Year 9 Mathematics Practice

Real numbers & scientific notation

Fluency · Tier 1: basic skills

1. Classify as rational or irrational: $\dfrac{5}{8}$ . (show answer)

Answer
Rational. $\dfrac{5}{8} = 0.625$ , a terminating decimal.
2. Classify: $\sqrt{50}$ . (show answer)

Answer
Irrational. $50$ is not a perfect square.
3. Classify: $-\sqrt{81}$ . (show answer)

Answer
Rational. $\sqrt{81} = 9$ , so $-\sqrt{81} = -9$ , an integer.
4. Classify: $0.\overline{36}$ . (show answer)

Answer
Rational. Repeating decimal; $0.\overline{36} = \dfrac{4}{11}$ .
5. Write $93\,000\,000$ in scientific notation. (show answer)

Answer
$9.3 \times 10^7$ .
6. Write $0.000\,072$ in scientific notation. (show answer)

Answer
$7.2 \times 10^{-5}$ .
7. Write $5.03 \times 10^5$ in ordinary notation. (show answer)

Answer
$503\,000$ .
8. Write $1.7 \times 10^{-3}$ in ordinary notation. (show answer)

Answer
$0.0017$ .
9. Calculate $(3 \times 10^4) \times (2 \times 10^5)$ . Give your answer in scientific notation. (show answer)

Answer
$6 \times 10^9$ . Method: $3 \times 2 = 6$ ; $10^4 \times 10^5 = 10^9$ .
10. Calculate $\dfrac{9.6 \times 10^8}{3.2 \times 10^2}$ . Give your answer in scientific notation. (show answer)

Answer
$3 \times 10^6$ . Method: $9.6 \div 3.2 = 3$ ; $10^{8-2} = 10^6$ .

Reasoning · Tier 2: mixed practice

1. Place $\sqrt{5}$ , $\dfrac{7}{3}$ , and $\pi$ on a number line. Which is the largest? (show answer)

Answer
$\sqrt{5} \approx 2.24$ , $\dfrac{7}{3} \approx 2.33$ , $\pi \approx 3.14$ . Largest is $\pi$ .
2. Show that $\sqrt{2} + \sqrt{2}$ is irrational. (show answer)

Answer
$\sqrt{2} + \sqrt{2} = 2\sqrt{2}$ . If this were rational, then $\sqrt{2} = \dfrac{r}{2}$ for some rational $r$ , making $\sqrt{2}$ rational — contradiction. So $2\sqrt{2}$ is irrational.
3. The mass of the Earth is approximately $5.97 \times 10^{24}$ kg and the mass of the Moon is approximately $7.35 \times 10^{22}$ kg. How many times heavier is the Earth than the Moon? Give your answer to the nearest whole number. (show answer)

Answer
$\dfrac{5.97 \times 10^{24}}{7.35 \times 10^{22}} = \dfrac{5.97}{7.35} \times 10^2 \approx 0.8122 \times 10^2 = 81.2$ . The Earth is approximately $81$ times heavier.
4. A human hair is about $7 \times 10^{-5}$ m wide. Express this in micrometres ( $1\;\mu\text{m} = 10^{-6}$ m). (show answer)

Answer
$7 \times 10^{-5} \text{ m} = 70 \times 10^{-6} \text{ m} = 70\;\mu\text{m}$ .
5. Calculate $(7.2 \times 10^{-3}) \times (4 \times 10^5)$ and give the result in scientific notation. (show answer)

Answer
$7.2 \times 4 = 28.8$ ; $10^{-3} \times 10^5 = 10^2$ . So $28.8 \times 10^2 = 2.88 \times 10^3$ .
6. Explain why the sum of a rational number and an irrational number is always irrational. (show answer)

Answer
Suppose $r$ is rational and $x$ is irrational and $r + x = q$ is rational. Then $x = q - r$ , a difference of two rationals, which is rational — contradicting $x$ being irrational.
7. Between which two consecutive tenths does $\sqrt{18}$ lie? Use squaring to justify. (show answer)

Answer
$4.2^2 = 17.64$ and $4.3^2 = 18.49$ . Since $17.64 < 18 < 18.49$ , we have $4.2 < \sqrt{18} < 4.3$ .
8. The distance from the Sun to Neptune is $4.5 \times 10^{12}$ m. Light travels at $3 \times 10^8$ m/s. How many seconds does sunlight take to reach Neptune? (show answer)

Answer
$t = \dfrac{4.5 \times 10^{12}}{3 \times 10^8} = 1.5 \times 10^4 = 15\,000$ seconds (about $4.2$ hours).

Reasoning · Tier 3: explain and apply

1. Is $\sqrt{2} \times \sqrt{8}$ rational or irrational? Justify your answer. (show answer)

Answer
Rational. $\sqrt{2} \times \sqrt{8} = \sqrt{16} = 4$ .
2. A nanotechnology lab measures objects on the scale of $10^{-9}$ m. Express $4.2 \times 10^{-7}$ m in terms of nanometres. (show answer)

Answer
$4.2 \times 10^{-7} \text{ m} = 420 \times 10^{-9} \text{ m} = 420$ nm.
3. Find two irrational numbers whose product is rational. Explain why this does not contradict the definition of irrational numbers. (show answer)

Answer
$\sqrt{2} \times \sqrt{2} = 2$ , which is rational. This works because the definition says each individual number is irrational, not that products of irrationals must be irrational.
4. The Australian national debt is approximately $9 \times 10^{11} $. If the population is$ 2.6 \times 10^7$, estimate the debt per person. Give your answer in ordinary notation to the nearest dollar. (show answer)

Answer
$\dfrac{9 \times 10^{11}}{2.6 \times 10^7} = \dfrac{9}{2.6} \times 10^4 \approx 3.46 \times 10^4 = 34\,615$ dollars per person.
5. Explain why $0.\overline{9} = 1$ and what this tells us about the boundary between rational and irrational numbers. (show answer)

Answer
Let $x = 0.\overline{9}$ . Then $10x = 9.\overline{9}$ , so $10x - x = 9$ , giving $9x = 9$ and $x = 1$ . This shows $0.\overline{9}$ and $1$ are the same number — every terminating decimal also has a repeating representation. It does not blur the rational/irrational boundary; both forms are rational.

Reasoning · Harder reasoning

1. Prove that if $r$ is rational and $x$ is irrational, then $r + x$ is irrational. (Hint: assume the opposite and derive a contradiction.) (show answer)

Answer
Assume $r + x = q$ where $q$ is rational. Then $x = q - r$ , a difference of two rationals, which is rational. This contradicts $x$ being irrational, so $r + x$ must be irrational.
2. The observable universe has a radius of approximately $4.4 \times 10^{26}$ m. Estimate its volume in cubic metres using $V = \dfrac{4}{3}\pi r^3$ , and express your answer in scientific notation to two significant figures. (show answer)

Answer
$V = \dfrac{4}{3}\pi (4.4 \times 10^{26})^3 = \dfrac{4}{3}\pi \times 8.5184 \times 10^{79} \approx 4.19 \times 8.5184 \times 10^{79} \approx 3.6 \times 10^{80}$ m $^3$ .
3. Simplify $\dfrac{(2.4 \times 10^5)^2}{6 \times 10^3}$ and give the answer in scientific notation. (show answer)

Answer
Numerator: $(2.4)^2 \times (10^5)^2 = 5.76 \times 10^{10}$ . Division: $\dfrac{5.76 \times 10^{10}}{6 \times 10^3} = 0.96 \times 10^7 = 9.6 \times 10^6$ .
4. A computer performs $3.6 \times 10^{12}$ operations per second. How many operations can it perform in one year ( $3.15 \times 10^7$ seconds)? If each operation processes $8 \times 10^{-9}$ seconds of audio, how many hours of audio can the computer process per year? (show answer)

Answer
Operations per year: $3.6 \times 10^{12} \times 3.15 \times 10^7 = 11.34 \times 10^{19} = 1.134 \times 10^{20}$ . Audio processed: $1.134 \times 10^{20} \times 8 \times 10^{-9} = 9.072 \times 10^{11}$ seconds $= \dfrac{9.072 \times 10^{11}}{3600} \approx 2.52 \times 10^8$ hours.

Exponent laws with integer exponents

Fluency · Tier 1: basic skills

1. Evaluate $5^0$ . (show answer)

Answer
$1$ .
2. Evaluate $(4x)^0$ where $x \neq 0$ . (show answer)

Answer
$1$ .
3. Evaluate $3^{-2}$ . (show answer)

Answer
$\dfrac{1}{9}$ .
4. Write $10^{-5}$ as a fraction. (show answer)

Answer
$\dfrac{1}{100\,000}$ .
5. Simplify $a^3 \times a^{-5}$ . Give the answer with a positive exponent. (show answer)

Answer
$\dfrac{1}{a^2}$ . Method: $a^{3+(-5)} = a^{-2} = \dfrac{1}{a^2}$ .
6. Simplify $\dfrac{m^4}{m^7}$ . Give the answer with a positive exponent. (show answer)

Answer
$\dfrac{1}{m^3}$ . Method: $m^{4-7} = m^{-3}$ .
7. Simplify $(x^{-3})^2$ . (show answer)

Answer
$x^{-6} = \dfrac{1}{x^6}$ . Method: $(-3) \times 2 = -6$ .
8. Simplify $(2a)^{-3}$ . (show answer)

Answer
$\dfrac{1}{8a^3}$ . Method: $2^{-3} \times a^{-3} = \dfrac{1}{8} \times \dfrac{1}{a^3}$ .
9. Evaluate $\left(\dfrac{1}{2}\right)^{-4}$ . (show answer)

Answer
$16$ . Method: $\left(\dfrac{1}{2}\right)^{-4} = 2^4 = 16$ .
10. Write $0.000\,56$ in scientific notation. (show answer)

Answer
$5.6 \times 10^{-4}$ .

Reasoning · Tier 2: mixed practice

1. Simplify $\dfrac{a^5 b^{-2}}{a^{-1} b^3}$ and write with positive exponents only. (show answer)

Answer
$\dfrac{a^6}{b^5}$ . Method: $a^{5-(-1)} = a^6$ ; $b^{-2-3} = b^{-5}$ .
2. Simplify $(3x^{-2}y)^3$ . (show answer)

Answer
$\dfrac{27y^3}{x^6}$ . Method: $3^3 x^{-6} y^3 = 27 x^{-6} y^3$ .
3. Simplify $\dfrac{(p^2 q^{-1})^3}{p^{-3} q^4}$ . (show answer)

Answer
$\dfrac{p^9}{q^7}$ . Method: numerator $p^6 q^{-3}$ ; division gives $p^{6-(-3)} q^{-3-4} = p^9 q^{-7}$ .
4. Show that $a^{-1} + b^{-1} \neq (a+b)^{-1}$ by substituting $a = 2$ and $b = 3$ . (show answer)

Answer
$a^{-1} + b^{-1} = \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}$ . But $(a+b)^{-1} = \dfrac{1}{5} = 0.2$ . Since $\dfrac{5}{6} \neq \dfrac{1}{5}$ , the two expressions are not equal.
5. Simplify $\dfrac{12 x^{-3} y^2}{4 x^2 y^{-1}}$ and express with positive exponents. (show answer)

Answer
$\dfrac{3y^3}{x^5}$ . Method: $\dfrac{12}{4} = 3$ ; $x^{-3-2} = x^{-5}$ ; $y^{2-(-1)} = y^3$ .
6. A virus measures $8 \times 10^{-8}$ m across. Express this in nanometres. (show answer)

Answer
$80$ nm. Method: $8 \times 10^{-8} \text{ m} = 80 \times 10^{-9} \text{ m} = 80$ nm.
7. Simplify $\left(\dfrac{2a^3}{b^{-2}}\right)^{-1}$ . (show answer)

Answer
$\dfrac{b^{-2}}{2a^3} = \dfrac{1}{2a^3 b^2}$ . Method: flip and apply positive exponent.
8. If $x^{-2} = 9$ , find the value of $x^2$ . (show answer)

Answer
$x^2 = \dfrac{1}{9}$ . Since $x^{-2} = 9$ , take the reciprocal: $x^2 = \dfrac{1}{9}$ .

Reasoning · Tier 3: explain and apply

1. Explain why $a^{-n} = \dfrac{1}{a^n}$ must be true if the quotient rule is to remain valid for all integer exponents. (show answer)

Answer
The quotient rule says $\dfrac{a^m}{a^n} = a^{m-n}$ . When $m < n$ , we get $a^{m-n}$ with a negative exponent. But direct cancellation gives $\dfrac{1}{a^{n-m}}$ . For these to be equal, $a^{-(n-m)}$ must equal $\dfrac{1}{a^{n-m}}$ , which means $a^{-k} = \dfrac{1}{a^k}$ .
2. Simplify $\dfrac{(5a^2 b^{-3})^2}{(a^{-1} b^2)^3}$ completely. (show answer)

Answer
$\dfrac{25a^4 b^{-6}}{a^{-3} b^6} = 25 a^{4-(-3)} b^{-6-6} = 25 a^7 b^{-12} = \dfrac{25a^7}{b^{12}}$ .
3. The intensity of light decreases with the square of the distance: $I = I_0 \times d^{-2}$ . If you triple the distance, by what factor does the intensity change? (show answer)

Answer
Intensity at distance $3d$ : $I_0 \times (3d)^{-2} = I_0 \times \dfrac{1}{9d^2} = \dfrac{1}{9} \times I_0 d^{-2}$ . The intensity decreases by a factor of $9$ .
4. A student writes $(x + y)^{-1} = x^{-1} + y^{-1}$ . Disprove this with a counterexample and explain the error. (show answer)

Answer
Let $x = 1, y = 1$ . Then $(1+1)^{-1} = \dfrac{1}{2}$ , but $1^{-1} + 1^{-1} = 2$ . The error is distributing an exponent over addition — the power-of-a-product rule applies to multiplication, not addition.
5. Simplify $\dfrac{(2 \times 10^3)^{-2} \times (5 \times 10^{-1})}{10^{-4}}$ and give the answer in scientific notation. (show answer)

Answer
$(2 \times 10^3)^{-2} = 2^{-2} \times 10^{-6} = \dfrac{1}{4} \times 10^{-6}$ . Multiply by $5 \times 10^{-1}$ : $\dfrac{5}{4} \times 10^{-7} = 1.25 \times 10^{-7}$ . Divide by $10^{-4}$ : $1.25 \times 10^{-7-(-4)} = 1.25 \times 10^{-3}$ .

Reasoning · Harder reasoning

1. Simplify $\dfrac{a^{2n} \times a^{-n+1}}{(a^2)^{n-1}}$ where $n$ is a positive integer. (show answer)

Answer
$\dfrac{a^{2n} \times a^{-n+1}}{a^{2n-2}} = \dfrac{a^{2n + (-n+1)}}{a^{2n-2}} = \dfrac{a^{n+1}}{a^{2n-2}} = a^{(n+1)-(2n-2)} = a^{3-n}$ .
2. Prove that for any non-zero $a$ and integers $m, n$ : $(a^m)^{-n} = (a^{-m})^n = a^{-mn}$ . (show answer)

Answer
$(a^m)^{-n}$ : multiply exponents gives $a^{m \times (-n)} = a^{-mn}$ . $(a^{-m})^n$ : multiply exponents gives $a^{(-m) \times n} = a^{-mn}$ . Both equal $a^{-mn}$ .
3. A radioactive substance halves every year. Write the fraction remaining after $t$ years as a power of $2$ . After how many years is less than $\dfrac{1}{1000}$ of the substance left? (Hint: solve $2^{-t} < 10^{-3}$ .) (show answer)

Answer
Fraction remaining after $t$ years: $\left(\dfrac{1}{2}\right)^t = 2^{-t}$ . Need $2^{-t} < 10^{-3}$ , i.e. $2^t > 1000$ . Since $2^{10} = 1024 > 1000$ , after $10$ years less than $\dfrac{1}{1000}$ remains.
4. Simplify $\dfrac{(3x^{-2}y^3)^{-2} \times (9x^4 y^{-1})^2}{(xy)^{-4}}$ completely, writing the answer with positive exponents. (show answer)

Answer
$(3x^{-2}y^3)^{-2} = 3^{-2} x^4 y^{-6} = \dfrac{x^4}{9y^6}$ . $(9x^4 y^{-1})^2 = 81 x^8 y^{-2}$ . Product: $\dfrac{81 x^{12}}{9 y^8} = 9 x^{12} y^{-8}$ . Divide by $(xy)^{-4} = x^{-4} y^{-4}$ : $9 x^{12-(-4)} y^{-8-(-4)} = 9 x^{16} y^{-4} = \dfrac{9x^{16}}{y^4}$ .

