Year 10 Mathematics Practice

Approximation & accuracy

Fluency · Tier 1: representations and basic error

1. State whether each is exact or approximate: (a) $\sqrt{5}$ (b) $2.236$ (c) $\dfrac{22}{7}$ (d) $3.14$ . (show answer)

Answer
(a) Exact (b) Approximate (c) Exact (d) Approximate.
2. Round $7.3496$ to (a) 1 d.p., (b) 2 d.p., (c) 3 significant figures. (show answer)

Answer
(a) $7.3$ (b) $7.35$ (c) $7.35$ (3 s.f.).
3. Truncate $7.3496$ to (a) 1 d.p., (b) 2 d.p. (show answer)

Answer
(a) $7.3$ (b) $7.34$ .
4. Find the absolute error when $\pi$ is approximated by $3.14$ . (show answer)

Answer
$|\pi - 3.14| = 0.001\,592\ldots \approx 0.0016$ .
5. Find the relative (percentage) error when $\sqrt{3}$ is approximated by $1.73$ . (show answer)

Answer
$\sqrt{3} = 1.732\,050\ldots$ Absolute error $= |1.73 - 1.732\,050\ldots| = 0.002\,05\ldots$ Relative error $= \dfrac{0.002\,05}{1.732\,05} \times 100\% \approx 0.118\%$ .
6. A calculator shows $0.666\,666\,7$ for $\dfrac{2}{3}$ . What is the absolute error? (show answer)

Answer
$\dfrac{2}{3} = 0.666\,666\ldots$ repeating. Absolute error $= |0.666\,666\,7 - 0.666\,666\ldots| \approx 3.3 \times 10^{-8}$ .
7. Round $0.005\,049$ to 2 significant figures. (show answer)

Answer
$0.0050$ .
8. Express $\dfrac{5}{6}$ as a decimal rounded to 4 d.p. and state the absolute error. (show answer)

Answer
$\dfrac{5}{6} = 0.8333\ldots$ Rounded to 4 d.p.: $0.8333$ . Absolute error $= 0.0000\overline{3} \approx 3.3 \times 10^{-5}$ .
9. A length of $\dfrac{7}{3}$ m is recorded as $2.33$ m. Find the absolute and relative error. (show answer)

Answer
$\dfrac{7}{3} = 2.333\ldots$ Absolute error $= |2.33 - 2.333\ldots| = 0.003\ldots$ Relative error $= \dfrac{0.003\overline{3}}{2.333\ldots} \times 100\% \approx 0.143\%$ .
10. True or false: truncation always gives a smaller error than rounding. Justify your answer. (show answer)

Answer
False. For example, truncating $3.6478$ to 2 d.p. gives $3.64$ (error $0.0078$ ), while rounding gives $3.65$ (error $0.0022$ ). Truncation gave the larger error.

Reasoning · Tier 2: accumulated error

1. A square has side length $5.2$ cm (rounded to 1 d.p.). Find the range of possible values for the area. (show answer)

Answer
Side is between $5.15$ and $5.25$ . Area range: $5.15^2 = 26.5225$ to $5.25^2 = 27.5625$ . Range is approximately $26.52$ to $27.56$ cm $^2$ .
2. An investment of $1000 grows by a factor of $1.065$ each year. Compare the amount after 10 years using the exact multiplier vs the multiplier rounded to $1.07$ . (show answer)

Answer
Exact: $1000 \times 1.065^{10} = 1000 \times 1.877\,14\ldots = 1877.14$ . Rounded: $1000 \times 1.07^{10} = 1000 \times 1.967\,15\ldots = 1967.15$ . Difference $\approx 90.01$ .
3. A recipe calls for $\dfrac{2}{3}$ cup of sugar. A cook measures $0.67$ cups. If the recipe is tripled, what is the total error? (show answer)

Answer
Exact amount per batch: $\dfrac{2}{3} = 0.666\ldots$ cup. Measured: $0.67$ . Error per batch: $0.67 - 0.666\ldots = 0.003\overline{3}$ . Tripled: total error $= 3 \times 0.003\overline{3} = 0.01$ cup.
4. The radius of a circle is $3.4$ cm (rounded to 1 d.p.). Find the range of possible values for the circumference. Use $C = 2\pi r$ . (show answer)

Answer
Radius between $3.35$ and $3.45$ cm. Circumference range: $2\pi(3.35) = 6.7\pi \approx 21.05$ to $2\pi(3.45) = 6.9\pi \approx 21.68$ cm.
5. A student calculates $2.45 \times 3.15 \times 1.85$ by rounding each factor to 1 d.p. first. Find the error compared to the exact product. (show answer)

Answer
Exact: $2.45 \times 3.15 \times 1.85 = 14.276\,625$ . Rounded factors: $2.5 \times 3.2 \times 1.9 = 15.2$ . Error $= |15.2 - 14.276\,625| = 0.923\,375$ .
6. Explain why keeping values in surd form during intermediate steps gives a more accurate final answer. (show answer)

Answer
Surds are exact representations. Rounding introduces error that compounds through further operations. For example, $\sqrt{2} \times \sqrt{2} = 2$ exactly, but $1.414 \times 1.414 = 1.999\,396$ .
7. The value $x = 1.4$ is used to compute $x^5$ . If the true value is $1.41$ , find the absolute error in $x^5$ . (show answer)

Answer
$1.4^5 = 5.378\,24$ . $1.41^5 = 5.584\,059\ldots$ Absolute error $= |5.584\,06 - 5.378\,24| \approx 0.206$ .
8. A measurement of $12.0$ cm has a relative error of $0.5\%$ . Find the range of possible true values. (show answer)

Answer
$0.5\%$ of $12.0 = 0.06$ . True value is between $12.0 - 0.06 = 11.94$ and $12.0 + 0.06 = 12.06$ cm.

Reasoning · Tier 3: analysis and explanation

1. Prove that when a value rounded to $\pm\epsilon$ is squared, the maximum absolute error in the square is approximately $2x\epsilon$ (for small $\epsilon$ ). Hint: compare $(x + \epsilon)^2$ with $x^2$ . (show answer)

Answer
$(x + \epsilon)^2 = x^2 + 2x\epsilon + \epsilon^2$ . For small $\epsilon$ , $\epsilon^2$ is negligible, so the error $\approx 2x\epsilon$ . Similarly $(x - \epsilon)^2 \approx x^2 - 2x\epsilon$ , confirming maximum absolute error $\approx 2x\epsilon$ .
2. A GPS unit reports a distance of $142$ km, rounded to the nearest kilometre. This distance is used to calculate fuel needed at $8.3$ L per 100 km. Find the maximum error in the fuel estimate. (show answer)

Answer
Distance between $141.5$ and $142.5$ km. Fuel: $\dfrac{141.5 \times 8.3}{100} = 11.7445$ to $\dfrac{142.5 \times 8.3}{100} = 11.8275$ L. Using $142$ : $11.786$ L. Maximum error $\approx 0.083$ L.
3. A scientist measures the sides of a cuboid as $3.2$ cm, $4.5$ cm and $6.1$ cm (each to 1 d.p.). Calculate the maximum and minimum possible volumes and the percentage range. (show answer)

Answer
Min volume: $3.15 \times 4.45 \times 6.05 = 84.835\ldots$ Max volume: $3.25 \times 4.55 \times 6.15 = 90.966\ldots$ Using rounded values: $3.2 \times 4.5 \times 6.1 = 87.84$ . Percentage range $= \dfrac{90.97 - 84.84}{87.84} \times 100\% \approx 6.98\%$ .
4. Explain, with an example, why subtraction of nearly equal approximate numbers is particularly dangerous for accuracy. (show answer)

Answer
Catastrophic cancellation: if $a \approx 1000.3$ and $b \approx 1000.1$ (each accurate to 4 s.f.), then $a - b \approx 0.2$ , which has only 1 significant figure. The relative error jumps from $\sim 0.01\%$ in each value to potentially $50\%$ in their difference.

