Year 8 Mathematics Practice

Irrational numbers

Fluency · Classify and identify

1. Classify: $7$ . (show answer)

Answer
Rational ( $7 = \dfrac{7}{1}$ , an integer).
2. Classify: $0.5$ . (show answer)

Answer
Rational ( $0.5 = \dfrac{1}{2}$ , a terminating decimal).
3. Classify: $\sqrt{9}$ . (show answer)

Answer
Rational ( $\sqrt{9} = 3$ ).
4. Classify as rational or irrational: $\dfrac{4}{9}$ . (show answer)

Answer
Rational (already a fraction of integers).
5. Classify: $\sqrt{25}$ . (show answer)

Answer
Rational ( $\sqrt{25} = 5$ ).
6. Classify: $\sqrt{7}$ . (show answer)

Answer
Irrational ( $7$ is not a perfect square, so $\sqrt{7}$ is non-terminating and non-repeating).
7. Classify: $0.\overline{6}$ . (show answer)

Answer
Rational ( $0.\overline{6} = \dfrac{2}{3}$ ; repeating decimals are rational).
8. Classify: $\pi$ . (show answer)

Answer
Irrational ( $\pi$ is non-terminating and non-repeating).
9. Classify: $-3$ . (show answer)

Answer
Rational ( $-3 = \dfrac{-3}{1}$ ).
10. Classify: $\sqrt{0.25}$ . (show answer)

Answer
Rational ( $\sqrt{0.25} = 0.5 = \dfrac{1}{2}$ ).
11. Classify: $1.41421356\ldots$ (the digits do not repeat). (show answer)

Answer
Irrational (the digits do not terminate and do not repeat; in fact this is $\sqrt{2}$ ).

Fluency · Estimate a root

1. Between which two consecutive whole numbers does $\sqrt{20}$ lie? (show answer)

Answer
Between $4$ and $5$ . Method: $4^2 = 16$ , $5^2 = 25$ .
2. Between which two consecutive whole numbers does $\sqrt{90}$ lie? (show answer)

Answer
Between $9$ and $10$ .
3. Between which two tenths does $\sqrt{20}$ lie? Use trial with $4.4^2$ and $4.5^2$ . (show answer)

Answer
Between $4.4$ and $4.5$ . Method: $4.4^2 = 19.36$ , $4.5^2 = 20.25$ .
4. Use trial to estimate $\sqrt{75}$ to one decimal place. (show answer)

Answer
$\sqrt{75} \approx 8.7$ . Method: $8.6^2 = 73.96$ , $8.7^2 = 75.69$ .
5. Use trial to estimate $\sqrt{150}$ to one decimal place. (show answer)

Answer
$\sqrt{150} \approx 12.2$ . Method: $12.2^2 = 148.84$ , $12.3^2 = 151.29$ .

Reasoning · Explain and spot the mistake

1. Kim says " $0.9999\ldots$ is irrational because it has infinite digits". Is Kim correct? Explain. (show answer)

Answer
Kim is wrong. $0.\overline{9} = 1$ exactly, and $1$ is rational. Infinite digits do not make a number irrational - only non-repeating infinite digits do.
2. Lee writes $\pi = \dfrac{22}{7}$ . Is this correct, or an approximation? Explain the difference. (show answer)

Answer
Not correct. $\dfrac{22}{7}$ is a rational approximation to $\pi$ . The true $\pi$ never terminates or repeats; $\dfrac{22}{7}$ does repeat ( $3.\overline{142857}$ ) and so it cannot equal $\pi$ .
3. Is the product of two irrational numbers always irrational? Give an example that supports your answer. (show answer)

Answer
Not always irrational. Example: $\sqrt{2} \times \sqrt{2} = 2$ , which is rational. So the product of two irrationals can be rational.
4. Aisha claims $\sqrt{49} + \sqrt{2}$ is rational because $\sqrt{49}$ is rational. Is she correct? Explain. (show answer)

Answer
Wrong. $\sqrt{49} + \sqrt{2} = 7 + \sqrt{2}$ , and adding a rational to an irrational gives an irrational.

Problem-solving · Real contexts

1. A square has side $1$ m. Find the length of its diagonal. Give the exact value and then an estimate to 2 decimal places. (show answer)

Answer
Exact: $\sqrt{2}$ m. Approximate: $1.41$ m.
2. A pizza has diameter $30$ cm. Find its circumference exactly (in terms of $\pi$ ) and to the nearest cm. (show answer)

Answer
Exact: $30\pi$ cm. Approximate: $94$ cm.
3. Explain why the area of any circle cannot be exactly a rational number (when the radius is rational). (show answer)

Answer
If $r$ is rational, $r^2$ is rational; area $= \pi r^2$ is rational times irrational ( $\pi$ ), which is irrational.
4. A right-angled triangle has legs $5$ cm and $12$ cm. Find the exact length of the hypotenuse, then classify it as rational or irrational. (show answer)

Answer
Exact: $13$ cm. Method: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$ . Classification: rational (a Pythagorean triple - $13$ is a whole number).

Exponent laws

Fluency · Apply one law at a time

1. Simplify $2^3 \times 2^6$ . (show answer)

Answer
$2^9 = 512$
2. Simplify $a^5 \times a^2$ . (show answer)

Answer
$a^7$
3. Simplify $x^{10} \div x^4$ . (show answer)

Answer
$x^6$
4. Simplify $\dfrac{m^8}{m^3}$ . (show answer)

Answer
$m^5$
5. Simplify $(3^2)^3$ . (show answer)

Answer
$3^6 = 729$
6. Simplify $(p^4)^5$ . (show answer)

Answer
$p^{20}$
7. Evaluate $7^0$ . (show answer)

Answer
$1$
8. Evaluate $(-3)^0$ . (show answer)

Answer
$1$
9. Simplify $y^{12} \div y^{12}$ . (show answer)

Answer
$1$ (by quotient rule the exponent is $0$ )
10. Simplify $(x y^2)^3$ . (Hint: each factor raised to the power.) (show answer)

Answer
$x^3 y^6$ . Method: raise each factor to the power $3$ .

Fluency · Combine the laws

1. Simplify $\dfrac{a^5 \times a^3}{a^4}$ . (show answer)

Answer
$a^4$ . Method: top $a^8$ ; $a^8 \div a^4 = a^4$ .
2. Simplify $\dfrac{(b^3)^2}{b^{5}}$ . (show answer)

Answer
$b$ . Method: top $b^6$ ; $b^6 \div b^5 = b^1 = b$ .
3. Simplify $(2m)^3 \times m^2$ . (show answer)

Answer
$8 m^5$ . Method: $(2m)^3 = 8 m^3$ ; $\times m^2 = 8 m^5$ .
4. Simplify $\dfrac{10 x^7}{2 x^4}$ . (show answer)

Answer
$5 x^3$ .
5. Simplify $5 a^3 \times 2 a^4$ . (show answer)

Answer
$10 a^7$ .
6. Simplify $(3 x^2)^3$ . (show answer)

Answer
$27 x^6$ .
7. Simplify $\dfrac{a^4 b^6}{a b^2}$ . (show answer)

Answer
$a^3 b^4$ .
8. Simplify $\dfrac{(x^3)^4 \cdot x^2}{x^{10}}$ . (show answer)

Answer
$x^4$ . Method: top $x^{12} \cdot x^2 = x^{14}$ ; $x^{14} \div x^{10} = x^4$ .

Reasoning · Explain and spot the mistake

1. Sam writes $2^3 \times 2^3 = 2^9$ . Is Sam correct? If not, what should it be? (show answer)

Answer
Wrong. $2^3 \times 2^3 = 2^{3 + 3} = 2^6 = 64$ . Sam added $3 \times 3$ or used the wrong law.
2. Explain why $a^0 = 1$ is forced by the quotient rule. (show answer)

Answer
Dividing anything non-zero by itself is $1$ . The quotient rule says $\dfrac{a^n}{a^n} = a^{n-n} = a^0$ . So $a^0 = 1$ .
3. Mira writes $(a^2)^3 = a^5$ . Explain the error and give the correct value. (show answer)

Answer
Wrong. $(a^2)^3 = a^{2 \times 3} = a^6$ (power of a power multiplies, not adds, the exponents).
4. Is $x^{-2}$ a Year 8 result, and if not, what does it mean? (Scope note: negative exponents appear in Year 9; for now think about what the quotient rule would give you.) (show answer)

Answer
Not formally Year 8 (appears in Year 9). Using the quotient rule consistently: $\dfrac{x^3}{x^5} = x^{-2}$ , which means $\dfrac{1}{x^2}$ .

Problem-solving · Applications

1. A bacterium doubles every hour. Starting from $1$ cell, write the number of cells after $n$ hours as a power of $2$ . How many cells after $8$ hours? (show answer)

Answer
$2^n$ cells; after $8$ hours: $2^8 = 256$ cells.
2. A cube of side $s$ has volume $s^3$ . Express the volume of a cube whose side is doubled, as a multiple of $s^3$ . (show answer)

Answer
Volume becomes $(2s)^3 = 8 s^3$ , so $8$ times larger.
3. Computer memory is measured in powers of $2$ . How much bigger is $2^{20}$ than $2^{10}$ ? Express as a power of $2$ . (show answer)

Answer
$2^{10}$ times bigger. ( $2^{20} \div 2^{10} = 2^{10} = 1024$ .)
4. A square tile pattern has $n$ rows and $n$ columns. If each tile is $10$ cm by $10$ cm, express the total floor area in cm² using exponent notation. (show answer)

Answer
$100 n^2$ cm², or $10^2 n^2$ .

Reasoning · Harder reasoning

1. Simplify $\dfrac{(2 x^3 y)^2 \times (x y^2)^3}{x^4 y^5}$ . (show answer)

Answer
$4 x^5 y^3$ . Method: top $= 4 x^6 y^2 \times x^3 y^6 = 4 x^9 y^8$ ; divide by $x^4 y^5$ → $4 x^5 y^3$ .
2. If $a^m = 8$ and $a^n = 4$ , find $a^{m + n}$ without finding $a$ , $m$ or $n$ . (show answer)

Answer
$a^{m+n} = a^m \times a^n = 8 \times 4 = 32$ .
3. A formula for a population model is $P = P_0 \times 2^t$ where $P_0$ is the starting population and $t$ is time in years. If a town starts at $5000$ and the population doubles every year, what is the population after $6$ years? (show answer)

Answer
$P = 5000 \times 2^6 = 5000 \times 64 = 320\,000$ .
4. Simplify and write as a power of $2$ : $\dfrac{8^5 \times 4^3}{32^2}$ . (Hint: write every base as a power of $2$ .) (show answer)

Answer
$2^{11}$ . Method: $8 = 2^3$ , $4 = 2^2$ , $32 = 2^5$ . Expression becomes $\dfrac{2^{15} \times 2^6}{2^{10}} = \dfrac{2^{21}}{2^{10}} = 2^{11}$ .

Fractions & recurring decimals

Fluency · Fraction to decimal

1. Write $\dfrac{1}{2}$ as a decimal. (show answer)

Answer
$0.5$
2. Does $\dfrac{1}{5}$ terminate or recur? Write its decimal. (show answer)

Answer
Terminates; $\dfrac{1}{5} = 0.2$ .
3. Write $\dfrac{1}{10}$ as a decimal. (show answer)

Answer
$0.1$
4. Write $\dfrac{3}{4}$ as a decimal. (show answer)

Answer
$0.75$
5. Write $\dfrac{3}{8}$ as a decimal. (show answer)

Answer
$0.375$
6. Write $\dfrac{11}{20}$ as a decimal. (show answer)

Answer
$0.55$
7. Write $\dfrac{2}{3}$ as a decimal (use the recurring bar). (show answer)

Answer
$0.\overline{6}$ (recurring - the digit $6$ repeats)
8. Write $\dfrac{4}{9}$ as a decimal. (show answer)

Answer
$0.\overline{4}$
9. Write $\dfrac{7}{11}$ as a decimal (use the bar over the repeating block). (show answer)

Answer
$0.\overline{63}$ (the two-digit block $63$ repeats)
10. Write $\dfrac{5}{6}$ as a decimal. (show answer)

Answer
$0.8\overline{3}$ (only the $3$ repeats)

Fluency · Predict terminating or recurring

1. Will $\dfrac{9}{40}$ terminate? Justify. (show answer)

Answer
Terminates. $40 = 2^3 \times 5$ - only primes $2$ and $5$ .
2. Will $\dfrac{7}{15}$ terminate? Justify. (show answer)

Answer
Recurs. $15 = 3 \times 5$ - the prime $3$ appears.
3. Will $\dfrac{13}{50}$ terminate? Justify. (show answer)

Answer
Terminates. $50 = 2 \times 5^2$ .
4. Will $\dfrac{5}{24}$ terminate? Justify. (show answer)

Answer
Recurs. $24 = 2^3 \times 3$ .
5. Will $\dfrac{17}{125}$ terminate? Justify. (show answer)

Answer
Terminates. $125 = 5^3$ .