Expanding binomials & factorising quadratics

Fluency · Tier 1: basic skills

1. Expand $(x + 2)(x + 6)$ . (show answer)

Answer
$x^2 + 8x + 12$ .
2. Expand $(x + 5)(x - 3)$ . (show answer)

Answer
$x^2 + 2x - 15$ .
3. Expand $(x - 4)(x - 7)$ . (show answer)

Answer
$x^2 - 11x + 28$ .
4. Expand $(x + 8)^2$ . (show answer)

Answer
$x^2 + 16x + 64$ .
5. Expand $(x - 3)^2$ . (show answer)

Answer
$x^2 - 6x + 9$ .
6. Expand $(x + 11)(x - 11)$ . (show answer)

Answer
$x^2 - 121$ .
7. Factorise $x^2 + 8x + 15$ . (show answer)

Answer
$(x + 3)(x + 5)$ . Method: $3 + 5 = 8$ and $3 \times 5 = 15$ .
8. Factorise $x^2 - 5x + 6$ . (show answer)

Answer
$(x - 2)(x - 3)$ . Method: $(-2) + (-3) = -5$ and $(-2)(-3) = 6$ .
9. Factorise $x^2 - 36$ . (show answer)

Answer
$(x + 6)(x - 6)$ . Difference of squares: $36 = 6^2$ .
10. Factorise $x^2 + 12x + 36$ . (show answer)

Answer
$(x + 6)^2$ . Perfect square: $12 = 2 \times 6$ and $36 = 6^2$ .

Reasoning · Tier 2: mixed practice

1. Expand and simplify $(x + 3)(x + 4) - (x + 1)(x + 2)$ . (show answer)

Answer
$(x^2 + 7x + 12) - (x^2 + 3x + 2) = 4x + 10$ . The $x^2$ terms cancel.
2. Factorise $x^2 + x - 20$ . (show answer)

Answer
$(x + 5)(x - 4)$ . Method: $5 + (-4) = 1$ and $5 \times (-4) = -20$ .
3. Factorise $x^2 - 3x - 28$ . (show answer)

Answer
$(x - 7)(x + 4)$ . Method: $(-7) + 4 = -3$ and $(-7)(4) = -28$ .
4. A square garden has side $(x + 5)$ m. A path of width $1$ m surrounds it. Find the area of the path in expanded form. (show answer)

Answer
Outer square side: $(x + 5 + 2) = (x + 7)$ m. Path area $= (x+7)^2 - (x+5)^2 = (x^2 + 14x + 49) - (x^2 + 10x + 25) = 4x + 24$ m $^2$ .
5. Factorise $x^2 - 100$ and hence evaluate $998^2 - 1000^2$ mentally. (show answer)

Answer
$x^2 - 100 = (x+10)(x-10)$ . So $998^2 - 1000^2 = (998+1000)(998-1000) = 1998 \times (-2) = -3996$ .
6. Show that $(x + a)^2 - (x - a)^2 = 4ax$ by expanding both sides. (show answer)

Answer
$(x+a)^2 = x^2 + 2ax + a^2$ . $(x-a)^2 = x^2 - 2ax + a^2$ . Subtracting: $(x^2 + 2ax + a^2) - (x^2 - 2ax + a^2) = 4ax$ .
7. Factorise $x^2 - 14x + 49$ . (show answer)

Answer
$(x - 7)^2$ . Perfect square: $14 = 2 \times 7$ and $49 = 7^2$ .
8. Factorise $2x^2 + 14x + 24$ completely. (show answer)

Answer
$2(x^2 + 7x + 12) = 2(x + 3)(x + 4)$ . Factor out $2$ first, then factorise the trinomial.

Reasoning · Tier 3: explain and apply

1. Explain why $(x + a)(x - a)$ has no $x$ term. Use the area model or algebra to justify. (show answer)

Answer
When expanding $(x+a)(x-a)$ , the middle terms are $+ax$ and $-ax$ , which sum to zero. Using the area model: the two rectangular strips ( $a \times x$ and $x \times a$ ) have opposite signs and cancel, leaving only $x^2 - a^2$ .
2. A rectangle has area $x^2 + 9x + 18$ cm $^2$ . Find expressions for its length and width. (show answer)

Answer
$x^2 + 9x + 18 = (x + 3)(x + 6)$ . Method: $3 + 6 = 9$ and $3 \times 6 = 18$ . So the length is $(x + 6)$ cm and the width is $(x + 3)$ cm (or vice versa).
3. Without expanding, decide whether $(x + 3)^2$ and $x^2 + 9$ are equivalent. Explain. (show answer)

Answer
They are not equivalent. $(x+3)^2 = x^2 + 6x + 9$ , which has a $6x$ term that $x^2 + 9$ lacks. For example, at $x = 1$ : $(1+3)^2 = 16$ but $1 + 9 = 10$ .
4. Factorise $x^2 - 2x - 35$ and use your factorisation to solve $x^2 - 2x - 35 = 0$ . (show answer)

Answer
$x^2 - 2x - 35 = (x - 7)(x + 5) = 0$ . So $x = 7$ or $x = -5$ .
5. Explain the connection between expanding and factorising using the analogy of multiplication and division of numbers. (show answer)

Answer
Expanding is like multiplication: $3 \times 4 = 12$ breaks a product into a single value. Factorising is like finding factors: $12 = 3 \times 4$ rewrites a value as a product. They undo each other. In algebra, expanding turns $(x+3)(x+4)$ into $x^2 + 7x + 12$ , and factorising reverses the process.

Reasoning · Harder reasoning

1. Expand $(2x + 3)(x - 4)$ . (Note: this is a non-monic product — the coefficient of $x^2$ is not $1$ .) (show answer)

Answer
$(2x+3)(x-4) = 2x^2 - 8x + 3x - 12 = 2x^2 - 5x - 12$ .
2. Factorise $x^4 - 16$ completely. (Hint: treat $x^4$ as $(x^2)^2$ and apply difference of squares twice.) (show answer)

Answer
$x^4 - 16 = (x^2)^2 - 4^2 = (x^2 + 4)(x^2 - 4)$ . Then $x^2 - 4 = (x+2)(x-2)$ , so $x^4 - 16 = (x^2 + 4)(x + 2)(x - 2)$ . The factor $(x^2 + 4)$ does not factorise further over the reals.
3. Prove that the sum of any two consecutive odd numbers is divisible by $4$ . (Hint: let the odd numbers be $2n - 1$ and $2n + 1$ , and use difference of squares.) (show answer)

Answer
Product of two consecutive odd numbers: $(2n-1)(2n+1) = 4n^2 - 1$ . Their sum is $(2n-1) + (2n+1) = 4n$ , which is divisible by $4$ . (Note: the question asks about the sum, not the product.)
4. If $x + \dfrac{1}{x} = 5$ , find the value of $x^2 + \dfrac{1}{x^2}$ . (Hint: square both sides and simplify.) (show answer)

Answer
Square both sides of $x + \dfrac{1}{x} = 5$ : $x^2 + 2 + \dfrac{1}{x^2} = 25$ . So $x^2 + \dfrac{1}{x^2} = 25 - 2 = 23$ .

Linear graphs, gradient & midpoint

Fluency · Tier 1: gradient, intercept and sketching

1. State the gradient and $y$ -intercept of $y = 5x - 2$ . (show answer)

Answer
Gradient $= 5$ , $y$ -intercept $= -2$ .
2. State the gradient and $y$ -intercept of $y = -3x + 7$ . (show answer)

Answer
Gradient $= -3$ , $y$ -intercept $= 7$ .
3. Rewrite $2x + y = 10$ in the form $y = mx + c$ . (show answer)

Answer
$y = -2x + 10$ .
4. Rewrite $6x - 3y = 9$ in gradient-intercept form. (show answer)

Answer
$y = 2x - 3$ .
5. Find the $x$ - and $y$ -intercepts of $5x + 2y = 20$ . (show answer)

Answer
$x$ -intercept: $(4, 0)$ ; $y$ -intercept: $(0, 10)$ .
6. Find the gradient of the line through $(1, 2)$ and $(4, 11)$ . (show answer)

Answer
$m = \dfrac{11 - 2}{4 - 1} = \dfrac{9}{3} = 3$ .
7. Find the gradient of the line through $(-3, 5)$ and $(1, -7)$ . (show answer)

Answer
$m = \dfrac{-7 - 5}{1 - (-3)} = \dfrac{-12}{4} = -3$ .
8. Find the midpoint of $(2, 6)$ and $(10, 14)$ . (show answer)

Answer
$M = \left(\dfrac{2+10}{2},\;\dfrac{6+14}{2}\right) = (6, 10)$ .
9. Find the distance between $(0, 0)$ and $(5, 12)$ . (show answer)

Answer
$d = \sqrt{25 + 144} = \sqrt{169} = 13$ .
10. Find the distance between $(-1, 3)$ and $(2, 7)$ . (show answer)

Answer
$d = \sqrt{9 + 16} = \sqrt{25} = 5$ .

Reasoning · Tier 2: mixed practice

1. A line passes through $(0, -4)$ with gradient $\dfrac{2}{3}$ . Write its equation. (show answer)

Answer
$y = \dfrac{2}{3}x - 4$ .
2. Find the equation of the line through $(1, 5)$ and $(3, 11)$ . (show answer)

Answer
Gradient: $m = \dfrac{11-5}{3-1} = 3$ . Using $(1, 5)$ : $5 = 3(1) + c$ , $c = 2$ . Equation: $y = 3x + 2$ .
3. Determine whether the lines $y = 2x + 3$ and $4x - 2y = 7$ are parallel, perpendicular, or neither. (show answer)

Answer
Rearrange $4x - 2y = 7$ : $y = 2x - \dfrac{7}{2}$ . Gradient $= 2$ . Same gradient as $y = 2x + 3$ , so the lines are parallel.
4. Find the midpoint and length of the segment joining $A(-2, 3)$ and $B(6, -1)$ . (show answer)

Answer
Midpoint: $\left(\dfrac{-2+6}{2},\;\dfrac{3+(-1)}{2}\right) = (2, 1)$ . Length: $\sqrt{8^2 + 4^2} = \sqrt{80} = 4\sqrt{5}$ .
5. A line has equation $3x + 4y = 24$ . Find its gradient, $x$ -intercept and $y$ -intercept. (show answer)

Answer
Rearrange: $y = -\dfrac{3}{4}x + 6$ . Gradient $= -\dfrac{3}{4}$ . $y$ -intercept: $(0, 6)$ . $x$ -intercept: $(8, 0)$ .
6. Find the equation of the line perpendicular to $y = -2x + 1$ that passes through $(4, 3)$ . (show answer)

Answer
Gradient of perpendicular: $m = \dfrac{1}{2}$ . Using $(4, 3)$ : $3 = \dfrac{1}{2}(4) + c$ , $c = 1$ . Equation: $y = \dfrac{1}{2}x + 1$ .
7. Show that the triangle with vertices $A(0, 0)$ , $B(4, 0)$ and $C(4, 3)$ is right-angled by calculating all three side lengths. (show answer)

Answer
$AB = 4$ , $BC = 3$ , $AC = \sqrt{16 + 9} = 5$ . Check: $3^2 + 4^2 = 9 + 16 = 25 = 5^2$ . Right-angled at $B$ .
8. The midpoint of $P(a, 3)$ and $Q(5, 9)$ is $(4, 6)$ . Find the value of $a$ . (show answer)

Answer
Midpoint $x$ -coordinate: $\dfrac{a + 5}{2} = 4$ , so $a + 5 = 8$ , $a = 3$ .

Reasoning · Tier 3: explain and apply

1. Explain why a vertical line cannot be written in the form $y = mx + c$ . What happens to the gradient formula when $x_1 = x_2$ ? (show answer)

Answer
A vertical line has the form $x = k$ . In the gradient formula, $x_1 = x_2$ makes the denominator zero, so $m$ is undefined. Since $y = mx + c$ requires a defined $m$ , vertical lines cannot be written in this form.
2. Two hikers start at point $A(1, 2)$ on a grid map (km units). Hiker 1 walks to $B(7, 10)$ . Hiker 2 walks to $C(9, 5)$ . Who walks further, and by how much? (show answer)

Answer
Hiker 1: $d = \sqrt{6^2 + 8^2} = \sqrt{100} = 10$ km. Hiker 2: $d = \sqrt{8^2 + 3^2} = \sqrt{73} \approx 8.54$ km. Hiker 1 walks further by $(10 - \sqrt{73}) \approx 1.46$ km.
3. A quadrilateral has vertices $P(0, 0)$ , $Q(6, 0)$ , $R(8, 4)$ , $S(2, 4)$ . By calculating gradients, show that $PQRS$ is a parallelogram. (show answer)

Answer
Gradient of $PQ$ : $\dfrac{0}{6} = 0$ . Gradient of $SR$ : $\dfrac{0}{6} = 0$ . Gradient of $PS$ : $\dfrac{4}{2} = 2$ . Gradient of $QR$ : $\dfrac{4}{2} = 2$ . Opposite sides have equal gradients, so $PQRS$ is a parallelogram.
4. The line $y = kx + 2$ is perpendicular to $y = 4x - 1$ . Find $k$ . (show answer)

Answer
Perpendicular gradients: $k \times 4 = -1$ , so $k = -\dfrac{1}{4}$ .
5. Point $M(3, 5)$ is the midpoint of $A(1, 2)$ and $B$ . Find the coordinates of $B$ . (show answer)

Answer
Midpoint formula: $3 = \dfrac{1 + x_B}{2}$ gives $x_B = 5$ ; $5 = \dfrac{2 + y_B}{2}$ gives $y_B = 8$ . So $B = (5, 8)$ .

Reasoning · Harder reasoning

1. Prove that the diagonals of the rectangle with vertices $A(0, 0)$ , $B(8, 0)$ , $C(8, 6)$ , $D(0, 6)$ bisect each other by finding both midpoints. (show answer)

Answer
Midpoint of $AC$ : $\left(\dfrac{0+8}{2},\;\dfrac{0+6}{2}\right) = (4, 3)$ . Midpoint of $BD$ : $\left(\dfrac{8+0}{2},\;\dfrac{0+6}{2}\right) = (4, 3)$ . Both midpoints are the same, so the diagonals bisect each other.
2. A line passes through $(2, -1)$ and is perpendicular to the line $3x - 5y = 10$ . Find its equation in general form $ax + by = c$ . (show answer)

Answer
Gradient of $3x - 5y = 10$ is $\dfrac{3}{5}$ . Perpendicular gradient: $-\dfrac{5}{3}$ . Through $(2, -1)$ : $y + 1 = -\dfrac{5}{3}(x - 2)$ . Multiply by $3$ : $3y + 3 = -5x + 10$ . Rearrange: $5x + 3y = 7$ .
3. Three points are $A(1, 1)$ , $B(5, 3)$ , $C(3, 5)$ . Find the perimeter of triangle $ABC$ , giving your answer in exact (surd) form. (show answer)

Answer
$AB = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$ . $BC = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$ . $AC = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$ . Perimeter $= 4\sqrt{5} + 2\sqrt{2}$ .
4. The vertices of a triangle are $P(-1, 4)$ , $Q(5, 0)$ and $R(3, 8)$ . Find the length of the median from $P$ to the midpoint of $QR$ . (show answer)

Answer
Midpoint of $QR$ : $\left(\dfrac{5+3}{2},\;\dfrac{0+8}{2}\right) = (4, 4)$ . Distance from $P(-1, 4)$ to $(4, 4)$ : $\sqrt{25 + 0} = 5$ .

Quadratic functions & equations

Fluency · Tier 1: identify and factorise

1. For $y = 2x^2 - 4x + 1$ , state $a$ , $b$ , $c$ and whether the parabola opens up or down. (show answer)

Answer
$a = 2$ , $b = -4$ , $c = 1$ . Since $a = 2 > 0$ , the parabola opens upward.
2. Find the $y$ -intercept of $y = -x^2 + 3x - 5$ . (show answer)

Answer
$y$ -intercept: $(0, -5)$ .
3. Factorise $x^2 + 7x + 12$ . (show answer)

Answer
$(x + 3)(x + 4)$ .
4. Factorise $x^2 - 9x + 20$ . (show answer)

Answer
$(x - 4)(x - 5)$ .
5. Factorise $x^2 + 2x - 15$ . (show answer)

Answer
$(x + 5)(x - 3)$ .
6. Solve $(x - 3)(x + 7) = 0$ . (show answer)

Answer
$x = 3$ or $x = -7$ .
7. Solve $x^2 - 5x + 6 = 0$ . (show answer)

Answer
$(x - 2)(x - 3) = 0$ : $x = 2$ or $x = 3$ .
8. Solve $x^2 + x - 12 = 0$ . (show answer)

Answer
$(x + 4)(x - 3) = 0$ : $x = -4$ or $x = 3$ .
9. Solve $x^2 - 16 = 0$ . (show answer)

Answer
$(x - 4)(x + 4) = 0$ : $x = 4$ or $x = -4$ .
10. Solve $3x^2 - 12x = 0$ . (show answer)

Answer
$3x(x - 4) = 0$ : $x = 0$ or $x = 4$ .

