Chapter 2 — Number Sense

Answer: D

Explanation

Irrational means: cannot be written as a fraction $\frac{p}{q}$. Test each one.

• A $\sqrt{9} = 3 = \frac31$ — rational.
• B $\frac{22}{7}$ is already a fraction — rational. (It is close to $\pi$, but it is not $\pi$.)
• C $0.\overline{3} = \frac13$ — a repeating decimal is always rational.
• D $\sqrt{5}$ — 5 is not a perfect square, and $\sqrt5$ cannot be written as any fraction. Irrational ✓

Takeaway: $\sqrt{n}$ is rational only when $n$ is a perfect square. And $\frac{22}{7}$ is an approximation of $\pi$, not $\pi$ itself — either way it is rational.

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Answer: B

Explanation

$-7$ is an integer, and $-7 = \frac{-7}{1}$ makes it rational too. Both parts of B are true.

Why the others are wrong:

• A — natural numbers are $\{1,2,3,\ldots\}$, all positive. $-7$ is not one.
• C — gets "integer" right but says "not rational". Every integer is rational: write it over 1.
• D — whole numbers are $\{0,1,2,\ldots\}$, so $-7$ is not whole; and it is an integer. Wrong twice.

Takeaway: Every integer is automatically rational. The nesting $\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q}$ means anything in a smaller set is also in every larger one.

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Answer: A

Explanation

In CAPS, $\mathbb{N} = \{1,2,3,\ldots\}$ — zero is not a natural number.

$\mathbb{Z} = \{\ldots,-2,-1,0,1,2,\ldots\}$ lists zero explicitly ✓

Why the others are wrong:

• B — $0 \notin \mathbb{N}$ under CAPS.
• C — $0 = \frac01$, so zero is rational, not irrational.
• D — zero is both an integer and rational.

Takeaway: Zero is inside $\mathbb{Z}$ and $\mathbb{Q}$, but outside $\mathbb{N}$. Some countries count 0 as natural; the NBT follows CAPS, which does not.

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Answer: D

Explanation

Simplify each one before judging it.

• A $\pi = 3.14159\ldots$ — never repeats, never ends. Irrational.
• B $\sqrt2 = 1.41421\ldots$ — 2 is not a perfect square. Irrational.
• C $e = 2.71828\ldots$ — irrational.
• D $\sqrt[3]{8} = 2$, since $2^3=8$. Just the integer 2 — rational ✓

Takeaway: A root only looks irrational. $\sqrt[3]{8}$ is really 2. Always simplify first — any root of a perfect power is rational.

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Answer: D

Explanation

Step 1 — Simplify. $\sqrt{12} = \sqrt{4\times3} = 2\sqrt3$

Step 2 — Add. $\sqrt3 + 2\sqrt3 = 3\sqrt3$

Step 3 — Classify. $\sqrt3$ is irrational, and a non-zero rational times an irrational is still irrational. So $3\sqrt3$ is irrational.

Why the others are wrong: A, B and C are nested (every natural is an integer, every integer is rational), so they all stand or fall on one question — can $3\sqrt3$ be written as a fraction? It cannot, so all three fail together. They tempt you if the neat coefficients make the answer feel like it should tidy into a whole number.

Takeaway: Simplify before classifying. Two irrationals can add to a rational — but only if they cancel, like $\sqrt3 + (-\sqrt3) = 0$. Here they reinforce instead.

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Section 2 — Ordering and Comparing Numbers

When comparing numbers in different forms (fractions, decimals, surds), convert everything to the same form — usually decimals — or use squaring to compare positive surds without a calculator.

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Answer: A

Explanation

Convert everything to decimals, then order.

$$0.6 = 0.600 \qquad \frac{5}{8} = 0.625 \qquad \frac{2}{3} = 0.6\overline{6} \approx 0.667$$

$$0.600 < 0.625 < 0.667 \quad\Rightarrow\quad 0.6 < \frac{5}{8} < \frac{2}{3}$$

Why the others are wrong: each wrong order embodies a specific habit of judging fractions by their looks —

• D swaps the two fractions: the "8 is bigger than 3, so $\frac{5}{8}$ beats $\frac{2}{3}$" trap. Larger denominator = smaller pieces.
• B puts $\frac{2}{3}$ first — the error of reading a fraction as its digits ("two point three-ish") instead of dividing.
• C parks the plain decimal 0.6 at the top, as if a clean decimal must outrank "messy" fractions. Only conversion settles rank.

Takeaway: Convert everything to decimals (to three places if needed) and then order. Every wrong option here comes from ranking by appearance instead of by value.

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Answer: A

Explanation

Step 1 — Square both (both are positive, so squaring keeps the inequality the right way round).

$$(\sqrt{5})^2 = 5 \qquad (\sqrt{3}+0.5)^2 = 3 + \sqrt{3} + 0.25 = 3.25 + \sqrt{3}$$

Step 2 — Compare 5 with $3.25+\sqrt3$. That means comparing $\sqrt3$ with $1.75$. Square again: $3$ versus $3.0625$.