Reasoning · Harder reasoning

1. A bank compounds interest monthly at a nominal rate of $4.8\%$ per annum. Compare the balance after 20 years on a $50 000 deposit using (a) the exact monthly multiplier $1 + \dfrac{0.048}{12}$ and (b) the multiplier rounded to 4 d.p. How large is the discrepancy? (show answer)

Answer
Exact monthly rate: $r = 1 + \dfrac{0.048}{12} = 1.004$ . After 240 months: $50\,000 \times 1.004^{240} = 50\,000 \times 2.6051\ldots = 130\,255.20$ . Rounded to 4 d.p. the multiplier is already $1.0040$ , so there is no rounding error at 4 d.p. in this case. With a cruder rounding (e.g. 3 d.p. giving $1.004$ ) the result is the same. If rounded to 2 d.p. as $1.00$ , the balance would be $50 000 -- a discrepancy of roughly $80 255.
2. The golden ratio is $\phi = \dfrac{1 + \sqrt{5}}{2}$ . A student approximates $\phi \approx 1.618$ and calculates $\phi^{10}$ . Find the percentage error compared to the exact value of $\phi^{10}$ . (show answer)

Answer
$\phi^{10} = \left(\dfrac{1+\sqrt{5}}{2}\right)^{10} = \dfrac{123 + 55\sqrt{5}}{2^{10}} \cdot 2^{10}$ . Exact value $= 122.991\,869\ldots$ Using $1.618^{10} = 122.966\ldots$ Percentage error $= \dfrac{|122.992 - 122.966|}{122.992} \times 100\% \approx 0.021\%$ .
3. Two measurements are $a = 10.0 \pm 0.05$ and $b = 9.9 \pm 0.05$ . Show that the relative error of $a - b$ can exceed $100\%$ while the relative errors of $a$ and $b$ individually are each under $1\%$ . (show answer)

Answer
$a - b$ ranges from $(10.0 - 0.05) - (9.9 + 0.05) = 0.0$ to $(10.0 + 0.05) - (9.9 - 0.05) = 0.2$ . Best estimate of $a - b = 0.1$ . Error up to $\pm 0.1$ , relative error $= \dfrac{0.1}{0.1} = 100\%$ , while individual relative errors are $\dfrac{0.05}{10.0} = 0.5\%$ and $\dfrac{0.05}{9.9} \approx 0.505\%$ .
4. A computer stores numbers in floating point with 7 significant digits. Explain how computing $\sqrt{10\,000\,001} - \sqrt{10\,000\,000}$ could lose almost all significant figures, and describe a rearrangement that avoids this problem. (show answer)

Answer
With 7 significant digits: $\sqrt{10\,000\,001} = 3162.277\,8\ldots$ and $\sqrt{10\,000\,000} = 3162.277\,7\ldots$ The difference $\approx 0.000\,016$ , but both stored values agree in their first 7 digits, so the subtraction leaves at most 1-2 correct digits. Rearrangement: multiply by the conjugate: $\sqrt{10\,000\,001} - \sqrt{10\,000\,000} = \dfrac{1}{\sqrt{10\,000\,001} + \sqrt{10\,000\,000}} \approx \dfrac{1}{6324.555} \approx 1.581 \times 10^{-4}$ , which retains full precision.

Algebraic techniques (factorise, fractions, exponents)

Fluency · Tier 1: core skills

1. Factorise $8x^2 - 12x$ . (show answer)

Answer
$4x(2x - 3)$ .
2. Factorise $5a^2b + 10ab^2 - 15ab$ . (show answer)

Answer
$5ab(a + 2b - 3)$ .
3. Simplify $\dfrac{18x^4 y^2}{6x^2 y^5}$ . (show answer)

Answer
$\dfrac{3x^2}{y^3}$ .
4. Simplify $(3a^2)^3 \times 2a^{-4}$ . (show answer)

Answer
$(3a^2)^3 = 27a^6$ . Then $27a^6 \times 2a^{-4} = 54a^2$ .
5. Expand $(x + 4)(x - 7)$ . (show answer)

Answer
$x^2 - 3x - 28$ .
6. Expand $(2x - 3)^2$ . (show answer)

Answer
$4x^2 - 12x + 9$ .
7. Factorise $x^2 - 49$ . (show answer)

Answer
$(x - 7)(x + 7)$ .
8. Factorise $x^2 + 3x - 18$ . (show answer)

Answer
$(x + 6)(x - 3)$ .
9. Simplify $\dfrac{4}{x} + \dfrac{3}{2x}$ . (show answer)

Answer
LCD $= 2x$ : $\dfrac{8}{2x} + \dfrac{3}{2x} = \dfrac{11}{2x}$ .
10. Make $t$ the subject of $s = ut + \dfrac{1}{2}at^2$ when $u = 0$ . (show answer)

Answer
When $u = 0$ : $s = \dfrac{1}{2}at^2$ , so $t^2 = \dfrac{2s}{a}$ , $t = \sqrt{\dfrac{2s}{a}}$ (taking the positive root).

Reasoning · Tier 2: multi-step problems

1. Factorise $3x^2 + 10x - 8$ using grouping. (show answer)

Answer
Product $= 3 \times (-8) = -24$ . Numbers: $12$ and $-2$ . Split: $3x^2 + 12x - 2x - 8 = 3x(x + 4) - 2(x + 4) = (x + 4)(3x - 2)$ .
2. Write $x^2 - 6x + 1$ in the form $(x - p)^2 + q$ . (show answer)

Answer
$x^2 - 6x + 1 = (x - 3)^2 - 9 + 1 = (x - 3)^2 - 8$ .
3. Simplify $\dfrac{x^2 - 4}{x^2 + 5x + 6}$ . (show answer)

Answer
$\dfrac{(x-2)(x+2)}{(x+2)(x+3)} = \dfrac{x - 2}{x + 3}$ (for $x \neq -2$ ).
4. Simplify $\dfrac{2}{x - 1} - \dfrac{3}{x + 2}$ . (show answer)

Answer
LCD $= (x-1)(x+2)$ : $\dfrac{2(x+2) - 3(x-1)}{(x-1)(x+2)} = \dfrac{2x + 4 - 3x + 3}{(x-1)(x+2)} = \dfrac{-x + 7}{(x-1)(x+2)}$ .
5. Make $v$ the subject of $E = \dfrac{1}{2}mv^2$ . (show answer)

Answer
$2E = mv^2$ , $v^2 = \dfrac{2E}{m}$ , $v = \sqrt{\dfrac{2E}{m}}$ .
6. Make $x$ the subject of $y = \dfrac{3x + 1}{x - 2}$ . (show answer)

Answer
$y(x - 2) = 3x + 1$ , $xy - 2y = 3x + 1$ , $xy - 3x = 2y + 1$ , $x(y - 3) = 2y + 1$ , $x = \dfrac{2y + 1}{y - 3}$ .
7. Simplify $\dfrac{x^2 + 6x + 9}{x^2 - 9}$ . (show answer)

Answer
$\dfrac{(x+3)^2}{(x-3)(x+3)} = \dfrac{x + 3}{x - 3}$ (for $x \neq -3$ ).
8. Factorise $2x^2 - 18$ completely. (show answer)

Answer
$2(x^2 - 9) = 2(x - 3)(x + 3)$ .