Reasoning · Tier 2: features and graphs

1. Find the axis of symmetry and turning point of $y = x^2 - 4x + 3$ . (show answer)

Answer
$x = -\dfrac{-4}{2(1)} = 2$ . Turning point: $y = 4 - 8 + 3 = -1$ . Turning point is $(2, -1)$ .
2. Find the $x$ -intercepts, turning point and $y$ -intercept of $y = x^2 - 2x - 8$ . Sketch the parabola. (show answer)

Answer
$x$ -intercepts: $x^2 - 2x - 8 = (x - 4)(x + 2) = 0$ , so $x = 4$ or $x = -2$ . Axis of symmetry: $x = 1$ . Turning point: $y = 1 - 2 - 8 = -9$ , so $(1, -9)$ . $y$ -intercept: $(0, -8)$ . Parabola opens upward.
3. A parabola has $x$ -intercepts at $x = -3$ and $x = 1$ . Find the axis of symmetry and the turning point if the equation is $y = x^2 + 2x - 3$ . (show answer)

Answer
Axis of symmetry: $x = \dfrac{-3 + 1}{2} = -1$ . Turning point: $y = 1 - 2 - 3 = -4$ , so $(-1, -4)$ .
4. Solve $x^2 = 7x - 10$ by first rearranging to standard form. (show answer)

Answer
$x^2 - 7x + 10 = 0$ . $(x - 2)(x - 5) = 0$ : $x = 2$ or $x = 5$ .
5. Solve $2x^2 - 14x + 20 = 0$ by first taking out a common factor. (show answer)

Answer
$2(x^2 - 7x + 10) = 0$ . $2(x - 2)(x - 5) = 0$ : $x = 2$ or $x = 5$ .
6. Explain why $x^2 + 4 = 0$ has no real solutions. What does this mean for the graph? (show answer)

Answer
$x^2 = -4$ has no real solution because a square is never negative. The parabola $y = x^2 + 4$ has its turning point at $(0, 4)$ , entirely above the $x$ -axis, so it never crosses it.
7. A rectangle has length $(x + 3)$ cm and width $(x - 1)$ cm. Its area is $21$ cm $^2$ . Find $x$ . (show answer)

Answer
$(x + 3)(x - 1) = 21$ . Expand: $x^2 + 2x - 3 = 21$ , so $x^2 + 2x - 24 = 0$ . $(x + 6)(x - 4) = 0$ : $x = -6$ or $x = 4$ . Since $x - 1 > 0$ , we need $x > 1$ , so $x = 4$ . The rectangle is $7$ cm by $3$ cm.
8. The product of two consecutive integers is $72$ . Find the integers. (show answer)

Answer
Let the integers be $n$ and $n + 1$ . $n(n + 1) = 72$ , so $n^2 + n - 72 = 0$ . $(n + 9)(n - 8) = 0$ : $n = 8$ or $n = -9$ . The consecutive integers are $8$ and $9$ (or $-9$ and $-8$ ).

Reasoning · Tier 3: explain and apply

1. Explain the connection between the factorised form $y = (x - p)(x - q)$ and the $x$ -intercepts. How do you find the axis of symmetry from $p$ and $q$ ? (show answer)

Answer
In $y = (x - p)(x - q)$ , the $x$ -intercepts are $x = p$ and $x = q$ (by the null factor law). The axis of symmetry is $x = \dfrac{p + q}{2}$ (midpoint of the intercepts).
2. A ball is launched upward with height $h = 24t - 4t^2$ metres after $t$ seconds. Find when it hits the ground and its maximum height. (show answer)

Answer
Ground: $24t - 4t^2 = 0$ , $4t(6 - t) = 0$ , so $t = 0$ or $t = 6$ s. Axis of symmetry: $t = 3$ . Max height: $h = 24(3) - 4(9) = 72 - 36 = 36$ m.
3. Two numbers add to $12$ and their product is $32$ . Set up and solve a quadratic equation to find them. (show answer)

Answer
Let the numbers be $x$ and $12 - x$ . Product: $x(12 - x) = 32$ , so $12x - x^2 = 32$ , $x^2 - 12x + 32 = 0$ . $(x - 4)(x - 8) = 0$ : $x = 4$ or $x = 8$ . The numbers are $4$ and $8$ .
4. The parabola $y = x^2 + bx + 9$ has only one $x$ -intercept. Find the two possible values of $b$ . (show answer)

Answer
One $x$ -intercept means the discriminant is zero: $b^2 - 4(1)(9) = 0$ , so $b^2 = 36$ , $b = 6$ or $b = -6$ .

Reasoning · Harder reasoning

1. A farmer has $60$ m of fencing to enclose a rectangular paddock against a wall (only three sides need fencing). If the width is $x$ m, show that the area is $A = 60x - 2x^2$ and find the dimensions that maximise the area. (show answer)

Answer
Width $= x$ , length $= 60 - 2x$ . Area $= x(60 - 2x) = 60x - 2x^2$ . This is a downward parabola. Axis of symmetry: $x = -\dfrac{60}{2(-2)} = 15$ . Maximum area: $A = 60(15) - 2(225) = 900 - 450 = 450$ m $^2$ . Dimensions: $15$ m wide, $30$ m long.
2. The parabola $y = x^2 - 6x + k$ passes through the point $(1, 2)$ . Find $k$ , then find the turning point and $x$ -intercepts. (show answer)

Answer
Substitute $(1, 2)$ : $2 = 1 - 6 + k$ , so $k = 7$ . Equation: $y = x^2 - 6x + 7$ . Turning point: $x = 3$ , $y = 9 - 18 + 7 = -2$ , so $(3, -2)$ . $x$ -intercepts: $x^2 - 6x + 7 = 0$ . Using the quadratic formula (or completing the square): $x = \dfrac{6 \pm \sqrt{36 - 28}}{2} = \dfrac{6 \pm 2\sqrt{2}}{2} = 3 \pm \sqrt{2}$ .
3. Show that if a monic quadratic $x^2 + bx + c = 0$ has solutions $p$ and $q$ , then $p + q = -b$ and $pq = c$ . (show answer)

Answer
$(x - p)(x - q) = x^2 - (p+q)x + pq$ . Comparing with $x^2 + bx + c$ : $-(p+q) = b$ so $p + q = -b$ , and $pq = c$ .
4. The sum of the squares of two consecutive positive odd numbers is $130$ . Find the numbers. (show answer)

Answer
Let the numbers be $n$ and $n + 2$ . $n^2 + (n+2)^2 = 130$ . $2n^2 + 4n + 4 = 130$ . $n^2 + 2n - 63 = 0$ . $(n + 9)(n - 7) = 0$ : $n = 7$ (positive). The numbers are $7$ and $9$ .

Linear modelling & simple interest

Fluency · Tier 1: basic skills

1. A taxi charges a $4.50 flag-fall plus $2.20 per km. Write a linear model for the fare $F$ after $d$ km. (show answer)

Answer
$F = 2.2d + 4.5$ .
2. Using your model from Q1, find the fare for a $12$ km trip. (show answer)

Answer
$F = 2.2(12) + 4.5 = 26.4 + 4.5 = 30.90$ dollars.
3. Calculate the simple interest on $3000 at $5\%$ p.a. for $2$ years. (show answer)

Answer
$I = 3000 \times 0.05 \times 2 = 300$ dollars.
4. Find the total amount when $8000 is invested at $3.5\%$ p.a. simple interest for $4$ years. (show answer)

Answer
$I = 8000 \times 0.035 \times 4 = 1120$ dollars. Total: $A = 8000 + 1120 = 9120$ dollars.
5. A phone battery starts at $100\%$ and drains at $8\%$ per hour. Write a linear model for battery percentage $B$ after $t$ hours. (show answer)

Answer
$B = -8t + 100$ .
6. Using your model from Q5, after how many hours will the battery reach $20\%$ ? (show answer)

Answer
$20 = -8t + 100$ , so $8t = 80$ , $t = 10$ hours.
7. A pool contains $12\,000$ litres. Water drains at $500$ litres per hour. Write a model for the volume $V$ after $t$ hours. (show answer)

Answer
$V = -500t + 12000$ .
8. Gemma earns $18.50 per hour plus a $25 daily transport allowance. Write a model for her daily earnings $E$ after $h$ hours. (show answer)

Answer
$E = 18.5h + 25$ .

Reasoning · Tier 2: mixed practice

1. Plan X charges $30 per month plus $0.10 per text. Plan Y charges $15 per month plus $0.40 per text. Find the break-even number of texts and state which plan is cheaper for $60$ texts per month. (show answer)

Answer
$30 + 0.1n = 15 + 0.4n$ , so $15 = 0.3n$ , $n = 50$ texts. At $60$ texts: Plan X $= 36$ , Plan Y $= 39$ . Plan X is cheaper.
2. Omar invests $4000 at simple interest and after $5$ years has $4800. Find the annual interest rate. (show answer)

Answer
$I = 4800 - 4000 = 800$ . $800 = 4000 \times r \times 5$ . $r = 0.04 = 4\%$ p.a.
3. A car is worth $25000 new and depreciates by $3000 per year (linear model). Write the model for its value $V$ after $t$ years and find when it will be worth $7000. (show answer)

Answer
$V = -3000t + 25000$ . Set $V = 7000$ : $7000 = -3000t + 25000$ , $3000t = 18000$ , $t = 6$ years.
4. Two runners start a race. Runner A starts $20$ m ahead and runs at $6$ m/s. Runner B starts at the start line and runs at $8$ m/s. Write models for their positions and find when Runner B catches Runner A. (show answer)

Answer
Runner A: $d_A = 6t + 20$ . Runner B: $d_B = 8t$ . Set equal: $8t = 6t + 20$ , $2t = 20$ , $t = 10$ s. Position: $80$ m from the start.
5. The cost of hiring a marquee is modelled by $C = 150 + 45h$ , where $h$ is the number of hours. Interpret the $150$ and the $45$ in context. (show answer)

Answer
The $150 is the fixed hire cost (charged regardless of time). The $45 is the hourly rate (cost per additional hour).
6. A swimming pool is being filled. After $3$ hours it has $2700$ litres; after $7$ hours it has $5100$ litres. Find the linear model and state how much water was in the pool initially. (show answer)

Answer
Gradient: $m = \dfrac{5100 - 2700}{7 - 3} = \dfrac{2400}{4} = 600$ L/h. Using $(3, 2700)$ : $2700 = 600(3) + c$ , $c = 900$ . Model: $W = 600t + 900$ . The pool initially had $900$ litres.

Reasoning · Tier 3: explain and apply

1. Explain why the simple interest formula $A = P(1 + rT)$ is linear in $T$ . What is the gradient of the graph of $A$ vs $T$ ? (show answer)

Answer
$A = P(1 + rT) = PrT + P$ . This has the form $y = mx + c$ with $m = Pr$ (constant interest per year) and $c = P$ (initial deposit). The gradient is $Pr$ .
2. Maria has $10000 to invest. Bank A offers $4\%$ p.a. simple interest. Bank B offers $3\%$ p.a. simple interest plus a one-off $200 bonus at account opening. Which bank gives more money after $8$ years? (show answer)

Answer
Bank A after $8$ years: $A = 10000(1 + 0.04 \times 8) = 10000 \times 1.32 = 13200$ dollars. Bank B after $8$ years: $A = 10000(1 + 0.03 \times 8) + 200 = 10000 \times 1.24 + 200 = 12600$ dollars. Bank A gives $600 more.
3. A hire company charges $120 for the first day and $80 for each additional day. Is this a linear model from day $1$ ? Write the model for total cost $C$ after $d$ days ( $d \geq 1$ ) and explain. (show answer)

Answer
For $d \geq 1$ : first day costs $120, each additional day costs $80. Total: $C = 80(d - 1) + 120 = 80d + 40$ . Yes, this is linear in $d$ : gradient $= 80$ (daily rate), $y$ -intercept $= 40$ (but note the model only applies for $d \geq 1$ , so the actual starting cost is $C(1) = 120$ ).
4. Two towns are connected by road ( $120$ km). Car A leaves Town X at $60$ km/h. Car B leaves Town Y at the same time at $40$ km/h, driving toward Town X. Write position models for each car (from Town X) and find when and where they meet. (show answer)

Answer
Car A from Town X: $d_A = 60t$ . Car B from Town X: $d_B = 120 - 40t$ . Set equal: $60t = 120 - 40t$ , $100t = 120$ , $t = 1.2$ hours ( $1$ h $12$ min). Position: $60 \times 1.2 = 72$ km from Town X.

Reasoning · Harder reasoning

1. Aisha borrows $15000 at $6\%$ p.a. simple interest. She repays $500 at the end of each year (after interest is calculated on the original principal). Write a model for the amount still owed after $n$ years and find after how many years the loan is fully repaid. (show answer)

Answer
Interest per year: $15000 \times 0.06 = 900$ dollars. Net reduction per year: $500 - 900 = -400$ dollars (the debt grows by $400 each year because repayments do not cover interest). Actually, since the question states interest is on the original principal: amount owed after $n$ years $= 15000 + 900n - 500n = 15000 + 400n$ . The debt increases, so the loan is never repaid under these terms. Alternatively, if we interpret "repays $500 after interest" as reducing the balance: total interest over $n$ years is $900n$ . Total repaid: $500n$ . Amount owed: $15000 + 900n - 500n = 15000 + 400n$ . The repayments ($500/year) are less than the annual interest ($900), so the loan balance grows. The repayment amount would need to exceed $900 per year to reduce the debt.
2. A phone company offers three plans. Plan A: $0 monthly, $0.80 per minute. Plan B: $20 monthly, $0.30 per minute. Plan C: $50 monthly, unlimited calls. Find the usage ranges (in minutes per month) for which each plan is cheapest. (show answer)

Answer
Plan A vs B: $0.8m = 20 + 0.3m$ , $0.5m = 20$ , $m = 40$ min. Plan B vs C: $20 + 0.3m = 50$ , $0.3m = 30$ , $m = 100$ min. Plan A cheapest: $0$ to $40$ min. Plan B cheapest: $40$ to $100$ min. Plan C cheapest: over $100$ min.
3. A candle and a sparkler are lit at the same time. The candle is $20$ cm tall and burns down $2$ cm per minute. The sparkler is $12$ cm tall and burns down $4$ cm per minute. Find when they are the same height and when each burns out completely. (show answer)

Answer
Candle: $h_c = -2t + 20$ , burns out at $t = 10$ min. Sparkler: $h_s = -4t + 12$ , burns out at $t = 3$ min. Same height: $-2t + 20 = -4t + 12$ , $2t = -8$ , $t = -4$ . Since $t = -4$ is negative, they are never the same height during the time both are burning (the sparkler starts shorter and burns faster, so the candle is always taller while both exist).

Transformations of functions

Fluency · Tier 1: identify the transformation

1. State the vertex of $y = x^2 + 7$ . (show answer)

Answer
$(0, 7)$
2. State the vertex of $y = x^2 - 4$ . (show answer)

Answer
$(0, -4)$
3. Is $y = 5x^2$ narrower or wider than $y = x^2$ ? (show answer)

Answer
Narrower (since $5 > 1$ ).
4. Is $y = 0.3x^2$ narrower or wider than $y = x^2$ ? (show answer)

Answer
Wider (since $0.3 < 1$ ).
5. Does $y = -x^2$ open upward or downward? (show answer)

Answer
Downward.
6. State the vertex of $y = (x - 6)^2$ . (show answer)

Answer
$(6, 0)$
7. State the vertex of $y = (x + 2)^2 - 1$ . (show answer)

Answer
$(-2, -1)$
8. Write the equation of $y = x^2$ shifted $5$ units down. (show answer)

Answer
$y = x^2 - 5$
9. Write the equation of $y = x^2$ shifted $3$ units right and $4$ units up. (show answer)

Answer
$y = (x - 3)^2 + 4$

Reasoning · Tier 2: mixed practice

1. Match each equation to its vertex: (a) $y = (x - 1)^2 + 2$ , (b) $y = (x + 3)^2 - 5$ , (c) $y = -x^2 + 4$ . Vertices: $(0, 4)$ , $(1, 2)$ , $(-3, -5)$ . (show answer)

Answer
(a) $(1, 2)$ , (b) $(-3, -5)$ , (c) $(0, 4)$ .
2. A parabola has vertex $(2, -3)$ and opens upward with the same width as $y = x^2$ . Write its equation. (show answer)

Answer
$y = (x - 2)^2 - 3$ .
3. Describe two different transformations that could move the vertex of $y = x^2$ to $(0, 9)$ . (show answer)

Answer
Method 1: vertical translation $y = x^2 + 9$ . Method 2: vertical stretch $y = 9x^2$ (passes through $(1, 9)$ but vertex is still at origin, so only the translation gives vertex $(0, 9)$ ). Accept $y = x^2 + 9$ and $y = -(x)^2 + 9$ (reflection plus shift).
4. The graph of $y = ax^2$ passes through $(1, 6)$ . Find $a$ . (show answer)

Answer
$a = 6$ . Method: $6 = a \times 1^2$ .
5. Arrange from widest to narrowest: $y = 4x^2$ , $y = x^2$ , $y = \tfrac{1}{4}x^2$ , $y = 2x^2$ . (show answer)

Answer
Widest to narrowest: $y = \tfrac{1}{4}x^2$ , $y = x^2$ , $y = 2x^2$ , $y = 4x^2$ .
6. A parabola opens downward, is narrower than $y = x^2$ , and has vertex at $(1, 5)$ . Write a possible equation. (show answer)

Answer
One possible answer: $y = -2(x - 1)^2 + 5$ . Accept any equation with $a < -1$ , $h = 1$ , $k = 5$ .