Since $3 < 3.0625$, we get $\sqrt3 < 1.75$, so $3.25+\sqrt3 < 5$, and therefore $\sqrt5$ is larger.

Why the others are wrong: the gap here is razor-thin — $2.232$ versus $2.236$ —

• B is what estimation gives: round $\sqrt3$ up to 1.75 and the inequality flips. A 0.004 gap is far smaller than rounding error.
• C ("equal") is the same story at 2 decimal places.
• D ("impossible without a calculator") — the squaring argument above is the calculator-free proof.

Takeaway: Square both sides to remove surds (valid when both are positive). When values are this close, approximation cannot be trusted, but exact comparison can.

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Answer: A

Explanation

Step 1 — Convert to decimals.

$1.4 = 1.400$

$\dfrac{3}{2} = 1.500$

$\sqrt{2} \approx 1.4142\ldots$

Step 2 — Compare.

$$1.400 < 1.414\ldots < 1.500 \implies 1.4 < \sqrt{2} < \frac{3}{2}$$

Why the others are wrong:

• B puts $\sqrt{2}$ below 1.4 — the misremembered benchmark ($\sqrt{2}$ "about 1.2"). In fact $1.4^2 = 1.96 < 2$, so $\sqrt{2} > 1.4$, guaranteed.
• C puts $\sqrt{2}$ above $\frac{3}{2}$ — usually $\sqrt{2}$ confused with $\sqrt{3} \approx 1.732$. Check: $1.5^2 = 2.25 > 2$, so $\sqrt{2} < 1.5$.
• D is internally impossible: it places $\frac{3}{2} = 1.5$ below 1.4. One glance at the two rational values kills it before $\sqrt{2}$ even enters the picture.

Takeaway: $\sqrt{2} \approx 1.414$ is a benchmark worth memorising — and when memory wavers, squaring the boundaries ($1.4^2 = 1.96$, $1.5^2 = 2.25$) pins $\sqrt{2}$ between them with certainty.

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Answer: D

Explanation

Step 1 — Bound $\sqrt{7}$ and $\sqrt{11}$ using perfect squares.

$7 > (2.6)^2 = 6.76$, so $\sqrt{7} > 2.6$.

$3 < \sqrt{11} < 4$ since $9 < 11 < 16$.

Step 2 — Test option D: $3$.

Is $\sqrt{7} < 3 < \sqrt{11}$? Squaring: $7 < 9 < 11$. Yes — $\sqrt{7} < \sqrt{9} = 3 < \sqrt{11}$. ✓

Step 3 — Check the others are outside the range.

A) $\dfrac{5}{2} = 2.5 < 2.6 < \sqrt{7}$ — too small.

B) $\dfrac{10}{3} \approx 3.33$ and $(10/3)^2 = 100/9 \approx 11.11 > 11$, confirming $\frac{10}{3} > \sqrt{11}$ — too large.

C) $3\sqrt{2} = \sqrt{18} > \sqrt{11}$ — too large.

Takeaway: To check whether a value $v$ satisfies $\sqrt{a} < v < \sqrt{b}$, just square everything: check $a < v^2 < b$. No calculator needed.

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Answer: D

Explanation

Step 1 — Rewrite both with the same exponent.

$$2^{30} = (2^3)^{10} = 8^{10}$$

$$3^{20} = (3^2)^{10} = 9^{10}$$

Step 2 — Compare.

Since $8 < 9$ and the exponent 10 is the same (and positive):

$$8^{10} < 9^{10} \implies 2^{30} < 3^{20}$$

Why the others are wrong:

• A is the "bigger exponent wins" instinct — 30 beats 20, surely? But the bases differ, and base strength can outweigh exponent count: rewritten with a common exponent, the fight is $8^{10}$ vs $9^{10}$, and 9 wins.
• B ("equal") mistakes matching exponents for matching values: both sides become something$^{10}$, but the somethings are 8 and 9.
• C is the surrender option — the factoring trick above is precisely the no-calculator method the NBT expects you to find.

Takeaway: To compare $a^m$ with $b^n$, force a common exponent using $\gcd(m,n)$: $2^{30} = (2^3)^{10}$, $3^{20} = (3^2)^{10}$. The comparison collapses to the bases, 8 vs 9.

