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DSC1630 Assignment 1 Semester 1 2026 (Answer Guide) - DUE 13 March 2026

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DSC1630 Assignment 1 Semester 1 2026 (Answer Guide) - DUE 13 March 2026 VERIFIED AND CERTIFIED ANSWERS. WRITTEN IN REQUIRED FORMAT AND WITHIN GIVEN GUIDELINES. IT IS GOOD TO USE AS A GUIDE AND FOR REFERENCE, NEVER PLAGARIZE. Thank you and success in your academics. UNISA, 2026

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DSC1630 Assignment 1 Semester 1 2026 (Answer Guide) - DUE 13 March
2026
VERIFIED AND CERTIFIED ANSWERS. WRITTEN IN REQUIRED FORMAT AND WITHIN GIVEN
GUIDELINES. IT IS GOOD TO USE AS A GUIDE AND FOR REFERENCE, NEVER PLAGARIZE.
Thank you and success in your academics.
UNISA, 2026


QUESTION 1
Given
Principal deposited:P = R900
Nominal interest rate: j = 6.5% p.a.
Compounding: quarterly → m = 4
Time before withdrawal: t1 = 3.5 years
Withdrawal amount: R1 000
Remaining time after withdrawal: t2 = 2 years




Step 1: Accumulated amount after 3.5 years
tm
j
S1 = P (1 + )
m
0.065 3.5×4


S1 = 900 (1 + )
4
S1 = 900(1.01625)14

S1 ≈ 900(1.2539) = R1 128.51

,Step 2: Amount remaining after withdrawal

Remaining balance = 1 128.51 − 1 000 = R128.51



Step 3: Accumulated amount after further 2 years
0.065 2×4


S2 = 128.51 (1 + )
4
S2 = 128.51(1.01625)8

S2 ≈ 128.51(1.1377) = R146.25



Final Answer (Q1)

R146



QUESTION 2
Given
Amount invested in second account: P = R1 000
r = 11% = 0.11
Simple interest rate:
Accumulated amount: S = R1 605




Formula (Simple Interest)

S = P (1 + rt)



Step 1: Substitute values

1 605 = 1 000(1 + 0.11t)

, Step 2: Solve for t
1.605 = 1 + 0.11t

0.605 = 0.11t
0.605
t=
0.11
t = 5.50 years



Final Answer (Q2)

5.50 years

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