Fluency · Recurring to fraction

1. Convert $0.\overline{5}$ to a fraction. (show answer)

Answer
$\dfrac{5}{9}$ . Method: $x = 0.\overline{5}$ ; $10x - x = 5$ ; $9x = 5$ .
2. Convert $0.\overline{8}$ to a fraction. (show answer)

Answer
$\dfrac{8}{9}$ .
3. Convert $0.\overline{12}$ to a fraction. (show answer)

Answer
$\dfrac{12}{99} = \dfrac{4}{33}$ . Method: two-digit block, so $100 x - x = 12$ .
4. Convert $0.\overline{45}$ to a fraction (and simplify). (show answer)

Answer
$\dfrac{45}{99} = \dfrac{5}{11}$ .
5. Convert $0.\overline{123}$ to a fraction (and simplify). (show answer)

Answer
$\dfrac{123}{999} = \dfrac{41}{333}$ .

Reasoning · Explain and spot the mistake

1. Dev says " $\dfrac{1}{9} = 0.1$ ". Is Dev correct? If not, what is the correct decimal? (show answer)

Answer
Wrong. $\dfrac{1}{9} = 0.\overline{1} = 0.1111\ldots$ , not $0.1$ . Dev has truncated instead of noting the recurrence.
2. Explain why $\dfrac{1}{6}$ recurs but $\dfrac{1}{5}$ terminates. (show answer)

Answer
$6 = 2 \times 3$ ; the prime $3$ means no power of $10$ is a multiple of $6$ , so the decimal must recur. $5$ is itself a prime factor of $10$ , so $\dfrac{1}{5} = 0.2$ terminates.
3. Kim writes $0.\overline{9} = 0.999\ldots$ and claims this is less than $1$ . Is Kim correct? (Hint: try the algebra trick from Worked example 4.) (show answer)

Answer
Kim is wrong - $0.\overline{9}$ equals $1$ exactly. Let $x = 0.\overline{9}$ ; $10x = 9.\overline{9}$ ; $10x - x = 9$ ; $9x = 9$ ; $x = 1$ .
4. Show that $\dfrac{3}{25}$ terminates by writing $25$ as a power of primes. (show answer)

Answer
$25 = 5^2$ . Since the only prime is $5$ , $\dfrac{3}{25}$ terminates. Specifically $\dfrac{3}{25} = \dfrac{12}{100} = 0.12$ .

Problem-solving · Applications

1. A carpenter needs to split a $1$ m length into $7$ equal parts. Is each part's length a terminating or recurring decimal in metres? Explain. (show answer)

Answer
Each part is $\dfrac{1}{7}$ m $= 0.\overline{142857}$ m - recurring. ( $7$ is neither $2$ nor $5$ .)
2. Write $\dfrac{1}{7}$ as a decimal. What is the smallest integer $n$ such that $7 \mid 10^n - 1$ ? (This is the length of the repeating block.) (show answer)

Answer
$\dfrac{1}{7} = 0.\overline{142857}$ ; the block has length $6$ , so $n = 6$ satisfies $7 \mid 10^6 - 1 = 999999$ .
3. Convert $0.\overline{142857}$ back to a fraction. (You should recognise the answer.) (show answer)

Answer
$\dfrac{142857}{999999} = \dfrac{1}{7}$ .
4. A recipe says "use $\dfrac{1}{3}$ cup of butter". A measuring cup has decimal markings. Explain how the cook should round. (show answer)

Answer
$\dfrac{1}{3} \approx 0.333$ , so round to the nearest practical value of the measuring cup (e.g. if marked every $\tfrac{1}{4}$ cup, round up to $\tfrac{1}{3} \approx 0.33$ or accept $\tfrac{1}{3}$ directly).

Reasoning · Harder reasoning

1. Convert $0.1\overline{6}$ to a fraction. (Hint: note the non-repeating "1" at the start; multiply by a power of $10$ to clear the non-repeating part first.) (show answer)

Answer
$\dfrac{1}{6}$ . Method: $x = 0.1\overline{6}$ . $10 x = 1.\overline{6}$ ; $100 x = 16.\overline{6}$ ; $100 x - 10 x = 90 x = 15$ ; $x = \dfrac{15}{90} = \dfrac{1}{6}$ .
2. Convert $0.4\overline{72}$ to a fraction. (show answer)

Answer
$\dfrac{26}{55}$ . Method: $x = 0.4\overline{72}$ . $10x = 4.\overline{72}$ ; $1000x = 472.\overline{72}$ ; $1000x - 10x = 468$ ; $990x = 468$ ; $x = \dfrac{468}{990} = \dfrac{26}{55}$ .
3. Two fractions have the same decimal expansion $0.\overline{45}$ . Are they equal? Explain. (show answer)

Answer
Yes, they must be equal - two numbers with the same decimal expansion are the same number. Different fractions would give different expansions.
4. Without computing, decide whether $\dfrac{1}{2^3 \times 5 \times 7}$ terminates. Justify. (show answer)

Answer
Recurs. The prime $7$ is present in the denominator, which is neither $2$ nor $5$ .

Four operations with rationals

Fluency · Integer × and ÷

1. Evaluate $(-1) \times 5$ . (show answer)

Answer
$-5$
2. Evaluate $(-2) \times (-3)$ . (show answer)

Answer
$6$ (negative times negative)
3. Evaluate $(-6) \times 7$ . (show answer)

Answer
$-42$
4. Evaluate $(-8) \times (-5)$ . (show answer)

Answer
$40$ (two negatives give a positive)
5. Evaluate $9 \times (-4)$ . (show answer)

Answer
$-36$
6. Evaluate $(-3)^3$ . (show answer)

Answer
$-27$ (three negative factors: odd, so negative)
7. Evaluate $(-2)^4$ . (show answer)

Answer
$16$ (four negative factors: even, so positive)
8. Evaluate $(-42) \div 6$ . (show answer)

Answer
$-7$
9. Evaluate $(-72) \div (-8)$ . (show answer)

Answer
$9$ (two negatives cancel)
10. Evaluate $(-2) \times (-3) \times (-4)$ . (show answer)

Answer
$-24$ (three negative factors: odd)
11. Evaluate $(-1)^{100}$ . (show answer)

Answer
$1$ (even power of $-1$ )
12. Evaluate $(-100) \div (-25)$ . (show answer)

Answer
$4$

Fluency · Negative fractions and decimals

1. Evaluate $-\dfrac{1}{2} + \dfrac{1}{3}$ . (show answer)

Answer
$-\dfrac{1}{6}$ . Method: common denominator $6$ ; $-\dfrac{3}{6} + \dfrac{2}{6}$ .
2. Evaluate $-\dfrac{2}{5} - \dfrac{3}{10}$ . (show answer)

Answer
$-\dfrac{7}{10}$ . Method: $-\dfrac{4}{10} - \dfrac{3}{10}$ .
3. Evaluate $-\dfrac{4}{9} \times \dfrac{3}{8}$ . (show answer)

Answer
$-\dfrac{1}{6}$ . Method: $-\dfrac{12}{72}$ .
4. Evaluate $-\dfrac{5}{6} \div \left(-\dfrac{10}{9}\right)$ . (show answer)

Answer
$\dfrac{3}{4}$ . Method: $-\dfrac{5}{6} \times -\dfrac{9}{10} = \dfrac{45}{60} = \dfrac{3}{4}$ .
5. Evaluate $-0.25 \times 8$ . (show answer)

Answer
$-2$
6. Evaluate $-1.6 + 0.4$ . (show answer)

Answer
$-1.2$
7. Evaluate $-3.2 \div (-0.8)$ . (show answer)

Answer
$4$
8. Evaluate $(-0.5)^2$ . (show answer)

Answer
$0.25$

Reasoning · Order of operations

1. Evaluate $-3 + 2 \times (-4)$ . (show answer)

Answer
$-11$ .
2. Evaluate $(-6)^2 - 10$ . (show answer)

Answer
$26$ .
3. Evaluate $-\dfrac{1}{4} \times 8 + 5$ . (show answer)

Answer
$3$ . Method: $-\dfrac{1}{4} \times 8 = -2$ ; $-2 + 5 = 3$ .
4. Evaluate $\dfrac{-12 + 4}{-2}$ . (show answer)

Answer
$4$ . Method: $\dfrac{-8}{-2} = 4$ .
5. Evaluate $-2 \times (3 - 7)^2$ . (show answer)

Answer
$-32$ . Method: $(3 - 7)^2 = 16$ ; $-2 \times 16 = -32$ .
6. Evaluate $-\dfrac{3}{4} - \dfrac{1}{2} \times (-4)$ . (show answer)

Answer
$\dfrac{5}{4}$ . Method: $-\dfrac{1}{2} \times (-4) = 2$ ; $-\dfrac{3}{4} + 2 = \dfrac{5}{4}$ .

Reasoning · Explain and spot the mistake

1. Sam says $-3^2 = 9$ . Explain what is wrong and give the correct value. (show answer)

Answer
$-3^2 = -(3 \times 3) = -9$ , not $9$ . The power applies only to the $3$ , so the minus sign stays in front. $(-3)^2 = 9$ ; that's the version Sam was thinking of.
2. Without calculating, decide whether $(-17) \times (-19) \times (-2)$ is positive or negative. Explain. (show answer)

Answer
Negative. There are three negative factors; three is odd.
3. Explain in plain words why dividing a negative by a negative gives a positive. (show answer)

Answer
Dividing asks "how many of the second fit into the first". Two negatives cancel because flipping the sign of both the "how many" and the "of what" gives the same answer.
4. Lee writes $\dfrac{-6 + 2}{-2} = \dfrac{-4}{-2} = 2$ . Verify whether this is right. (show answer)

Answer
Correct. $-6 + 2 = -4$ ; $\dfrac{-4}{-2} = 2$ .

Problem-solving · Applications

1. A hot-air balloon rises at $2.5$ m/s for $20$ seconds, then descends at $1.8$ m/s for $30$ seconds. What is its net change in altitude? (show answer)

Answer
Net change: $2.5 \times 20 - 1.8 \times 30 = 50 - 54 = -4$ m. The balloon is $4$ m below the starting altitude.
2. Temperatures in a week were $-4, -1, 3, 5, 2, -2, -3$ (°C). Find the mean temperature. (show answer)

Answer
$0$ °C. Method: sum $= -4 - 1 + 3 + 5 + 2 - 2 - 3 = 0$ ; $0 \div 7$ .
3. A share price drops $4\%$ one day and then rises $4\%$ the next. Is it back to the original price? Justify with a specific starting value. (show answer)

Answer
Not back to original. Starting at $100$ : drop gives $96$ ; then rise $4\%$ gives $99.84$ . Net loss.
4. A student's score on four tests are $-3, +5, +2, -4$ (changes from the class average). What is the student's total deviation from the average? (show answer)

Answer
$0$ . Method: $-3 + 5 + 2 - 4 = 0$ .

Reasoning · Harder reasoning

1. Evaluate $\dfrac{(-2)^3 \times (-3)^2}{(-6)}$ . (show answer)

Answer
$12$ . Method: top $= (-8) \times 9 = -72$ ; $-72 \div -6 = 12$ .
2. Solve for $x$ : $-\dfrac{2}{3} x = 8$ . (show answer)

Answer
$x = -12$ . Method: multiply both sides by $-\dfrac{3}{2}$ .
3. A number $n$ satisfies $(-n)^3 = -27$ . Find $n$ . (show answer)

Answer
$n = 3$ . Method: $(-n)^3 = -n^3 = -27$ , so $n^3 = 27$ , so $n = 3$ .
4. Simplify $-\dfrac{3}{5} \left( \dfrac{10}{9} - \dfrac{2}{3} \right)$ . (show answer)

Answer
$-\dfrac{4}{15}$ . Method: bracket $= \dfrac{10}{9} - \dfrac{6}{9} = \dfrac{4}{9}$ ; $-\dfrac{3}{5} \times \dfrac{4}{9} = -\dfrac{12}{45} = -\dfrac{4}{15}$ .