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Answer: C

Explanation

Square each one (all three are positive):

$$\left(\frac{\sqrt{5}}{2}\right)^2 = 1.25 \qquad \left(\sqrt{\frac{3}{2}}\right)^2 = 1.5 \qquad \left(\frac{\sqrt{3}+1}{2}\right)^2 = \frac{4+2\sqrt3}{4} = 1 + \frac{\sqrt3}{2} \approx 1.866$$

Order them: $1.25 < 1.5 < 1.866$, so $$\frac{\sqrt{5}}{2} < \sqrt{\frac{3}{2}} < \frac{\sqrt{3}+1}{2}$$

Why the others are wrong:

• A and B both drop the cross term when squaring: $(\sqrt3+1)^2$ taken as $3+1=4$, making the third expression square to 1 and look smallest. The full expansion is $4+2\sqrt3$.
• D compares the raw numbers under the roots, forgetting that the denominator in $\frac{\sqrt5}{2}$ quarters its square.

Takeaway: Squaring turns a surd comparison into plain numbers — but $(a+b)^2 = a^2+2ab+b^2$, cross term included, every time.

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Section 3 — Arithmetic: BODMAS and Fractions

BODMAS: Brackets, Orders (powers/roots), Division, Multiplication (left to right), Addition, Subtraction (left to right). Division and multiplication have equal priority — work left to right. Same for addition and subtraction.

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Answer: B

Explanation

Step 1 — Multiplication before addition/subtraction.

$$3 + \underbrace{2 \times 4}_{= 8} - 1 = 3 + 8 - 1$$

Step 2 — Left to right.

$$3 + 8 - 1 = 11 - 1 = 10$$

Why the others are wrong:

• A (19) is pure left-to-right reading: $3 + 2 = 5$, then $5 \times 4 = 20$, then $-1$. The most common BODMAS failure there is.
• D (12) does the multiplication correctly but then adds the 1 instead of subtracting: $3 + 8 + 1$. Sign slips love to hide behind a correctly-handled multiplication.
• C (18) survives no consistent order of operations at all — it is what appears when brackets get imagined mid-calculation. Writing the multiplication's result (8) down before touching the additions leaves it nowhere to come from.

Takeaway: Multiplication before addition and subtraction, always — then strictly left to right among equals. Write the intermediate product down; most BODMAS errors happen in the head, not on the page.

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Answer: D

Explanation

Step 1 — Multiplication first.

$$\frac{2}{3} \times \frac{3}{4} = \frac{2 \times 3}{3 \times 4} = \frac{6}{12} = \frac{1}{2}$$

Step 2 — Addition.

$$\frac{1}{2} + \frac{1}{2} = 1$$

Why the others are wrong:

• C ($\frac{7}{8}$) is the left-to-right error in fraction costume: $\left(\frac{1}{2} + \frac{2}{3}\right) \times \frac{3}{4} = \frac{7}{6} \times \frac{3}{4} = \frac{7}{8}$. BODMAS doesn't switch off because the numbers grew fraction bars.
• B ($\frac{5}{6}$) is $\frac{1}{2} + \frac{1}{3}$ — the product $\frac{6}{12}$ mis-simplified to $\frac{1}{3}$ instead of $\frac{1}{2}$. Simplify by dividing top and bottom by the same number.
• A ($\frac{3}{4}$) merely echoes the last fraction in the question — a no-work option.

Takeaway: Multiplication before addition, even when everything is a fraction. Multiply tops, multiply bottoms, simplify carefully — $\frac{6}{12}$ is $\frac{1}{2}$, and the whole expression collapses to $\frac12 + \frac12 = 1$.

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Answer: D

Explanation

Step 1 — Division first (same priority as multiplication, apply before subtraction).

$$\frac{1}{3} \div \frac{2}{3} = \frac{1}{3} \times \frac{3}{2} = \frac{3}{6} = \frac{1}{2}$$

Step 2 — Subtraction.

$$\frac{3}{4} - \frac{1}{2} = \frac{3}{4} - \frac{2}{4} = \frac{1}{4}$$

Why the others are wrong:

• A ($\frac{5}{12}$) is $\frac{3}{4} - \frac{1}{3}$ — the division vanished entirely; only the two visible fractions got processed.
• B ($\frac{7}{12}$) treats "$\div \frac{2}{3}$" as "$\div 2$": $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$, then $\frac{3}{4} - \frac{1}{6} = \frac{7}{12}$. The reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$, not $\frac{1}{2}$.
• C ($\frac{1}{2}$) is the intermediate result $\frac{1}{3} \div \frac{2}{3}$ reported as the final answer — the subtraction step was never completed.

Takeaway: Dividing by a fraction means multiplying by its reciprocal: $\div \frac{a}{b} = \times \frac{b}{a}$. Do the division first (it outranks subtraction), write its result down, and then finish the expression.

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Answer: A

Explanation

Step 1 — Powers first. $2^3 = 8$

Step 2 — First fraction. $\dfrac{8-1}{8+1} = \dfrac{7}{9}$

Step 3 — Add (both are already over 9): $$\frac{7}{9} + \frac{3}{9} = \frac{10}{9}$$

Why the others are wrong:

• B ($\frac89$) mangled $\frac39$ into $\frac19$. Simplifying must not change the value.
• C ($\frac43$) built the first fraction upside down.
• D ($\frac{7}{12}$) mashed everything into $\frac{8-1}{8+1+3}$, absorbing one fraction's numerator into the other's denominator.