Reasoning · Tier 3: explain and extend

1. Explain why $x^2 + 4$ cannot be factorised over the real numbers but $x^2 - 4$ can. (show answer)

Answer
$x^2 - 4 = (x-2)(x+2)$ uses the difference of two squares. For $x^2 + 4$ , there are no real numbers $a, b$ with $a^2 - b^2 = x^2 + 4$ in the required form. Since $x^2 \geq 0$ , $x^2 + 4 \geq 4 > 0$ , so it never equals zero and cannot be split into real linear factors.
2. By completing the square, show that $x^2 + 6x + 11 > 0$ for all real $x$ . (show answer)

Answer
$x^2 + 6x + 11 = (x + 3)^2 - 9 + 11 = (x + 3)^2 + 2$ . Since $(x+3)^2 \geq 0$ , the expression $\geq 2 > 0$ for all real $x$ .
3. Simplify $\dfrac{1}{x+1} + \dfrac{1}{x+2} + \dfrac{1}{x+3}$ as a single fraction. (show answer)

Answer
LCD $= (x+1)(x+2)(x+3)$ . Numerator: $(x+2)(x+3) + (x+1)(x+3) + (x+1)(x+2)$ $= (x^2 + 5x + 6) + (x^2 + 4x + 3) + (x^2 + 3x + 2)$ $= 3x^2 + 12x + 11$ . Result: $\dfrac{3x^2 + 12x + 11}{(x+1)(x+2)(x+3)}$ .
4. The surface area of a cylinder is $S = 2\pi r^2 + 2\pi rh$ . Factorise the right-hand side, then rearrange for $h$ . (show answer)

Answer
$S = 2\pi r(r + h)$ . Rearranging: $r + h = \dfrac{S}{2\pi r}$ , so $h = \dfrac{S}{2\pi r} - r = \dfrac{S - 2\pi r^2}{2\pi r}$ .
5. Factorise $x^4 - 16$ completely. (show answer)

Answer
$x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)$ .

Reasoning · Harder reasoning

1. Factorise $6x^2 - 7xy - 3y^2$ and verify by expanding. (show answer)

Answer
Product $= 6 \times (-3) = -18$ . Numbers: $-9$ and $2$ . Split: $6x^2 - 9xy + 2xy - 3y^2 = 3x(2x - 3y) + y(2x - 3y) = (2x - 3y)(3x + y)$ . Check: $(2x - 3y)(3x + y) = 6x^2 + 2xy - 9xy - 3y^2 = 6x^2 - 7xy - 3y^2$ . Correct.
2. If $a + \dfrac{1}{a} = 5$ , find the value of $a^2 + \dfrac{1}{a^2}$ . (show answer)

Answer
$\left(a + \dfrac{1}{a}\right)^2 = a^2 + 2 + \dfrac{1}{a^2}$ . So $25 = a^2 + 2 + \dfrac{1}{a^2}$ , hence $a^2 + \dfrac{1}{a^2} = 23$ .
3. Show that $\dfrac{1}{n} - \dfrac{1}{n+1} = \dfrac{1}{n(n+1)}$ and hence find $\dfrac{1}{1 \times 2} + \dfrac{1}{2 \times 3} + \dfrac{1}{3 \times 4} + \cdots + \dfrac{1}{99 \times 100}$ . (show answer)

Answer
$\dfrac{1}{n} - \dfrac{1}{n+1} = \dfrac{(n+1) - n}{n(n+1)} = \dfrac{1}{n(n+1)}$ . The sum telescopes: $\left(\dfrac{1}{1} - \dfrac{1}{2}\right) + \left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \cdots + \left(\dfrac{1}{99} - \dfrac{1}{100}\right) = 1 - \dfrac{1}{100} = \dfrac{99}{100}$ .
4. A rectangle has area $x^2 + 7x + 10$ and length $x + 5$ . Find the width, perimeter (in terms of $x$ ), and the value of $x$ if the perimeter is $26$ . (show answer)

Answer
Width $= \dfrac{x^2 + 7x + 10}{x + 5} = \dfrac{(x+5)(x+2)}{x+5} = x + 2$ . Perimeter $= 2(x + 5 + x + 2) = 2(2x + 7) = 4x + 14$ . If perimeter $= 26$ : $4x + 14 = 26$ , $x = 3$ .

Quadratic equations

Fluency · Tier 1: solve by factorisation and formula

1. Solve $x^2 - 7x + 10 = 0$ by factorisation. (show answer)

Answer
$(x - 2)(x - 5) = 0$ : $x = 2$ or $x = 5$ .
2. Solve $x^2 + 2x - 15 = 0$ by factorisation. (show answer)

Answer
$(x + 5)(x - 3) = 0$ : $x = -5$ or $x = 3$ .
3. Solve $3x^2 - 12x = 0$ . (show answer)

Answer
$3x(x - 4) = 0$ : $x = 0$ or $x = 4$ .
4. Solve $x^2 - 25 = 0$ . (show answer)

Answer
$(x - 5)(x + 5) = 0$ : $x = 5$ or $x = -5$ .
5. Use the quadratic formula to solve $x^2 + 4x + 1 = 0$ . Give exact answers. (show answer)

Answer
$x = \dfrac{-4 \pm \sqrt{16 - 4}}{2} = \dfrac{-4 \pm \sqrt{12}}{2} = -2 \pm \sqrt{3}$ .
6. Use the quadratic formula to solve $2x^2 - 3x - 1 = 0$ . Give exact answers. (show answer)

Answer
$x = \dfrac{3 \pm \sqrt{9 + 8}}{4} = \dfrac{3 \pm \sqrt{17}}{4}$ .
7. Find the discriminant of $x^2 + 6x + 9 = 0$ and state the number of solutions. (show answer)

Answer
$\Delta = 36 - 36 = 0$ . One repeated solution: $x = -3$ .
8. Find the discriminant of $2x^2 - x + 3 = 0$ and state the number of solutions. (show answer)

Answer
$\Delta = 1 - 24 = -23 < 0$ . No real solutions.
9. Solve $x^2 + 8x + 12 = 0$ by completing the square. (show answer)

Answer
$x^2 + 8x = -12$ . $(x + 4)^2 = 16 - 12 = 4$ . $x + 4 = \pm 2$ . $x = -2$ or $x = -6$ .
10. Solve $(x - 1)(x + 5) = 7$ . (show answer)

Answer
$x^2 + 4x - 5 = 7$ , so $x^2 + 4x - 12 = 0$ . $(x + 6)(x - 2) = 0$ : $x = -6$ or $x = 2$ .

Reasoning · Tier 2: applications and analysis

1. A rectangle has length $(x + 4)$ cm and width $(x - 2)$ cm. Its area is $32$ cm $^2$ . Find $x$ . (show answer)

Answer
$(x + 4)(x - 2) = 32$ , $x^2 + 2x - 8 = 32$ , $x^2 + 2x - 40 = 0$ . $x = \dfrac{-2 \pm \sqrt{4 + 160}}{2} = \dfrac{-2 \pm \sqrt{164}}{2} = -1 \pm \sqrt{41}$ . Since $x > 2$ : $x = -1 + \sqrt{41} \approx 5.40$ cm.
2. Solve $4x^2 - 12x + 9 = 0$ and explain why there is only one solution. (show answer)

Answer
$\Delta = 144 - 144 = 0$ . $(2x - 3)^2 = 0$ , so $x = \dfrac{3}{2}$ . There is one repeated solution because the parabola touches the $x$ -axis at exactly one point.
3. Solve $x^2 - 6x + 3 = 0$ by (a) completing the square and (b) the quadratic formula. Verify the answers agree. (show answer)

Answer
(a) $x^2 - 6x = -3$ . $(x - 3)^2 = 9 - 3 = 6$ . $x = 3 \pm \sqrt{6}$ . (b) $x = \dfrac{6 \pm \sqrt{36 - 12}}{2} = \dfrac{6 \pm \sqrt{24}}{2} = 3 \pm \sqrt{6}$ . Both methods give the same answer.
4. For what values of $k$ does $x^2 + 4x + k = 0$ have two distinct real solutions? (show answer)

Answer
Two distinct real solutions requires $\Delta > 0$ : $16 - 4k > 0$ , so $k < 4$ .
5. The height of a ball is $h = 25t - 5t^2$ metres after $t$ seconds. When is the ball at a height of $20$ m? (show answer)

