Question 1 (2 Marks ) (i) Vincent invested a sum of $5,000 in a bank account earning 5½% p.a simple interest on 1 September 2003. On what date will Vincent’s bank account reach an accumulated balance of $5,039.18? 23 October 2003

(ii)

A 90-day note is to mature for $2,000 plus simple interest at 7% p.a. Calculate the maturity payment (correct to the nearest cent). $2,034.52

Parts (iii) to (iv) are based on the following information: A 180-day promissory note will mature for $10,000. Sixty days after issue it was purchased by Quentin for $9,786.30. (iii) Calculate (as a percentage, correct to 2 decimal places) the rate of simple discount p.a. used in calculating the purchase price paid by Quentin. 6.50% p.a.

(iv) Quentin held the note for 100 days and then sold it for a price of $9,930.28. Calculate the rate of simple interest p.a. (as a percentage, correct to 2 decimal places) earned by Quentin on his investment. 5.37% p.a.

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Question 2. (2 Marks) (i) If j1 = 6.8%, calculate j4 (as a percentage, correct to 2 decimal places). j4 = 6.63%

(ii) How long (in years, correct to two decimal places) will it take for $10,000 to accumulate to $16,000 at j4 = 5.6%. 8.45 years

(iii)

If an investment of $1,000 grows by 50% in 5 years, what rate of interest, j2 , is earned (as a percentage, correct to two decimal places)? j2 = 8.28%

(iv)

A sum of $2,000 is due in 10 years time. Calculate the present value if interest is at j2 = 5% for the first 3 years and j4 = 4% thereafter. (Correct to the nearest cent). $1,305.23

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Question 3 (3 marks) (i) A promissory note is priced as follows: at $970.41 when a discount rate of 6% p.a. is used, but at $964.00 when a discount rate of 7.3% p.a. is used. Use linear interpolation to estimate the discount rate if the price is $968.00 (as a percentage, correct to 2 decimal places). [1 mark] at 6% at i% at 7.3% RHS = 970.41 RHS = 968.00 RHS = 964.00

i% = 0.06 + 968.00 - 970.41 * (0.073 - 0.06) 964.00 – 970.41 i% = 6.49%

(ii)

A savings account pays simple interest at 4% p.a. Calculate the interest earned on the following account for the period 1 October 2003 to 31 December 2003 (inclusive), using the daily balance method (correct to the nearest cent). Date Deposit Withdrawal Balance 25 / 10 / 2003 $450 $450 17 / 11 / 2003 $200 $250 23 / 12 / 2003 $400 $650 [1 mark]

Interest from 25/10 to 16/11 = 450 * 0.04 * 23/365 = 1.1342 Interest from 17/11 to 22/12 = 250 * 0.04 * 36/365 = 0.9863 Interest from 23/12 to 31/12 = 650 * 0.04 * 9/365 = 0.64109 Total interest = 1.1342 + 0.9863 + 0.64109 = $2.76

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(iii)

Sam purchases a television worth $3,000 on 1 October 2003 and arranges to pay the amount with the following three payments: Time of payment 1 October 2003 1 November 2003 1 December 2003 Size of payment $X $X + 500 $1,000

(a) Draw a time line, clearly showing the timing of the 3 repayments and the original debt. [½ mark] Original Loan: Time: Replacement Payments: $3,000 Oct 03 $X Nov 03 $X + 500 Dec 03 $1,000

(b) Using a focal date of 1 October 2003 write an equation of value for the loan transaction, assuming compound interest is at i per month. [½ mark] 3,000 = X +[( X + 500)* (1+i)-1 ] + 1,000(1+i)-2

(c) Solve the equation of value to find the size of X (correct to the nearest cent), if the rate of interest charged is j12 = 12%. [½ mark] 3,000 = X +[( X + 500)* (1.01)-1 ] + 1,000(1.01)-2 3,000 = X + X*0.990099009 + 495.049505 + 980.2960494 1,524.6544 = X*1.990099009 X = 766.12

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Question 4. (3 marks) (i) Answer the following questions. (a)

[1/4 mark for each answer]

A negotiable certificate of...