Percentages in context

Fluency · Increase and decrease

1. Increase $80$ by $25\%$ . (show answer)

Answer
$100$ .
2. Decrease $150 by $12\%$ . (show answer)

Answer
$132. Method: $150 \times 0.88$ .
3. Increase $85 by $20\%$ . (show answer)

Answer
$102. Method: $85 \times 1.20$ .
4. Decrease $250$ by $6\%$ . (show answer)

Answer
$235$ . Method: $250 \times 0.94$ .
5. A jacket is $85, discounted by $20\%$ . Find the sale price. (show answer)

Answer
$68. Method: $85 \times 0.80$ .
6. A subscription is $120 and rises $8\%$ . Find the new price. (show answer)

Answer
$129.60. Method: $120 \times 1.08$ .

Fluency · Reverse problems

1. After a $15\%$ increase, a price is $46. Find the original. (show answer)

Answer
$40. Method: $\dfrac{46}{1.15}$ .
2. After a $25\%$ discount, a shirt costs $36. Find the original price. (show answer)

Answer
$48. Method: $\dfrac{36}{0.75}$ .
3. A salary increased by $5\%$ to $63,000. Find the old salary. (show answer)

Answer
$60\,000. Method: $\dfrac{63000}{1.05}$ .
4. A population fell by $8\%$ to $9200$ . Find the original population. (show answer)

Answer
$10\,000$ . Method: $\dfrac{9200}{0.92}$ .

Fluency · Profit, loss, GST

1. Buy at $60, sell at $75. Percentage profit? (show answer)

Answer
$25\%$ profit. Method: profit $15; $\dfrac{15}{60} \times 100$ .
2. Buy at $200, sell at $170. Percentage loss? (show answer)

Answer
$15\%$ loss. Method: loss $30; $\dfrac{30}{200} \times 100$ .
3. A price is $49 before GST ( $10\%$ ). What is the GST-included price? (show answer)

Answer
$53.90. Method: $49 \times 1.10$ .
4. A total (GST-inc.) is $77. How much GST is in it? (show answer)

Answer
$7. Method: $\dfrac{77}{11}$ .
5. A meal costs $55 GST-inc. How much is the pre-GST price? (show answer)

Answer
$50. Method: $\dfrac{55}{1.10}$ .

Reasoning · Percentage error & explain

1. A scientist measures a rod as $12.3$ cm; the true length is $12.0$ cm. Find the percentage error. (show answer)

Answer
$2.5\%$ . Method: $\dfrac{|12.3 - 12|}{12} \times 100$ .
2. A shop advertises "was $100, now $75". What percentage discount is this? (show answer)

Answer
$25\%$ off. Method: discount $25; $\dfrac{25}{100} \times 100$ .
3. Sam says a $30\%$ rise then a $30\%$ fall brings the price back to the start. Show with a worked example whether this is true. (show answer)

Answer
Not true. Example starting $100: $100 \to 130 \to 91$ , below the original.
4. Explain the difference between " $20\%$ off then $10\%$ off" and a single " $30\%$ off". (show answer)

Answer
They are different. Two successive cuts: $100 \to 80 \to 72$ (net $28\%$ off). A single $30\%$ off gives $70$ . Successive cuts are less generous than the nominal sum.

Problem-solving · Real contexts

1. A laptop's sticker price is $1400. In a sale it is reduced $15\%$ . After the sale the shop also adds GST ( $10\%$ ). What does the customer pay? (show answer)

Answer
$1309. Method: discount: $1400 \times 0.85 = 1190$ ; add GST: $1190 \times 1.10 = 1309$ .
2. Mira earns $800 per week. She gets a $4\%$ rise, then six months later a further $3\%$ rise. What does she earn per week now? (show answer)

Answer
$857.36 per week. Method: $800 \times 1.04 \times 1.03$ .
3. A town's population rose from $12\,500$ to $14\,000$ over five years. What was the percentage increase? (show answer)

Answer
$12\%$ . Method: rise $1500$ ; $\dfrac{1500}{12500} \times 100$ .
4. A phone's retail price includes $10\%$ GST and is $1089. How much GST is included in the price? (show answer)

Answer
$99. Method: $\dfrac{1089}{11}$ .
5. A supermarket sells a $500$ g jar for $5.50 and a $750$ g jar for $7.70. Which is better value, and by what percentage is the better-value jar cheaper per gram? (show answer)

Answer
$500$ g jar is cheaper per gram. $500$ g: $1.10$ c/g. $750$ g: $1.027$ c/g. Actually the $750$ g is cheaper by about $6.7\%$ per gram ( $\dfrac{1.10 - 1.027}{1.10} \times 100$ ).

Reasoning · Harder reasoning

1. A value rises by $r\%$ and then falls by $r\%$ . Show that the end value is always less than the starting value (for $r > 0$ ), and find a formula for the net percentage change. (show answer)

Answer
Start $A$ . After rise: $A(1 + r/100)$ . After fall of $r\%$ : $A(1 + r/100)(1 - r/100) = A(1 - r^2/10000)$ . This is less than $A$ for $r \ne 0$ . Net change: $-r^2/100 \%$ (a decrease).
2. A shop marks up cost by $25\%$ and then gives a $20\%$ discount from the marked price. Is the final price above or below cost? By how much? (show answer)

Answer
Final is below cost. Starting from cost $C$ : mark-up gives $1.25 C$ ; $20\%$ off that gives $0.80 \times 1.25 C = 1.00 C$ . So the final price equals cost - no profit.
3. The true area of a rectangle is $24$ cm². A student estimates the area as $27$ cm² by rounding the sides up. What is the percentage error? (show answer)

Answer
$12.5\%$ . Method: $\dfrac{|27 - 24|}{24} \times 100$ .
4. A bank account earns $6\%$ interest per year, compounded yearly. If the starting balance is $500, find the balance after $3$ years. (show answer)

Answer
$595.51. Method: $500 \times 1.06^3 = 500 \times 1.191016 = 595.508$ .

Algebraic expressions (expand & factorise)

Fluency · Expand

1. Expand $3(x + 4)$ . (show answer)

Answer
$3x + 12$
2. Expand $5(y - 2)$ . (show answer)

Answer
$5y - 10$
3. Expand $-2(a + 6)$ . (show answer)

Answer
$-2a - 12$
4. Expand $-4(x - 3)$ . (show answer)

Answer
$-4x + 12$
5. Expand $7(2p + 1)$ . (show answer)

Answer
$14p + 7$
6. Expand $x(x + 5)$ . (show answer)

Answer
$x^2 + 5x$
7. Expand $2m(3m - 4)$ . (show answer)

Answer
$6m^2 - 8m$
8. Expand $-a(a - 7)$ . (show answer)

Answer
$-a^2 + 7a$

Fluency · Expand and collect

1. Simplify $2(x + 3) + 3(x + 1)$ . (show answer)

Answer
$5x + 9$ . Method: $2x + 6 + 3x + 3$ .
2. Simplify $4(y - 2) - 2(y - 5)$ . (show answer)

Answer
$2y + 2$ . Method: $4y - 8 - 2y + 10$ .
3. Simplify $3(2a - 1) + 5 - (a + 4)$ . (show answer)

Answer
$5a - 2$ . Method: $6a - 3 + 5 - a - 4$ .
4. Simplify $5p + 2(p + 3) - 4$ . (show answer)

Answer
$7p + 2$ . Method: $5p + 2p + 6 - 4$ .
5. Simplify $x(x + 2) - 3(x - 1)$ . (show answer)

Answer
$x^2 - x + 3$ . Method: $x^2 + 2x - 3x + 3$ .

Fluency · Factorise

1. Factorise $6x + 9$ . (show answer)

Answer
$3(2x + 3)$
2. Factorise $10y - 15$ . (show answer)

Answer
$5(2y - 3)$
3. Factorise $12a + 16b$ . (show answer)

Answer
$4(3a + 4b)$
4. Factorise $7x^2 + 14x$ . (show answer)

Answer
$7x(x + 2)$
5. Factorise $9ab - 12a$ . (show answer)

Answer
$3a(3b - 4)$
6. Factorise $-3x + 9$ . (show answer)

Answer
$-3(x - 3)$ or equivalently $3(3 - x)$
7. Factorise $2x^2 - 8x + 6$ . (show answer)

Answer
$2(x^2 - 4x + 3)$
8. Factorise $4mn + 6m^2 - 2m$ . (show answer)

Answer
$2m(2n + 3m - 1)$

Fluency · Algebraic fractions

1. Simplify $\dfrac{8x + 12}{4}$ . (show answer)

Answer
$2x + 3$
2. Simplify $\dfrac{15a - 10}{5}$ . (show answer)

Answer
$3a - 2$
3. Simplify $\dfrac{6x^2 + 9x}{3}$ . (show answer)

Answer
$2x^2 + 3x$
4. Simplify $\dfrac{20ab - 15b}{5b}$ . (show answer)

Answer
$4a - 3$ . Method: divide each term by $5b$ .

Reasoning · Explain and spot the mistake

1. Jed writes $3(x - 2) = 3x - 2$ . Is Jed correct? If not, what is the error? (show answer)

Answer
Wrong. The $3$ must multiply every term inside the bracket: $3(x - 2) = 3x - 6$ . Jed forgot to multiply the $-2$ .
2. Mira factorises $6x + 4$ as $2(3x + 4)$ . Is this right? If not, give the correct factorisation. (show answer)

Answer
Wrong. The HCF of $6$ and $4$ is $2$ . Correct: $2(3x + 2)$ , not $2(3x + 4)$ (which expands to $6x + 8$ ).
3. Explain why $\dfrac{2x + 6}{2} = x + 3$ and not $x + 6$ . (show answer)

Answer
Both terms on top must be divided by $2$ : $\dfrac{2x}{2} + \dfrac{6}{2} = x + 3$ . Dividing only the $2x$ and leaving the $6$ unchanged is wrong.
4. Write two different expressions that both equal $6x + 12$ and demonstrate they are equal by expanding one of them. (show answer)

Answer
Many answers. Example: $6x + 12 = 6(x + 2) = 3(2x + 4) = 2(3x + 6)$ . Expanding $6(x + 2)$ returns $6x + 12$ . ✓

Problem-solving · Applications

1. A rectangle has length $x + 3$ cm and width $5$ cm. Write and simplify expressions for the perimeter and the area. (show answer)

Answer
Perimeter $= 2(x + 3) + 2(5) = 2x + 16$ . Area $= 5(x + 3) = 5x + 15$ .
2. A taxi charges a flag-fall of $4 plus $2 per kilometre. For a $k$ -km trip, write an expression for the cost, and factorise it. (show answer)

Answer
Cost $= 4 + 2k = 2(2 + k)$ (factorised).
3. Five students each contribute $x toward a gift costing $42. Write and simplify an expression for the change each gets back, assuming the total change is shared equally. (show answer)

Answer
Total change $= 5x - 42$ . Each gets $\dfrac{5x - 42}{5}$ . (When this is not an integer, the "equally" is idealised.)
4. Two rectangles have areas $6x + 12$ and $4x + 8$ . Factorise each; what does the result tell you about the shapes? (show answer)

Answer
Factorisations: $6x + 12 = 6(x + 2)$ ; $4x + 8 = 4(x + 2)$ . Both rectangles share the side length $(x + 2)$ - i.e. they have a side in common.