Answer
$25t - 5t^2 = 20$ . $5t^2 - 25t + 20 = 0$ . $t^2 - 5t + 4 = 0$ . $(t - 1)(t - 4) = 0$ : $t = 1$ s or $t = 4$ s.
6. Solve $\dfrac{6}{x} + x = 5$ by first multiplying through by $x$ . (show answer)

Answer
$\dfrac{6}{x} + x = 5$ . Multiply by $x$ : $6 + x^2 = 5x$ . $x^2 - 5x + 6 = 0$ . $(x - 2)(x - 3) = 0$ : $x = 2$ or $x = 3$ .
7. The sum of a number and its reciprocal is $\dfrac{10}{3}$ . Find the number. (show answer)

Answer
Let the number be $x$ . $x + \dfrac{1}{x} = \dfrac{10}{3}$ . Multiply by $3x$ : $3x^2 + 3 = 10x$ . $3x^2 - 10x + 3 = 0$ . $(3x - 1)(x - 3) = 0$ : $x = \dfrac{1}{3}$ or $x = 3$ .
8. Show that $x^2 + 2x + 5 = 0$ has no real solutions using (a) the discriminant and (b) completing the square. (show answer)

Answer
(a) $\Delta = 4 - 20 = -16 < 0$ , so no real solutions. (b) $x^2 + 2x + 5 = (x + 1)^2 + 4$ . Since $(x+1)^2 \geq 0$ , the expression $\geq 4 > 0$ , so it can never equal zero.

Reasoning · Tier 3: extended reasoning

1. Prove that the quadratic formula follows from completing the square on $ax^2 + bx + c = 0$ . (show answer)

Answer
$ax^2 + bx + c = 0$ . Divide by $a$ : $x^2 + \dfrac{b}{a}x + \dfrac{c}{a} = 0$ . Complete the square: $\left(x + \dfrac{b}{2a}\right)^2 = \dfrac{b^2}{4a^2} - \dfrac{c}{a} = \dfrac{b^2 - 4ac}{4a^2}$ . Square root: $x + \dfrac{b}{2a} = \pm\dfrac{\sqrt{b^2 - 4ac}}{2a}$ . Therefore $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .
2. A farmer encloses a rectangular area using $80$ m of fencing against a river (three sides). Find the maximum area and the dimensions that achieve it. (show answer)

Answer
Width $= x$ , length $= 80 - 2x$ . Area $= x(80 - 2x) = 80x - 2x^2$ . Axis of symmetry: $x = \dfrac{80}{4} = 20$ . Maximum area: $80(20) - 2(400) = 1600 - 800 = 800$ m $^2$ . Dimensions: $20$ m by $40$ m.
3. The parabola $y = 2x^2 + bx + c$ passes through $(1, 3)$ and $(3, 7)$ . Find $b$ and $c$ . (show answer)

Answer
$(1, 3)$ : $2 + b + c = 3$ , so $b + c = 1$ . $(3, 7)$ : $18 + 3b + c = 7$ , so $3b + c = -11$ . Subtract: $2b = -12$ , $b = -6$ . $c = 1 - (-6) = 7$ .
4. For what values of $m$ does the line $y = mx + 1$ intersect the parabola $y = x^2$ at exactly one point? (show answer)

Answer
$x^2 = mx + 1$ , so $x^2 - mx - 1 = 0$ . Exactly one intersection: $\Delta = 0$ . $m^2 + 4 = 0$ . Since $m^2 \geq 0$ , $m^2 + 4 \geq 4 > 0$ for all real $m$ . So $\Delta > 0$ always, meaning the line always intersects the parabola at two points -- there is no value of $m$ giving exactly one intersection.

Reasoning · Harder reasoning

1. Solve $\dfrac{x}{x+1} + \dfrac{x+1}{x} = \dfrac{5}{2}$ by letting $u = \dfrac{x}{x+1}$ . (show answer)

Answer
Let $u = \dfrac{x}{x+1}$ , so $\dfrac{x+1}{x} = \dfrac{1}{u}$ . The equation becomes $u + \dfrac{1}{u} = \dfrac{5}{2}$ . Multiply by $2u$ : $2u^2 + 2 = 5u$ , $2u^2 - 5u + 2 = 0$ , $(2u - 1)(u - 2) = 0$ , $u = \dfrac{1}{2}$ or $u = 2$ . If $u = \dfrac{1}{2}$ : $\dfrac{x}{x+1} = \dfrac{1}{2}$ , $2x = x + 1$ , $x = 1$ . If $u = 2$ : $\dfrac{x}{x+1} = 2$ , $x = 2x + 2$ , $x = -2$ . Solutions: $x = 1$ or $x = -2$ .
2. The roots of $x^2 - 5x + 3 = 0$ are $\alpha$ and $\beta$ . Without finding $\alpha$ and $\beta$ , evaluate $\alpha^2 + \beta^2$ and $\dfrac{1}{\alpha} + \dfrac{1}{\beta}$ . (show answer)

Answer
By Vieta's formulas: $\alpha + \beta = 5$ and $\alpha\beta = 3$ . $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 25 - 6 = 19$ . $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{5}{3}$ .
3. Find all values of $k$ such that $kx^2 + (k+2)x + 1 = 0$ has a repeated root. State any restrictions on $k$ . (show answer)

Answer
$\Delta = (k+2)^2 - 4k = k^2 + 4k + 4 - 4k = k^2 + 4$ . For a repeated root: $\Delta = 0$ , so $k^2 + 4 = 0$ . Since $k^2 \geq 0$ , $k^2 + 4 \geq 4 > 0$ for all real $k$ . There is no real value of $k$ that gives a repeated root (restriction: $k \neq 0$ since the equation must be quadratic).
4. A ball is thrown upward from a $15$ m platform with initial velocity $20$ m/s. Its height is $h = 15 + 20t - 5t^2$ . Find the times when it is at $30$ m height, and the time when it hits the ground. Give exact answers. (show answer)

Answer
$15 + 20t - 5t^2 = 30$ : $5t^2 - 20t + 15 = 0$ , $t^2 - 4t + 3 = 0$ , $(t-1)(t-3) = 0$ : $t = 1$ or $t = 3$ s. Hits ground: $15 + 20t - 5t^2 = 0$ , $t^2 - 4t - 3 = 0$ , $t = \dfrac{4 \pm \sqrt{16 + 12}}{2} = 2 \pm \sqrt{7}$ . Taking the positive root: $t = 2 + \sqrt{7} \approx 4.65$ s.

Simultaneous linear equations

Fluency · Tier 1: basic solving

1. Solve by substitution: $y = x + 3$ and $y = 2x - 1$ . (show answer)

Answer
$x + 3 = 2x - 1$ , $x = 4$ , $y = 7$ .
2. Solve by substitution: $y = 4x$ and $3x + y = 14$ . (show answer)

Answer
$3x + 4x = 14$ , $7x = 14$ , $x = 2$ , $y = 8$ .
3. Solve by elimination: $x + y = 10$ and $x - y = 4$ . (show answer)

Answer
Add: $2x = 14$ , $x = 7$ , $y = 3$ .
4. Solve by elimination: $3x + y = 11$ and $x + y = 5$ . (show answer)

Answer
Subtract: $2x = 6$ , $x = 3$ , $y = 2$ .
5. Solve: $2x + y = 7$ and $x - y = 2$ . (show answer)

Answer
Add: $3x = 9$ , $x = 3$ , $y = 1$ .
6. Solve: $y = 2x + 3$ and $3x - y = -1$ . (show answer)

Answer
Substitute: $3x - (2x + 3) = -1$ , $x - 3 = -1$ , $x = 2$ , $y = 7$ .
7. Solve: $4x + 3y = 18$ and $2x + 3y = 12$ . (show answer)

Answer
Subtract: $2x = 6$ , $x = 3$ , $y = 2$ .
8. Solve: $x + 2y = 8$ and $3x + 2y = 12$ . (show answer)

Answer
Subtract: $2x = 4$ , $x = 2$ , $y = 3$ .
9. Write the equations for: "Two numbers add to 20 and differ by 6." Solve them. (show answer)

Answer
$x + y = 20$ and $x - y = 6$ . Add: $2x = 26$ , $x = 13$ , $y = 7$ .
10. Write the equations for: "A pen costs $2 more than a pencil. Three pens and two pencils cost $21." Solve them. (show answer)

Answer
Let pencil $= p$ , pen $= p + 2$ . $3(p + 2) + 2p = 21$ , $5p + 6 = 21$ , $p = 3$ . Pencil $3, pen $5.