Reasoning · Tier 3: explain and apply

1. Explain why replacing $x$ with $(x - h)$ shifts the graph to the right rather than the left. (show answer)

Answer
Replacing $x$ with $(x - h)$ means we need a larger $x$ -value to produce the same output. For example, the vertex of $y = x^2$ is at $x = 0$ ; for $y = (x - 3)^2$ the output is $0$ when $x - 3 = 0$ , i.e. $x = 3$ . Every point shifts right by $h$ .
2. Sam says " $y = -(x - 2)^2$ has vertex at $(-2, 0)$ ." Identify and correct the error. (show answer)

Answer
The vertex is at $(2, 0)$ , not $(-2, 0)$ . Sam confused the sign: $(x - 2)$ means shift right $2$ .
3. A ball's height is modelled by $y = -5(x - 1)^2 + 8$ , where $x$ is time in seconds. State the maximum height and when it occurs. (show answer)

Answer
Maximum height is $8$ m, occurring at $x = 1$ s. The vertex $(1, 8)$ gives the peak because $a = -5 < 0$ (opens downward).
4. Two parabolas have equations $y = 2(x - 3)^2 + 1$ and $y = 2(x - 3)^2 - 4$ . How are they related? What is the vertical distance between their vertices? (show answer)

Answer
Both have the same shape ( $a = 2$ ) and the same axis of symmetry ( $x = 3$ ). The second is a vertical translation of the first, shifted $5$ units down. Vertical distance between vertices: $1 - (-4) = 5$ units.

Reasoning · Harder reasoning

1. Find the equation in vertex form of a parabola that opens downward, passes through $(0, 0)$ and $(4, 0)$ , and has a maximum value of $8$ . (show answer)

Answer
$y = -2(x - 2)^2 + 8$ . Method: axis of symmetry at $x = \tfrac{0+4}{2} = 2$ ; vertex $(2, 8)$ ; sub $(0, 0)$ : $0 = a(0-2)^2 + 8$ , so $4a = -8$ , $a = -2$ .
2. The graph of $y = a(x - h)^2 + k$ passes through $(0, 5)$ and $(6, 5)$ and has a minimum value of $-4$ . Find $a$ , $h$ , and $k$ . (show answer)

Answer
$a = 1$ , $h = 3$ , $k = -4$ . Method: axis of symmetry at $x = \tfrac{0+6}{2} = 3$ ; minimum is $k = -4$ ; vertex $(3, -4)$ ; sub $(0, 5)$ : $5 = a(9) - 4$ , $9a = 9$ , $a = 1$ .
3. A parabola $y = a(x - h)^2 + k$ has vertex $(3, 7)$ and passes through $(5, -1)$ . Find the values of $a$ , $h$ , and $k$ , and state whether the parabola opens upward or downward. (show answer)

Answer
$a = -2$ , $h = 3$ , $k = 7$ . Method: vertex gives $h = 3$ , $k = 7$ ; sub $(5, -1)$ : $-1 = a(5-3)^2 + 7$ , $4a = -8$ , $a = -2$ . Opens downward since $a < 0$ .
4. Explain why the graph of $y = (x - p)(x - q)$ has its axis of symmetry at $x = \dfrac{p + q}{2}$ , and find the vertex coordinates in terms of $p$ and $q$ . (show answer)

Answer
The axis of symmetry is the midpoint of the roots $p$ and $q$ , so $x = \tfrac{p+q}{2}$ . Substituting: $y = \bigl(\tfrac{p+q}{2} - p\bigr)\bigl(\tfrac{p+q}{2} - q\bigr) = \bigl(\tfrac{q-p}{2}\bigr)\bigl(\tfrac{p-q}{2}\bigr) = -\tfrac{(p-q)^2}{4}$ . Vertex: $\bigl(\tfrac{p+q}{2},\; -\tfrac{(p-q)^2}{4}\bigr)$ .

Surface area & volume (prisms & cylinders)

Fluency · Tier 1: basic calculations

1. Find the surface area of a cuboid $10 \times 6 \times 4$ cm. (show answer)

Answer
$248$ cm $^2$ . Method: $2(60 + 40 + 24)$ .
2. Find the surface area of a cube of side $7$ cm. (show answer)

Answer
$294$ cm $^2$ . Method: $6 \times 49$ .
3. A triangular prism has an equilateral triangle base of side $5$ cm (height $\approx 4.33$ cm) and length $10$ cm. Find its surface area. (show answer)

Answer
$171.65$ cm $^2$ (approx). Method: two triangles $2 \times \tfrac{1}{2}(5)(4.33) = 21.65$ ; three rectangles $3 \times 5 \times 10 = 150$ ; total $\approx 171.65$ .
4. Find the total surface area of a cylinder with $r = 3$ cm and $h = 10$ cm. Give your answer in terms of $\pi$ and as a decimal. (show answer)

Answer
$78\pi \approx 245.0$ cm $^2$ . Method: $2\pi(9) + 2\pi(3)(10) = 18\pi + 60\pi$ .
5. Find the volume of a cylinder with $r = 6$ cm and $h = 20$ cm. Leave your answer in terms of $\pi$ . (show answer)

Answer
$720\pi$ cm $^3$ . Method: $\pi(36)(20)$ .
6. A cylinder has $r = 4$ cm and $h = 8$ cm. Find (a) the curved surface area, (b) the total surface area, (c) the volume. (show answer)

Answer
(a) $64\pi \approx 201.1$ cm $^2$ . (b) $96\pi \approx 301.6$ cm $^2$ . (c) $128\pi \approx 402.1$ cm $^3$ .
7. Convert a cylinder volume of $2500$ cm $^3$ to litres. (show answer)

Answer
$2.5$ L. Method: $2500 \div 1000$ .
8. Find the surface area of a cylinder with diameter $14$ cm and height $9$ cm. (show answer)

Answer
$\approx 703.7$ cm $^2$ . Method: $r = 7$ ; $2\pi(49) + 2\pi(7)(9) = 98\pi + 126\pi = 224\pi$ .

Reasoning · Tier 2: mixed practice

1. A cylindrical can has volume $1000$ cm $^3$ and height $12$ cm. Find its radius to one decimal place. (show answer)

Answer
$r \approx 5.2$ cm. Method: $\pi r^2 \times 12 = 1000$ ; $r^2 = \tfrac{1000}{12\pi} \approx 26.53$ ; $r \approx 5.2$ .
2. Two cylinders have the same volume. Cylinder A has $r = 5$ cm. Cylinder B has $r = 10$ cm. If A has height $20$ cm, find the height of B. (show answer)

Answer
$h = 5$ cm. Method: $\pi(25)(20) = \pi(100)h$ ; $h = \tfrac{500}{100} = 5$ .
3. A closed cylinder uses $600\pi$ cm $^2$ of sheet metal. If $r = 10$ cm, find $h$ . (show answer)

Answer
$h = 20$ cm. Method: $2\pi(100) + 2\pi(10)h = 600\pi$ ; $200\pi + 20\pi h = 600\pi$ ; $20\pi h = 400\pi$ .
4. A rectangular prism $20 \times 10 \times 8$ cm has a cylindrical hole of radius $3$ cm drilled through its length. Find the remaining volume. (show answer)

Answer
$\approx 1034.5$ cm $^3$ . Method: prism $= 1600$ ; cylinder hole $= \pi(9)(20) = 180\pi \approx 565.5$ ; $1600 - 565.5$ .
5. Which has the greater surface area: a cube of side $10$ cm or a cylinder with $r = 5$ cm and $h = 10$ cm? Justify. (show answer)

Answer
Cube SA $= 600$ cm $^2$ . Cylinder SA $= 2\pi(25) + 2\pi(5)(10) = 50\pi + 100\pi = 150\pi \approx 471.2$ cm $^2$ . The cube has greater surface area.
6. A prism has a cross-section that is a right trapezium with parallel sides $6$ cm and $10$ cm and height $4$ cm. Its length is $15$ cm. Find the surface area. (The non-parallel sides are $4$ cm and $\approx 5.66$ cm.) (show answer)

Answer
$\approx 445.8$ cm $^2$ . Method: two trapezium ends $= 2 \times \tfrac{1}{2}(6+10)(4) = 64$ ; four rectangles: $6 \times 15 = 90$ , $10 \times 15 = 150$ , $4 \times 15 = 60$ , $5.66 \times 15 \approx 84.9$ ; total $\approx 64 + 384.9 \approx 448.9$ . (Accept minor rounding differences.)

Reasoning · Tier 3: explain and apply

1. Explain why the formula $\text{SA} = 2\pi r^2 + 2\pi r h$ has two separate terms. What does each represent physically? (show answer)

Answer
The first term $2\pi r^2$ is the area of the two circular ends (top and bottom). The second term $2\pi r h$ is the curved lateral surface -- the rectangle you get when you unroll the cylinder, whose width is the circumference $2\pi r$ and whose height is $h$ .
2. A manufacturer wants to double the volume of a cylindrical can without changing the radius. By what factor must the height change? (show answer)

Answer
The height must double. Since $V = \pi r^2 h$ and $r$ is fixed, doubling $V$ requires doubling $h$ .
3. If you double the radius of a cylinder but keep the height the same, by what factor does the volume increase? Explain why. (show answer)

Answer
Volume increases by a factor of $4$ . Since $V = \pi r^2 h$ , replacing $r$ with $2r$ gives $\pi(2r)^2 h = 4\pi r^2 h$ . The $r^2$ term means radius has a squared effect on volume.
4. A composite solid is a cylinder ( $r = 3$ cm, $h = 10$ cm) with a hemisphere ( $r = 3$ cm) on top. Find the total surface area. (Hemisphere curved SA $= 2\pi r^2$ .) (show answer)

Answer
$\approx 226.2$ cm $^2$ . Method: cylinder curved SA $= 2\pi(3)(10) = 60\pi$ ; cylinder base circle $= \pi(9) = 9\pi$ (only the bottom; the top is covered by the hemisphere); hemisphere curved SA $= 2\pi(9) = 18\pi$ ; total $= (60 + 9 + 18)\pi = 87\pi \approx 273.3$ cm $^2$ . (Accept equivalent working.)
5. A water pipe is $50$ m long with an inner radius of $5$ cm. Find the volume of water it can hold, in litres. (show answer)

Answer
$\approx 392.7$ L. Method: $r = 0.05$ m; $V = \pi(0.05)^2(50) = \pi(0.0025)(50) = 0.125\pi \approx 0.3927$ m $^3$ ; $\times 1000 = 392.7$ L.

Reasoning · Harder reasoning

1. A cylinder and a cube have the same volume. The cylinder has $r = 5$ cm and $h = 10$ cm. Find the side length of the cube to one decimal place. (show answer)

Answer
Side $\approx 9.3$ cm. Method: cylinder $V = \pi(25)(10) = 250\pi \approx 785.4$ cm $^3$ ; cube side $= \sqrt[3]{785.4} \approx 9.3$ .
2. A closed cylinder has total surface area $200\pi$ cm $^2$ . Express $h$ in terms of $r$ , and find the radius that maximises the volume. (Hint: substitute into $V = \pi r^2 h$ and look for the maximum.) (show answer)

Answer
Express $h$ : from $2\pi r^2 + 2\pi r h = 200\pi$ , divide through by $2\pi$ : $r^2 + r h = 100$ , so $h = \dfrac{100 - r^2}{r} = \dfrac{100}{r} - r$ .

Substitute into $V$ : $V = \pi r^2 h = \pi r^2 \left(\dfrac{100}{r} - r\right) = 100\pi r - \pi r^3$ .

Find the maximum by table of values (trial and improvement). Evaluate $V = \pi(100r - r^3)$ at whole-number $r$ :

| $r$ | $100r - r^3$ | $V \approx$ | |-----|--------------|-------------| | 3 | $300 - 27 = 273$ | $857.7$ | | 4 | $400 - 64 = 336$ | $1055.6$ | | 5 | $500 - 125 = 375$ | $1178.1$ | | 6 | $600 - 216 = 384$ | $\mathbf{1206.4}$ | | 7 | $700 - 343 = 357$ | $1121.5$ | | 8 | $800 - 512 = 288$ | $904.8$ |

The maximum is near $r = 6$ . Refine with decimals:

| $r$ | $100r - r^3$ | $V \approx$ | |-----|--------------|-------------| | 5.7 | $570 - 185.193 = 384.807$ | $1208.9$ | | 5.8 | $580 - 195.112 = 384.888$ | $\mathbf{1208.9}$ | | 5.9 | $590 - 205.379 = 384.621$ | $1208.0$ |

The radius that maximises the volume is $r \approx 5.8$ cm. (Then $h = \dfrac{100}{5.8} - 5.8 \approx 11.5$ cm, and $V \approx 1209$ cm $^3$ .)
3. A composite solid is a rectangular prism $10 \times 8 \times 6$ cm with a half-cylinder (radius $4$ cm, length $10$ cm) sitting on its top face. Find (a) the total volume, and (b) the total exposed surface area. (show answer)

Answer
(a) Volume: prism $= 480$ ; half-cylinder $= \tfrac{1}{2}\pi(16)(10) = 80\pi \approx 251.3$ ; total $\approx 731.3$ cm $^3$ . (b) SA: prism base $= 10 \times 8 = 80$ ; two prism ends $= 2(8 \times 6) = 96$ ; two prism long sides $= 2(10 \times 6) = 120$ ; prism top has rectangle $10 \times 8$ minus half-cylinder footprint (the diameter strip is part of the prism top, but the half-cylinder sits on it): exposed top $= 80 - 8 \times 10 = 0$ (the half-cylinder covers the entire top, so no exposed top); half-cylinder curved $= \tfrac{1}{2}(2\pi)(4)(10) = 40\pi \approx 125.7$ ; two half-circle ends $= 2 \times \tfrac{1}{2}\pi(16) = 16\pi \approx 50.3$ ; total SA $\approx 80 + 96 + 120 + 125.7 + 50.3 = 472.0$ cm $^2$ . (Accept reasonable variations depending on which faces are considered exposed.)
4. A cylindrical tank of radius $0.5$ m and height $1.2$ m is lying on its side. When it is half full, what volume of water does it contain? Give your answer in litres. (show answer)

Answer
$V = \tfrac{1}{2}\pi r^2 L = \tfrac{1}{2}\pi(0.5)^2(1.2) = 0.15\pi \approx 0.4712$ m $^3 \approx 471.2$ L.

Trigonometry in right-angled triangles

Fluency · Tier 1: basic skills

1. In a right-angled triangle, angle $A = 30°$ . The opposite side is $5$ cm and the hypotenuse is $10$ cm. Find $\sin 30°$ . (show answer)

Answer
$\sin 30° = \dfrac{5}{10} = 0.5$ .
2. In a right-angled triangle with $\theta = 45°$ , the adjacent and opposite sides are both $7$ cm. Find $\tan 45°$ . (show answer)

Answer
$\tan 45° = \dfrac{7}{7} = 1$ .
3. Find the opposite side if $\theta = 40°$ and $H = 20$ cm. (show answer)

Answer
$O = 20 \sin 40° \approx 20 \times 0.6428 = 12.86$ cm.
4. Find the adjacent side if $\theta = 55°$ and $H = 18$ cm. (show answer)

Answer
$A = 18 \cos 55° \approx 18 \times 0.5736 = 10.32$ cm.
5. Find the hypotenuse if $\theta = 33°$ and $O = 10$ cm. (show answer)

Answer
$H = \dfrac{10}{\sin 33°} = \dfrac{10}{0.5446} \approx 18.36$ cm.
6. Find $\theta$ if $O = 6$ cm and $H = 13$ cm. (show answer)

Answer
$\theta = \sin^{-1}\!\left(\dfrac{6}{13}\right) \approx 27.5°$ .
7. Find $\theta$ if $A = 9$ cm and $H = 15$ cm. (show answer)

Answer
$\theta = \cos^{-1}\!\left(\dfrac{9}{15}\right) = \cos^{-1}(0.6) \approx 53.1°$ .
8. Find $\theta$ if $O = 12$ cm and $A = 5$ cm. (show answer)

Answer
$\theta = \tan^{-1}\!\left(\dfrac{12}{5}\right) = \tan^{-1}(2.4) \approx 67.4°$ .
9. A kite string is $50$ m long and makes an angle of $60°$ with the ground. How high is the kite? (show answer)

Answer
Height $= 50 \sin 60° = 50 \times 0.8660 \approx 43.3$ m.
10. A $3$ m ladder leans against a wall at $72°$ to the ground. How far up the wall does it reach? (show answer)

Answer
Height up wall $= 3 \sin 72° \approx 3 \times 0.9511 = 2.85$ m.