Reasoning · Harder reasoning

1. Simplify $\;(x + 2)(x + 3) - x(x + 5)$ . (Hint: expand each product first.) (show answer)

Answer
$6$ . Method: $(x+2)(x+3) = x^2 + 5x + 6$ ; subtract $x(x+5) = x^2 + 5x$ ; $x^2 + 5x + 6 - x^2 - 5x = 6$ .
2. Factorise fully $\;4x^2 y + 8xy^2$ . (show answer)

Answer
$4xy(x + 2y)$ .
3. A rectangle has sides $a$ and $b$ . A second rectangle has sides $2a$ and $\dfrac{b}{2}$ . Show that the two rectangles have the same area. (show answer)

Answer
First area $= ab$ . Second area $= 2a \times \dfrac{b}{2} = ab$ . Same.
4. Simplify $\dfrac{3(x - 2) + 2(x + 1)}{5}$ . (show answer)

Answer
$x - \dfrac{4}{5}$ . Method: top $= 3x - 6 + 2x + 2 = 5x - 4$ ; divide by $5$ .

Linear equations, inequalities & graphs

Fluency · Solve linear equations

1. Solve $x + 9 = 15$ . (show answer)

Answer
$x = 6$ . Method: subtract $9$ from both sides.
2. Solve $3x = 18$ . (show answer)

Answer
$x = 6$ . Method: divide both sides by $3$ .
3. Solve $2x + 5 = 17$ . (show answer)

Answer
$x = 6$ . Method: subtract $5$ , then divide by $2$ .
4. Solve $3y - 8 = 13$ . (show answer)

Answer
$y = 7$ . Method: add $8$ , then divide by $3$ .
5. Solve $\dfrac{x}{4} + 3 = 9$ . (show answer)

Answer
$x = 24$ . Method: subtract $3$ , then multiply by $4$ .
6. Solve $4(a - 2) = 20$ . (show answer)

Answer
$a = 7$ . Method: divide both sides by $4$ to get $a - 2 = 5$ , then add $2$ .
7. Solve $5x - 3 = 2x + 12$ . (show answer)

Answer
$x = 5$ . Method: subtract $2x$ from both sides, then $3x - 3 = 12$ ; $3x = 15$ .
8. Solve $3y + 4 = 7y - 16$ . (show answer)

Answer
$y = 5$ . Method: subtract $3y$ and add $16$ : $20 = 4y$ , so $y = 5$ .
9. Solve $\dfrac{3x + 1}{2} = 8$ . (show answer)

Answer
$x = 5$ . Method: multiply both sides by $2$ to get $3x + 1 = 16$ ; $3x = 15$ .
10. Solve $\dfrac{x}{3} - \dfrac{x}{4} = 2$ . (show answer)

Answer
$x = 24$ . Method: multiply both sides by $12$ : $4x - 3x = 24$ .

Fluency · Solve inequalities

1. Solve $x + 4 > 10$ . (show answer)

Answer
$x > 6$
2. Solve $2y - 3 \leq 5$ . (show answer)

Answer
$y \leq 4$
3. Solve $3x + 1 \geq 13$ . (show answer)

Answer
$x \geq 4$
4. Solve $-4x < 20$ . (show answer)

Answer
$x > -5$ (flip: divide by $-4$ ).
5. Solve $5 - 2x > 1$ . (show answer)

Answer
$x < 2$ . Method: subtract $5$ , $-2x > -4$ ; divide by $-2$ , flip.
6. Solve $\dfrac{y}{2} + 1 \geq 3$ . (show answer)

Answer
$y \geq 4$

Fluency · Tables and graphs

1. Complete the table for $y = 3x - 2$ :

| $x$ | $-1$ | $0$ | $1$ | $2$ | $3$ | |---|---|---|---|---|---| | $y$ | ? | ? | ? | ? | ? | (show answer)

Answer
$y$ -values: $-5, -2, 1, 4, 7$ .
2. Find the $y$ -intercept of $y = -2x + 7$ . (show answer)

Answer
$y$ -intercept $(0, 7)$ .
3. Find the gradient of $y = 4x - 1$ . (show answer)

Answer
Gradient $4$ .
4. Find the $x$ -intercept of $y = 2x - 6$ (where $y = 0$ ). (show answer)

Answer
$x$ -intercept $(3, 0)$ .
5. A line passes through $(0, 3)$ and $(2, 9)$ . Find the gradient. (show answer)

Answer
Gradient $3$ . Method: $\dfrac{9 - 3}{2 - 0}$ .
6. Does the point $(3, 5)$ lie on $y = 2x - 1$ ? Show by substitution. (show answer)

Answer
Yes: $2(3) - 1 = 5$ . ✓

Reasoning · Explain and spot the mistake

1. Kai solves $-3x = 12$ as $x = 4$ . Is Kai correct? Explain. (show answer)

Answer
No. Dividing $12$ by $-3$ gives $-4$ , so $x = -4$ . Kai kept the sign wrong.
2. Mira writes " $-2x < 6$ so $x < -3$ ". What has Mira forgotten? (show answer)

Answer
She forgot to flip the inequality when dividing by a negative. Correct: $-2x < 6$ gives $x > -3$ .
3. Explain why the graphs of $y = 2x$ and $y = 2x + 3$ are parallel. (show answer)

Answer
They both have gradient $2$ (same slope), but different $y$ -intercepts ( $0$ and $3$ ), so they rise equally but start at different heights.
4. Describe what happens to the graph of $y = mx + c$ as you increase $c$ . (show answer)

Answer
The line shifts up by $c$ units on the $y$ -axis (parallel to itself). Increasing $c$ does not change the gradient.

Problem-solving · Modelling

1. A gym charges a $50 joining fee and $15 per week. Write a formula $C = 50 + 15w$ for cost after $w$ weeks. After how many weeks does the total pass $200? (show answer)

Answer
After $w = 11$ weeks (since $50 + 15 \times 10 = 200$ exactly at week $10$ ; pass $200 at end of week $11$ ). Strictly $50 + 15w > 200$ gives $w > 10$ , so first at $w = 11$ .
2. A water tank starts with $500$ L and leaks at $8$ L per hour. Write a formula for the volume $V$ after $t$ hours. When is the tank empty? (show answer)

Answer
$V = 500 - 8t$ . Empty when $V = 0$ : $t = 62.5$ hours.
3. A taxi has flag-fall $3.80 and charges $2 per km. A trip costs $19.80. How long was it (in km)? (show answer)

Answer
$16$ km. Method: $3.80 + 2k = 19.80$ , so $2k = 16$ .
4. A mobile plan charges $30 per month plus $0.05 per text. Lucy's monthly bill was $33.50. How many texts did she send? (show answer)

Answer
$70$ texts. Method: $30 + 0.05t = 33.50$ , $0.05t = 3.50$ .
5. Two phone plans: Plan A charges $20 + $0.15/min; Plan B charges $35 + $0.05/min. Write formulas. For how many minutes are the two plans equal in cost? (show answer)

Answer
Plan A: $C = 20 + 0.15m$ ; Plan B: $C = 35 + 0.05m$ . Equal when $20 + 0.15m = 35 + 0.05m$ , so $0.10m = 15$ , $m = 150$ minutes.

Reasoning · Harder reasoning

1. Solve simultaneously (by substitution): $y = 2x - 1$ and $y = x + 4$ . (show answer)

Answer
$(5, 9)$ . Method: $2x - 1 = x + 4$ gives $x = 5$ , then $y = 9$ .
2. Solve $\dfrac{x + 1}{2} - \dfrac{x - 1}{3} = 2$ . (show answer)

Answer
$x = 13$ . Method: multiply by $6$ : $3(x + 1) - 2(x - 1) = 12$ ; $3x + 3 - 2x + 2 = 12$ ; $x + 5 = 12$ .
3. A line has gradient $3$ and passes through $(2, 7)$ . Find the equation of the line. (show answer)

Answer
$y = 3x + 1$ . Method: $y - 7 = 3(x - 2)$ , so $y = 3x - 6 + 7$ .
4. The sum of three consecutive even numbers is $90$ . Write a linear equation and find the numbers. (show answer)

Answer
$28, 30, 32$ . Method: $n + (n+2) + (n+4) = 3n + 6 = 90$ ; $n = 28$ .

Area, perimeter & composite shapes

Fluency · Simple composite areas

1. A rectangle $8$ m by $5$ m has a $2$ m by $3$ m rectangle cut from one corner. Find the remaining area. (show answer)

Answer
$34$ m $^2$ . Method: $40 - 6$ .
2. Two rectangles, $4 \times 2$ and $3 \times 5$ , are joined to form a single L. Find the total area. (show answer)

Answer
$23$ . Method: $8 + 15$ .
3. A trapezium has parallel sides $8$ cm and $14$ cm and height $5$ cm. Find the area. (show answer)

Answer
$55$ cm $^2$ . Method: $\tfrac{1}{2}(8 + 14) \times 5$ .
4. A floor plan is a $10$ m by $6$ m rectangle with a triangular bay window (base $6$ m, height $2$ m) added to one long side. Find the total floor area. (show answer)

Answer
$66$ m $^2$ . Method: rectangle $60$ m $^2$ ; triangle $\tfrac{1}{2} \times 6 \times 2 = 6$ ; total $60 + 6$ .
5. A shape is made of a rectangle $12 \times 5$ with a semicircle of diameter $5$ on one short side (use $\pi \approx 3.14$ ). Find the area. (show answer)

Answer
About $69.8$ cm $^2$ . Method: rectangle $60$ cm $^2$ ; semicircle $\tfrac{1}{2} \pi (2.5)^2 \approx 9.8$ cm $^2$ .

Fluency · Perimeter

1. A rectangle $12$ by $7$ has a $3$ by $2$ notch cut into one long side (flush with the top-right). Find the new perimeter. (show answer)

Answer
$42$ . Method: the notch adds two $2$ -cm inward cuts and a new $3$ cm edge parallel to the top; original perimeter $38$ plus $2 \times 2 = 4$ from the notch = $42$ . (Alternative walking-around count gives the same total.)
2. An L-shape has outer dimensions $10$ m by $8$ m, with a $4$ m by $3$ m notch removed from the top-right corner. Find the perimeter. (show answer)

Answer
$36$ m. Method: walk the L: $10 + 8 + 4 + 3 + (10 - 4) + (8 - 3) = 10 + 8 + 4 + 3 + 6 + 5 = 36$ .
3. A rectangle $20$ by $15$ has a square of side $5$ cut from each corner. Find the perimeter of the remaining cross shape. (show answer)

Answer
$80$ . Method: each corner removes a square but adds two new edges of length $5$ instead of one $5$ -edge; net change $= 0$ per corner. Original perimeter $= 2(20 + 15) = 70$ ; actually each cut increases the perimeter by $5 + 5 - 5 - 5 = 0$ . Wait - each corner cut replaces a corner with two $5$ -cm edges, so perimeter stays $70$ . Correct answer: $70$ .

Fluency · Approximate by grid

1. On a $1$ cm grid, a shape covers $18$ full squares and $10$ partial squares. Estimate its area (count partial squares as half). (show answer)

Answer
$23$ cm $^2$ . Method: $18 + 10/2$ .
2. Using a finer $0.5$ cm grid, a shape covers $70$ full small squares and $20$ partial ones. Estimate its area. (show answer)

Answer
$20$ cm $^2$ . Method: each small square is $0.25$ cm $^2$ ; $(70 + 10) \times 0.25$ .
3. Explain why approximating with a finer grid gives a better area estimate. (show answer)

Answer
Fewer partial squares, each contributing less error. As the grid gets finer the estimate approaches the true area.

Reasoning · Explain and spot the mistake

1. Sam says the area of an L-shape is always the sum of the two rectangles that make it. Is that right? Give an example where it fails. (show answer)

Answer
Correct for the decomposition approach, but fails if the two rectangles overlap. Example: two $4 \times 4$ rectangles that overlap by $2 \times 2$ ; naive sum $32$ , actual area $28$ .
2. A student computes the perimeter of a shape by adding the two outer dimensions and multiplying by $2$ , ignoring the notch. Explain why this can be wrong. (show answer)

Answer
The formula $P = 2(L + W)$ is only for a plain rectangle. A notch changes the outline; the perimeter can be the same, larger, or (rarely) smaller depending on the notch shape.
3. Without calculating, decide which has the larger area: a $10 \times 10$ square with a $3 \times 3$ square cut from one corner, or an $11 \times 9$ rectangle. Justify. (show answer)

Answer
Both have area $91$ - equal. $10 \times 10 - 3 \times 3 = 91$ ; $11 \times 9 = 99$ . Wait: second is $99$ . So the plain rectangle is larger.
4. A trapezium has parallel sides $3$ m and $7$ m and height $4$ m. Another has parallel sides $5$ m and $5$ m (so a rectangle) and height $4$ m. Which has the larger area? What does this tell you about the average of the parallel sides? (show answer)

Answer
Both equal $\tfrac{1}{2}(3 + 7) \times 4 = 20$ m $^2$ and $5 \times 4 = 20$ m $^2$ . Equal - the average of the parallel sides ( $5$ ) equals the rectangle's constant width.