Reasoning · Tier 2: multi-step and applications

1. Solve $\dfrac{x}{2} + \dfrac{y}{3} = 4$ and $x - y = 3$ . (show answer)

Answer
From eq 2: $x = y + 3$ . Substitute: $\dfrac{y+3}{2} + \dfrac{y}{3} = 4$ . Multiply by $6$ : $3(y+3) + 2y = 24$ , $5y + 9 = 24$ , $y = 3$ , $x = 6$ .
2. Solve $5x + 2y = 1$ and $3x - 4y = 11$ . (show answer)

Answer
Multiply eq 1 by $2$ : $10x + 4y = 2$ . Add to eq 2: $13x = 13$ , $x = 1$ , $y = -2$ .
3. Determine whether $2x - 6y = 4$ and $x - 3y = 7$ have no solution, one solution, or infinitely many solutions. (show answer)

Answer
Multiply eq 2 by $2$ : $2x - 6y = 14$ . But eq 1 is $2x - 6y = 4$ . Since $14 \neq 4$ , there is no solution (parallel lines).
4. A fruit shop sells apples at $3 per kg and bananas at $2 per kg. A customer buys 5 kg of fruit for $12. How many kg of each? (show answer)

Answer
$a + b = 5$ and $3a + 2b = 12$ . From eq 1: $b = 5 - a$ . Substitute: $3a + 2(5-a) = 12$ , $a + 10 = 12$ , $a = 2$ kg apples, $b = 3$ kg bananas.
5. Two cars leave the same point. Car A travels north at $80$ km/h and car B travels north at $60$ km/h but left $1$ hour earlier. When and where does car A overtake car B? (show answer)

Answer
Car B position: $60t$ (where $t$ is hours after B left). Car A position: $80(t - 1)$ (A left 1 hour later). Overtake: $80(t-1) = 60t$ , $80t - 80 = 60t$ , $20t = 80$ , $t = 4$ hours after B left. Position: $60 \times 4 = 240$ km from start.
6. The perimeter of a rectangle is $36$ cm and the length is $4$ cm more than the width. Find the dimensions. (show answer)

Answer
$2l + 2w = 36$ and $l = w + 4$ . Substitute: $2(w+4) + 2w = 36$ , $4w + 8 = 36$ , $w = 7$ cm, $l = 11$ cm.
7. Solve $2x + 3y = 1$ and $4x + 6y = 2$ . How many solutions are there? Explain. (show answer)

Answer
Eq 2 is exactly $2 \times$ eq 1: $4x + 6y = 2(2x + 3y) = 2(1) = 2$ . The equations are identical, so there are infinitely many solutions.
8. A test has $30$ questions. Correct answers score $4$ marks; wrong answers lose $1$ mark. A student scores $75$ . How many correct answers? (show answer)

Answer
Let $c$ = correct, $w$ = wrong. $c + w = 30$ and $4c - w = 75$ . Add: $5c = 105$ , $c = 21$ correct answers.

Reasoning · Tier 3: extended problems

1. Three friends buy cinema tickets and snacks. Use the information below to set up and solve simultaneous equations: 2 tickets and 1 snack cost $35; 1 ticket and 2 snacks cost $25. Find the cost of a ticket and a snack. (show answer)

Answer
$2t + s = 35$ and $t + 2s = 25$ . Multiply eq 2 by $2$ : $2t + 4s = 50$ . Subtract eq 1: $3s = 15$ , $s = 5$ . $t = \dfrac{35 - 5}{2} = 15$ . Ticket $15, snack $5.
2. A boat travels $30$ km upstream in $3$ hours and $30$ km downstream in $2$ hours. Find the speed of the boat in still water and the speed of the current. (show answer)

Answer
Let boat speed $= b$ , current $= c$ . Upstream: $\dfrac{30}{b - c} = 3$ , so $b - c = 10$ . Downstream: $\dfrac{30}{b + c} = 2$ , so $b + c = 15$ . Add: $2b = 25$ , $b = 12.5$ km/h, $c = 2.5$ km/h.
3. A company's cost function is $C = 500 + 8n$ and revenue function is $R = 20n - 0.05n^2$ . Find the break-even point(s). (show answer)

Answer
$500 + 8n = 20n - 0.05n^2$ . $0.05n^2 - 12n + 500 = 0$ . $n^2 - 240n + 10\,000 = 0$ . $n = \dfrac{240 \pm \sqrt{57\,600 - 40\,000}}{2} = \dfrac{240 \pm \sqrt{17\,600}}{2} = \dfrac{240 \pm 40\sqrt{11}}{2} = 120 \pm 20\sqrt{11}$ . So $n \approx 120 - 66.3 = 53.7$ or $n \approx 120 + 66.3 = 186.3$ . Break-even at approximately $54$ and $186$ units.
4. The line $ax + by = 1$ passes through $(2, 1)$ and $(4, -1)$ . Find $a$ and $b$ , then write the equation in the form $y = mx + c$ . (show answer)

Answer
Through $(2,1)$ : $2a + b = 1$ . Through $(4,-1)$ : $4a - b = 1$ . Add: $6a = 2$ , $a = \dfrac{1}{3}$ , $b = 1 - \dfrac{2}{3} = \dfrac{1}{3}$ . Equation: $\dfrac{x}{3} + \dfrac{y}{3} = 1$ , so $x + y = 3$ , $y = -x + 3$ .

Reasoning · Harder reasoning

1. A two-digit number has digits that add to $9$ . If the digits are reversed, the new number is $27$ less than the original. Find the number. (Hint: let the tens digit be $t$ and units digit be $u$ , so the number is $10t + u$ .) (show answer)

Answer
Number $= 10t + u$ . $t + u = 9$ and $10t + u - (10u + t) = 27$ , so $9t - 9u = 27$ , $t - u = 3$ . Add: $2t = 12$ , $t = 6$ , $u = 3$ . The number is $63$ .
2. Solve the system $\dfrac{2}{x} + \dfrac{3}{y} = 7$ and $\dfrac{5}{x} - \dfrac{1}{y} = 9$ by letting $u = \dfrac{1}{x}$ and $v = \dfrac{1}{y}$ . (show answer)

Answer
Let $u = \dfrac{1}{x}$ , $v = \dfrac{1}{y}$ . $2u + 3v = 7$ and $5u - v = 9$ . From eq 2: $v = 5u - 9$ . Substitute: $2u + 3(5u - 9) = 7$ , $17u - 27 = 7$ , $u = 2$ , $v = 1$ . So $x = \dfrac{1}{2}$ , $y = 1$ .
3. Find the equation of the line passing through the intersection of $2x + y = 5$ and $x - y = 1$ that also passes through $(0, 4)$ . (show answer)

Answer
Solve $2x + y = 5$ and $x - y = 1$ : add, $3x = 6$ , $x = 2$ , $y = 1$ . The intersection is $(2, 1)$ . Line through $(2, 1)$ and $(0, 4)$ : gradient $= \dfrac{4 - 1}{0 - 2} = -\dfrac{3}{2}$ . Equation: $y = -\dfrac{3}{2}x + 4$ .
4. A shop sells two sizes of coffee. On Monday, 40 small and 25 large coffees earned $285. On Tuesday, 30 small and 35 large coffees earned $295. Find the price of each size, then find the day's revenue if 50 small and 50 large are sold. (show answer)

Answer
$40s + 25l = 285$ and $30s + 35l = 295$ . Multiply eq 1 by $7$ and eq 2 by $5$ : $280s + 175l = 1995$ and $150s + 175l = 1475$ . Subtract: $130s = 520$ , $s = 4$ . $40(4) + 25l = 285$ , $25l = 125$ , $l = 5$ . Small $4, large $5. Revenue for 50 of each: $50(4) + 50(5) = 200 + 250 = 450$ .