Reasoning · Tier 2: mixed practice

1. From the top of a $25$ m cliff, the angle of depression to a boat is $38°$ . How far is the boat from the base of the cliff? (show answer)

Answer
$\approx 32.0$ m. Method: the angle of depression equals the angle at the boat. $\tan 38° = \dfrac{25}{d}$ ; $d = \dfrac{25}{\tan 38°} = \dfrac{25}{0.7813} \approx 32.0$ .
2. A road rises $1$ m for every $8$ m of horizontal distance. Find the angle of incline. (show answer)

Answer
$\approx 7.1°$ . Method: $\tan \theta = \dfrac{1}{8} = 0.125$ ; $\theta = \tan^{-1}(0.125)$ .
3. A rectangular gate is $1.8$ m wide and $1.2$ m tall. Find the angle its diagonal makes with the bottom edge. (show answer)

Answer
$\approx 33.7°$ . Method: $\tan \theta = \dfrac{1.2}{1.8} = 0.6\overline{6}$ ; $\theta = \tan^{-1}(0.667)$ .
4. An isosceles triangle has equal sides of $10$ cm and a base of $12$ cm. Find the base angles. (Hint: split it into two right-angled triangles.) (show answer)

Answer
$\approx 53.1°$ . Method: half-base $= 6$ cm; $\cos \theta = \dfrac{6}{10} = 0.6$ ; $\theta = \cos^{-1}(0.6)$ .
5. A ship sails $15$ km due east then $9$ km due north. Find the bearing from the starting point to the ship's final position. (show answer)

Answer
Bearing $\approx 031°$ . Method: $\tan \alpha = \dfrac{9}{15} = 0.6$ ; $\alpha = \tan^{-1}(0.6) \approx 31.0°$ . Bearing from east is $90° - 31° = 59°$ ... Correction: bearing is measured clockwise from north. The ship is east then north, so the angle from north $= 90° - \tan^{-1}(\tfrac{9}{15})$ . Actually: from start, east $= 15$ , north $= 9$ . Bearing $= \tan^{-1}(\tfrac{15}{9}) = \tan^{-1}(1.667) \approx 59.0°$ . Bearing $\approx 059°$ .
6. Two buildings are $30$ m apart. From the roof of the shorter building ( $20$ m tall), the angle of elevation to the top of the taller building is $35°$ . Find the height of the taller building. (show answer)

Answer
$41.0$ m. Method: let extra height above the shorter building $= x$ . $\tan 35° = \dfrac{x}{30}$ ; $x = 30 \tan 35° \approx 21.0$ m. Total height $= 20 + 21.0 = 41.0$ m.

Reasoning · Tier 3: explain and apply

1. Explain why $\sin \theta$ can never be greater than $1$ . (show answer)

Answer
$\sin \theta = \dfrac{O}{H}$ . In a right-angled triangle, the opposite side is always shorter than the hypotenuse (the hypotenuse is the longest side). Therefore $O < H$ , so $\dfrac{O}{H} < 1$ , meaning $\sin \theta < 1$ for acute angles.
2. Show that for any acute angle $\theta$ , $\tan \theta = \dfrac{\sin \theta}{\cos \theta}$ . (show answer)

Answer
$\dfrac{\sin \theta}{\cos \theta} = \dfrac{O/H}{A/H} = \dfrac{O}{H} \times \dfrac{H}{A} = \dfrac{O}{A} = \tan \theta$ .
3. A surveyor needs to find the width of a river. She stands at point A on one bank and sights a tree at point B directly opposite. She then walks $40$ m along the bank to point C and measures angle ACB as $56°$ . Find the width of the river. (show answer)

Answer
$\approx 59.3$ m. Method: the river width $w$ is opposite the $56°$ angle; the $40$ m walk is adjacent. $\tan 56° = \dfrac{w}{40}$ ; $w = 40 \tan 56° \approx 59.3$ m.
4. Without a calculator, explain why $\sin 30° = \cos 60°$ . Use a diagram if helpful. (show answer)

Answer
In a 30-60-90 triangle, the side opposite $30°$ is half the hypotenuse, and the side adjacent to $30°$ is $\dfrac{\sqrt{3}}{2}$ times the hypotenuse. Now $\sin 30° = \dfrac{\text{opp}_{30}}{H} = \dfrac{1}{2}$ . For $60°$ , the opposite and adjacent sides swap: $\cos 60° = \dfrac{\text{adj}_{60}}{H} = \dfrac{\text{opp}_{30}}{H} = \dfrac{1}{2}$ . So $\sin 30° = \cos 60°$ .
5. A ski slope is $200$ m long (measured along the slope) and drops $80$ m vertically. Find (a) the angle of the slope, and (b) the horizontal distance covered. (show answer)

Answer
(a) $\sin \theta = \dfrac{80}{200} = 0.4$ ; $\theta = \sin^{-1}(0.4) \approx 23.6°$ . (b) Horizontal distance $= \sqrt{200^2 - 80^2} = \sqrt{33600} \approx 183.3$ m. Or: $200 \cos 23.6° \approx 183.3$ m.

Reasoning · Harder reasoning

1. From a point on level ground, the angle of elevation to the top of a building is $28°$ . Moving $20$ m closer, the angle becomes $42°$ . Find the height of the building. (show answer)

Answer
$\approx 24.9$ m. Method: let $h$ = height, $d$ = original distance. From the first position: $\tan 28° = \dfrac{h}{d}$ , so $d = \dfrac{h}{\tan 28°}$ . From the closer position: $\tan 42° = \dfrac{h}{d - 20}$ , so $d - 20 = \dfrac{h}{\tan 42°}$ . Substituting: $\dfrac{h}{\tan 28°} - 20 = \dfrac{h}{\tan 42°}$ ; $h\!\left(\dfrac{1}{\tan 28°} - \dfrac{1}{\tan 42°}\right) = 20$ ; $h\!\left(\dfrac{1}{0.5317} - \dfrac{1}{0.9004}\right) = 20$ ; $h(1.8807 - 1.1106) = 20$ ; $h \times 0.7701 = 20$ ; $h \approx 26.0$ m. (Accept $25-26$ m depending on rounding.)
2. A regular hexagon has side length $8$ cm. Using trigonometry, find (a) the distance from the centre to a vertex, and (b) the area of the hexagon. (show answer)

Answer
(a) In a regular hexagon, the distance from centre to vertex equals the side length, so $8$ cm. (b) The hexagon splits into $6$ equilateral triangles of side $8$ cm. Area of each $= \dfrac{\sqrt{3}}{4}(8^2) = 16\sqrt{3}$ . Total $= 96\sqrt{3} \approx 166.3$ cm $^2$ .
3. Prove that in any right-angled triangle, $\sin^2\theta + \cos^2\theta = 1$ . (Hint: use the definitions and Pythagoras' theorem.) (show answer)

Answer
$\sin^2\theta + \cos^2\theta = \left(\dfrac{O}{H}\right)^2 + \left(\dfrac{A}{H}\right)^2 = \dfrac{O^2 + A^2}{H^2}$ . By Pythagoras, $O^2 + A^2 = H^2$ . Therefore $\dfrac{H^2}{H^2} = 1$ .
4. A plane takes off at an angle of $15°$ to the horizontal and climbs at a constant speed of $250$ km/h. After $2$ minutes, find (a) the plane's altitude, and (b) the horizontal distance it has covered from the runway. (show answer)

Answer
Distance travelled in $2$ min $= 250 \times \dfrac{2}{60} = 8.33$ km. (a) Altitude $= 8.33 \sin 15° \approx 8.33 \times 0.2588 \approx 2.16$ km. (b) Horizontal distance $= 8.33 \cos 15° \approx 8.33 \times 0.9659 \approx 8.05$ km.

Pythagoras & trigonometry applications

Fluency · Tier 1: basic skills

1. A right-angled triangle has sides 6 cm and 8 cm. Find the hypotenuse. (show answer)

Answer
$c = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ cm.
2. The hypotenuse of a right-angled triangle is 13 m and one side is 5 m. Find the other side. (show answer)

Answer
$a = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$ m.
3. A rectangular room is 4 m by 3 m. Find the length of the diagonal of the floor. (show answer)

Answer
$d = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ m.
4. Find $\sin 40°$ , $\cos 40°$ , and $\tan 40°$ using a calculator (2 d.p.). (show answer)

Answer
$\sin 40° \approx 0.64$ , $\cos 40° \approx 0.77$ , $\tan 40° \approx 0.84$ .
5. In a right-angled triangle with hypotenuse 20 cm and an angle of $50°$ , find the opposite side. (show answer)

Answer
$\text{opp} = 20 \sin 50° \approx 20 \times 0.7660 = 15.32$ cm.
6. Find the angle whose tangent is $\dfrac{3}{4}$ . (show answer)

Answer
$\theta = \tan^{-1}\!\left(\dfrac{3}{4}\right) = \tan^{-1}(0.75) \approx 36.87° \approx 36.9°$ .
7. Find the distance between the points $(1, 3)$ and $(4, 7)$ . (show answer)

Answer
$d = \sqrt{(4-1)^2 + (7-3)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ units.
8. Find the distance between $(-2, 5)$ and $(3, -7)$ . (show answer)

Answer
$d = \sqrt{(3-(-2))^2 + (-7-5)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ units.
9. A 6 m ladder reaches 5.5 m up a wall. Find the angle the ladder makes with the ground. (show answer)

Answer
$\sin\theta = \dfrac{5.5}{6}$ , so $\theta = \sin^{-1}(0.9167) \approx 66.4°$ .
10. A box has dimensions 2 m $\times$ 3 m $\times$ 6 m. Find the space diagonal. (show answer)

Answer
$d = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ m.

Reasoning · Tier 2: mixed practice

1. A kite is flying at a height of 40 m. The string is 65 m long. What angle does the string make with the ground? (show answer)

Answer
$\sin\theta = \dfrac{40}{65} = \dfrac{8}{13}$ , so $\theta = \sin^{-1}(0.6154) \approx 38.0°$ .
2. From the top of a 50 m cliff, the angle of depression to a boat is $18°$ . How far is the boat from the base of the cliff? (show answer)

Answer
$\tan 18° = \dfrac{50}{d}$ , so $d = \dfrac{50}{\tan 18°} \approx \dfrac{50}{0.3249} \approx 153.9$ m.
3. A hiker walks 8 km on a bearing of $040°$ and then 6 km due east. How far north and how far east is the hiker from the starting point? (show answer)

Answer
North: $8 \cos 40° \approx 8 \times 0.7660 = 6.13$ km. East from first leg: $8 \sin 40° \approx 8 \times 0.6428 = 5.14$ km. Total east: $5.14 + 6 = 11.14$ km. Total north: $6.13$ km.
4. Points $A(0, 0)$ , $B(6, 0)$ , and $C(3, 5)$ form a triangle. Find the perimeter. (show answer)

Answer
$AB = 6$ , $BC = \sqrt{(6-3)^2 + (0-5)^2} = \sqrt{9 + 25} = \sqrt{34} \approx 5.83$ , $AC = \sqrt{9 + 25} = \sqrt{34} \approx 5.83$ . Perimeter $\approx 6 + 5.83 + 5.83 = 17.66$ units.
5. A tent pole 2.4 m tall is supported by a rope pegged 1.8 m from the base. Find the angle the rope makes with the ground. (show answer)

Answer
$\tan\theta = \dfrac{2.4}{1.8} = \dfrac{4}{3}$ , so $\theta = \tan^{-1}(1.333) \approx 53.1°$ .
6. A rectangular prism has a base of 5 cm $\times$ 12 cm and a space diagonal of 15 cm. Find the height of the prism. (show answer)

Answer
Space diagonal: $d^2 = l^2 + w^2 + h^2$ . So $15^2 = 5^2 + 12^2 + h^2$ , giving $225 = 25 + 144 + h^2$ , thus $h^2 = 56$ and $h = \sqrt{56} \approx 7.48$ cm.
7. A ship sails 20 km on a bearing of $120°$ . How far south and how far east is it from its starting point? (show answer)

Answer
South: $20 \cos 60° = 20 \times 0.5 = 10$ km (bearing $120°$ is $60°$ from south). East: $20 \sin 60° = 20 \times 0.8660 \approx 17.32$ km.
8. Two buildings are 30 m apart. From the top of the shorter building (20 m tall), the angle of elevation to the top of the taller building is $35°$ . Find the height of the taller building. (show answer)

Answer
Let extra height above the shorter building be $x$ . $\tan 35° = \dfrac{x}{30}$ , so $x = 30 \tan 35° \approx 30 \times 0.7002 = 21.0$ m. Total height $= 20 + 21.0 = 41.0$ m.

Reasoning · Tier 3: explain and apply

1. Explain why the distance formula is a direct application of Pythagoras' theorem. Use a diagram to support your answer. (show answer)

Answer
On the Cartesian plane, the horizontal distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $|x_2 - x_1|$ and the vertical distance is $|y_2 - y_1|$ . These form the two shorter sides of a right-angled triangle. Applying Pythagoras' theorem: $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$ , giving $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ .
2. A helicopter is directly above a point on the ground. Observer A is 500 m due north of that point and measures the angle of elevation as $28°$ . Observer B is 700 m due east and measures the angle of elevation as $20°$ . Are both observers looking at the same helicopter height? Show your working. (show answer)

Answer
Observer A: height $= 500 \tan 28° \approx 500 \times 0.5317 = 265.9$ m. Observer B: height $= 700 \tan 20° \approx 700 \times 0.3640 = 254.8$ m. The heights differ ( $265.9 \neq 254.8$ ), so either the measurements are imprecise or the helicopter is not directly above the assumed point.
3. A surveyor needs to find the width of a river. She stands at point $A$ on one bank and sights a tree at point $B$ directly across the river. She then walks 50 m along the bank to point $C$ and measures $\angle ACB = 62°$ . Find the width of the river. (show answer)

Answer
$\tan 62° = \dfrac{w}{50}$ , where $w$ is the river width. $w = 50 \tan 62° \approx 50 \times 1.8807 = 94.0$ m.
4. Prove that for any three points forming a triangle on the coordinate plane, the triangle inequality holds: the sum of any two side lengths exceeds the third. (show answer)

Answer
This is the standard triangle inequality. For any triangle with vertices on the coordinate plane, the shortest path between two points is the straight line (the side). Going via a third point is longer, so $AB + BC > AC$ (and cyclic permutations). A rigorous proof uses the Cauchy--Schwarz inequality or the properties of the Euclidean metric.
5. A rescue helicopter is at coordinates $(3, 7)$ and must reach a boat at $(-5, 1)$ . If each grid unit represents 2 km, find the actual distance the helicopter must fly. (show answer)

Answer
$d = \sqrt{(-5-3)^2 + (1-7)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$ grid units. Actual distance $= 10 \times 2 = 20$ km.

Reasoning · Harder reasoning

1. A cone has a slant height of 13 cm and a base radius of 5 cm. Find the height of the cone and then calculate the angle between the slant surface and the base. (show answer)

Answer
Height: $h = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$ cm. Angle between slant and base: $\cos\alpha = \dfrac{5}{13}$ , so $\alpha = \cos^{-1}(0.3846) \approx 67.4°$ .
2. Two lighthouses are 10 km apart on a straight coastline running east--west. A ship at sea measures the bearing to the western lighthouse as $320°$ and the bearing to the eastern lighthouse as $050°$ . Find the ship's perpendicular distance from the coastline. (show answer)

Answer
The western lighthouse is at $W$ and the eastern at $E$ , 10 km apart. Bearing $320°$ to $W$ means the angle from north is $320°$ , so the line from the ship to $W$ makes $40°$ west of north. Bearing $050°$ to $E$ means $50°$ east of north. The ship is south of the coastline. At the ship, the angle $\angle WSE = 360° - 320° + 50° = 90°$ . Using the right triangle: let perpendicular distance $= h$ . $\tan 40° = \dfrac{d_W}{h}$ and $\tan 50° = \dfrac{d_E}{h}$ where $d_W + d_E = 10$ . So $h(\tan 40° + \tan 50°) = 10$ , giving $h = \dfrac{10}{\tan 40° + \tan 50°} = \dfrac{10}{0.8391 + 1.1918} \approx \dfrac{10}{2.0309} \approx 4.92$ km.
3. A cube has side length $s$ . Show that the space diagonal has length $s\sqrt{3}$ and find the angle the space diagonal makes with the base of the cube. (show answer)

Answer
Space diagonal: $d = \sqrt{s^2 + s^2 + s^2} = \sqrt{3s^2} = s\sqrt{3}$ . The base diagonal is $s\sqrt{2}$ . The angle $\alpha$ between the space diagonal and the base satisfies $\tan\alpha = \dfrac{s}{s\sqrt{2}} = \dfrac{1}{\sqrt{2}}$ , so $\alpha = \tan^{-1}\!\left(\dfrac{1}{\sqrt{2}}\right) \approx 35.3°$ .
4. Three mobile phone towers are at positions $A(0, 0)$ , $B(8, 0)$ , and $C(3, 6)$ on a coordinate grid (units in km). A phone receives signal from a tower only if it is within 7 km. Determine whether a phone at position $P(5, 4)$ can receive signal from all three towers. (show answer)

Answer
$PA = \sqrt{5^2 + 4^2} = \sqrt{41} \approx 6.40$ km (within 7 km — yes). $PB = \sqrt{(5-8)^2 + 4^2} = \sqrt{9 + 16} = 5$ km (within 7 km — yes). $PC = \sqrt{(5-3)^2 + (4-6)^2} = \sqrt{4 + 4} = \sqrt{8} \approx 2.83$ km (within 7 km — yes). The phone can receive signal from all three towers.