Takeaway: Powers first ($2^3 = 8$, not 6, and not $2^{3-1}$), simplify each fraction on its own, then add over a common denominator. Fractions never merge by proximity.

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Answer: D

Explanation

Top (LCD 6): $\ \frac12 + \frac13 = \frac56$

Bottom (LCD 12): $\ \frac14 + \frac16 = \frac{5}{12}$

Divide — multiply by the reciprocal: $$\frac{5/6}{5/12} = \frac{5}{6} \times \frac{12}{5} = 2$$

Why the others are wrong:

• B (1) cancelled the matching 5s and stopped, ignoring that $\frac16 \div \frac1{12} = 2$ remains.
• C ($\frac32$) paired the fractions wrongly across the bar. Numerator terms stay in the numerator.
• A ($\frac54$) flipped the wrong fraction. Write "$\times \frac{12}{5}$" out before multiplying.

Takeaway: A fraction bar is a division sign. Finish the top, finish the bottom, then multiply by the reciprocal. Expect clean collapses — but earn them step by step.

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Answer: A

Explanation

Step 1 — Split the general term. $$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$$

Step 2 — Write the sum out and watch it collapse. $$\left(1 - \frac12\right) + \left(\frac12 - \frac13\right) + \cdots + \left(\frac19 - \frac1{10}\right)$$

Everything in the middle cancels, leaving first minus last: $$1 - \frac{1}{10} = \frac{9}{10}$$

Why the others are wrong — all three mis-collapse the telescope:

• B ($\frac1{10}$) reports the piece being subtracted, not the result.
• C ($\frac{10}{11}$) ran one term too far.
• D ($\frac89$) stopped one term early.

Takeaway: After any telescoping collapse, check the two endpoints — once the middle cancels, they are the only places left to go wrong.

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Section 4 — Percentages, Ratios, and Proportions

Percentage questions on the NBT almost always involve one of three things: finding a percentage of a quantity, reversing a percentage (finding the original), or combining ratios. Master the equation $\text{part} = \% \times \text{whole}$.

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Answer: A

Explanation

Step 1 — Use the multiplier method.

A 15% increase means multiplying by $1 + 0.15 = 1.15$.

$$240 \times 1.15 = 276$$

Why the others are wrong:

• B (R255) adds fifteen rand, not fifteen per cent — R255 = 240 + 15. Percentages scale with the base: 15 % of R240 is R36.
• C (R288) applies a 20 % increase ($240 \times 1.2$) — a slipped rate.
• D (R264) applies 10 % ($240 \times 1.1$) — the "round rate" reflex overriding the stated 15 %.

Takeaway: A 15 % increase means multiplying by 1.15 (a decrease, by 0.85). One multiplier, one multiplication — and check the size: the increase should be R36, visibly more than B's R15 and less than C's R48.

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Answer: C

Explanation

Step 1 — Set up the equation.

$$0.30 \times x = 45$$

Step 2 — Solve.

$$x = \frac{45}{0.30} = \frac{45}{\frac{3}{10}} = 45 \times \frac{10}{3} = 150$$

Why the others are wrong:

• A (13.5) multiplies instead of divides: $45 \times 0.30$ finds 30 % of 45 — the question already told you 45 is the 30 %.
• B (135) divides by $\frac{1}{3}$, quietly replacing 30 % with "a third". Close, but 30 % is $\frac{3}{10}$, not $\frac{1}{3}$ — the R15 gap between 135 and 150 is the cost of that approximation.
• D (15) takes a third of the part ($45 \div 3$) — dividing in the right spirit but by the wrong reading entirely.

Sanity check the winner: 30 % of 150 = 45 ✓.

Takeaway: "Part = % × whole" rearranges to "whole = part ÷ %". And treat "30 % ≈ a third" as estimation only — the NBT prices the difference into the options.

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Answer: A

Explanation

Step 1 — Find the total number of parts.

$$3 + 5 = 8 \text{ parts}$$

Step 2 — Find one part.

$$\text{One part} = \frac{40}{8} = 5$$

Step 3 — Find the boys.

$$\text{Boys} = 3 \times 5 = 15$$

Why the others are wrong:

• D (25) is the number of girls — the right method answering the wrong question. Underline what is being asked before you finish.
• C (20) splits the class in half, ignoring the ratio altogether.
• B (10) fails the reconstruction test instantly: 10 boys means 30 girls, a ratio of 1 : 3, not 3 : 5.

