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 $3! = 6$.

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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).

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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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Answer: B

Explanation

Required total for a mean of 75 over 4 tests: $75 \times 4 = 300$

Current total: $70 + 82 + 65 = 217$

Fourth score: $300 - 217 = 83$

Why the others are wrong: C (80) is the round guess; A (78) and D (85) bracket the truth to punish mental arithmetic. Only 83 passes the check: $\frac{300}{4} = 75$ ✓

Takeaway: Required score = (target mean × tests) − (current total). Or think in deficits: the three scores are 8 short of $3 \times 75$, so the fourth must be 8 above 75. Same answer, built-in check.

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Answer: B

Explanation

Girls: $30 - 18 = 12$

Totals: boys $18 \times 72 = 1296$; girls $12 \times 80 = 960$; together 2256

Class average: $\dfrac{2256}{30} = 75.2$

Why the others are wrong:

• C (76.0) is the plain average of 72 and 80 — valid only if the groups were equal size.
• D (77.0) swaps the weights, giving the smaller group the bigger pull. More boys must drag the average below 76, so anything above 76 is impossible.
• A (74.4) used a 30 % girl-weight instead of 40 %.

Takeaway: Predict the side before computing: the bigger group pulls the average toward itself. That one prediction eliminates half the options.

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Answer: D

Explanation

The data (already sorted): 6, 8, 10, 14, 22 — five values.

Median: the 3rd value, 10

Mean: $\dfrac{60}{5} = 12$

Difference: $12 - 10 = 2$

Why the others are wrong:

• A (0) assumes mean = median. True for symmetric data, but the lone 22 drags the mean up while the median stays put.
• B (4) mis-totalled the sum.
• C (6) misread the diagram, treating leaves as whole values. The stems carry the tens.

Takeaway: Read stem-and-leaf as (stem × 10) + leaf, and the data arrive pre-sorted. Mean above median is the fingerprint of right-skewed data — here caused by that 22.

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Section 9 — Compound Probability

Independent events: $P(A \cap B) = P(A) \times P(B)$.

Union: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

Conditional probability: $P(A | B) = \dfrac{P(A \cap B)}{P(B)}$.

Without replacement: the second draw depends on the first — adjust the denominator.

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Answer: A

Explanation

Step 1 — List sample space.

HH, HT, TH, TT — four equally likely outcomes.

Step 2 — Count favourable outcomes.

Only HH. $P(\text{HH}) = \dfrac{1}{4}$.

Alternatively: tosses are independent, so $P(\text{H}) \times P(\text{H}) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$.

Why the others are wrong:

• B ($\frac12$) is the single-toss probability doing double duty — two tosses require multiplying, not reusing.
• C ($\frac13$) is the seductive "three outcomes: no heads, one head, two heads — each equally likely" model. The outcomes exist but their probabilities differ ($\frac14, \frac12, \frac14$): "one head" happens two ways (HT, TH).
• D ($\frac34$) is $P(\text{at least one head})$ — the complement of the wanted event's opposite, a mis-aimed complement.

Takeaway: Independent events multiply: $\frac12 \times \frac12 = \frac14$. When in doubt, write the sample space (HH, HT, TH, TT) — four equally likely outcomes settle every one of these disputes at a glance.

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Answer: B

Explanation

Step 1 — Find each probability.

$P(\text{red}) = \dfrac{3}{5}$, $P(\text{blue}) = \dfrac{2}{5}$.

Since the marble is replaced, the second draw is independent of the first.

Step 2 — Multiply.

$$P(\text{red then blue}) = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}$$

Why the others are wrong:

• C ($\frac{3}{10}$) is the without-replacement answer: $\frac{3}{5} \times \frac{2}{4}$. The word "replaced" is the hinge of the whole question — the bag resets, so the second denominator stays 5.
• D ($\frac{2}{5}$) is $P(\text{blue})$ alone — the first draw never entered the calculation.
• A ($\frac{1}{5}$) is a near-miss simplification of the right idea ($\frac{6}{25}$ carelessly reduced). $\frac{6}{25}$ is already in lowest terms — 6 and 25 share no factor.

Takeaway: "With replacement" ⇒ independent draws ⇒ multiply with the same denominators: $\frac35 \times \frac25 = \frac{6}{25}$. Before multiplying anything, circle whether the draw is with or without replacement — it changes exactly one number, and the examiner always offers both versions.

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Answer: C

Explanation

Step 1 — Total outcomes. $6 \times 6 = 36$.

Step 2 — Pairs that sum to 7.

$(1,6),\ (2,5),\ (3,4),\ (4,3),\ (5,2),\ (6,1)$ — exactly 6 pairs.