Measurement errors

Fluency · Tier 1: basic skills

1. The actual mass of a parcel is 3.2 kg. A scale reads 3.35 kg. Find the absolute error. (show answer)

Answer
Absolute error $= |3.35 - 3.2| = 0.15$ kg.
2. Find the relative error for the measurement in Q1. (show answer)

Answer
Relative error $= \dfrac{0.15}{3.2} = 0.046875$ .
3. Find the percentage error for the measurement in Q1. (show answer)

Answer
Percentage error $= 0.046875 \times 100\% \approx 4.7\%$ .
4. A length is 15.8 cm, measured to the nearest 0.1 cm. State the upper and lower bounds. (show answer)

Answer
Lower bound $= 15.75$ cm, upper bound $= 15.85$ cm.
5. A time is recorded as 24 seconds, to the nearest second. State the upper and lower bounds. (show answer)

Answer
Lower bound $= 23.5$ s, upper bound $= 24.5$ s.
6. A mass is given as 500 g, to the nearest 10 g. What is the maximum possible error? (show answer)

Answer
Maximum possible error $= \dfrac{10}{2} = 5$ g.
7. A stick is measured as 1.2 m to the nearest 0.1 m, and another as 0.8 m to the nearest 0.1 m. Find the bounds of their combined length. (show answer)

Answer
Combined stated length $= 2.0$ m. Max error $= 0.05 + 0.05 = 0.10$ m. Lower bound $= 1.90$ m, upper bound $= 2.10$ m.
8. A student estimates $\pi$ as 3.14. Find the percentage error (use $\pi = 3.14159\ldots$ ). (show answer)

Answer
Absolute error $= |3.14 - 3.14159| = 0.00159$ . Percentage error $= \dfrac{0.00159}{3.14159} \times 100\% \approx 0.051\%$ .
9. The temperature is recorded as $22°$ C to the nearest degree. What are the upper and lower bounds? (show answer)

Answer
Lower bound $= 21.5°$ C, upper bound $= 22.5°$ C.
10. A car's odometer reads 45 230 km, rounded to the nearest km. State the bounds. (show answer)

Answer
Lower bound $= 45\,229.5$ km, upper bound $= 45\,230.5$ km.

Reasoning · Tier 2: mixed practice

1. Two students measure the same length. Student A gets 12.4 cm (actual 12.0 cm) and Student B gets 47.2 cm (actual 46.0 cm). Who has the smaller percentage error? (show answer)

Answer
Student A: $\dfrac{0.4}{12.0} \times 100\% = 3.33\%$ . Student B: $\dfrac{1.2}{46.0} \times 100\% = 2.61\%$ . Student B has the smaller percentage error.
2. A rectangle is measured as 20 cm $\times$ 15 cm, each to the nearest cm. Find the upper and lower bounds of the area. (show answer)

Answer
Length bounds: $19.5$ to $20.5$ cm. Width bounds: $14.5$ to $15.5$ cm. Lower area $= 19.5 \times 14.5 = 282.75$ cm $^2$ . Upper area $= 20.5 \times 15.5 = 317.75$ cm $^2$ .
3. A runner completes 400 m (to the nearest metre) in 52.3 s (to the nearest 0.1 s). Calculate the speed and find the maximum percentage error in the speed. (show answer)

Answer
Speed $= \dfrac{400}{52.3} \approx 7.65$ m/s. Distance \% error $= \dfrac{0.5}{400} \times 100\% = 0.125\%$ . Time \% error $= \dfrac{0.05}{52.3} \times 100\% \approx 0.096\%$ . Max \% error in speed $\approx 0.125\% + 0.096\% = 0.22\%$ .
4. A square has a side measured as 8.0 cm to the nearest mm. Find the upper and lower bounds of its area. (show answer)

Answer
Side bounds: $7.95$ cm to $8.05$ cm. Lower area $= 7.95^2 = 63.2025$ cm $^2$ . Upper area $= 8.05^2 = 64.8025$ cm $^2$ .
5. Explain why percentage error is more useful than absolute error when comparing measurements of different magnitudes. (show answer)

Answer
Percentage error scales the error to the size of the quantity, enabling fair comparison. A 1 cm error on a 10 cm measurement (10%) is far more significant than a 1 cm error on a 1000 cm measurement (0.1%), but the absolute errors are identical.
6. A cylindrical tank has radius 1.2 m and height 3.0 m, each to the nearest 0.1 m. Calculate the volume and find the upper and lower bounds. (show answer)

Answer
Volume $= \pi r^2 h = \pi \times 1.2^2 \times 3.0 \approx 13.57$ m $^3$ . Lower: $\pi \times 1.15^2 \times 2.95 \approx 12.26$ m $^3$ . Upper: $\pi \times 1.25^2 \times 3.05 \approx 14.97$ m $^3$ .
7. A thermometer has a precision of $0.5°$ C. If it reads $37.0°$ C, state the error bounds and express the maximum percentage error. (show answer)

Answer
Bounds: $36.75°$ C to $37.25°$ C. Max error $= 0.25°$ C. Percentage error $= \dfrac{0.25}{37.0} \times 100\% \approx 0.68\%$ .

Reasoning · Tier 3: explain and apply

1. A surveyor measures the distance between two points as 84.3 m. The true distance is 85.0 m. She then uses this measurement to calculate the area of a square plot. Find the percentage error in the distance and in the area. (show answer)

Answer
Distance percentage error $= \dfrac{|84.3 - 85.0|}{85.0} \times 100\% = \dfrac{0.7}{85.0} \times 100\% \approx 0.82\%$ . Area using measured distance $= 84.3^2 = 7106.49$ m $^2$ . Actual area $= 85.0^2 = 7225$ m $^2$ . Area percentage error $= \dfrac{|7106.49 - 7225|}{7225} \times 100\% \approx 1.64\%$ , which is approximately double the distance error (since area $\propto d^2$ ).
2. The density of an object is calculated using $\rho = \dfrac{m}{V}$ . The mass is $120 \pm 0.5$ g and the volume is $50 \pm 1$ cm $^3$ . Find the density and the maximum percentage error. (show answer)

Answer
Density $= \dfrac{120}{50} = 2.4$ g/cm $^3$ . Mass \% error $= \dfrac{0.5}{120} \times 100\% \approx 0.42\%$ . Volume \% error $= \dfrac{1}{50} \times 100\% = 2.0\%$ . Max \% error in density $\approx 0.42\% + 2.0\% = 2.42\%$ . Density $= 2.4 \pm 0.06$ g/cm $^3$ .
3. Explain why the maximum error of a difference $A - B$ uses the sum of the individual errors, not the difference. (show answer)

Answer
The error in $A - B$ is maximised when $A$ is at its upper bound and $B$ is at its lower bound (or vice versa). This gives the largest possible spread: $(A + \delta_A) - (B - \delta_B) = (A - B) + (\delta_A + \delta_B)$ . The individual errors add regardless of whether we add or subtract the measurements.
4. A GPS device is accurate to $\pm 3$ m. Two points are recorded 200 m apart. What is the maximum percentage error in this distance? If the points were only 10 m apart, how would the percentage error change? (show answer)

Answer
Max error $= 3 + 3 = 6$ m. For 200 m: \% error $= \dfrac{6}{200} \times 100\% = 3\%$ . For 10 m: \% error $= \dfrac{6}{10} \times 100\% = 60\%$ . The GPS is unreliable for short distances.

Reasoning · Harder reasoning

1. The area of a circle is calculated from a measured diameter of $14.0 \pm 0.1$ cm. Find the percentage error in the area. (Hint: area depends on $d^2$ .) (show answer)

Answer
Radius $= 7.0$ cm, so area $= \pi \times 7.0^2 = 49\pi \approx 153.94$ cm $^2$ . Diameter \% error $= \dfrac{0.1}{14.0} \times 100\% \approx 0.714\%$ . Since area $\propto d^2$ , area \% error $\approx 2 \times 0.714\% = 1.43\%$ .
2. A student measures the acceleration due to gravity using $g = \dfrac{2s}{t^2}$ , where $s = 1.200 \pm 0.005$ m and $t = 0.495 \pm 0.005$ s. Calculate $g$ and the maximum percentage error. Comment on whether this is an acceptable result given that the accepted value is $9.81$ m/s $^2$ . (show answer)

Answer
$g = \dfrac{2 \times 1.200}{0.495^2} = \dfrac{2.400}{0.245025} \approx 9.795$ m/s $^2$ . Distance \% error $= \dfrac{0.005}{1.200} \times 100\% \approx 0.42\%$ . Time \% error $= \dfrac{0.005}{0.495} \times 100\% \approx 1.01\%$ . Since $g \propto t^{-2}$ , time contributes $2 \times 1.01\% = 2.02\%$ . Total \% error $\approx 0.42\% + 2.02\% = 2.44\%$ . This gives $g = 9.80 \pm 0.24$ m/s $^2$ . The accepted value $9.81$ falls within this range, so the result is acceptable.
3. Two measurements are subtracted: $A = 50.0 \pm 0.5$ and $B = 49.0 \pm 0.5$ . Find the result and its percentage error. Explain why subtraction of nearly equal quantities leads to a large relative error (this is called catastrophic cancellation). (show answer)

Answer
Result $= 50.0 - 49.0 = 1.0$ . Max error $= 0.5 + 0.5 = 1.0$ . Percentage error $= \dfrac{1.0}{1.0} \times 100\% = 100\%$ . When two nearly equal quantities are subtracted, the result is small but the absolute error remains the sum of the original errors, leading to a very large relative error. This catastrophic cancellation should be avoided in practice by redesigning the measurement approach.
4. A map has a scale of 1 : 25 000. A distance on the map is measured as $4.2 \pm 0.1$ cm. Find the actual distance and the bounds (in metres) of the actual distance. (show answer)

Answer
Map distance $= 4.2$ cm, so actual distance $= 4.2 \times 25\,000 = 105\,000$ cm $= 1050$ m. Lower bound: $(4.2 - 0.1) \times 25\,000 = 4.1 \times 25\,000 = 102\,500$ cm $= 1025$ m. Upper bound: $(4.2 + 0.1) \times 25\,000 = 4.3 \times 25\,000 = 107\,500$ cm $= 1075$ m. Actual distance is between 1025 m and 1075 m.

Direct proportion, ratio & scale

Fluency · Tier 1: basic skills

1. $y$ is directly proportional to $x$ . When $x = 4$ , $y = 20$ . Find $k$ and write the equation. (show answer)

Answer
$k = \dfrac{20}{4} = 5$ . Equation: $y = 5x$ .
2. Using the equation from Q1, find $y$ when $x = 7$ . (show answer)

Answer
$y = 5 \times 7 = 35$ .
3. Does this table show direct proportion? $x$ : 3, 6, 9, 12; $y$ : 9, 18, 27, 36. (show answer)

Answer
Yes. $\dfrac{y}{x} = 3$ for every pair, so $y = 3x$ .
4. Does this table show direct proportion? $x$ : 2, 4, 6, 8; $y$ : 5, 9, 13, 17. (show answer)

Answer
No. $\dfrac{5}{2} = 2.5$ , $\dfrac{9}{4} = 2.25$ , $\dfrac{13}{6} \approx 2.17$ , $\dfrac{17}{8} = 2.125$ . The ratio is not constant (this is $y = 2x + 1$ ).
5. A map has a scale of 1 : 25 000. Two points are 6.4 cm apart on the map. Find the actual distance in metres. (show answer)

Answer
Actual distance $= 6.4 \times 25\,000 = 160\,000$ cm $= 1\,600$ m.
6. An actual length of 3 km needs to be drawn on a map at 1 : 50 000 scale. What is the drawing length in cm? (show answer)

Answer
$3$ km $= 300\,000$ cm. Drawing length $= \dfrac{300\,000}{50\,000} = 6$ cm.
7. A car travels at a constant speed of 80 km/h. How far does it travel in 2.5 hours? (show answer)

Answer
Distance $= 80 \times 2.5 = 200$ km.
8. A liquid flows at 5 litres per minute. How long does it take to fill a 120-litre container? (show answer)

Answer
Time $= \dfrac{120}{5} = 24$ minutes.
9. Convert 1 AUD $= 0.70$ USD. How many USD do you get for 250 AUD? (show answer)

Answer
$250 \times 0.70 = 175$ USD.
10. A block of iron has density 7.87 g/cm $^3$ . Find the mass of a 15 cm $^3$ piece. (show answer)

Answer
Mass $= 7.87 \times 15 = 118.05$ g.

Reasoning · Tier 2: mixed practice

1. The graph of $y$ against $x$ is a straight line through the origin with gradient 2.5. Write the equation and state whether $y$ is directly proportional to $x$ . (show answer)

Answer
$y = 2.5x$ . Yes, $y$ is directly proportional to $x$ because the line passes through the origin and $k = 2.5$ .
2. A recipe for 6 serves needs 450 g of flour. How much flour is needed for 10 serves? What assumption are you making? (show answer)

Answer
Flour per serve $= \dfrac{450}{6} = 75$ g. For 10 serves: $75 \times 10 = 750$ g. Assumption: the amount of flour is directly proportional to the number of serves (the recipe scales linearly).
3. On a 1 : 200 floor plan, a room measures 3.5 cm by 2.8 cm. Find the actual area of the room in square metres. (show answer)

Answer
Drawing dimensions: $3.5$ cm $\times$ $2.8$ cm. Actual: $3.5 \times 200 = 700$ cm $= 7$ m and $2.8 \times 200 = 560$ cm $= 5.6$ m. Area $= 7 \times 5.6 = 39.2$ m $^2$ .
4. A cyclist covers 36 km in 1.5 hours. A runner covers 15 km in 1.25 hours. Express each as a rate in km/h and determine who is faster. (show answer)

Answer
Cyclist: $\dfrac{36}{1.5} = 24$ km/h. Runner: $\dfrac{15}{1.25} = 12$ km/h. The cyclist is faster.
5. Gold has a density of 19.3 g/cm $^3$ . A gold bar has dimensions 25 cm $\times$ 5 cm $\times$ 2 cm. Find its mass in kilograms. (show answer)

Answer
Volume $= 25 \times 5 \times 2 = 250$ cm $^3$ . Mass $= 19.3 \times 250 = 4\,825$ g $= 4.825$ kg.
6. The extension of a spring is directly proportional to the load. A 6 kg load produces an extension of 9 cm. Find the extension for a 10 kg load and the load needed for a 15 cm extension. (show answer)

Answer
$k = \dfrac{9}{6} = 1.5$ cm/kg. Extension for 10 kg: $1.5 \times 10 = 15$ cm. Load for 15 cm: $\dfrac{15}{1.5} = 10$ kg.
7. Water flows into a pool at 40 litres per minute and drains out at 15 litres per minute simultaneously. How long until the pool, which holds 5000 litres, is full? (show answer)

Answer
Net flow rate $= 40 - 15 = 25$ litres per minute. Time $= \dfrac{5000}{25} = 200$ minutes $= 3$ hours $20$ minutes.
8. On a map at 1 : 10 000 scale, a park has an area of 12 cm $^2$ . What is the actual area in square metres? (show answer)

Answer
Scale factor for length $= 10\,000$ . Area scale factor $= 10\,000^2 = 10^8$ . Actual area $= 12 \times 10^8$ cm $^2 = 12 \times 10^4$ m $^2 = 120\,000$ m $^2$ (or $0.12$ km $^2$ ).

Reasoning · Tier 3: explain and apply

1. Explain how you can tell from a table of values whether two quantities are directly proportional, and give an example of a table that almost looks proportional but is not. (show answer)

Answer
Calculate $\dfrac{y}{x}$ for every pair. If the ratio is the same each time, the relationship is directly proportional. Example of a near-miss: $x$ : 1, 2, 3, 4; $y$ : 3, 6, 9, 13. The first three ratios are 3, but the last is 3.25, so it is not proportional.
2. A delivery company charges a $5 base fee plus $2 per kilogram. Is the total cost directly proportional to the weight? Justify your answer and sketch the graph. (show answer)

Answer
No. The total cost $C = 5 + 2w$ has a non-zero $y$ -intercept ($5 base fee), so $C$ is not directly proportional to $w$ . The graph is a straight line crossing the $C$ -axis at 5, not through the origin. Doubling the weight does not double the cost.
3. Two maps of the same region have scales 1 : 20 000 and 1 : 50 000. A road is 8 cm long on the first map. How long is it on the second map? (show answer)

Answer
Road actual length $= 8 \times 20\,000 = 160\,000$ cm. On the second map: $\dfrac{160\,000}{50\,000} = 3.2$ cm.
4. The power output of a solar panel is directly proportional to its area. A 2 m $^2$ panel produces 600 W. A rooftop can fit 14 m $^2$ of panels. What is the maximum power output? Discuss one real-world factor that might cause the actual output to differ. (show answer)

Answer
$k = \dfrac{600}{2} = 300$ W/m $^2$ . Maximum output $= 300 \times 14 = 4\,200$ W. Real-world factors: panels may not all face the sun at the optimal angle; shading, temperature, and panel efficiency losses reduce actual output.
5. A student claims that since $y = 3x + 1$ has a constant rate of change (gradient 3), it must be a proportional relationship. Identify and explain the student's error. (show answer)

Answer
The student confuses a constant rate of change (gradient) with direct proportionality. Direct proportion requires $y = kx$ (the line passes through the origin). Since $y = 3x + 1$ has $y = 1$ when $x = 0$ , it does not pass through the origin and is not a proportional relationship.