Takeaway: Add the parts (8), divide the total to get one part (5), multiply by the boys' share (3). Then reconstruct: girls $= 25$, and $15 + 25 = 40$ ✓ with ratio $15{:}25 = 3{:}5$ ✓ — a five-second check that also catches "answered the wrong person" errors like D.

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Answer: B

Explanation

After a 20 % discount the price is 80 % of the original, so:

$$0.80 \times \text{original} = 360 \quad\Rightarrow\quad \text{original} = \frac{360}{0.80} = \text{R}450$$

Check: $450 \times 0.8 = 360$ ✓

Why the others are wrong:

• A (R432) adds 20 % onto the sale price. The 20 % came off the bigger original, so adding 20 % of the smaller price cannot restore it.
• D (R288) discounts again — running the machine forwards instead of backwards.
• C (R480) divides by 0.75, turning "20 % off" into "a quarter off".

Takeaway: The original was multiplied by 0.80, so recover it by dividing by 0.80. Never add the percentage back — and always re-apply the discount to check.

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Answer: B

Explanation

Step 1 — Set up.

Let the three numbers be $2k$, $3k$, $5k$.

$$2k + 3k + 5k = 200 \implies 10k = 200 \implies k = 20$$

Step 2 — Find the largest.

$$5k = 5 \times 20 = 100$$

Why the others are wrong:

• A (60) is the middle number ($3k$) and D (40) is the smallest ($2k$) — both are correct values from the correct working, attached to the wrong word. "Largest" is doing the real work in this question.
• C (80) takes $\frac{2}{5}$ of the total — a fraction built from the ratio's outer digits (2 and 5) instead of from part-over-sum ($\frac{5}{10}$).

Takeaway: With three ratio parts, one variable: $2k + 3k + 5k = 200$ gives $k = 20$, so the parts are 40, 60, 100. Each part's share of the whole is (its digits)/(sum of digits) — the largest is $\frac{5}{10} = 50\%$, and half of 200 is 100 ✓.

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Answer: D

Explanation

Step 1 — Make $y$ match in both ratios. It is 4 in the first and 2 in the second, so scale the second up: $$y:z = 2:5 = 4:10$$

Step 2 — Link them. $x:y:z = 3:4:10$

Step 3 — Read off. $x:z = 3:10$

Why the others are wrong:

• A ($3:5$) takes the outer numbers as they stand. But $x$'s 3 is measured against $y=4$ and $z$'s 5 against $y=2$ — two different yardsticks.
• C ($15:8$) multiplies crosswise, the same yardstick error.
• B ($6:5$) shuffles numbers with no shared $y$ at all.

Takeaway: Scale both ratios until the shared middle term matches, then read off the ends. Or multiply the fractions: $\frac{x}{z} = \frac34 \cdot \frac25 = \frac{3}{10}$.

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Answer: D

Explanation

Step 1 — Find the acid. It never changes when water is added. $$40\% \times 50 = 20 \text{ litres}$$

Step 2 — Set up the new concentration with $w$ litres of water added: $$\frac{20}{50+w} = 0.25$$

Step 3 — Solve. $20 = 12.5 + 0.25w \Rightarrow w = 30$

Check: $\frac{20}{80} = 25\%$ ✓

Why the others are wrong — all three are numbers lifted from the question:

• A (20) is the amount of acid, the one thing that never changes.
• B (25) and C (40) echo the two percentages. Percentages are not volumes.

Takeaway: Anchor on what stays constant (the acid) and let the total change. Then check your answer as a concentration — echo-options cannot survive that.

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Section 5 — Parity: Odd and Even Expressions

A number is even if it is divisible by 2, odd if not. Key rules: - even ± even = even; odd ± odd = even; even ± odd = odd - even × anything = even; odd × odd = odd

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Answer: B

Explanation

Step 1 — Test each with $n = 4$ (even).

A) $4 + 2 = 6$ — even.

B) $4 + 1 = 5$ — odd. ✓

C) $2(4) = 8$ — even.

D) $4^2 = 16$ — even.

Step 2 — Prove B is always odd.

$n$ is even means $n = 2k$ for some integer $k$. Then $n + 1 = 2k + 1$ — always odd.

Takeaway: Even + odd = odd. $n$ even, $1$ odd → $n + 1$ odd. All the other options add or multiply even numbers, keeping the result even.

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Answer: C

Explanation

Step 1 — Apply parity rules.

A) $p + q$: odd + even = odd. Not always even.

B) $p - q$: odd − even = odd. Not always even.

C) $p \cdot q$: odd × even = even. ✓ (Any multiple of an even number is even.)

D) $p^2 + q$: $(p^2)$ — odd² = odd. odd + even = odd. Not always even.

Takeaway: Multiplying by an even number always gives an even result — the factor of 2 in $q$ makes the product divisible by 2, regardless of $p$.

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Answer: C

Explanation

Step 1 — Test each.

A) $m^2 + n$: odd² = odd; odd + odd = even. Always even, not odd.

