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DSC1630 Assignment 02 2021 AS PER UPDATED TUTORIAL LETTER
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This document includes the assignment questions, answers and workings. (Including calculator workings - Sharp EL-738F)

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  • April 15, 2021
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DSC1630

Introductory Financial Mathematics

Department of Decision Sciences

Assignment 02 for 2021 AS PER UPDATED TUTORIAL
LETTER

Unique assignment number: 370781
Due Date: 15 June 2021




Questions 1 and 2 relate to the following situation:

Thando invested R10 000 in a special savings account on 15 May at an interest rate of 15%
per year, compounded every three months for seven months. Interest is calculated on 1
January, 1 April, 1 July and 1 October of every year.

Preview of Question 1

Question 1

If simple interest is used for the odd periods and compound interest for the rest of the term,
the amount of interest received by Thando after seven months is

[1] R901,35.
[2] R1 644,57.
[3] R896,95.

,[4] R665,54.
[5] none of the above.

Answer:

Odd period: Time period is not a full compounding period but smaller. Use simple interest
𝑗𝑚 𝑡𝑚
for odd periods 𝑆 = 𝑃(1 + 𝑟𝑡) and compound interest for the full period 𝑆 = 𝑃 (1 + )
𝑚


Simple interest
𝑆 = 𝑃(1 + 𝑟𝑡)
𝑆 ≡ 𝑎𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑒𝑑 𝑎𝑚𝑜𝑢𝑛𝑡 / 𝑓𝑢𝑡𝑢𝑟𝑒 𝑣𝑎𝑙𝑢𝑒
𝑃 ≡ 𝑝𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙 / 𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑣𝑎𝑙𝑢𝑒
𝑟 ≡ 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑟𝑎𝑡𝑒 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟
𝑡 ≡ 𝑡𝑖𝑚𝑒 𝑖𝑛 𝑦𝑒𝑎𝑟𝑠

Using the Sharp EL-738F calculator:

10 000(1 + 0,15 × 47 ÷ 365) = 10 193,15




Simple interest:

10 575,39 × (1 + 0,15 × 75 ÷ 365) = 10 901,35
10 901,35 − 10 000 = 901,35

OR:

Draw a time line:




There are two odd periods. (One from 15 May to 1 July) (One from 1 October to 15
December). If we use the days table in the Study guide, then the number of days between:

, 15 May and 1 July is 182 − 135 = 47 days,
1 October and 15 December is 349 − 274 = 75 days.

We use simple interest for odd periods and compound interest for full periods. Therefore,
the accumulated value at 15 Dec is:

47 0,15 1 75
10 000 (1 + 0,15 × ) (1 + ) (1 + 0,15 × 365)
365 4


10 901,34776

The interest is equal to:

𝐼 = 10 901,34776 − 10 000

= 901,34776

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