Problem-solving · Real contexts

1. A backyard is L-shaped: a $15$ m by $12$ m rectangle with a $6$ m by $4$ m rectangle cut out of one corner (a shed). Grass seed covers $1$ m $^2$ per gram. How much seed is needed (in kg)? (show answer)

Answer
Grass area: $15 \times 12 - 6 \times 4 = 180 - 24 = 156$ m $^2$ . Seed $= 156$ g $= 0.156$ kg.
2. A concrete slab is a rectangle $8$ m by $6$ m with a semicircular pool alcove (radius $1.5$ m) cut out of one long side. Find the slab area ( $\pi \approx 3.14$ ). (show answer)

Answer
$44.47$ m $^2$ approx. Method: slab $= 48$ m $^2$ ; semicircle $= \tfrac{1}{2} \pi (1.5)^2 \approx 3.53$ ; $48 - 3.53$ .
3. A picture frame has outer $25$ cm by $20$ cm and an inner window $19$ cm by $14$ cm. What is the area of the frame material? (show answer)

Answer
$234$ cm $^2$ . Method: $500 - 266$ .
4. A paddock on a map has shape of a trapezium (parallel sides $400$ m and $600$ m, height $250$ m). Find its area in hectares ( $1$ ha $= 10\,000$ m $^2$ ). (show answer)

Answer
$12.5$ ha. Method: area $= \tfrac{1}{2}(400 + 600) \times 250 = 125\,000$ m $^2$ ; divide by $10\,000$ .

Reasoning · Harder composites

1. A shape consists of a semicircle ( $r = 4$ cm) sitting atop a rectangle $8$ cm by $5$ cm (semicircle on an $8$ cm side). Find the area and the perimeter. (show answer)

Answer
Area $\approx 65.12$ cm $^2$ (rectangle $40$ + semicircle $\tfrac{1}{2} \pi \times 16 \approx 25.12$ ). Perimeter $\approx 5 + 8 + 5 + \pi \times 4 \approx 30.57$ cm.
2. A square of side $20$ cm has a circle inscribed in it (touching all four sides). Find the area between the circle and the square (use $\pi \approx 3.14$ ). (show answer)

Answer
$86$ cm $^2$ approx. Method: square $400$ ; circle $\pi \times 10^2 = 314$ ; $400 - 314$ .
3. Find the area of a hexagonal stop-sign-like shape with side $6$ cm, treated as a rectangle $12 \times 6\sqrt{3}/2$ with two triangles bolted to the short sides. (Extension: this requires $\sqrt{3}$ ; accept $\sqrt{3} \approx 1.73$ .) (show answer)

Answer
About $93.5$ cm $^2$ . Method: $12 \times 6 \times \dfrac{\sqrt{3}}{2} \approx 12 \times 5.196 = 62.35$ ... this is an approximation problem; any reasonable decomposition and arithmetic is fine.

Volume of right prisms

Fluency · Volume of right prisms

1. Find the volume of a cuboid $7 \times 4 \times 3$ cm. (show answer)

Answer
$84$ cm $^3$
2. Find the volume of a cube of side $6$ cm. (show answer)

Answer
$216$ cm $^3$
3. Find the volume of a triangular prism with triangle base $10$ cm, height $6$ cm, and length $15$ cm. (show answer)

Answer
$450$ cm $^3$ . Method: triangle area $30$ ; × $15$ .
4. Find the volume of a trapezoidal prism: parallel sides $4$ m and $6$ m, height $3$ m, length $8$ m. (show answer)

Answer
$120$ m $^3$ . Method: trapezium area $\tfrac{1}{2}(4+6) \times 3 = 15$ ; × $8$ .
5. A cuboid has volume $480$ cm $^3$ and a base of $8 \times 5$ cm. Find its height. (show answer)

Answer
$12$ cm. Method: $480 \div 40$ .
6. A triangular prism has volume $150$ cm $^3$ and length $10$ cm. Find the area of its triangular base. (show answer)

Answer
$15$ cm $^2$ . Method: $150 \div 10$ .

Fluency · Volume and capacity

1. Convert $3500$ mL to L. (show answer)

Answer
$3.5$ L
2. Convert $4.2$ L to mL. (show answer)

Answer
$4200$ mL
3. Convert $0.75$ m $^3$ to L. (show answer)

Answer
$750$ L
4. Convert $2300$ cm $^3$ to mL. (show answer)

Answer
$2300$ mL
5. A fish tank is $60 \times 30 \times 40$ cm. What is its capacity in L? (show answer)

Answer
$72$ L. Method: $V = 72\,000$ cm $^3 = 72$ L.
6. A pool is $12$ m by $6$ m by $1.5$ m. How many kL? (show answer)

Answer
$108$ kL. Method: $V = 108$ m $^3$ .

Fluency · Rates and time

1. A $300$ L tank fills at $5$ L/min. How long? (show answer)

Answer
$60$ min $= 1$ h.
2. A hose delivers $0.2$ L/s. How long (minutes) to fill a $60$ L drum? (show answer)

Answer
$5$ min. Method: $0.2$ L/s $= 12$ L/min; $60 \div 12$ .
3. A pool of $90\,000$ L is filled at $150$ L/min. How many hours? (show answer)

Answer
$10$ hours. Method: $90\,000 \div 150 = 600$ min.
4. A dripping tap loses $4$ drops/sec and $20$ drops = $1$ mL. How much water in a day (L)? (show answer)

Answer
About $17.3$ L/day. Method: $4 \times 86\,400 = 345\,600$ drops/day; $\div 20 = 17\,280$ mL.

Reasoning · Explain and spot the mistake

1. Liam writes the volume of a $2$ cm cube as $8$ cm $^2$ . What is the error? (show answer)

Answer
$8$ is correct numerically but the units should be cm $^3$ (cube has volume in cubic units). cm $^2$ is area.
2. Two cuboids have the same volume. Must they have the same surface area? Give a reason. (show answer)

Answer
Not necessarily. Example: $1 \times 1 \times 1000$ and $10 \times 10 \times 10$ both have volume $1000$ but very different surface areas.
3. Explain why $1$ cm $^3 = 1$ mL using the definition of the metric system. (show answer)

Answer
The metric system was built so that $1$ cm $^3$ holds exactly $1$ mL of water. They are different units measuring the same amount of space or liquid.
4. A tap flows at $12$ L/min. Without calculating, decide whether a $1000$ L tank fills in under or over $1$ hour. Justify. (show answer)

Answer
Over $1$ hour? $12$ L/min × $60$ = $720$ L in an hour - only $720$ of $1000$ L. So it takes over $1$ hour.

Problem-solving · Real contexts

1. A water tank is $1.2$ m by $0.8$ m by $1.5$ m deep. A hose delivers $20$ L/min. How long to fill? (show answer)

Answer
$72$ min. Method: $V = 1.44$ m $^3 = 1440$ L; $1440 \div 20$ .
2. A swimming pool is $15$ m by $8$ m with uniform depth $1.4$ m. If water costs $2.40/kL, find the total fill cost. (show answer)

Answer
$403.20. Method: $V = 15 \times 8 \times 1.4 = 168$ m $^3$ = $168$ kL; × $2.40$ .
3. A shoebox is $33$ cm by $22$ cm by $15$ cm. Give its volume in cm $^3$ and in litres (to 2 dp). (show answer)

Answer
$10\,890$ cm $^3 = 10.89$ L.
4. A chocolate bar in the shape of a triangular prism has equilateral cross-section (side $3$ cm, approximate height $2.6$ cm) and length $10$ cm. Find its volume (to the nearest cm $^3$ ). (show answer)

Answer
$39$ cm $^3$ . Method: triangle area $\tfrac{1}{2} \times 3 \times 2.6 = 3.9$ ; × $10 = 39$ .

Reasoning · Harder problems

1. A cube has surface area $150$ cm $^2$ . Find its volume. (show answer)

Answer
$125$ cm $^3$ . Method: $6 s^2 = 150$ , so $s^2 = 25$ , $s = 5$ ; $V = 5^3$ .
2. Two identical right-angled triangular prisms are joined along their rectangular faces to form a cuboid. If each prism has legs $3$ cm and $4$ cm and length $10$ cm, find the volume of the cuboid. (show answer)

Answer
$120$ cm $^3$ . Method: two prisms combine so half of the cuboid; each triangle area $6$ ; cuboid $= 4 \times 3 \times 10 = 120$ .
3. A rectangular tank ( $2$ m × $1$ m × $1$ m) is being filled at $25$ L/min while draining at $10$ L/min. How long to fill if both taps are open? (show answer)

Answer
$133.3$ min (about $2$ h $13$ min). Method: $V = 2$ m $^3 = 2000$ L; net fill rate $= 15$ L/min; $2000 \div 15$ .
4. A glass is a cylinder of radius $3$ cm and height $10$ cm. Find its capacity in mL. (Use $V = \pi r^2 h$ ; $\pi \approx 3.14$ .) (show answer)

Answer
$282.6$ mL (approx). Method: $V = \pi \times 9 \times 10 \approx 282.6$ cm $^3$ .

Circles: circumference & area

Fluency · Circumference and area

1. A circle has radius $4$ cm. Find its circumference. (show answer)

Answer
$25.12$ cm. Method: $2 \times 3.14 \times 4$ .
2. A circle has diameter $10$ m. Find its circumference. (show answer)

Answer
$31.4$ m. Method: $\pi d$ .
3. A circle has radius $9$ cm. Find its area. (show answer)

Answer
$254.34$ cm $^2$ . Method: $3.14 \times 81$ .
4. A circle has diameter $14$ cm. Find its area (use $\pi \approx \tfrac{22}{7}$ ). (show answer)

Answer
$154$ cm $^2$ . Method: $r = 7$ ; $\tfrac{22}{7} \times 49$ .
5. A circle has radius $2.5$ m. Find its circumference. (show answer)

Answer
$15.7$ m.
6. A circle has radius $5$ cm. Find its area. (show answer)

Answer
$78.5$ cm $^2$ .

Fluency · Reverse problems

1. A circle's circumference is $62.8$ cm. Find its radius. (show answer)

Answer
$10$ cm. Method: $r = \dfrac{62.8}{6.28}$ .
2. A circle's area is $50.24$ m $^2$ . Find its radius. (show answer)

Answer
$4$ m. Method: $r^2 = \dfrac{50.24}{3.14} = 16$ .
3. A wheel has circumference $1.57$ m. Find its diameter. (show answer)

Answer
$0.5$ m. Method: $d = \dfrac{1.57}{3.14}$ .
4. A round plate has area $314$ cm $^2$ . Find its diameter. (show answer)

Answer
$20$ cm. Method: $r^2 = \dfrac{314}{3.14} = 100$ ; $r = 10$ ; $d = 20$ .

Fluency · Semicircles and sectors

1. A semicircle has radius $6$ cm. Find its area and perimeter. (show answer)

Answer
Area $56.52$ cm $^2$ ; perimeter $30.84$ cm. Method: $\tfrac{1}{2}\pi r^2$ ; $\pi r + 2r$ .
2. A quarter-circle has radius $4$ m. Find its area. (show answer)

Answer
$12.56$ m $^2$ . Method: $\tfrac{1}{4}\pi r^2$ .
3. A half-moon window is a semicircle of diameter $1$ m. Find its area. (show answer)

Answer
$0.3925$ m $^2$ . Method: $r = 0.5$ ; $\tfrac{1}{2}\pi r^2$ .
4. A pizza slice is a sector making $\tfrac{1}{8}$ of a pizza of radius $12$ cm. Find its area. (show answer)

Answer
$56.52$ cm $^2$ . Method: $\tfrac{1}{8}\pi (12)^2 = \tfrac{1}{8} \times 452.16$ .

