Question

2. (3 marks The amount invested in the term deposit was $2295. On what date was the investment made if the rate of return was 3.2% per annum? )A term deposit reached maturity on July 2,2018. The maturity value was $2345.25. (4 marks) Two payments were scheduled to be made on March 10, 2017 and December 5, 2017 of $700 and $1250 respectively. What single payment on August 15, 2017 would be equivalent to these two payments if the interest agreed upon was 6.25% per annum? 3. It 0.0 b is boter Fer deu

Answer 1

Answer 2)
We have to provide the date of term deposit, so first
we get the total period for which interest is paid.
Thus, P (1 + r)^n = MV where r = rate of interest, n=period in years
2295 (1 + 0.032)^n = 2345.25
(1.032)^n = 2345.25/2295 = 1.021895
taking log on both side, we get
n log(1.032) = log (1.021895)
n = log (1.021895) / log(1.032)
n = 0.00940627421 / 0.0136797
n= 0.6876 year
n = 365 * 0.6876 = 251 days
Days count backward:
july	1
june	30
may	31
apr	30
march	31
feb	28
jan	31
dec	31
nov	30
oct	8
TOTAL DAYS	251
Start of deposit = 8th day of Oct, 2017 = 24th of October, 2017