Reasoning · Harder reasoning

1. A model car is built at a scale of 1 : 18. The model weighs 1.2 kg. If the model and the real car are made of materials with the same density, estimate the mass of the real car. (Hint: mass scales with volume, which scales as the cube of the linear scale factor.) (show answer)

Answer
Linear scale factor $= 18$ . Volume scale factor $= 18^3 = 5\,832$ . Estimated real car mass $= 1.2 \times 5\,832 = 6\,998.4$ kg $\approx 7\,000$ kg. (In practice, real cars are not solid like models, so the actual mass would be much less — around 1\,200--1\,800 kg. The calculation shows what would happen if density were identical throughout.)
2. Two quantities are related by $y = kx^2$ . Is $y$ directly proportional to $x$ ? Is $y$ directly proportional to $x^2$ ? A ball is dropped and falls $d = 4.9t^2$ metres in $t$ seconds. Find the distance fallen in 3 seconds and the time to fall 100 m. (show answer)

Answer
$y$ is not directly proportional to $x$ (doubling $x$ quadruples $y$ ). However, $y$ is directly proportional to $x^2$ with constant $k$ . Distance in 3 s: $d = 4.9 \times 9 = 44.1$ m. Time to fall 100 m: $100 = 4.9t^2$ , so $t^2 = \dfrac{100}{4.9} \approx 20.41$ and $t \approx 4.52$ s.
3. Water leaks from a tank at a rate proportional to the depth of water. When the depth is 2 m, the leak rate is 0.5 litres per minute. Write the rate as an equation and find the leak rate when the depth is 3.5 m. (show answer)

Answer
Let rate $= k \times \text{depth}$ . When depth $= 2$ : $0.5 = k \times 2$ , so $k = 0.25$ litres per minute per metre. Equation: rate $= 0.25d$ . When $d = 3.5$ : rate $= 0.25 \times 3.5 = 0.875$ litres per minute.
4. A photographer enlarges a 10 cm $\times$ 15 cm photo to fit inside a frame that is 35 cm wide. What scale factor is used? What is the height of the enlarged photo? If the print costs $0.08 per cm $^2$ , find the printing cost of the enlargement. (show answer)

Answer
Scale factor $= \dfrac{35}{15} = \dfrac{7}{3} \approx 2.333$ . Height $= 10 \times \dfrac{7}{3} = \dfrac{70}{3} \approx 23.3$ cm. Enlarged area $= 35 \times \dfrac{70}{3} = \dfrac{2450}{3} \approx 816.7$ cm $^2$ . Cost $= 816.7 \times 0.08 \approx 65.33$ dollars.

Enlargement & similarity

Fluency · Tier 1: basic skills

1. A triangle has vertices $A(2,1)$ , $B(4,1)$ , $C(2,3)$ . Enlarge with centre $O(0,0)$ and $k=3$ . State the image vertices. (show answer)

Answer
$A'(6,3)$ , $B'(12,3)$ , $C'(6,9)$ .
2. A rectangle is enlarged by scale factor $k = 2$ . If the original has length $5$ cm and width $3$ cm, find the image length and width. (show answer)

Answer
Image length $= 10$ cm, image width $= 6$ cm.
3. The original perimeter of a square is $20$ cm. What is the perimeter after enlargement with $k = 4$ ? (show answer)

Answer
$20 \times 4 = 80$ cm.
4. A shape has area $18$ cm $^2$ . After enlargement with $k = 3$ , what is the new area? (show answer)

Answer
$18 \times 3^2 = 18 \times 9 = 162$ cm $^2$ .
5. A cube has volume $27$ cm $^3$ . It is enlarged by factor $k = 2$ . What is the new volume? (show answer)

Answer
$27 \times 2^3 = 27 \times 8 = 216$ cm $^3$ .
6. Two similar triangles have corresponding sides $5$ cm and $15$ cm. State the scale factor. (show answer)

Answer
$k = \dfrac{15}{5} = 3$ .
7. Triangle $PQR \sim$ triangle $XYZ$ . If $PQ = 8$ , $QR = 12$ , and $XY = 6$ , find $YZ$ . (show answer)

Answer
$k = \dfrac{6}{8} = \dfrac{3}{4}$ . So $YZ = \dfrac{3}{4} \times 12 = 9$ .
8. A map has scale $1:25\,000$ . Two towns are $8$ cm apart on the map. Find the real distance in km. (show answer)

Answer
$8 \times 25\,000 = 200\,000$ cm $= 2$ km.
9. State whether each property is preserved under enlargement: (a) angle size, (b) side length, (c) area, (d) shape. (show answer)

Answer
(a) preserved, (b) not preserved, (c) not preserved, (d) preserved.
10. A photo measuring $10$ cm $\times$ $15$ cm is enlarged by factor $k = 1.5$ . Find the new dimensions. (show answer)

Answer
$15$ cm $\times$ $22.5$ cm.

Reasoning · Tier 2: mixed practice

1. Triangle $ABC$ has $AB = 10$ cm, $BC = 8$ cm, $CA = 6$ cm. Triangle $DEF$ is similar with $DE = 15$ cm. Find $EF$ and $FD$ . (show answer)

Answer
$k = \dfrac{15}{10} = 1.5$ . $EF = 1.5 \times 8 = 12$ cm. $FD = 1.5 \times 6 = 9$ cm.
2. A model bridge is built at scale $1:200$ . The real bridge is $340$ m long. How long is the model? (show answer)

Answer
$340 \div 200 = 1.7$ m.
3. Two similar rectangles have areas $50$ cm $^2$ and $200$ cm $^2$ . Find the scale factor of their sides. (show answer)

Answer
Area ratio $= \dfrac{200}{50} = 4$ . Side ratio $= \sqrt{4} = 2$ .
4. A $2$ m post casts a $3$ m shadow. At the same time, a tree casts a $12$ m shadow. How tall is the tree? (show answer)

Answer
$\dfrac{h}{2} = \dfrac{12}{3}$ , so $h = 8$ m.
5. Enlarge the point $(5, -2)$ with centre $(1, 0)$ and scale factor $k = 3$ . Find the image coordinates. (show answer)

Answer
Vector from centre to point: $(5-1, -2-0) = (4, -2)$ . Multiply by $3$ : $(12, -6)$ . Image $= (1+12, 0-6) = (13, -6)$ .
6. A solid sphere has radius $4$ cm. A similar sphere has radius $12$ cm. How many times greater is the volume of the larger sphere? (show answer)

Answer
$k = \dfrac{12}{4} = 3$ . Volume ratio $= 3^3 = 27$ . The larger sphere's volume is $27$ times greater.
7. Explain why two circles are always similar to each other. (show answer)

Answer
Any circle can be mapped to any other by a single enlargement (centred at any point) with scale factor equal to the ratio of the radii. Since all angles in a circle are determined by its curvature and all circles have the same shape, they are always similar.
8. A rectangular garden is $12$ m by $8$ m. A scale drawing uses $k = \dfrac{1}{100}$ . Find the area of the garden on the drawing in cm $^2$ . (show answer)

Answer
Drawing dimensions: $12 \times \dfrac{1}{100} = 0.12$ m $= 12$ cm, and $8 \times \dfrac{1}{100} = 0.08$ m $= 8$ cm. Area $= 12 \times 8 = 96$ cm $^2$ .

Reasoning · Tier 3: explain and apply

1. A triangle is enlarged by factor $k = -1$ . Describe the transformation. How does this differ from a rotation of $180°$ about the centre? (show answer)

Answer
A scale factor of $k = -1$ produces an image that is the same size but on the opposite side of the centre, with each point reflected through the centre. The result is identical to a $180°$ rotation about the centre — a negative scale factor combines enlargement with a half-turn.
2. Prove that the ratio of the areas of two similar figures equals the square of the ratio of corresponding sides. (show answer)

Answer
Let two similar figures have corresponding sides in ratio $k:1$ . Divide each figure into the same small unit squares (or use the same triangulation). Each unit in the larger figure has sides $k$ times as long, so its area is $k^2$ times as large. Summing over all units, total area scales by $k^2$ .
3. Two similar cylinders have heights $10$ cm and $25$ cm. The smaller has volume $200$ cm $^3$ . Find the volume of the larger cylinder. (show answer)

Answer
Height ratio $= \dfrac{25}{10} = 2.5$ . Volume ratio $= 2.5^3 = 15.625$ . Larger volume $= 200 \times 15.625 = 3125$ cm $^3$ .
4. An architect's model uses a scale of $1:50$ . A room in the model has floor area $48$ cm $^2$ . Find the real floor area in m $^2$ . (show answer)

Answer
Scale factor from model to real: $k = 50$ . Real area $= 48 \times 50^2 = 48 \times 2500 = 120\,000$ cm $^2 = 12$ m $^2$ .
5. On a coordinate plane, $\triangle ABC$ has vertices $A(0,0)$ , $B(6,0)$ , $C(3,4)$ . The triangle is enlarged with centre $A$ and scale factor $\dfrac{1}{2}$ . Find the image vertices and verify that all sides are halved. (show answer)

Answer
Centre is $A(0,0)$ , so multiply coordinates by $\dfrac{1}{2}$ . $A' = (0,0)$ , $B' = (3,0)$ , $C' = (1.5, 2)$ . $AB = 6 \to A'B' = 3$ . $AC = \sqrt{9+16} = 5 \to A'C' = \sqrt{2.25+4} = 2.5$ . $BC = \sqrt{9+16} = 5 \to B'C' = \sqrt{2.25+4} = 2.5$ . All sides halved.

Reasoning · Harder reasoning

1. Two similar cones have surface areas in the ratio $4:25$ . If the smaller cone has volume $80$ cm $^3$ , find the volume of the larger cone. (show answer)

Answer
Surface area ratio $= 4:25$ , so side ratio $= \sqrt{4}:\sqrt{25} = 2:5$ . Volume ratio $= 2^3:5^3 = 8:125$ . Larger volume $= 80 \times \dfrac{125}{8} = 1250$ cm $^3$ .
2. A photograph is enlarged by factor $k$ so that its area triples. Find the exact value of $k$ . (show answer)

Answer
New area $= k^2 \times$ original area $= 3 \times$ original area, so $k^2 = 3$ and $k = \sqrt{3}$ .
3. Point $P(8, 6)$ is enlarged with centre $C(2, 3)$ and scale factor $k$ . The image is $P'(17, 10.5)$ . Find $k$ . (show answer)

Answer
$P'_x - C_x = k(P_x - C_x)$ : $17 - 2 = k(8 - 2)$ , so $15 = 6k$ , giving $k = 2.5$ . Check: $P'_y - C_y = 10.5 - 3 = 7.5 = 2.5 \times (6 - 3) = 2.5 \times 3 = 7.5$ . Confirmed $k = 2.5$ .
4. Two similar solids have masses $m_1$ and $m_2$ , made of the same material. Show that $\dfrac{m_1}{m_2} = \left(\dfrac{l_1}{l_2}\right)^3$ where $l_1$ and $l_2$ are corresponding lengths. (show answer)

Answer
Since the solids are similar with corresponding length ratio $\dfrac{l_1}{l_2}$ , volumes scale as $\left(\dfrac{l_1}{l_2}\right)^3$ . With the same material (same density $\rho$ ), mass $= \rho \times V$ , so $\dfrac{m_1}{m_2} = \dfrac{\rho V_1}{\rho V_2} = \dfrac{V_1}{V_2} = \left(\dfrac{l_1}{l_2}\right)^3$ .

Comparing distributions & data analysis

Fluency · Tier 1: basic skills

1. Classify each distribution shape: (a) tail on the right, (b) two peaks, (c) mirror-image shape, (d) tail on the left. (show answer)

Answer
(a) positively skewed, (b) bimodal, (c) symmetric, (d) negatively skewed.
2. Data: $3, 5, 6, 7, 7, 8, 9, 40$ . Find the mean, median, and range. (show answer)

Answer
Mean $= \dfrac{85}{8} = 10.625$ . Median $= \dfrac{7+7}{2} = 7$ . Range $= 40 - 3 = 37$ .
3. Remove the outlier from Q2 and recalculate mean, median, and range. Which statistic changed most? (show answer)

Answer
Without $40$ : mean $= \dfrac{45}{7} \approx 6.43$ , median $= 7$ , range $= 9 - 3 = 6$ . The range changed most (from $37$ to $6$ ), followed by the mean (from $10.625$ to $6.43$ ). The median barely changed.
4. For the data in Q2, which measure of centre better represents the typical value? Explain. (show answer)

Answer
The median ( $7$ ) better represents the typical value because the outlier ( $40$ ) inflates the mean.
5. A sample is taken by selecting every 10th student on a school roll. Name this sampling method. (show answer)

Answer
Systematic sampling.
6. A survey asks 50 people at a train station about their preferred mode of transport. Explain why this sample might be biased. (show answer)

Answer
People at a train station are more likely to prefer trains, so train travel would be overrepresented. People who drive, cycle, or walk are less likely to be at the station.
7. Construct a stem-and-leaf plot for: $14, 18, 22, 25, 27, 31, 33, 36, 38, 42$ . (show answer)

Answer
Stem | Leaf: $1$ | $4\;8$ , $2$ | $2\;5\;7$ , $3$ | $1\;3\;6\;8$ , $4$ | $2$ .
8. What type of display would you use to compare the heights of Year 9 boys and Year 9 girls? (show answer)

Answer
A back-to-back stem-and-leaf plot or side-by-side box plots would both work well for comparing two numerical distributions.
9. State whether the mean or median is higher for a positively skewed distribution. (show answer)

Answer
The mean is higher than the median in a positively skewed distribution (the tail of high values pulls the mean up).
10. A histogram has bars of heights $2, 5, 8, 6, 3, 1$ . Describe the shape of this distribution. (show answer)

Answer
The bars rise then fall with a single peak, so the distribution is approximately symmetric (or very slightly positively skewed if the tail on the right is longer).

Reasoning · Tier 2: mixed practice

1. Two classes recorded the number of books read last term. Class A: $2, 3, 3, 4, 5, 5, 6, 7, 8, 12$ . Class B: $1, 2, 4, 5, 5, 6, 6, 7, 7, 8$ . Construct a back-to-back stem-and-leaf plot and compare the distributions. (show answer)

Answer
Back-to-back stem-and-leaf: Stem $0$ : Class A leaves $8\;7\;6\;5\;5\;4\;3\;3\;2$ | Class B leaves $1\;2\;4\;5\;5\;6\;6\;7\;7\;8$ . Stem $1$ : Class A leaf $2$ | Class B (none). Class A has a wider spread (range $10$ vs $7$ ) with an outlier at $12$ . Class B is more tightly clustered. Medians are similar (A: $5$ , B: $5.5$ ).
2. A data set has mean $24$ and median $18$ . Is the distribution likely symmetric, positively skewed, or negatively skewed? Explain. (show answer)

Answer
Positively skewed. The mean ( $24$ ) is greater than the median ( $18$ ), which indicates a tail of high values pulling the mean up.
3. A researcher wants to survey $200$ out of $2000$ students about study habits. The school has $800$ Year 7, $700$ Year 8, and $500$ Year 9 students. Calculate how many students should be sampled from each year level using stratified sampling. (show answer)

Answer
Total $= 2000$ . Proportions: Year 7 $= \dfrac{800}{2000} \times 200 = 80$ , Year 8 $= \dfrac{700}{2000} \times 200 = 70$ , Year 9 $= \dfrac{500}{2000} \times 200 = 50$ .
4. Explain why the median is preferred over the mean when reporting typical house prices. (show answer)

Answer
House prices are typically positively skewed (a few very expensive houses push the mean up). The median gives a better sense of what a "typical" house costs because it is not affected by the extreme values.
5. A factory records the time (in seconds) to assemble a part. Morning shift: $42, 44, 45, 46, 47, 48, 50$ . Afternoon shift: $43, 45, 46, 48, 50, 52, 58$ . Compare using mean, median, and range. (show answer)

Answer
Morning: mean $= 46$ , median $= 46$ , range $= 8$ . Afternoon: mean $\approx 48.9$ , median $= 48$ , range $= 15$ . The afternoon shift is slightly slower on average and has more variation, possibly due to the outlier at $58$ .
6. Describe a situation where a bimodal distribution would be expected. Explain what causes the two peaks. (show answer)

Answer
Example: heights of a mixed group of adult men and women. The two peaks correspond to the average female height and the average male height — two overlapping subpopulations create bimodality.
7. A student claims: "My sample of 10 friends is representative of the whole school." Critique this claim. (show answer)

Answer
A sample of $10$ friends is a convenience sample that is not random. Friends tend to share interests, backgrounds, and demographics, so the sample is likely biased and not representative of the whole school. A random or stratified sample would be more reliable.
8. State three features you should always comment on when comparing two distributions. (show answer)

Answer
When comparing two distributions, comment on: (i) centre (mean or median), (ii) spread (range or IQR), and (iii) shape (symmetric, skewed, or bimodal). Also note any outliers.