B) $m + n$: odd + odd = even. Always even.

C) $m \cdot n$: odd × odd = odd. Always odd. ✓

D) $m - n$: odd − odd = even. Always even.

Step 2 — Formal proof for C.

$m = 2a+1$, $n = 2b+1$ for integers $a, b$.

$mn = (2a+1)(2b+1) = 4ab + 2a + 2b + 1 = 2(2ab + a + b) + 1$

This is of the form $2k + 1$ — always odd.

Takeaway: The product of two odd numbers is always odd. Options A, B, and D are all always even when both inputs are odd — a result that surprises many students.

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Answer: B

Explanation

Step 1 — Write as $n$, $n+1$, $n+2$.

Test option D: $p + q = n + (n+1) = 2n + 1$ — always odd. Eliminate.

Test option A: $p + q + r = 3n + 3 = 3(n+1)$. If $n = 2$: $3(3) = 9$ (odd). Not always even. Eliminate.

Test option C: If $n = 2$: $4 + 9 + 16 = 29$ (odd). Eliminate.

Test option B: $pqr = n(n+1)(n+2)$. Among any 3 consecutive integers, at least one is even. An even factor makes the whole product even. ✓

Takeaway: Among any set of consecutive integers, at least one is even. Therefore their product is always even. In fact, $n(n+1)(n+2)$ is always divisible by $3! = 6$.

Why the others are wrong: A is $3q$, which is even only when $q$ is. D is always ODD — two consecutive integers are one even and one odd. C works for $1, 2, 3$ but not for $2, 3, 4$: $4 + 9 + 16 = 29$. One counter-example is enough to kill an "always".

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Answer: C

Explanation

Step 1 — Recall when a sum is odd.

$x + y$ is odd ⟺ exactly one of $x$, $y$ is odd (different parities).

Step 2 — Apply to $a + b$.

$a + b$ is odd → $a$ and $b$ have different parities. So one is even and the other is odd.

Step 3 — Confirm with $a - b$.

$a - b = a + (-b)$. The parity of $-b$ equals the parity of $b$. So $a - b$ odd means $a$ and $b$ also have different parities — consistent.

Step 4 — Check option D.

If $a$ even, $b$ odd: $a^2$ even, $b^2$ odd, $a^2 + b^2$ = odd. So D is false (it would be odd, not even).

Takeaway: Both $a + b$ and $a - b$ being odd simultaneously forces different parities. Check option D carefully — $a^2 + b^2$ being "even" sounds plausible but is actually false when parities differ.

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Section 6 — Last Digit of Large Powers

The units digit of $n^k$ follows a repeating cycle. To find the units digit of $n^k$: 1. Find the cycle of units digits for powers of $n$ (cycles of length 1, 2, or 4). 2. Find $k \bmod \text{cycle length}$. 3. The remainder gives the position in the cycle (remainder 0 means the last position).

Why the others are wrong: A and B both give an EVEN sum — even + even and odd + odd. D fails for the pair that does work: an odd square plus an even square is odd.

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Answer: A

Explanation

Step 1 — Find the cycle of units digits.

$7^1 = 7$, $7^2 = 49$, $7^3 = 343$, $7^4 = 2401$ — so the cycle is 7, 9, 3, 1, length 4.

Step 2 — Locate $7^{20}$. $20 \div 4 = 5$ remainder 0, which means the last position: digit 1.

Why the others are wrong: C (7), D (9) and B (3) are positions 1, 2 and 3 of the same cycle. The whole question turns on one convention: remainder 0 means the last position, because $7^4, 7^8, 7^{12}$ all end where $7^4$ ends.

Takeaway: Powers of 7 (and 3) cycle every 4. Any exponent divisible by 4 lands on 1. Anchor it with $7^4 = 2401$ whenever you doubt the convention.

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Answer: B

Explanation

Step 1 — Find the cycle.

$3^1 = 3$ → 3; $3^2 = 9$ → 9; $3^3 = 27$ → 7; $3^4 = 81$ → 1.

Cycle: 3, 9, 7, 1 (length 4).

Step 2 — Find position of $3^{100}$.

$$100 \div 4 = 25 \text{ remainder } 0$$

Remainder 0 → last position in cycle → units digit 1.

Why the others are wrong: A (3), D (9) and C (7) are cycle positions 1, 2 and 3 — the full menu of remainder-handling mistakes. The most seductive is A: "powers of 3 end in 3." They do — once every four exponents. 100 is a multiple of 4, so $3^{100}$ sits at the cycle's end, not its start.

Takeaway: Powers of 3 and 7 share cycle length 4 and both land on 1 whenever the exponent is a multiple of 4. The bulletproof form: $3^{100} = (3^4)^{25} = 81^{25}$, and a number ending in 1 raised to any power still ends in 1 ✓.