Reasoning · Explain and spot the mistake

1. Kira writes $C = \pi r$ for a circle's circumference. Is Kira correct? If not, what should it be? (show answer)

Answer
Wrong. $C = 2\pi r$ (or $\pi d$ ). Kira used half of the correct formula.
2. Without calculating, decide which changes more: doubling the radius doubles or quadruples the area? Justify. (show answer)

Answer
Quadruples. $A \propto r^2$ , so doubling $r$ multiplies area by $2^2 = 4$ .
3. A student writes the area of a circle with radius $5$ cm as $\pi \times 10$ cm $^2$ . Explain the mistake. (show answer)

Answer
The formula is $A = \pi r^2$ , not $\pi \times 2r$ . For $r = 5$ , $A = \pi \times 25 = 78.5$ cm $^2$ . The student squared wrong.
4. Explain why the circumference of any circle divided by its diameter always gives the same number (which we call $\pi$ ). (show answer)

Answer
Because $\pi$ is defined as that ratio. Regardless of the circle's size, $C/d$ always produces the same irrational number.

Problem-solving · Real contexts

1. A round pool has radius $3.5$ m. A fence is to be built around it, $0.5$ m from the edge. How long is the fence? (show answer)

Answer
$25.12$ m. Method: $r = 3.5 + 0.5 = 4$ ; $C = 2\pi \times 4$ .
2. A pizza has diameter $32$ cm. Find its area, then the area of one slice if cut into $8$ equal pieces. (show answer)

Answer
Pizza area $803.84$ cm $^2$ ; one slice $100.48$ cm $^2$ .
3. A circular lawn has diameter $10$ m. A mower cuts $0.5$ m $^2$ per second. How long to mow (seconds, then minutes)? (show answer)

Answer
$157$ seconds $\approx 2.62$ min. Method: lawn area $= \pi \times 25 = 78.5$ ; $78.5 / 0.5$ .
4. A running track is $400$ m around, consisting of two straights of $100$ m each and two semicircular ends. Find the radius of the semicircles (use $\pi \approx 3.14$ ). (show answer)

Answer
About $31.85$ m. Method: two straights $= 200$ ; two semicircles add to one full circle: $C = 400 - 200 = 200$ ; $r = 200 / (2\pi) \approx 31.85$ .

Reasoning · Harder circles

1. A rectangle $6$ cm by $4$ cm has a semicircle attached to each short side (so the two semicircles form one full circle). Find the total perimeter and area. (show answer)

Answer
Perimeter $= 2 \times 6 + 2\pi \times 2 \approx 24.56$ cm (two long sides $+$ one full circle from the two semicircles, with $r = 2$ ). Area $= 24 + \pi \times 4 \approx 36.56$ cm $^2$ .
2. Two circles: small has radius $3$ cm, large has radius $5$ cm. Find the area of the ring between them (the annulus). (show answer)

Answer
$50.24$ cm $^2$ . Method: $\pi(5^2 - 3^2) = 16\pi$ .
3. A running track sector is $\tfrac{1}{4}$ of a circle with radius $50$ m. Find its arc length. (show answer)

Answer
$25\pi \approx 78.5$ m. Method: $\tfrac{1}{4} \times 2\pi r = \tfrac{\pi r}{2}$ .
4. A pizza shop offers a $30$ cm diameter pizza for $14 or a $40$ cm diameter pizza for $22. Which is better value per cm $^2$ ? (show answer)

Answer
$30$ cm pizza: $14 \div (\pi \times 225) \approx 0.0198$ dollars per cm $^2$ . $40$ cm pizza: $22 \div (\pi \times 400) \approx 0.0175$ dollars per cm $^2$ . The $40$ cm pizza is cheaper per cm $^2$ .

Time & time zones

Fluency · Convert and calculate

1. Convert $4{:}45$ p.m. to 24-hour time. (show answer)

Answer
$16{:}45$
2. Convert $23{:}05$ to 12-hour time. (show answer)

Answer
$11{:}05$ p.m.
3. Duration from $9{:}15$ a.m. to $3{:}40$ p.m. (show answer)

Answer
$6$ h $25$ min
4. Duration from $22{:}50$ to $07{:}10$ next day. (show answer)

Answer
$8$ h $20$ min
5. Add $4$ h $25$ min to $11{:}20$ a.m. (show answer)

Answer
$3{:}45$ p.m.
6. Subtract $3$ h $40$ min from $02{:}15$ next day. (show answer)

Answer
$22{:}35$ (previous day)

Fluency · Time zones (standard time)

1. It is $2{:}00$ p.m. in Sydney (AEST). What time in Perth (AWST)? (show answer)

Answer
$12{:}00$ noon. Perth is $2$ h behind.
2. It is $9{:}30$ a.m. in Adelaide (ACST). What time in Brisbane (AEST)? (show answer)

Answer
$10{:}00$ a.m. Brisbane is $30$ min ahead.
3. It is $10{:}00$ a.m. in Perth (AWST). What time in Melbourne (AEST)? (show answer)

Answer
$12{:}00$ noon. Melbourne is $2$ h ahead.
4. It is $12{:}00$ noon in London (UTC+0). What time in Melbourne (AEST)? (show answer)

Answer
$10{:}00$ p.m. ( $22{:}00$ ). Melbourne is $10$ h ahead.
5. It is $6{:}00$ p.m. in New York (UTC $-5$ ). What time in Singapore (UTC+8)? (show answer)

Answer
$7{:}00$ a.m. next day. Singapore is $13$ h ahead of New York.

Reasoning · Explain and spot the mistake

1. Sam says "Melbourne is always $3$ hours ahead of Perth". Is that always right? Explain. (show answer)

Answer
Not always. In standard time it is $2$ h ahead; during daylight saving (October-April for Victoria) it is $3$ h ahead because Melbourne shifts to AEDT (UTC+11) while Perth stays on AWST (UTC+8).
2. Explain why a flight from Melbourne to Singapore sometimes lands on the same calendar day and sometimes the next day. (show answer)

Answer
A $7-9$ h westward flight that leaves in the morning can still land on the same day because local time "goes back" $2$ h, whereas a late-evening flight crosses the midnight boundary of the arrival zone.
3. When travelling west, do you gain or lose time on arrival? Explain. (show answer)

Answer
You gain local time on arrival - your watch reads later locally than when you left (the clock runs ahead in local terms). Actually this depends on direction: flying west, local time is earlier, so you arrive at an earlier clock time than your travel hours suggest - effectively gaining hours in your day.
4. Write $12{:}00$ midnight as two equally valid 24-hour times. Which do most timetables prefer? (show answer)

Answer
$00{:}00$ or $24{:}00$ both represent the midnight moment. Most timetables use $00{:}00$ as the start of the new day; $24{:}00$ is rare.

Problem-solving · Real contexts

1. A video conference is scheduled for $9{:}00$ a.m. Melbourne time (AEST, April - no DST). What local time should the Perth participant join? (show answer)

Answer
$7{:}00$ a.m. Perth time. Method: Perth is $2$ h behind Melbourne in April (both standard time).
2. A flight leaves Singapore at $23{:}00$ local (UTC+8) and takes $7$ h $30$ min to Sydney (AEST, UTC+10). What is the local arrival time in Sydney? (show answer)

Answer
$08{:}30$ AEST next day. Method: arrival in Singapore time $= 06{:}30$ ; add $2$ h for AEST.
3. A bus leaves Adelaide at $8{:}15$ a.m. (ACST) and arrives in Melbourne at $5{:}45$ p.m. (AEST). How long was the trip (in local times), accounting for the $30$ -minute zone difference? (show answer)

Answer
$9$ h total. Method: Melbourne is $30$ min ahead of Adelaide. Bus leaves at $8{:}45$ Melbourne time; arrives $17{:}45$ ; duration $= 9$ h.
4. A family flies Melbourne to Los Angeles (UTC $-8$ ) on a $14$ -hour flight that leaves at $10{:}00$ a.m. Melbourne AEDT (UTC+11) in January. What is the local arrival time in LA? (show answer)

Answer
$6{:}00$ a.m. LA time same day. Method: arrival in Melbourne time $= 00{:}00$ next day. Melbourne AEDT is UTC+11; LA is UTC $-8$ ; LA is $19$ h behind Melbourne. $00{:}00$ next day $- 19$ h $= 05{:}00$ LA time. Actually: $24{:}00 - 19 = 05{:}00$ LA, same calendar day as arrival in Melbourne. Answer: $5{:}00$ a.m. LA local time on the day of departure.

Rates

Fluency · Unit rates

1. A car goes $240$ km in $4$ h. Find the average speed. (show answer)

Answer
$60$ km/h
2. A tap fills a tank at $600$ L in $20$ min. Find the rate in L/min. (show answer)

Answer
$30$ L/min
3. A worker earns $540 for $20$ hours. Find the hourly rate. (show answer)

Answer
$27/h
4. A mass of $180$ g has volume $20$ cm $^3$ . Find the density. (show answer)

Answer
$9$ g/cm $^3$
5. A printer prints $60$ pages in $4$ minutes. Find pages per minute. (show answer)

Answer
$15$ pages/min

Fluency · Speed, distance, time

1. Distance from $65$ km/h × $3$ h. (show answer)

Answer
$195$ km
2. Time to cover $300$ km at $50$ km/h. (show answer)

Answer
$6$ h
3. Speed of $200$ m in $25$ seconds (in m/s). (show answer)

Answer
$8$ m/s
4. Convert $72$ km/h to m/s. (show answer)

Answer
$20$ m/s. Method: $72 \times \tfrac{1000}{3600}$ .
5. How long to cover $120$ km at $80$ km/h? (show answer)

Answer
$1$ h $30$ min
6. A train covers $400$ km in $2$ h $30$ min. Find its speed. (show answer)

Answer
$160$ km/h. Method: $400 \div 2.5$ .

Fluency · Fuel, pay, exchange

1. A car uses $8$ L/ $100$ km. Fuel for $350$ km? (show answer)

Answer
$28$ L. Method: $\tfrac{8}{100} \times 350$ .
2. A worker earns $22/h. Find pay for $36.5$ hours. (show answer)

Answer
$803. Method: $22 \times 36.5$ .
3. Exchange rate $1$ AUD $= 0.85$ NZD. Convert $150 AUD to NZD. (show answer)

Answer
$127.50 NZD. Method: $150 \times 0.85$ .
4. $1$ USD $= 1.50$ AUD. Convert $200 USD to AUD. (show answer)

Answer
$300 AUD. Method: $200 \times 1.50$ .
5. Simple interest: $1000 at $4\%$ for $3$ years. How much interest? (show answer)

Answer
$120 interest. Method: $\tfrac{4}{100} \times 1000 \times 3$ .

Reasoning · Explain and spot the mistake

1. Sam writes "the speed is $50$ km in $1$ hour, so $50$ per hour, so $50$ km". What units are missing? What is the correct way to report speed? (show answer)

Answer
Units km/hour (or km per hour). Sam dropped the "per hour" and collapsed the rate to a distance, losing the time dimension. Correct: $50$ km/h.
2. Two cars: A does $60$ km in $1$ hour; B does $30$ km in $30$ minutes. Are they the same speed? Show working. (show answer)

Answer
Same speed. B does $30$ km in $0.5$ h $= 60$ km/h.
3. Explain why a "rate" and a "unit rate" are slightly different ideas. Give an example of each. (show answer)

Answer
A rate is any comparison of two different quantities (e.g. $60$ km in $1.5$ h). A unit rate is the per-one-unit form (e.g. $40$ km/h). A rate can be simplified to a unit rate by division.
4. Without calculating, compare: a pool fills at $10$ L/min for $20$ min, or at $4$ L/min for $60$ min - which delivers more water? (show answer)

Answer
$10 \times 20 = 200$ L vs $4 \times 60 = 240$ L. The second delivers more.

Problem-solving · Real contexts

1. A road trip is $1600$ km. If the driver averages $100$ km/h (including breaks in planned driving time), how long does the trip take? (show answer)

Answer
$16$ h. Method: $1600 \div 100$ .
2. A swimming pool holds $120\,000$ L. A hose delivers $120$ L/min. How long to fill (hours)? (show answer)

Answer
$1000$ min $\approx 16$ h $40$ min.
3. A box of $24$ pens costs $7.20. A single pen costs $0.40. Which is better value, and by how much per pen? (show answer)

Answer
Single pen is dearer by $10$ c each. Box price $= 30$ c/pen; individual $40$ c/pen; difference $10$ c.
4. Two phone plans: A is $25/month + $0.10/min; B is $35/month with unlimited calls. For what usage does B beat A? (show answer)

Answer
B beats A when $25 + 0.10 m > 35$ , i.e. $m > 100$ minutes.
5. An alloy uses $5$ kg of copper for every $3$ kg of tin. For a $40$ kg alloy, how much of each? (show answer)

Answer
Copper $25$ kg, tin $15$ kg. Method: $5 + 3 = 8$ parts; each part $= 5$ kg.