Step 3 — Probability.

$$P(\text{sum} = 7) = \frac{6}{36} = \frac{1}{6}$$

Why the others are wrong:

• D ($\frac{1}{12} = \frac{3}{36}$) counts unordered pairs — {1,6}, {2,5}, {3,4} — forgetting that (1,6) and (6,1) are different rolls. The dice are distinct objects even if they look identical.
• A ($\frac{5}{36}$) is the count for the neighbouring sums: both 6 and 8 can be made exactly 5 ways. One pair short of 7's six.
• B ($\frac{7}{36}$) puts the 7 from the question straight into the numerator — a surface-number trap.

Takeaway: With two dice, always count ordered pairs out of 36. Sum 7 is the unique total achievable 6 ways — one for every value of the first die — which is why $\frac{6}{36} = \frac16$ and why 7 is the most likely sum of all.

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Answer: C

Explanation

Addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Independence gives the intersection: $0.4 \times 0.3 = 0.12$

$$P(A \cup B) = 0.4 + 0.3 - 0.12 = 0.58$$

Why the others are wrong:

• B (0.70) adds and stops, double-counting the overlap. That only works for mutually exclusive events — and independent events are never mutually exclusive.
• A (0.12) reports the intersection as the union.
• D (0.28) is $P(A) - P(A\cap B)$, the probability of "A but not B".

Takeaway: Label every number you compute (∩, ∪, "only A") — three of the four options here are just each other's intermediate steps.

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Answer: C

Explanation

Step 1 — Find $P(\text{sport or music})$.

$$P(S \cup M) = P(S) + P(M) - P(S \cap M) = 0.60 + 0.40 - 0.25 = 0.75$$

Step 2 — Find $P(\text{neither})$.

$$P(\text{neither}) = 1 - P(S \cup M) = 1 - 0.75 = 0.25$$

Why the others are wrong: every wrong option is a different region of the Venn diagram, correctly computed and wrongly labelled —

• A (0.15) is "music only" ($0.40 - 0.25$).
• D (0.35) is "sport only" ($0.60 - 0.25$).
• B (0.20) is the bare difference between the two percentages ($0.60 - 0.40$) — a region of nothing at all.

Check the full picture: sport-only 0.35 + music-only 0.15 + both 0.25 + neither 0.25 = 1.00 ✓.

Takeaway: $P(\text{neither}) = 1 - P(A \cup B)$, with the union built as $0.60 + 0.40 - 0.25 = 0.75$. Sketch the Venn regions and make them sum to 1 — that closing check catches region-mislabeling instantly.

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Section 10 — Bounds and Estimation

When a measurement is rounded to a given place, the true value lies in a symmetric interval around the rounded value. When a measurement is truncated (digits simply dropped), the true value lies in a one-sided interval.

Rounded to nearest unit $u$: $\text{stated} - \frac{u}{2} \leq x < \text{stated} + \frac{u}{2}$

Truncated: $\text{stated} \leq x < \text{stated} + u$

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Answer: A

Explanation

Lower bound: the smallest value that rounds up to 80 is 75 — and 75 itself rounds to 80, so it is included.

Upper bound: anything below 85 rounds to 80, but 85 rounds to 90 — so 85 is excluded.

$$75 \leq x < 85$$

Why the others are wrong:

• B used the full unit (±10) instead of half — twice too wide. It admits 85, which rounds to 90.
• C flips both endpoint conventions — excluding 75 and including 85, backwards on both.
• D is one-sided, which is truncation (see Q55), not rounding.

Takeaway: "Rounded to nearest $u$" gives stated $\pm u/2$, lower inclusive and upper exclusive. Test your endpoints — "does 75 round to 80? does 85?" — and the conventions sort themselves out.

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Answer: C

Explanation

Step 1 — Identify the rounding unit. Rounded to 1 d.p. means rounded to the nearest 0.1.

Half-unit = $0.05$.

Step 2 — Upper bound = $7.4 + 0.05 = 7.45$ cm.

True value: $7.35 \leq x < 7.45$.

Why the others are wrong:

• A (7.40) is the stated measurement itself — "upper bound" asks how far the true value could reach, not what was written down.
• B (7.44) thinks digit-by-digit: "7.44 still shows 7.4". True, but so does 7.449, 7.4499… The supremum of "rounds to 7.4" is 7.45, approached but never reached.
• D (7.49) is the truncation instinct — anything 7.4-something. But 7.47, say, rounds to 7.5. Rounding halves the window that truncation allows.

Takeaway: For rounding to 1 d.p., the bound is stated ± 0.05: here $7.35 \leq x < 7.45$. The upper bound is exclusive — exactly 7.45 would round up to 7.5. State bounds as half the rounding unit, never by fiddling with the next digit.