Linear relationships & inequalities

Fluency · Tier 1: basic skills

1. Solve $5x + 3 = 28$ . (show answer)

Answer
$5x = 25$ , so $x = 5$ .
2. Make $w$ the subject of $P = 2l + 2w$ . (show answer)

Answer
$2w = P - 2l$ , so $w = \dfrac{P - 2l}{2}$ .
3. Solve $4x - 7 > 9$ and state the solution as an inequality. (show answer)

Answer
$4x > 16$ , so $x > 4$ .
4. Solve $-3x \leq 12$ and show the solution on a number line. (show answer)

Answer
Divide by $-3$ and flip: $x \geq -4$ . Closed circle at $-4$ , shade right.
5. State whether $y = 2x + 1$ and $y = 2x - 5$ are parallel, perpendicular, or neither. (show answer)

Answer
Both have gradient $2$ , so parallel.
6. State whether $y = 4x + 3$ and $y = -\tfrac{1}{4}x + 7$ are parallel, perpendicular, or neither. (show answer)

Answer
$m_1 = 4$ , $m_2 = -\tfrac{1}{4}$ . Product $= 4 \times (-\tfrac{1}{4}) = -1$ , so perpendicular.
7. Find the gradient of a line perpendicular to a line with gradient $5$ . (show answer)

Answer
$m = -\tfrac{1}{5}$ .
8. For the inequality $y > x + 2$ , state whether the boundary line is solid or dashed, and whether you shade above or below. (show answer)

Answer
Dashed line (strict inequality). Shade above (since $y >$ ).

Reasoning · Tier 2: mixed practice

1. The formula $C = \tfrac{5}{9}(F - 32)$ converts Fahrenheit to Celsius. Find $F$ when $C = 20$ . (show answer)

Answer
$20 = \tfrac{5}{9}(F - 32)$ . Multiply both sides by $\tfrac{9}{5}$ : $36 = F - 32$ . $F = 68$ .
2. Solve $7 - 3x \geq 2x + 22$ and graph the solution on a number line. (show answer)

Answer
$7 - 3x \geq 2x + 22$ . $-5x \geq 15$ . Divide by $-5$ , flip: $x \leq -3$ . Closed circle at $-3$ , shade left.
3. Find the equation of the line parallel to $y = -2x + 6$ passing through $(3, 1)$ . (show answer)

Answer
Gradient $= -2$ . Through $(3, 1)$ : $y - 1 = -2(x - 3)$ , so $y = -2x + 7$ .
4. Find the equation of the line perpendicular to $y = \tfrac{1}{2}x - 3$ passing through $(6, 4)$ . (show answer)

Answer
Given gradient $\tfrac{1}{2}$ , perpendicular gradient $= -2$ . Through $(6, 4)$ : $y - 4 = -2(x - 6)$ , so $y = -2x + 16$ .
5. Graph the region satisfying $2x + y \leq 8$ in the first quadrant (where $x \geq 0$ , $y \geq 0$ ). (show answer)

Answer
Boundary: $y = -2x + 8$ . Solid line. Intercepts $(4, 0)$ and $(0, 8)$ . Test $(0,0)$ : $0 \leq 8$ true, shade below (toward origin). Region is the triangle with vertices $(0, 0)$ , $(4, 0)$ , $(0, 8)$ .
6. A rectangle has perimeter $P = 2l + 2w$ . If $P = 40$ and $l$ must be at least twice $w$ , write two inequalities and find the range of possible values for $w$ . (show answer)

Answer
$2l + 2w = 40$ gives $l = 20 - w$ . Also $l \geq 2w$ : $20 - w \geq 2w$ , so $20 \geq 3w$ , $w \leq \tfrac{20}{3} \approx 6.67$ . Since $w > 0$ and $l > 0$ (so $w < 20$ ), the range is $0 < w \leq \tfrac{20}{3}$ .

Reasoning · Tier 3: explain and apply

1. Prove that the triangle with vertices $A(0, 0)$ , $B(4, 2)$ , and $C(2, -4)$ contains a right angle by checking perpendicular gradients. State which angle is $90°$ . (show answer)

Answer
$m_{AB} = \tfrac{2}{4} = \tfrac{1}{2}$ . $m_{AC} = \tfrac{-4}{2} = -2$ . Product: $\tfrac{1}{2} \times (-2) = -1$ . So $AB \perp AC$ , meaning the right angle is at $A$ .
2. The lines $y = kx + 3$ and $y = (2k - 5)x + 1$ are parallel. Find $k$ . (show answer)

Answer
Parallel means equal gradients: $k = 2k - 5$ , so $k = 5$ .
3. Find the equation of the perpendicular bisector of the segment from $A(2, 6)$ to $B(8, 2)$ . (show answer)

Answer
Midpoint: $\left(\tfrac{2+8}{2}, \tfrac{6+2}{2}\right) = (5, 4)$ . Gradient of $AB$ : $\tfrac{2-6}{8-2} = -\tfrac{2}{3}$ . Perpendicular gradient: $\tfrac{3}{2}$ . Equation: $y - 4 = \tfrac{3}{2}(x - 5)$ , so $y = \tfrac{3}{2}x - \tfrac{7}{2}$ .
4. A factory produces $x$ standard items and $y$ premium items. Each standard item needs $2$ hours; each premium item needs $5$ hours. The factory has at most $100$ hours. Write and graph the inequality, then find three integer combinations that use all $100$ hours. (show answer)

Answer
$2x + 5y \leq 100$ , with $x \geq 0$ , $y \geq 0$ . Boundary intercepts: $(50, 0)$ and $(0, 20)$ . Three integer solutions using exactly $100$ hours: $(0, 20)$ , $(25, 10)$ , $(50, 0)$ .

Reasoning · Harder reasoning

1. Show that the quadrilateral with vertices $P(1, 1)$ , $Q(5, 3)$ , $R(4, 5)$ , and $S(0, 3)$ is a parallelogram by proving both pairs of opposite sides are parallel. Is it also a rectangle? Justify using gradients. (show answer)

Answer
$m_{PQ} = \tfrac{3-1}{5-1} = \tfrac{1}{2}$ . $m_{SR} = \tfrac{5-3}{4-0} = \tfrac{1}{2}$ . So $PQ \parallel SR$ . $m_{QR} = \tfrac{5-3}{4-5} = -2$ . $m_{PS} = \tfrac{3-1}{0-1} = -2$ . So $QR \parallel PS$ . Both pairs of opposite sides are parallel, confirming a parallelogram. Check rectangle: $m_{PQ} \times m_{QR} = \tfrac{1}{2} \times (-2) = -1$ . Adjacent sides are perpendicular, so it is a rectangle.
2. Two inequality constraints are $x + 2y \leq 10$ and $3x + y \leq 12$ , with $x \geq 0$ and $y \geq 0$ . Find the vertices of the feasible region and determine which vertex maximises $P = 4x + 3y$ . (show answer)

Answer
Vertices of feasible region: $(0, 0)$ , $(4, 0)$ , $(\tfrac{14}{5}, \tfrac{18}{5})$ , $(0, 5)$ . Intersection of $x + 2y = 10$ and $3x + y = 12$ : solve to get $x = \tfrac{14}{5}$ , $y = \tfrac{18}{5}$ . Evaluate $P$ : at $(0,0)$ : $0$ ; at $(4,0)$ : $16$ ; at $(\tfrac{14}{5}, \tfrac{18}{5})$ : $\tfrac{56+54}{5} = \tfrac{110}{5} = 22$ ; at $(0,5)$ : $15$ . Maximum $P = 22$ at $(\tfrac{14}{5}, \tfrac{18}{5})$ .
3. The line $L_1$ passes through $(a, 2a)$ and $(3, 7)$ , and the line $L_2$ passes through $(1, -1)$ and $(4, 5)$ . If $L_1$ is perpendicular to $L_2$ , find $a$ . (show answer)

Answer
$m_{L_2} = \tfrac{5-(-1)}{4-1} = 2$ . Perpendicular: $m_{L_1} = -\tfrac{1}{2}$ . $m_{L_1} = \tfrac{7-2a}{3-a}$ . Set equal: $\tfrac{7-2a}{3-a} = -\tfrac{1}{2}$ . Cross-multiply: $2(7-2a) = -(3-a)$ . $14 - 4a = -3 + a$ . $17 = 5a$ . $a = \tfrac{17}{5}$ .

