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Grade 8 — Term 1 & 2 Revision draws its questions at random from a bank of 18 questions, with full worked solutions for every one.
A few of the question types, with full solutions, so you know what to expect.
Determine the highest common factor (HCF) of \(48\) and \(60\).
Answer: \(12\)
Write each number as a product of its prime factors:
\(48 = 2^4 \times 3\)
\(60 = 2^2 \times 3 \times 5\)
The HCF takes the lowest power of each common prime factor:
\(\text{HCF} = 2^2 \times 3 = 12\)
Share \(R450\) between two people in the ratio \(4:5\). How much does the person with the larger share receive?
Total number of parts: \(4 + 5 = 9\)
Value of one part: \(R450 \div 9 = R50\)
The larger share is \(5\) parts:
\(5 \times R50 = R250\)
A jacket costs \(R800\). In a sale it is discounted by \(15\%\). Calculate the sale price.
Calculate the discount amount:
\(15\% \text{ of } R800 = \dfrac{15}{100} \times 800 = R120\)
Subtract the discount from the original price:
\(R800 - R120 = R680\)
Calculate: \(-6 \times (-4) + (-10)\)
Multiply first — a negative times a negative is positive:
\(-6 \times (-4) = 24\)
Then add \(-10\):
\(24 + (-10) = 14\)
Calculate: \((-3)^2 - 4 \times (-2)\)
Apply BODMAS — powers and multiplication before subtraction.
\((-3)^2 = 9\)
\(4 \times (-2) = -8\)
\(9 - (-8) = 9 + 8 = 17\)