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Answer: C

Explanation

Lower bounds (half a unit down): $l = 11.5$ cm and $w = 7.5$ cm.

Minimum area = min length × min width: $$11.5 \times 7.5 = 86.25 \text{ cm}^2$$

Why the others are wrong:

• D (96) is $12 \times 8$ — the uncertainty ignored entirely.
• B (90) took only one dimension to its minimum. You need both at once.
• A (84) knocked a whole centimetre off. Bounds move by half a unit.

Takeaway: Extreme-value questions push every measurement to the extreme that serves the goal — and bounds move by half the rounding unit, not a full one.

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Answer: C

Explanation

Step 1 — Maximise $a - b$: use maximum $a$ and minimum $b$.

$$a_{\max} = 5.0 + 0.1 = 5.1$$

$$b_{\min} = 3.0 - 0.1 = 2.9$$

Step 2 — Calculate.

$$a_{\max} - b_{\min} = 5.1 - 2.9 = 2.2$$

Why the others are wrong:

• A (2.0) subtracts the nominal values ($5.0 - 3.0$) — the uncertainty never entered the calculation.
• B (2.1) applies the error to only one of the two measurements. Both quantities can sit at their unfavourable extremes at the same time, and "maximum possible" means assuming they do.
• D (2.3) over-applies the errors ($\pm 0.1$ counted three times) — enthusiasm without bookkeeping.

Takeaway: Max of $a - b$ = (max $a$) − (min $b$); min of $a - b$ = (min $a$) − (max $b$). Under subtraction the error margins add: total swing here is $\pm 0.2$ around 2.0. Write each extreme down explicitly (5.1 and 2.9) before subtracting.

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Answer: B

Explanation

Truncating to 2 d.p. means digits after the second place are simply dropped, not rounded. So anything reading $3.47\ldots$ truncates to 3.47.

• Lower: $3.47000\ldots$ truncates to 3.47, so 3.47 is included.
• Upper: $3.4799\ldots$ still truncates to 3.47, but 3.48 truncates to itself — so 3.48 is excluded.

$$3.47 \leq x < 3.48$$

Why the others are wrong:

• A and D are the rounding intervals — the reflex when "truncated" goes unread. Truncation never reaches below the stated value: 3.468 truncates to 3.46.
• C closes the wrong end. 3.48 truncates to 3.48, so it must be excluded.

Takeaway: Rounding gives a symmetric interval (± half a unit). Truncation gives a one-sided one, $[\text{stated},\ \text{stated}+\text{unit})$. That one word moves the whole interval — read the verb first.

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Section 11 — Scientific Notation, Rates and Speed

Scientific notation writes a number as $a \times 10^n$ where $1 \leq a < 10$ and $n$ is an integer. The exponent counts how many places the decimal point moves: positive for large numbers, negative for small ones.

$$4\,500\,000 = 4.5 \times 10^6 \qquad 0.00045 = 4.5 \times 10^{-4}$$

To multiply or divide, handle the coefficients and the powers of ten separately, using the exponent laws:

$$(a \times 10^m)(b \times 10^n) = (a \times b) \times 10^{m+n} \qquad \frac{a \times 10^m}{b \times 10^n} = \frac{a}{b} \times 10^{m-n}$$

If the coefficient ends up outside the range $1 \leq a < 10$, renormalise: $37 \times 10^4 = 3.7 \times 10^5$.

Speed, distance and time are one relationship read three ways:

$$\text{distance} = \text{speed} \times \text{time} \qquad \text{speed} = \frac{\text{distance}}{\text{time}} \qquad \text{time} = \frac{\text{distance}}{\text{speed}}$$

Two rules prevent almost every error here. First, units must agree — 30 minutes is $0.5$ hours, not $30$, when the speed is in km/h. Second, average speed is always total distance over total time, never the average of the speeds.

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Answer: A

Explanation

Move the decimal point until one non-zero digit sits in front of it:

$$0.00045 \rightarrow 4.5$$

That took 4 places to the right, which makes the coefficient bigger — so the power of ten must shrink it back. The exponent is negative:

$$0.00045 = 4.5 \times 10^{-4}$$

Check by reversing: $4.5 \times 10^{-4}$ moves the point 4 places left, back to $0.00045$ ✓

Why the others are wrong:

• B has the right digits but a positive exponent, giving $45\,000$.
• C equals the answer numerically, but $45$ is not between 1 and 10 — so it is not scientific notation.
• D counted the three zeros and forgot that the 4 occupies a place too.

Takeaway: Count the places the point moves and let the direction set the sign — left positive, right negative. Then check the coefficient really is between 1 and 10.