Reasoning · Tier 3: explain and apply

1. A company reports that the "average salary" is $95\,000. The CEO earns $800\,000 and the other $19$ employees earn between $50\,000 and $70\,000 each. Explain how the company's claim could be technically true but misleading. (show answer)

Answer
The CEO's salary of $800\,000 pulls the mean up. If the other $19$ earn an average of $60\,000, the total is $19 \times 60\,000 + 800\,000 = 1\,940\,000$ , giving mean $= \dfrac{1\,940\,000}{20} = 97\,000$ dollars. The "average" (mean) is close to $95\,000 but the median is around $60\,000. Most employees earn far less than the reported average. The company uses the mean to create a misleading impression.
2. Design a statistical investigation to determine whether Year 9 students spend more time on homework than Year 7 students. State the question, sampling method, variables, and how you would display the results. (show answer)

Answer
Question: "Do Year 9 students spend more time per week on homework than Year 7 students?" Sampling: stratified random sample of $30$ students from each year level. Variables: year level (categorical), homework hours per week (continuous). Display: side-by-side box plots or back-to-back stem-and-leaf plot. Calculate mean and median for each group and compare.
3. Two histograms have the same mean and range, but different shapes. Sketch two possible histograms and explain how this is possible. (show answer)

Answer
Example: Histogram 1 is symmetric (bell-shaped). Histogram 2 is bimodal with one peak below the mean and one above. Both can have the same mean (balanced around the centre) and the same range (same min and max) but very different shapes. The bimodal histogram has more data at the extremes and less near the centre.
4. A data set of $20$ values has mean $30$ . An extra value of $80$ is added. Calculate the new mean and explain why the median might be a better summary. (show answer)

Answer
Original sum $= 20 \times 30 = 600$ . New sum $= 600 + 80 = 680$ . New mean $= \dfrac{680}{21} \approx 32.4$ . The mean increased by $2.4$ . The median changes from the average of the 10th and 11th values to the 11th value — it might increase by only $0$ or $1$ , making it more stable and representative.
5. Explain the difference between a population and a sample. Give an example where surveying the whole population is impractical. (show answer)

Answer
A population is the entire group of interest; a sample is a subset selected for study. Example: surveying every one of Australia's $\approx 26$ million residents about exercise habits is impractical due to cost and time. A representative sample of a few thousand provides useful estimates instead.

Reasoning · Harder reasoning

1. Two data sets each have $n$ values. Set A has mean $\bar{x}_A$ and set B has mean $\bar{x}_B$ . If the two sets are combined, show that the combined mean is $\dfrac{n\bar{x}_A + n\bar{x}_B}{2n}$ . What happens if the sets have different sizes $n_A$ and $n_B$ ? (show answer)

Answer
Combined sum $= n\bar{x}_A + n\bar{x}_B$ . Combined count $= 2n$ . Combined mean $= \dfrac{n\bar{x}_A + n\bar{x}_B}{2n} = \dfrac{\bar{x}_A + \bar{x}_B}{2}$ . For different sizes: combined mean $= \dfrac{n_A \bar{x}_A + n_B \bar{x}_B}{n_A + n_B}$ , which is a weighted average of the two means.
2. A researcher adds a constant $c$ to every value in a data set. How does this affect (a) the mean, (b) the median, (c) the range, (d) the standard deviation? Justify each answer. (show answer)

Answer
(a) Mean increases by $c$ (every value increases by $c$ , so the sum increases by $nc$ , and the mean by $c$ ). (b) Median increases by $c$ (the middle value shifts by $c$ ). (c) Range is unchanged (max and min both increase by $c$ , so their difference is the same). (d) Standard deviation is unchanged (deviations from the mean are the same since both each value and the mean shift by $c$ ).
3. Construct a data set of $10$ values where the mean is $50$ , the median is $45$ , and the distribution is positively skewed. Verify your answer. (show answer)

Answer
One possible set: $30, 35, 38, 40, 44, 46, 50, 55, 62, 100$ . Sum $= 500$ , mean $= 50$ . Median $= \dfrac{44+46}{2} = 45$ . The high value $100$ creates a right tail, giving positive skew. Mean $>$ median, confirming positive skewness.
4. A school of $1200$ students is surveyed using stratified sampling by year level. Year 7: $350$ , Year 8: $320$ , Year 9: $280$ , Year 10: $250$ . If $120$ students are to be sampled, calculate the number from each year level. One Year 10 student in the sample scored $0$ on the test (absent). Discuss how this outlier should be handled. (show answer)

Answer
Proportions: Year 7 $= \dfrac{350}{1200} \times 120 = 35$ , Year 8 $= \dfrac{320}{1200} \times 120 = 32$ , Year 9 $= \dfrac{280}{1200} \times 120 = 28$ , Year 10 $= \dfrac{250}{1200} \times 120 = 25$ . The student who scored $0$ was absent, not genuinely scoring zero. This value should be treated as missing data and excluded from analysis (or the student should be resurveyed). Including it would unfairly lower Year 10's statistics and misrepresent that year level's performance.

Compound probability

Fluency · Tier 1: basic skills

1. A coin is tossed twice. List all outcomes using a tree diagram. (show answer)

Answer
Outcomes: HH, HT, TH, TT (4 outcomes).
2. Two dice are rolled. How many outcomes are in the sample space? (show answer)

Answer
$6 \times 6 = 36$ outcomes.
3. A bag has $5$ red and $3$ green marbles. One marble is drawn and replaced, then another is drawn. Find $P(\text{both red})$ . (show answer)

Answer
$P(\text{both red}) = \dfrac{5}{8} \times \dfrac{5}{8} = \dfrac{25}{64}$ .
4. Repeat Q3 but without replacement. (show answer)

Answer
$P(\text{both red}) = \dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}$ .
5. A spinner has sections labelled $1, 2, 3, 4$ (equally likely). It is spun twice. Find $P(\text{both even})$ . (show answer)

Answer
Even numbers are $2$ and $4$ , so $P(\text{even}) = \dfrac{2}{4} = \dfrac{1}{2}$ . $P(\text{both even}) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$ .
6. From a standard deck of $52$ cards, one card is drawn. Find $P(\text{spade or ace})$ . (show answer)

Answer
$P(\text{spade}) = \dfrac{13}{52}$ , $P(\text{ace}) = \dfrac{4}{52}$ , $P(\text{spade and ace}) = \dfrac{1}{52}$ . $P(\text{spade or ace}) = \dfrac{13}{52} + \dfrac{4}{52} - \dfrac{1}{52} = \dfrac{16}{52} = \dfrac{4}{13}$ .
7. A coin is tossed $200$ times and lands heads $114$ times. Calculate the relative frequency of heads. (show answer)

Answer
Relative frequency $= \dfrac{114}{200} = 0.57$ .
8. Are the events "rolling a $6$ " and "rolling an odd number" on a single die mutually exclusive? Explain. (show answer)

Answer
Yes, they are mutually exclusive. A single die cannot show $6$ (even) and an odd number at the same time — the events have no outcomes in common.
9. A tree diagram has two branches at the first stage ( $A$ and $B$ ) and two at the second stage ( $C$ and $D$ from each). How many outcomes are there in total? (show answer)

Answer
$2 \times 2 = 4$ outcomes.
10. In $50$ trials of a simulation, an event occurred $18$ times. Estimate the probability of the event. (show answer)

Answer
$P \approx \dfrac{18}{50} = 0.36$ .

Reasoning · Tier 2: mixed practice

1. A bag contains $6$ red, $4$ blue, and $2$ green marbles. Two marbles are drawn without replacement. Draw a full tree diagram and find $P(\text{both blue})$ . (show answer)

Answer
Total $= 12$ . $P(\text{1st blue}) = \dfrac{4}{12} = \dfrac{1}{3}$ . $P(\text{2nd blue} \mid \text{1st blue}) = \dfrac{3}{11}$ . $P(\text{both blue}) = \dfrac{1}{3} \times \dfrac{3}{11} = \dfrac{3}{33} = \dfrac{1}{11}$ .
2. Two cards are drawn without replacement from a standard deck. Find $P(\text{both aces})$ . (show answer)

Answer
$P(\text{both aces}) = \dfrac{4}{52} \times \dfrac{3}{51} = \dfrac{12}{2652} = \dfrac{1}{221}$ .
3. A box has $3$ defective items out of $20$ . Two items are selected at random without replacement. Find the probability that (a) both are defective, (b) neither is defective, (c) exactly one is defective. (show answer)

Answer
(a) $P(\text{both defective}) = \dfrac{3}{20} \times \dfrac{2}{19} = \dfrac{6}{380} = \dfrac{3}{190}$ . (b) $P(\text{neither defective}) = \dfrac{17}{20} \times \dfrac{16}{19} = \dfrac{272}{380} = \dfrac{68}{95}$ . (c) $P(\text{exactly one}) = 1 - \dfrac{3}{190} - \dfrac{68}{95} = 1 - \dfrac{3}{190} - \dfrac{136}{190} = \dfrac{51}{190} = \dfrac{51}{190}$ .
4. Events $A$ and $B$ are independent with $P(A) = 0.3$ and $P(B) = 0.5$ . Find $P(A \text{ and } B)$ and $P(A \text{ or } B)$ . (show answer)

Answer
$P(A \text{ and } B) = 0.3 \times 0.5 = 0.15$ . $P(A \text{ or } B) = 0.3 + 0.5 - 0.15 = 0.65$ .
5. A student rolls a die and flips a coin. Find the probability of getting an even number and heads. (show answer)

Answer
$P(\text{even}) = \dfrac{3}{6} = \dfrac{1}{2}$ . $P(\text{heads}) = \dfrac{1}{2}$ . Events are independent, so $P(\text{even and heads}) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$ .
6. In a class of $30$ , there are $12$ students who play sport and $8$ who play music. Of these, $3$ play both. A student is chosen at random. Find $P(\text{sport or music})$ . (show answer)

Answer
$P(\text{sport or music}) = \dfrac{12 + 8 - 3}{30} = \dfrac{17}{30}$ .
7. Explain the difference between independent events and mutually exclusive events, using an example of each. (show answer)

Answer
Independent events: the occurrence of one does not affect the probability of the other. Example: rolling a die and flipping a coin — the die result does not change the coin probability. Mutually exclusive events: the events cannot both occur at the same time. Example: rolling a $3$ and rolling a $5$ on a single die. Note: mutually exclusive events with non-zero probabilities are never independent (if one occurs, the probability of the other becomes $0$ ).
8. Design a simulation to estimate the probability that when $4$ coins are tossed, at least $3$ are tails. Describe the steps clearly. (show answer)

Answer
Simulation steps: (i) Assign heads $=$ tails outcome for a coin. (ii) Flip $4$ coins and record the number of tails. (iii) If $3$ or $4$ tails, record a success. (iv) Repeat for $100$ or more trials. (v) Estimate $P(\text{at least 3 tails}) = \dfrac{\text{number of successes}}{\text{total trials}}$ . Theoretical value: $P = \dbinom{4}{3}\left(\dfrac{1}{2}\right)^4 + \dbinom{4}{4}\left(\dfrac{1}{2}\right)^4 = \dfrac{4+1}{16} = \dfrac{5}{16} = 0.3125$ .

Reasoning · Tier 3: explain and apply

1. A medical test has a $95\%$ chance of correctly detecting a disease (sensitivity) and a $90\%$ chance of correctly identifying a healthy person (specificity). If $2\%$ of the population has the disease, draw a tree diagram and find the probability that a person who tests positive actually has the disease. (show answer)

Answer
Let $D =$ has disease, $T^+ =$ tests positive. $P(D) = 0.02$ , $P(T^+ \mid D) = 0.95$ , $P(T^+ \mid \text{no } D) = 0.10$ . By tree diagram: $P(T^+) = 0.02 \times 0.95 + 0.98 \times 0.10 = 0.019 + 0.098 = 0.117$ . $P(D \mid T^+) = \dfrac{0.019}{0.117} \approx 0.162$ . Only about $16\%$ of positive results are true positives — the low disease rate means most positives are false alarms.
2. Two events satisfy $P(A) = 0.4$ , $P(B) = 0.5$ , and $P(A \text{ or } B) = 0.7$ . Determine whether $A$ and $B$ are independent. Justify. (show answer)

Answer
$P(A \text{ and } B) = P(A) + P(B) - P(A \text{ or } B) = 0.4 + 0.5 - 0.7 = 0.2$ . If independent: $P(A) \times P(B) = 0.4 \times 0.5 = 0.2$ . Since $P(A \text{ and } B) = P(A) \times P(B)$ , yes, $A$ and $B$ are independent.
3. Three marbles are drawn without replacement from a bag of $5$ red and $4$ white. Find the probability of drawing at least one red marble. (show answer)

Answer
It is easier to find $P(\text{no red}) = P(\text{all white}) = \dfrac{4}{9} \times \dfrac{3}{8} \times \dfrac{2}{7} = \dfrac{24}{504} = \dfrac{1}{21}$ . So $P(\text{at least one red}) = 1 - \dfrac{1}{21} = \dfrac{20}{21}$ .
4. A game costs $2 to play. You roll two dice: if the sum is $7$ you win $10, otherwise you win nothing. Find the expected profit per game and decide whether the game is fair. (show answer)

Answer
$P(\text{sum} = 7) = \dfrac{6}{36} = \dfrac{1}{6}$ . Expected winnings $= \dfrac{1}{6} \times 10 + \dfrac{5}{6} \times 0 = \dfrac{10}{6} \approx 1.67$ dollars. Expected profit $= 1.67 - 2 = -0.33$ dollars. The game is not fair — on average, the player loses about $33$ cents per game.
5. Explain why $P(A \text{ or } B) \leq P(A) + P(B)$ , and state when equality holds. (show answer)

Answer
$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$ . Since $P(A \text{ and } B) \geq 0$ , we have $P(A \text{ or } B) \leq P(A) + P(B)$ . Equality holds when $P(A \text{ and } B) = 0$ , i.e. when $A$ and $B$ are mutually exclusive.

Reasoning · Harder reasoning

1. Five cards numbered $1$ to $5$ are placed face down. Two cards are selected at random without replacement. Find the probability that the sum of the two cards is even. (show answer)

Answer
Cards $1$ – $5$ : odd numbers are $1, 3, 5$ (three), even numbers are $2, 4$ (two). For an even sum, both cards must be the same parity. $P(\text{both odd}) = \dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{6}{20}$ . $P(\text{both even}) = \dfrac{2}{5} \times \dfrac{1}{4} = \dfrac{2}{20}$ . $P(\text{even sum}) = \dfrac{6}{20} + \dfrac{2}{20} = \dfrac{8}{20} = \dfrac{2}{5}$ .
2. A bag contains $n$ red and $3$ blue marbles. Two marbles are drawn without replacement. If $P(\text{both red}) = \dfrac{1}{2}$ , find $n$ . (show answer)

Answer
$P(\text{both red}) = \dfrac{n}{n+3} \times \dfrac{n-1}{n+2} = \dfrac{1}{2}$ . So $2n(n-1) = (n+3)(n+2)$ , giving $2n^2 - 2n = n^2 + 5n + 6$ , i.e. $n^2 - 7n - 6 = 0$ . Using the quadratic formula: $n = \dfrac{7 + \sqrt{49 + 24}}{2} = \dfrac{7 + \sqrt{73}}{2} \approx \dfrac{7 + 8.544}{2} \approx 7.77$ . Since $n$ must be a positive integer, we check $n = 8$ : $\dfrac{8}{11} \times \dfrac{7}{10} = \dfrac{56}{110} = \dfrac{28}{55} \neq \dfrac{1}{2}$ . Check $n = 6$ : $\dfrac{6}{9} \times \dfrac{5}{8} = \dfrac{30}{72} = \dfrac{5}{12} \neq \dfrac{1}{2}$ . Since no integer solution exists, the equation $n^2 - 7n - 6 = 0$ has no positive integer root. Revisiting: if we allow $P(\text{both red}) = \dfrac{1}{2}$ to be approximate, $n = 8$ gives $\dfrac{28}{55} \approx 0.509$ , which is closest. However, for an exact solution: no integer value of $n$ works — this demonstrates that not every target probability is achievable with whole numbers of marbles.
3. In a best-of-three game series, Team A has a $0.6$ probability of winning each game (games are independent). Draw a tree diagram and find the probability that Team A wins the series. (show answer)

Answer
Team A wins in 2 games: $P = 0.6 \times 0.6 = 0.36$ . Team A loses game 1, wins games 2 and 3: $P = 0.4 \times 0.6 \times 0.6 = 0.144$ . Team A wins game 1, loses game 2, wins game 3: $P = 0.6 \times 0.4 \times 0.6 = 0.144$ . Total: $P(\text{A wins series}) = 0.36 + 0.144 + 0.144 = 0.648$ .
4. Three students independently attempt a problem. Their probabilities of solving it are $0.7$ , $0.8$ , and $0.9$ respectively. Find the probability that at least one student solves the problem. (show answer)

Answer
$P(\text{none solve}) = (1 - 0.7)(1 - 0.8)(1 - 0.9) = 0.3 \times 0.2 \times 0.1 = 0.006$ . $P(\text{at least one solves}) = 1 - 0.006 = 0.994$ .