly; (b) Continuously? Give answers to the nearest dollar. 2. If an investment of $20000 is made in an account returning 3.2% then how long will it take for the account to have $24000 if (a) Interest is compounded quarterly? (four times a year) (b) Interest is compounded continuously. Give answers to two decimal places. 3. The demand function for a product is given by p = 10000e-9/4. (a) At what price per unit will the quantity demanded equal 20 units? (b) If the price is $1000 per unit then what will the demand be, to the nearest (whole) unit?

Answer 1

& compound interest bemula : F =P ( 14 ont where F = Futeure value p - Present value PAPR t = Number of year n = Number of compounding poriods per year. @ Here, R=4.8% = 0.098 t = 5 F = 10000 n = 12 Therefore 10000 = P(1 + 0.098 1285 or, P = 7870 $7870 would need to deposit now. 12 Henee, 6 Here, R=4.8% = 0.048 t=5 F = 10000 n: 365 Therefore, 365 XS sze, 10000 - P(14 0043 on, p = 7866 Henee, $ 7866 would need to deposit now.

@ @ Here, P-20000 R - 9.0% - 0.032 F - 29000 n = 4 Formula as bebre. 24000 - 20000(1+ 0,032 ) or, at ln (126) = ln (&) as, t in (45) on t 4 ln (126/125) Here it will take. 6 years. © Here, P = 20000 R = 3.2% = 0.032 F = 24000 n = 365 There bore. W 24000 - 20000 (1+0,4329991 os, /365.03213654 5. 9454 In 1965.642 ) = 2n13) og t= ln (6/8) 365 ln (9.65092) C 365 Hence, it will take 6 yea

③ Creven that, – 2/4 P = 10000 e @ Here, q=20 Now, p= 10000 e 4 ~67 Hence, the price is $67 per unit. 6 Here, p= 1000 100. 1000 - 10000 e 2/4 on, &- 4/4 = 70 or, - I line = ln (to) oh, q=g Hence, the demand will be o unit.