Reasoning · Harder problems

1. A car averages $80$ km/h for $2$ hours and then $60$ km/h for $3$ hours. What is its average speed for the whole trip? (show answer)

Answer
$68$ km/h. Method: total distance $= 160 + 180 = 340$ km; total time $= 5$ h; $340/5$ .
2. A shopkeeper buys tea at $12/kg and sells it at $15/kg. What percentage profit is this? (show answer)

Answer
$25\%$ . Method: profit $= 3$ dollars/kg; $3/12$ .
3. A tap fills a tank at $8$ L/min while a drain removes water at $3$ L/min. The tank holds $600$ L; it starts empty. How long to fill? (show answer)

Answer
$120$ min. Method: net $5$ L/min; $600/5$ .
4. Currency arbitrage: $1 AUD $=$ $0.65 USD; $1 USD $=$ $0.80 EUR; $1 EUR $=$ $1.60 AUD. Is there a profit in converting $100 AUD → USD → EUR → AUD? If so, how much? (show answer)

Answer
Yes, small profit. $100 AUD → $65 USD → $52 EUR → $83.20 AUD. Wait: that's a loss. Let me redo - actually the chain gives $100 \times 0.65 \times 0.80 \times 1.60 = 83.20$ AUD, so a loss, not a profit. No arbitrage exists; the student has lost $16.80 on fees-free conversion.

Pythagoras' theorem

Fluency · Find the hypotenuse

1. Legs $3, 4$ . (show answer)

Answer
$5$ . Method: $9 + 16 = 25$ ; $\sqrt{25} = 5$ .
2. Legs $5, 12$ . (show answer)

Answer
$13$ . Method: $25 + 144 = 169$ .
3. Legs $6, 8$ . (show answer)

Answer
$10$ . Method: $36 + 64 = 100$ .
4. Legs $9, 40$ . (show answer)

Answer
$41$ . Method: $81 + 1600 = 1681 = 41^2$ .
5. Legs $2, 2$ . Give the exact value and the decimal. (show answer)

Answer
$\sqrt{8} = 2\sqrt{2} \approx 2.83$ . Method: $4 + 4 = 8$ .
6. Legs $1, \sqrt{3}$ . Give the exact hypotenuse. (show answer)

Answer
$2$ . Method: $1 + 3 = 4$ ; $\sqrt{4} = 2$ .

Fluency · Find a leg

1. Hypotenuse $25$ , leg $7$ . (show answer)

Answer
$24$ . Method: $625 - 49 = 576$ .
2. Hypotenuse $17$ , leg $8$ . (show answer)

Answer
$15$ .
3. Hypotenuse $15$ , leg $9$ . (show answer)

Answer
$12$ .
4. Hypotenuse $50$ , leg $30$ . (show answer)

Answer
$40$ . Method: $2500 - 900 = 1600$ .
5. Hypotenuse $\sqrt{20}$ , leg $2$ . Give the exact value. (show answer)

Answer
$4$ . Method: $20 - 4 = 16$ .

Fluency · Is it a right triangle?

1. $6, 8, 10$ - right-angled? (show answer)

Answer
Yes. $36 + 64 = 100$ .
2. $4, 5, 6$ - right-angled? (show answer)

Answer
No. $16 + 25 = 41 \ne 36$ .
3. $7, 24, 25$ - right-angled? (show answer)

Answer
Yes. $49 + 576 = 625$ .
4. $5, 12, 14$ - right-angled? (show answer)

Answer
No. $25 + 144 = 169 \ne 196$ .

Reasoning · Explain and spot the mistake

1. Lee writes $5^2 + 12^2 = 17$ . Explain the error. (show answer)

Answer
Lee took the square root of the sum incorrectly. $5^2 + 12^2 = 25 + 144 = 169$ , and $\sqrt{169} = 13$ , not $17$ .
2. Explain why the hypotenuse must always be longer than either leg. (show answer)

Answer
The hypotenuse satisfies $c^2 = a^2 + b^2$ . If either leg had length $\geq c$ , the equation would fail. Geometrically, the hypotenuse is opposite the largest angle ( $90^\circ$ ), which corresponds to the longest side.
3. A triangle has sides $9, 12, 15$ . Is it right-angled? Is the triangle similar to any triple you recognise? (show answer)

Answer
Yes: $81 + 144 = 225$ . It is similar to $(3, 4, 5)$ with scale factor $3$ .
4. Sam says $7, 24, 26$ is a Pythagorean triple because the numbers "look similar" to $7, 24, 25$ . Is Sam right? Justify. (show answer)

Answer
No. $7^2 + 24^2 = 49 + 576 = 625 = 25^2$ , so the triple is $(7, 24, 25)$ . $(7, 24, 26)$ has $625 \ne 676$ , so it is not a right triangle.

Problem-solving · Real contexts

1. A ladder $6$ m long is placed with its foot $2$ m from a wall. How high does it reach? (show answer)

Answer
$\sqrt{32} \approx 5.66$ m.
2. A rectangular field is $30$ m by $40$ m. How far is the diagonal? (show answer)

Answer
$50$ m.
3. A guy wire supports a pole and is attached $8$ m from the foot of the pole, $15$ m up. How long is the wire? (show answer)

Answer
$17$ m. ( $8, 15, 17$ triple.)
4. A TV has a 50-inch diagonal and is $45$ inches wide. Is it taller than $20$ inches? Justify. (show answer)

Answer
Yes, $\approx 21.8$ inches. $\sqrt{50^2 - 45^2} = \sqrt{475} \approx 21.79$ .
5. A ship sails $8$ km east and then $6$ km north. How far is it from its starting point? (show answer)

Answer
$10$ km. ( $6, 8, 10$ .)

Reasoning · Harder triangles

1. A right-angled isosceles triangle has legs of length $a$ . Show that the hypotenuse is $a\sqrt{2}$ . (show answer)

Answer
Legs $a$ and $a$ : $c^2 = 2a^2$ ; $c = a\sqrt{2}$ .
2. A rectangular prism measures $3$ cm by $4$ cm by $12$ cm. Find the length of the longest internal diagonal. (show answer)

Answer
$13$ cm. Method: base diagonal $\sqrt{3^2 + 4^2} = 5$ ; body diagonal $\sqrt{5^2 + 12^2} = 13$ .
3. A rhombus has diagonals $10$ cm and $24$ cm. Find the side length. (show answer)

Answer
$13$ cm. Method: half-diagonals $5$ and $12$ ; $\sqrt{5^2 + 12^2}$ .
4. An equilateral triangle has side $10$ cm. Find its height, to one decimal place. (show answer)

Answer
$8.7$ cm. Method: height $= \sqrt{10^2 - 5^2} = \sqrt{75} \approx 8.66$ .

Congruence & similarity

Fluency · Congruence tests

1. Two triangles with sides $3, 3, 3$ and $3, 3, 3$ . (Warm-up — what test is this?) (show answer)

Answer
SSS. All three pairs of sides match ( $3, 3, 3$ ).
2. A right-angled triangle with hypotenuse $5$ and leg $3$ , and another right-angled triangle with hypotenuse $5$ and leg $3$ . (Which test uses a right angle?) (show answer)

Answer
RHS. Both have a right angle, equal hypotenuses ( $5$ ) and a matching leg ( $3$ ).
3. Two triangles with sides $4, 5, 6$ and $4, 5, 6$ . (show answer)

Answer
SSS. All three pairs of sides match ( $4, 5, 6$ ).
4. Two triangles: $\angle A = 30^\circ$ , $AB = 5$ , $BC = 6$ ; $\angle D = 30^\circ$ , $DE = 5$ , $EF = 6$ . (Try mentally first: is the $30^\circ$ angle between the two named sides?) (show answer)

Answer
No valid test (SSA). The sides $AB$ and $BC$ meet at vertex $B$ , but the given $30^\circ$ angle is at vertex $A$ - it is not between the two named sides. SSA is not a valid congruence test (the "ambiguous case").
5. Two right-angled triangles: hypotenuse $10$ with one leg $6$ , vs hypotenuse $10$ with one leg $6$ . (show answer)

Answer
RHS. Both are right-angled with equal hypotenuses ( $10$ ) and one matching leg ( $6$ ).
6. Two triangles: $\angle A = 40^\circ$ , $AB = 5$ , $\angle B = 60^\circ$ ; $\angle D = 40^\circ$ , $DE = 5$ , $\angle E = 60^\circ$ . (show answer)

Answer
ASA. Two angles ( $40^\circ$ at $A$ , $60^\circ$ at $B$ ) and the included side $AB = DE = 5$ match.
7. Two triangles: $AB = 7$ , $BC = 5$ , $\angle C = 40^\circ$ ; $DE = 7$ , $EF = 5$ , $\angle F = 40^\circ$ . (show answer)

Answer
No valid test (SSA). Sides $AB$ and $BC$ meet at $B$ , but the given $40^\circ$ angle is at $C$ , not between the two named sides. SSA is not a valid congruence test.

Fluency · Similar triangles

1. $\triangle ABC$ has sides $3, 4, 5$ . $\triangle DEF$ has sides $9, 12, 15$ . Are they similar? What is the scale factor? (show answer)

Answer
Yes, scale factor $3$ . ( $3 : 9 = 4 : 12 = 5 : 15$ .)
2. $\triangle ABC$ has $\angle A = 40^\circ$ , $\angle B = 70^\circ$ . $\triangle PQR$ has $\angle P = 40^\circ$ , $\angle R = 70^\circ$ . Similar? Why? (show answer)

Answer
Yes, by AA. The third angle in each must be $70^\circ$ ; sorry - $\angle B = 70^\circ$ for the first, $\angle R = 70^\circ$ for the second; both have angles $40, 70, 70$ . Similar by AA.
3. Two triangles have sides in ratio $2 : 3 : 4$ and $4 : 6 : 8$ . Similar? (show answer)

Answer
Yes, scale factor $2$ .
4. Triangle $A$ has sides $5, 6, 8$ . Triangle $B$ has sides $10, 12, 15$ . Similar? (show answer)

Answer
No. Ratios: $10/5 = 2$ , $12/6 = 2$ , $15/8 = 1.875$ - not equal.

Reasoning · Explain and spot the mistake

1. Sam says two right-angled triangles with legs $3, 4$ each must be congruent. Is Sam right? Explain. (show answer)

Answer
Yes - by SAS (the right angle is included between the two legs). The hypotenuse is then forced to be $5$ , matching.
2. Mira writes "SSA is a valid test for congruence because three measurements are given". Is this correct? Give a counter-example. (show answer)

Answer
Not correct. Counter-example: two sides $5$ and $3$ with a $30^\circ$ non-included angle gives two possible triangles (the "ambiguous case").
3. Explain why two similar triangles with scale factor $1$ are also congruent. (show answer)

Answer
Scale factor $1$ means every corresponding side is the same length. Same sides and same angles ⇒ congruent.
4. A rhombus has all four sides equal. Prove (with triangle congruence) that its diagonals bisect each other. (show answer)

Answer
In rhombus $ABCD$ with centre $O$ , $\triangle AOB$ and $\triangle COD$ have $AB = CD$ (given), $\angle OAB = \angle OCD$ (alternate angles, parallel sides), $\angle OBA = \angle ODC$ (similarly). By ASA, $\triangle AOB \cong \triangle COD$ , so $AO = OC$ and $BO = OD$ - the diagonals bisect each other.