Non-linear graphs (quadratic, exponential, circle)

Fluency · Tier 1: basic skills

1. State the vertex and direction (up/down) of $y = (x + 1)^2 - 9$ . (show answer)

Answer
Vertex $(-1, -9)$ . Opens upward ( $a = 1 > 0$ ).
2. Find the $x$ -intercepts of $y = x^2 - 4x - 5$ by factorising. (show answer)

Answer
$x^2 - 4x - 5 = (x - 5)(x + 1) = 0$ . $x$ -intercepts: $x = 5$ and $x = -1$ .
3. State the centre and radius of $(x - 3)^2 + (y + 2)^2 = 25$ . (show answer)

Answer
Centre $(3, -2)$ , radius $5$ .
4. Write the equation of a circle with centre $(0, 4)$ and radius $6$ . (show answer)

Answer
$x^2 + (y - 4)^2 = 36$ .
5. For $y = 3^x$ , find $y$ when $x = 0$ , $x = 2$ , and $x = -1$ . (show answer)

Answer
$x = 0$ : $y = 1$ . $x = 2$ : $y = 9$ . $x = -1$ : $y = \tfrac{1}{3}$ .
6. For $y = \dfrac{4}{x}$ , find $y$ when $x = 1$ , $x = 4$ , and $x = -2$ . (show answer)

Answer
$x = 1$ : $y = 4$ . $x = 4$ : $y = 1$ . $x = -2$ : $y = -2$ .
7. Identify whether each function is a parabola, circle, exponential, or hyperbola: (a) $y = 5^x$ , (b) $x^2 + y^2 = 16$ , (c) $y = \tfrac{3}{x}$ , (d) $y = -x^2 + 2$ . (show answer)

Answer
(a) Exponential. (b) Circle. (c) Hyperbola. (d) Parabola.
8. State the asymptote(s) of $y = \dfrac{-2}{x}$ . (show answer)

Answer
Horizontal asymptote: $y = 0$ . Vertical asymptote: $x = 0$ .

Reasoning · Tier 2: mixed practice

1. Write $y = x^2 + 8x + 12$ in vertex form by completing the square, then state the vertex. (show answer)

Answer
$y = (x^2 + 8x + 16) - 16 + 12 = (x + 4)^2 - 4$ . Vertex $(-4, -4)$ .
2. A ball is thrown upward with height $h = -4.9t^2 + 19.6t + 1.5$ . Find the maximum height and the time it is reached. (show answer)

Answer
$t = -\dfrac{19.6}{2(-4.9)} = 2$ seconds. $h = -4.9(4) + 19.6(2) + 1.5 = -19.6 + 39.2 + 1.5 = 21.1$ metres.
3. Determine whether the point $(3, 4)$ lies inside, on, or outside the circle $(x - 1)^2 + (y - 2)^2 = 9$ . (show answer)

Answer
Substitute: $(3-1)^2 + (4-2)^2 = 4 + 4 = 8$ . Since $8 < 9 = r^2$ , the point is inside the circle.
4. Sketch $y = 2^x$ and $y = \left(\tfrac{1}{2}\right)^x$ on the same axes. Describe the relationship between the two curves. (show answer)

Answer
$y = \left(\tfrac{1}{2}\right)^x = 2^{-x}$ , so the second curve is a reflection of $y = 2^x$ in the $y$ -axis. Both pass through $(0, 1)$ . One grows right, the other decays right.
5. A hyperbola passes through $(2, 5)$ and has the form $y = \tfrac{k}{x}$ . Find $k$ and state the equations of the asymptotes. (show answer)

Answer
$5 = \tfrac{k}{2}$ , so $k = 10$ . Asymptotes: $x = 0$ and $y = 0$ .
6. The parabola $y = a(x - 1)^2 + 3$ passes through $(3, 11)$ . Find $a$ . (show answer)

Answer
$11 = a(3-1)^2 + 3 = 4a + 3$ . $4a = 8$ . $a = 2$ .

Reasoning · Tier 3: explain and apply

1. A tunnel has a parabolic cross-section. At ground level it is $8$ m wide and the maximum height is $5$ m. Taking the origin at the centre of the base, find the equation of the parabola and determine whether a truck $3$ m wide and $4$ m tall can pass through. (show answer)

Answer
The base goes from $(-4, 0)$ to $(4, 0)$ . Maximum height at $(0, 5)$ . Equation: $y = a x^2 + 5$ . At $x = 4$ : $0 = 16a + 5$ , so $a = -\tfrac{5}{16}$ . Equation: $y = -\tfrac{5}{16}x^2 + 5$ . The truck is $3$ m wide, so its edges are at $x = \pm 1.5$ . Height at $x = 1.5$ : $y = -\tfrac{5}{16}(2.25) + 5 = -0.703 + 5 = 4.297$ m. Since $4.297 > 4$ , yes, the truck can pass through.
2. Show algebraically that the circle $x^2 + y^2 = 25$ and the line $y = x + 1$ intersect at two points. Find the coordinates. (show answer)

Answer
Substitute $y = x + 1$ into $x^2 + y^2 = 25$ : $x^2 + (x+1)^2 = 25$ . $2x^2 + 2x + 1 = 25$ . $2x^2 + 2x - 24 = 0$ . $x^2 + x - 12 = 0$ . $(x+4)(x-3) = 0$ . $x = -4$ or $x = 3$ . Points: $(-4, -3)$ and $(3, 4)$ .
3. Compare the graphs of $y = 2^x$ and $y = 3 \times 2^x$ . Explain how the multiplier affects the curve. (show answer)

Answer
$y = 3 \times 2^x$ has the same shape as $y = 2^x$ but every $y$ -value is multiplied by $3$ . It is a vertical stretch by factor $3$ . The $y$ -intercept moves from $(0, 1)$ to $(0, 3)$ . The asymptote remains $y = 0$ .
4. Explain why $y = \dfrac{k}{x}$ can never equal zero, no matter how large $x$ becomes. (show answer)

Answer
For $y = \tfrac{k}{x}$ to equal zero, we would need $\tfrac{k}{x} = 0$ , which means $k = 0 \times x = 0$ . But $k \neq 0$ (otherwise it is not a hyperbola), so no value of $x$ can make $y = 0$ . As $x$ grows larger, $y$ gets closer and closer to $0$ but never reaches it.

Reasoning · Harder reasoning

1. A quadratic $y = ax^2 + bx + c$ has vertex $(2, -3)$ and passes through $(0, 5)$ . Find $a$ , $b$ , and $c$ . (show answer)

Answer
Vertex form: $y = a(x - 2)^2 - 3$ . Through $(0, 5)$ : $5 = a(4) - 3$ , $a = 2$ . Expand: $y = 2(x^2 - 4x + 4) - 3 = 2x^2 - 8x + 5$ . So $a = 2$ , $b = -8$ , $c = 5$ .
2. Find the two points where the parabola $y = x^2$ and the circle $x^2 + y^2 = 2$ intersect (for $y \geq 0$ ). (show answer)

Answer
From $y = x^2$ , substitute into $x^2 + y^2 = 2$ : $y + y^2 = 2$ (since $x^2 = y$ ). $y^2 + y - 2 = 0$ . $(y+2)(y-1) = 0$ . $y = 1$ (take $y \geq 0$ ). Then $x^2 = 1$ , so $x = \pm 1$ . Points: $(1, 1)$ and $(-1, 1)$ .
3. The curve $y = 2^x$ is reflected in the $y$ -axis and then shifted up by $3$ units. Write the equation of the resulting curve and state its asymptote. (show answer)

Answer
Reflect in $y$ -axis: $y = 2^{-x}$ . Shift up $3$ : $y = 2^{-x} + 3$ . Asymptote: $y = 3$ .