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Answer: C

Explanation

Split the problem in two — coefficients together, powers of ten together:

$$(3 \times 2) \times (10^{5} \times 10^{-8}) = 6 \times 10^{5 + (-8)} = 6 \times 10^{-3}$$

When you multiply powers of the same base you add the exponents. Here $5 + (-8) = -3$.

The coefficient 6 already lies between 1 and 10, so no renormalising is needed.

Why the others are wrong:

• A multiplied the exponents ($5 \times -8 = -40$). Multiplying the powers means adding the exponents.
• B subtracted instead of adding: $5 - (-8) = 13$. Subtraction is the rule for division.
• D added the coefficients ($3+2$) and the exponents. The coefficients get multiplied, like any other numbers.

Takeaway: Multiply the coefficients, add the exponents; divide the coefficients, subtract the exponents. Then check the coefficient is still in $[1, 10)$ and fix it if not.

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Answer: B

Explanation

Convert the time to hours first, so it matches the km/h in the speed. $30$ minutes is half an hour:

$$2 \text{ h } 30 \text{ min} = 2.5 \text{ h}$$

Then:

$$\text{distance} = \text{speed} \times \text{time} = 60 \times 2.5 = 150 \text{ km}$$

Sense-check: 2 hours alone covers 120 km, and the extra half-hour adds 30 km. $120 + 30 = 150$ ✓

Why the others are wrong:

• A used only the whole 2 hours and discarded the 30 minutes.
• C treated 30 minutes as a whole extra hour, computing $60 \times 3$.
• D read the time as "2.30" and multiplied by 25 — a units slip. Minutes are sixtieths, not hundredths, so $30$ min $= 0.5$ h, never $0.30$ h.

Takeaway: Convert minutes to hours by dividing by 60 before you touch a km/h speed. Half an hour is $0.5$, twenty minutes is $\frac{1}{3}$, forty-five minutes is $0.75$.

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Answer: B

Explanation

Average speed is total distance ÷ total time — never the average of the two speeds, because more time is spent at the slower one.

Distance: $120 + 120 = 240$ km

Time: $\dfrac{120}{60} = 2$ h out, $\dfrac{120}{40} = 3$ h back, so $5$ h in total.

$$\text{average speed} = \frac{240}{5} = 48 \text{ km/h}$$

48 sits below the midpoint 50, as expected: 3 of the 5 hours are at the slower speed.

Why the others are wrong:

• A ($50$) is the trap — the plain mean $\frac{60+40}{2}$, correct only for equal times, not equal distances.
• C ($52$) is pulled above the midpoint, as if more time were spent at the faster speed.
• D ($45$) over-corrects downward with no calculation behind it.

Takeaway: For equal distances at two speeds, the average is always below the plain mean, and the distance cancels out entirely. Which side of the midpoint your answer falls on is an instant sanity check.

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Exam-Bank Extras — Question Types Confirmed in Recent Papers

The NBT MAT reuses question types from a stable bank year after year. The question below is modelled directly on a type repeatedly confirmed in recent papers and not yet represented in this chapter. The answer has been independently machine-verified.

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Answer: B

Explanation

Forty squares without a calculator is impossible — which is the signal that a structure is hiding. Pair the terms and use $a^2-b^2 = (a-b)(a+b)$:

$$40^2 - 39^2 = (1)(79) = 40 + 39$$ $$38^2 - 37^2 = (1)(75) = 38 + 37$$

Each pair differs by 1, so every difference of squares collapses to a plain sum. The whole monster is just:

$$40 + 39 + 38 + \cdots + 1 = \frac{40 \times 41}{2} = 820$$

Why the others are wrong:

• D ($1640$) is $40 \times 41$ — forgot to halve.
• C ($800$) estimated $40^2/2$ instead of doing the collapse.
• A ($780$) summed only to 39, dropping the 40.

Takeaway: $a^2-(a-1)^2 = a + (a-1)$, so an alternating sum of consecutive squares always collapses into a simple arithmetic series. On a no-calculator paper, "impossible arithmetic" is a promise that a shortcut exists.

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Mixed Practice — Chapter 2

These 30 questions mix all Number Sense topics. Work through them without looking back at the sections. Check your answers with the key on the following page — detailed explanations are not given here, so use the earlier sections to diagnose any errors.

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Answer: C

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Answer: D

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Answer: B

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Answer: C

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Answer: B

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Answer: C

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Answer: D

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Answer: D

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Answer: A

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Answer: B

---

Answer: A

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Answer: C

---

Answer: D

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Answer: A

---

Answer: A

---

Answer: A

---

Answer: A

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Answer: C

---

Answer: C

---

Answer: C

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Answer: D

---

Answer: D

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Answer: B

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Answer: C

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Answer: B

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Answer: D

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Answer: B

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Answer: B

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Answer: A

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Answer: D

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