MIP1501 Assignment 4 (COMPLETE ANSWERS) 2025 - DUE 31 August 2025

MIP1501 Assignment 4
(COMPLETE ANSWERS)
2025 - DUE 31 August 2025
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,Exam (elaborations)
MIP1501 Assignment 4 (COMPLETE
ANSWERS) 2025 - DUE 31 August 2025
Course

 Mathematics for the intermediate phase (MIP1501)
 Institution
 University Of South Africa (Unisa)
 Book
 New Successful Mathematics Intermediate Phase

MIP1501 Assignment 4 (COMPLETE ANSWERS) 2025 - DUE 31 August 2025;
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1.1. Set up one example question (also provide solutions) which falls under
the following: 1.1.1. Level 1: Mathematical knowledge (4) 1.1.2. Level
2: Routine Procedures (in a real-life context). (4) 1.1.3. Level 3:
Complex Procedures (in real life context) (4) 1.1.4. Level 4: Problem
Solving (in real life context) (4) 1.2. The photo below is 120 mm x 90
mm. It is enlarged to 360 mm x 270 mm. The elephant’s tail in the
smaller photo is 15 mm long. How long is its tail in the larger photo?
Show your calculations. (4) 1.3. Modelling refers to the idea that
learners build in their minds and play out with objects or in picture
form on paper sometimes to solve problems. Give at least three
benefits that you may achieve in your primary mathematics class by
employing modelling. (6) 1.4. Jabu had a few marbles. Today he played
and doubled his number of marbles. Then Thabo gave him three
marbles for free. Jabu has 21 marbles now. How many did he have
before he started playing? Solve this problem using a suitable model.
(5) 1.5. Discuss the steps that you can deploy in constructing number
problems. (10)



1.1 Example questions + solutions (by
cognitive level)
1.1.1 Level 1: Mathematical knowledge (4)
Question. Write 3 equivalent fractions for 25\tfrac{2}{5}52.
Solution. Multiply numerator & denominator by the same number:

, 25=410=615=820\tfrac{2}{5}=\tfrac{4}{10}=\tfrac{6}{15}=\tfrac{8}{20}52=104=156=208.
Notes. Recall facts/definitions; no context.




1.1.2 Level 2: Routine procedures (real-life) (4)
Question. A pencil costs R4.50. How much for 6 pencils?
Solution. 6×R4.50=R27.006\times \text{R}4.50=\text{R}27.006×R4.50=R27.00.
Notes. Single, familiar procedure (repeated addition/multiplication).




1.1.3 Level 3: Complex procedures (real-life) (4)
Question. A 2.4 m rope is cut into 3 equal skipping ropes. Each rope then has 15 cm tied off as a
handle (not used). What length of rope used for skipping is left on each?
Solution.

 Equal parts: 2.4 m÷3=0.8 m=80 cm2.4\text{ m} \div 3 = 0.8\text{ m} = 80\
text{ cm}2.4 m÷3=0.8 m=80 cm.
 Subtract handle: 80 cm−15 cm=65 cm80\text{ cm} - 15\text{ cm} = 65\
text{ cm}80 cm−15 cm=65 cm.
Answer: 65 cm per rope.
Notes. Multiple steps with unit conversion.




1.1.4 Level 4: Problem solving (real-life, non-routine) (4)
Question. A fruit seller sells apples at R6 each or a 4-pack for R20. Lerato has R50 and needs at
least 9 apples. What is the cheapest way to buy at least 9 apples, and what will it cost?
Solution.
Try combinations:

 Two 4-packs (8 apples) + 1 single = 2×20+6=R462\times20 + 6 = \
text{R}462×20+6=R46 → 9 apples.
 One 4-pack (4) + 5 singles = 20+5×6=R5020 + 5\times6 = \text{R}5020+5×6=R50 →
more expensive.
 Three 4-packs (12) = R60 → too much.
Cheapest: 2 four-packs + 1 single = R46 for 9 apples.
Notes. Requires strategic search/justification.