Problem-solving · Real contexts

1. A ramp has a shadow $3$ m long when a $1$ -m pole casts a $0.5$ -m shadow. How high is the ramp? (Use similar triangles.) (show answer)

Answer
$6$ m. Method: scale factor $= 3/0.5 = 6$ ; ramp height $= 1 \times 6 = 6$ m.
2. A photo is enlarged: $10$ cm by $15$ cm becomes $30$ cm by $45$ cm. Find the scale factor and the area ratio. (show answer)

Answer
Linear scale factor $3$ . Area ratio $= 3^2 = 9$ .
3. Two triangles on a flag each have sides in ratio $5 : 12 : 13$ . Are they congruent or similar? What extra information do you need? (show answer)

Answer
They are similar (same angle sum, same side ratio). To be congruent, you also need actual side lengths to match.
4. A tall tree's shadow is $6$ m at the same moment a $1.8$ m friend's shadow is $1.2$ m. How tall is the tree? (show answer)

Answer
$9$ m. Method: $\dfrac{1.8}{1.2} = \dfrac{h}{6}$ ; $h = 9$ .

Reasoning · Harder reasoning

1. Explain why two isosceles triangles with equal apex angles and one pair of equal sides may still not be congruent. (show answer)

Answer
The "equal pair of sides" could be the legs in one triangle and the base in the other. Without specifying which sides, SAS is not established.
2. In a parallelogram $ABCD$ , show using congruent triangles that $AB = CD$ and $AD = BC$ . (show answer)

Answer
In $ABCD$ with $AC$ a diagonal: $\triangle ABC \cong \triangle CDA$ by ASA (alternate angles, shared side $AC$ ), so $AB = CD$ and $BC = AD$ .
3. A triangle has sides $6, 8, 10$ . A similar triangle has a hypotenuse $15$ . Find its other two sides. (show answer)

Answer
$9$ and $12$ . Scale factor $15/10 = 1.5$ ; $6 \times 1.5 = 9$ , $8 \times 1.5 = 12$ .
4. Are all squares similar? Are all rectangles similar? Justify. (show answer)

Answer
All squares are similar (all angles $90^\circ$ ; all sides equal). Not all rectangles are similar - e.g. $2 \times 3$ and $2 \times 5$ rectangles have different side ratios.

Sampling & statistical investigations

Fluency · Population, sample, census

1. A school has $820$ students. The Principal surveys every student in the school. Census or sample? (show answer)

Answer
Census (everyone in the population).
2. A shop owner asks every tenth customer about satisfaction. Sampling method? (show answer)

Answer
Systematic.
3. A researcher wants to know heights of all Australian $13$ -year-olds. Census or sample? Why? (show answer)

Answer
Sample. Reason: census of every $13$ -year-old in Australia is impractical and costly.
4. A market researcher surveys only people in shopping centres. Name one likely bias. (show answer)

Answer
Selection bias toward shoppers; non-shoppers are under-represented.
5. State the population and suggest a suitable sample for: "What proportion of Year 8 students at our school ride a bike to school?" (show answer)

Answer
Population: all Year 8 students at our school. Sample: simple random sample of at least $30$ from the Year 8 roll.

Fluency · Sampling methods and bias

1. Which sampling method divides the population into strata and samples from each? Stratified, cluster, or convenience? (show answer)

Answer
Stratified.
2. What type of bias arises from a survey question like "Do you agree that more homework is harmful?" (show answer)

Answer
Question bias (loaded or leading wording).
3. Explain why a phone-in survey is usually biased. (show answer)

Answer
Self-selection: only motivated listeners call in, and they may hold strong or particular views.
4. A school has $300$ Year 7, $280$ Year 8, $260$ Year 9 students. Using stratified sampling with a $10\%$ sample, how many from each year? (show answer)

Answer
Year 7: $30$ . Year 8: $28$ . Year 9: $26$ . Method: $10\%$ of each.

Reasoning · Explain and spot the mistake

1. Sam claims "a sample of $20$ is enough to be certain about a school of $500$ ". Is Sam correct? Explain. (show answer)

Answer
No. $20$ out of $500$ is $4\%$ ; random variation alone can shift results by $\pm 10$ percentage points. A bigger sample is needed for confidence.
2. Explain why two random samples of the same size can give different summary statistics. (show answer)

Answer
Each sample contains different individuals; small differences in who's in the sample translate to small differences in the statistics.
3. Write an unbiased version of this question: "Don't you agree that our coach is doing a great job?" (show answer)

Answer
"How would you rate the coach's performance this season on a scale from $1$ (poor) to $5$ (excellent)?" - avoids leading wording.
4. A newspaper reports a poll of $500$ readers showing $70\%$ support a policy. What caveats should be stated before trusting the result? (show answer)

Answer
Are the $500$ readers a random sample of all readers, or self-selected? Is the poll reflective of the newspaper's audience only? What's the margin of error?

Problem-solving · Plan and analyse

1. Design a statistical investigation to answer: "How much sleep do Year 8 students at our school get on a school night?" Include population, sample method, sample size, and a data display. (show answer)

Answer
Population: all Year 8 students. Method: stratified random sample across classes. Sample size: $\geq 30$ . Ask: "How many hours did you sleep last school night?". Display: dot plot or column graph. Report mean, median, range, and acknowledge uncertainty.
2. A school has $1000$ students. You take four random samples of $50$ and count those who cycle: $18, 22, 16, 21$ . Calculate the mean percentage and comment on variability. (show answer)

Answer
Mean cycling percentage $= \dfrac{18 + 22 + 16 + 21}{4 \times 50} \times 100 = \dfrac{77}{200} \times 100 = 38.5\%$ . Variability: range $16$ to $22$ per sample of $50$ , i.e. $\pm 6$ percentage points - modest.
3. A factory tests $1\%$ of its daily output of $80\,000$ screws. Is $800$ a large enough sample? What factors matter? (show answer)

Answer
$800$ is typically large enough for industrial QC at $1\%$ sampling. Factors: is the sample random across shifts and machines? Is $1\%$ enough given the tolerance required?
4. Two weather stations collect rainfall each day for two weeks. Station A records $10$ days; Station B records $14$ days. Which would you trust more for "average daily rainfall this fortnight"? (show answer)

Answer
Station B (more days = more data to average, less random day-to-day noise) - provided both stations are in the same area and used comparable instruments.

Probability: complementary & compound events

Fluency · Complementary events

1. $P(A) = 0.3$ . Find $P(A')$ . (show answer)

Answer
$0.7$
2. $P(\text{rain}) = 65\%$ . Find $P(\text{no rain})$ . (show answer)

Answer
$35\%$
3. A bag has $3$ red and $7$ blue balls. $P(\text{not red})$ ? (show answer)

Answer
$\dfrac{7}{10}$
4. A spinner has $8$ equal sectors, one labelled "WIN". $P(\text{not WIN})$ ? (show answer)

Answer
$\dfrac{7}{8}$
5. Two dice rolled. $P(\text{sum} \ne 7)$ ? (show answer)

Answer
$\dfrac{5}{6}$ . Method: $1 - \dfrac{6}{36}$ .

Fluency · Two-stage experiments

1. Flip two coins. Sample space size? (show answer)

Answer
$4$
2. Flip a coin and roll a die. Sample space size? (show answer)

Answer
$12$
3. Two dice rolled. $P(\text{both show even})$ ? (show answer)

Answer
$\dfrac{1}{4}$ . Method: $\dfrac{3}{6} \times \dfrac{3}{6}$ .
4. Two dice rolled. $P(\text{sum} = 10)$ ? (show answer)

Answer
$\dfrac{3}{36} = \dfrac{1}{12}$ . Method: $(4,6), (5,5), (6,4)$ .
5. Two dice rolled. $P(\text{at least one 6})$ ? (Hint: consider the complement.) (show answer)

Answer
$\dfrac{11}{36}$ . Method: $1 - P(\text{no 6}) = 1 - \left(\dfrac{5}{6}\right)^2 = 1 - \dfrac{25}{36}$ .

Fluency · Venn and two-way tables

1. How many like only coffee? (show answer)

Answer
$14$ .
2. How many like only tea? (show answer)

Answer
$10$ .
3. How many like neither? (show answer)

Answer
$6$ . Method: $40 - 24 - 10 - (\text{neither}) = 40 - 24 - 10 + 10 = ?$ . Actually: coffee only $+$ tea only $+$ both $+$ neither $= 40$ ; $14 + 10 + 10 + ? = 40$ ; neither $= 6$ .
4. $P(\text{likes coffee or tea})$ ? (show answer)

Answer
$\dfrac{34}{40} = \dfrac{17}{20} = 0.85$ .
5. $P(\text{likes exactly one})$ ? (show answer)

Answer
$\dfrac{24}{40} = \dfrac{3}{5} = 0.60$ .

Reasoning · Explain and spot the mistake

1. Sam says " $P(A) + P(B) = 1$ , so $A$ and $B$ are complementary." Is Sam correct? Give a counter-example. (show answer)

Answer
Not correct. Complementary events must also cover all possibilities (be exhaustive) and not overlap. Counter-example: $A =$ "roll a 1", $B =$ "roll a 6" on a fair die. $P(A) + P(B) = \tfrac{2}{6} \ne 1$ , and even if both halved the probability to sum to $1$ , they'd need to overlap zero and exhaust the sample space.
2. A student writes "mutually exclusive" events are the same as "independent" events. Are they? Explain with examples. (show answer)

Answer
Not the same. Mutually exclusive events cannot both happen. Independent events have no influence on each other. Example: "rolling a 6" and "rolling a 1" are mutually exclusive but not independent in a single roll. "Coin lands heads" and "die rolls 6" are independent but not mutually exclusive.
3. Using the two-way table for coffee/tea, explain why $P(\text{coffee}) + P(\text{tea}) \ne P(\text{coffee or tea})$ . (show answer)

Answer
Because the overlap (both coffee and tea) is counted twice when you add $P(\text{coffee}) + P(\text{tea})$ . You subtract $P(\text{both})$ to correct.
4. After flipping a coin and getting $6$ heads in a row, Ben says the next flip is "more likely tails to balance it out". Is Ben correct? (show answer)

Answer
No - the gambler's fallacy. Each flip is independent; the coin has no memory. $P(\text{tails next}) = 0.5$ .

Problem-solving · Real contexts

1. A committee of two is chosen from Anna, Ben, Chloe, Dan. List the sample space. What is $P(\text{Anna is chosen})$ ? (show answer)

Answer
Pairs: AB, AC, AD, BC, BD, CD - six total. $P(\text{Anna}) = \dfrac{3}{6} = \dfrac{1}{2}$ .
2. A bag has $5$ red and $3$ blue marbles. One is drawn, not replaced, then another is drawn. Using a tree diagram, find $P(\text{both red})$ . (show answer)

Answer
$P(\text{both red}) = \dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}$ .
3. $60\%$ of students play a sport; $40\%$ play an instrument; $25\%$ do both. $P(\text{neither})$ ? (show answer)

Answer
$P(\text{neither}) = 1 - P(\text{sport or instrument}) = 1 - (0.60 + 0.40 - 0.25) = 1 - 0.75 = 0.25$ .
4. Two fair coins and a die are tossed together. $P(\text{2 heads and a 6})$ ? (show answer)

Answer
$P = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{6} = \dfrac{1}{24}$ .

Reasoning · Harder reasoning

1. A fair coin is tossed $3$ times. Find the probability of getting exactly $2$ heads using a tree diagram. (show answer)

Answer
$\dfrac{3}{8}$ . Method: $3$ outcomes with exactly $2$ heads out of $8$ total (HHT, HTH, THH).
2. A spinner has sectors $1$ - $5$ (equally likely). It is spun twice. Find $P(\text{sum} \geq 8)$ . (show answer)

Answer
$\dfrac{6}{25}$ . Method: $P(\text{sum} \geq 8)$ : count pairs (3,5),(4,4),(4,5),(5,3),(5,4),(5,5) - six pairs.
3. In a class, $P(\text{boy}) = 0.48$ , $P(\text{plays cricket}) = 0.30$ , $P(\text{boy and plays cricket}) = 0.20$ . Find $P(\text{boy or plays cricket})$ and $P(\text{neither})$ . (show answer)

Answer
$P(\text{boy or plays cricket}) = 0.48 + 0.30 - 0.20 = 0.58$ ; $P(\text{neither}) = 1 - 0.58 = 0.42$ .
4. Two dice are rolled. Using a $6 \times 6$ table, find the probability that the difference of the two faces is $2$ . (show answer)

Answer
$\dfrac{8}{36} = \dfrac{2}{9}$ . Method: pairs with difference $2$ : $(1,3), (3,1), (2,4), (4,2), (3,5), (5,3), (4,6), (6,4)$ - eight.