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Answer: A

Explanation

Both cycles have length 4, so one division serves both: $103 = 4 \times 25 + 3$, position 3.

• Cycle of 2 is 2, 4, 8, 6 → position 3 gives 8
• Cycle of 3 is 3, 9, 7, 1 → position 3 gives 7

Add: $8 + 7 = 15$, so the units digit is 5.

Why the others are wrong:

• C (3) read both cycles one position early.
• D (7) treated 103 as a multiple of 4, landing on both cycle ends.
• B (1) read one cycle right and reset the other to its start. Same exponent means the same position in both.

Takeaway: Find each units digit, add them, then take the units digit of that little sum. Compute the remainder once when the cycles share a length.

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Answer: B

Explanation

The cycle of 7 has length 4, so you need the top exponent $7^7$ modulo 4.

Since $7 \equiv 3 \pmod 4$ and $7^2 \equiv 1 \pmod 4$: $$7^7 = (7^2)^3 \times 7 \equiv 1 \times 3 = 3 \pmod 4$$

Position 3 in the cycle 7, 9, 3, 1 gives units digit 3.

Why the others are wrong:

• C (7) flattens the tower into $(7^7)^7 = 7^{49}$. A tower is read top-down: $7^{(7^7)}$ is a vastly bigger number.
• A (1) and D (9) come from garbled arithmetic on the top exponent. Do the small computation honestly.

Takeaway: A tower $a^{b^c}$ means $a^{(b^c)}$, never $(a^b)^c$. Reduce the top exponent modulo the cycle length, then read the cycle — two tiny calculations replace an astronomical number.

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Section 7 — Counting and Permutations

Fundamental rule: If event A can happen in $m$ ways and event B can happen in $n$ ways, then A then B can happen in $m \times n$ ways.

Permutation (order matters, no repetition): $n \times (n-1) \times \cdots$ for as many slots as needed.

Combination (order does not matter): $\dbinom{n}{r} = \dfrac{n!}{r!\,(n-r)!}$.

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Answer: C

Explanation

Step 1 — Count the choices for each position.

Position 1: 4 books to choose from. Position 2: 3 remaining books. Position 3: 2 remaining books. Position 4: 1 remaining book.

Step 2 — Multiply.

$$4 \times 3 \times 2 \times 1 = 4! = 24$$

Why the others are wrong:

• D (256) is $4^4$ — each shelf position given all four books, as if one book could stand in several places at once. Powers count with repetition; arrangements shrink the pool each time.
• B (16) is $4^2$ — a half-remembered "multiply the 4 by something" instinct.
• A (4) counts the books, not the arrangements — the no-work option.

Takeaway: Arranging $n$ distinct objects in a row gives $n! = n \times (n-1) \times \cdots \times 1$; here $4! = 24$. The shrinking factors (4, 3, 2, 1) are the signature of "no repetition" — if your factors don't shrink, you've allowed clones.

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Answer: A

Explanation

Step 1 — Count choices for each digit position.

Hundreds digit: 5 choices. Tens digit: 4 remaining choices. Units digit: 3 remaining choices.

Step 2 — Multiply.

$$5 \times 4 \times 3 = 60$$

Why the others are wrong:

• C (125) is $5^3$ — repetition allowed, five choices for every slot. The words "no digit may be repeated" are exactly what turns $5 \times 5 \times 5$ into $5 \times 4 \times 3$.
• D (120) is $5!$ — arranging all five digits, but the number only has three positions.
• B (15) is $5 \times 3$ — the two visible numbers multiplied, with no counting model behind it.

Takeaway: This is the permutation $P(5,3) = 5 \times 4 \times 3 = 60$: one factor per position, shrinking by one each time. Match the number of factors to the number of slots, and let "no repetition" shrink them.

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Answer: B

Explanation

Order matters — the two roles are different. President: 5 choices. Vice-president: 4 left.

$$5 \times 4 = 20$$

Why the others are wrong:

• A (10) is the committee count $\binom52$ — right only if the roles were interchangeable. President-Thabo/VP-Amina differs from the reverse, so each pair counts twice.
• C (25) is $5^2$, letting one student hold both offices.
• D (15) adds the choices. Successive choices always multiply.

Takeaway: Named roles → order matters → permutation. Unnamed committee → order does not → combination. The factor of 2 between 20 and 10 is exactly the two orderings of each pair.

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Answer: B

Explanation

Step 1 — Count the letters.

L, E, V, E, L — five letters: L appears twice, E appears twice, V appears once.

Step 2 — Use the formula for arrangements with repeats.

$$\frac{5!}{2! \times 2!} = \frac{120}{2 \times 2} = \frac{120}{4} = 30$$

Why the others are wrong:

• A (120) is $5!$ — treating all five letters as distinct, as if the two L's were different colours. Swapping identical letters produces the same word, so 120 counts every arrangement four times.
• C (60) divides by only one $2!$ — the repeated L's handled, the repeated E's forgotten. Every repeated letter needs its own factorial.
• D (20) divides by $3! = 6$ — lumping the four repeated letters (L, L, E, E) into one group of "repeats". The divisor is per-letter: $2! \times 2! = 4$, not $3!$.

Takeaway: Arrangements with repeats: $\frac{n!}{k_1!\,k_2!\cdots}$ — one factorial for each letter counted separately. LEVEL: $\frac{5!}{2!\,2!} = 30$. Tally the letters before touching the formula.

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Answer: D

Explanation

Reds: choose 2 from 4 → $\binom{4}{2} = 6$

Non-red: choose 1 from the other 5 (3 blue + 2 green) → $\binom{5}{1} = 5$

Multiply: $6 \times 5 = 30$

Why the others are wrong — each corrupts one factor:

• A (20) used $4 \times 5$: choosing a pair of reds is not the same as choosing one.
• B (24) used $6 \times 4$: the non-red pool is 5, greens included.
• C (36) used $6 \times 6$ — the first factor recycled.

Takeaway: For "exactly $k$ of type X": $\binom{|X|}{k} \times \binom{\text{others}}{\text{rest}}$. Label each factor as you write it — every wrong option here is a mislabelled factor.

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Answer: C

Explanation

Step 1 — Fill the restricted slot first. Even means the units digit is 2, 4 or 6 — 3 choices.

Step 2 — Fill the rest from the 4 digits left: $4 \times 3 \times 2 = 24$

Step 3 — Multiply: $3 \times 24 = 72$

Why the others are wrong:

• A (48) counted only 2 even digits, overlooking the 6.
• D (96) counted 4, sweeping an odd digit into the even pile.
• B (60) is a three-slot permutation that ignored the structure entirely.

Takeaway: Fill the most restricted slot first, then the free ones. When a restriction names a property, list the elements that actually have it — miscounting that set is the whole game.

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Section 8 — Mean, Median, and Weighted Average

Mean = sum of all values ÷ number of values. Median = middle value when data is sorted (for an even count, average the two middle values). Weighted mean = $\dfrac{\sum (\text{weight} \times \text{value})}{\sum \text{weights}}$.

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Answer: A

Explanation

Step 1 — Sum the values.

$$60 + 72 + 68 + 75 + 80 = 355$$

Step 2 — Divide by 5.

$$\text{Mean} = \frac{355}{5} = 71$$

Why the others are wrong: C (72) and D (75) are values sitting in the data set — echo bait for anyone who "eyeballs the middle". B (70) is the round number nearest the truth, the reward for estimating the sum instead of computing it. None of the three equals $355 \div 5$, and that division is the only authority.

Takeaway: Build the sum in a running chain — $60 + 72 = 132$; $+68 = 200$; $+75 = 275$; $+80 = 355$ — then divide once: $355 \div 5 = 71$. Chained addition with a checkpoint at each step is how you beat "clean-looking" wrong options without a calculator.

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Answer: C

Explanation

Step 1 — Sort the data in ascending order.

$$1, 3, 4, 6, 7, 8, 9$$

Step 2 — Find the middle value. There are 7 values; the median is the $\frac{7+1}{2} = 4$th value.

$$1, 3, 4, \underbrace{6}_{\text{4th}}, 7, 8, 9 \implies \text{median} = 6$$

Why the others are wrong:

• B (5) is the midrange $\frac{1+9}{2}$ — halfway between min and max. The median is the middle member, not the middle of the endpoints.
• D (7) echoes the count of values (seven data points) — and also happens to be a datum, doubling its bait value.
• A (4) confuses the middle position (4th) with the value 4 — position and value are different currencies.

Takeaway: Sort first, always: $1,3,4,\mathbf{6},7,8,9$ → the 4th of 7 values is the median, 6. Every wrong option here is a different "middle-ish" number — only sorting tells you which middle the median actually is.

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Answer: D

Explanation

Step 1 — Find the required total.

$$\text{Total} = \text{mean} \times n = 15 \times 4 = 60$$

Step 2 — Subtract the known values.

$$60 - 12 - 18 - 14 = 16$$

Why the others are wrong:

• B (15) echoes the mean itself — "if the mean is 15, the missing number is probably 15" is a guess, not a computation. (It would only be right if the other three averaged exactly 15; they average $\frac{44}{3} \approx 14.7$.)
• A (14) echoes a number already in the data.
• C (17) is a one-off arithmetic slip in $60 - 44$. The subtraction chain $60 - 12 - 18 - 14$ done one number at a time ($48, 30, 16$) leaves no room for it.

Takeaway: "Mean × count = total" is the master move: target total 60, known sum 44, missing value 16. Verify: $\frac{12+18+14+16}{4} = \frac{60}{4} = 15$ ✓.

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