Exponential equations & growth/decay

Fluency · Tier 1: basic skills

1. For $y = 7 \times 2^x$ , state the initial value and the base. (show answer)

Answer
Initial value $a = 7$ , base $b = 2$ .
2. Evaluate $y = 4^x$ when $x = 0$ , $x = 1$ , $x = 3$ . (show answer)

Answer
$x = 0$ : $y = 1$ . $x = 1$ : $y = 4$ . $x = 3$ : $y = 64$ .
3. State whether $y = 0.6^x$ represents growth or decay. (show answer)

Answer
Decay (since $0 < 0.6 < 1$ ).
4. A quantity starts at $100$ and increases by $20\%$ each year. Write the exponential model. (show answer)

Answer
$y = 100 \times 1.2^t$ .
5. Solve $3^x = 81$ . (show answer)

Answer
$81 = 3^4$ , so $x = 4$ .
6. Solve $2^x = \tfrac{1}{8}$ . (show answer)

Answer
$\tfrac{1}{8} = 2^{-3}$ , so $x = -3$ .
7. Solve $10^x = 10000$ . (show answer)

Answer
$10000 = 10^4$ , so $x = 4$ .
8. A substance has a half-life of $4$ hours. If you start with $160$ g, how much remains after $12$ hours? (show answer)

Answer
Number of half-lives: $\tfrac{12}{4} = 3$ . Mass: $160 \times \left(\tfrac{1}{2}\right)^3 = 160 \times \tfrac{1}{8} = 20$ g.

Reasoning · Tier 2: mixed practice

1. Solve $4^x = 64$ . (show answer)

Answer
$64 = 4^3$ , so $x = 3$ .
2. Solve $25^x = 125$ . (show answer)

Answer
$25 = 5^2$ , $125 = 5^3$ . $(5^2)^x = 5^3$ . $2x = 3$ . $x = \tfrac{3}{2}$ .
3. A population of $800$ insects triples every $5$ days. Write a model and find the population after $15$ days. (show answer)

Answer
Model: $N = 800 \times 3^{t/5}$ . After $15$ days: $N = 800 \times 3^3 = 800 \times 27 = 21600$ insects.
4. A sample of $500$ g has a half-life of $6$ years. Find the mass after $18$ years. Write the general model. (show answer)

Answer
Half-lives: $\tfrac{18}{6} = 3$ . Mass: $500 \times \left(\tfrac{1}{2}\right)^3 = 500 \times \tfrac{1}{8} = 62.5$ g. Model: $M = 500 \times \left(\tfrac{1}{2}\right)^{t/6}$ .
5. A savings account starts with $1000 and earns $8\%$ per year (compounded annually). Use the Rule of 70 to estimate the doubling time. Then calculate $1000 \times 1.08^9$ to check. (show answer)

Answer
Rule of 70: $\tfrac{70}{8} = 8.75$ years. Check: $1000 \times 1.08^9 = 1000 \times 1.999 \approx 1999$ dollars. Confirms doubling in about $9$ years.
6. Solve $2 \times 5^x = 250$ . (show answer)

Answer
$5^x = 125 = 5^3$ . $x = 3$ .

Reasoning · Tier 3: explain and apply

1. Explain why an exponential decay function $y = a \times b^x$ (with $0 < b < 1$ ) never reaches zero, no matter how large $x$ is. (show answer)

Answer
Since $b > 0$ , raising $b$ to any power gives a positive result: $b^x > 0$ for all $x$ . Multiplying by $a > 0$ keeps it positive. So $y = a \times b^x > 0$ always. The curve approaches the $x$ -axis (its asymptote) but never touches or crosses it.
2. Two bacteria colonies start at the same time. Colony A has $100$ bacteria and doubles every $2$ hours. Colony B has $3200$ bacteria and halves every $2$ hours. After how many hours do they have the same population? (show answer)

Answer
Colony A: $N_A = 100 \times 2^{t/2}$ . Colony B: $N_B = 3200 \times \left(\tfrac{1}{2}\right)^{t/2}$ . Set equal: $100 \times 2^{t/2} = 3200 \times 2^{-t/2}$ . Multiply both sides by $2^{t/2}$ : $100 \times 2^t = 3200$ . $2^t = 32 = 2^5$ . $t = 5$ . They are equal after $\mathbf{10}$ hours. Wait -- let $u = t/2$ : $100 \times 2^u = 3200 \times 2^{-u}$ , $100 \times 2^{2u} = 3200$ , $2^{2u} = 32 = 2^5$ , $2u = 5$ , $u = 2.5$ , $t = 5$ hours. Check: $N_A = 100 \times 2^{2.5} \approx 566$ , $N_B = 3200 \times 2^{-2.5} \approx 566$ . Equal after 5 hours.
3. A car worth $30000 depreciates at $12\%$ per year. After how many whole years is it first worth less than $10000? (show answer)

Answer
$V = 30000 \times 0.88^t$ . Need $V < 10000$ : $0.88^t < \tfrac{1}{3}$ . By trial: $0.88^8 \approx 0.3596$ , $0.88^9 \approx 0.3165$ , $0.88^{10} \approx 0.2785$ . After $9$ years: $V \approx 9494 < 10000$ . First worth less than $10000 after 9 whole years.
4. The mass of a radioactive isotope is modelled by $M = 800 \times \left(\tfrac{1}{2}\right)^{t/5}$ . Find the half-life, the mass after $20$ years, and the time when the mass first drops below $100$ g. (show answer)

Answer
Half-life $= 5$ years (the denominator in the exponent). After $20$ years: $M = 800 \times \left(\tfrac{1}{2}\right)^4 = 800 \times \tfrac{1}{16} = 50$ g. Below $100$ g: $\left(\tfrac{1}{2}\right)^{t/5} < \tfrac{1}{8} = \left(\tfrac{1}{2}\right)^3$ , so $\tfrac{t}{5} > 3$ , $t > 15$ years.

Reasoning · Harder reasoning

1. Solve $8^{2x+1} = 4^{3x-2}$ . (Hint: express both sides as powers of $2$ .) (show answer)

Answer
$8 = 2^3$ and $4 = 2^2$ . So $2^{3(2x+1)} = 2^{2(3x-2)}$ . $6x + 3 = 6x - 4$ . $3 = -4$ , which is false. No solution.
2. A lake contains $10000$ fish. Due to overfishing the population drops by $10\%$ each year, but each year $500$ fish are also added by restocking. Write a recurrence relation and find the population after $3$ years. Is the population stabilising, growing, or declining? (show answer)

Answer
Let $P_n$ be the population after year $n$ . $P_0 = 10000$ . $P_{n+1} = 0.9 P_n + 500$ . Year 1: $0.9(10000) + 500 = 9500$ . Year 2: $0.9(9500) + 500 = 9050$ . Year 3: $0.9(9050) + 500 = 8645$ . The population is declining. (It would stabilise at $P = 0.9P + 500$ , so $0.1P = 500$ , $P = 5000$ , but it has not reached that level yet.)
3. Two investments start at the same time. Investment A is $5000 growing at $6\%$ p.a. Investment B is $8000 growing at $3\%$ p.a. After how many whole years does Investment A first exceed Investment B? (Use trial and improvement.) (show answer)

Answer
Need $5000 \times 1.06^t > 8000 \times 1.03^t$ . $\left(\tfrac{1.06}{1.03}\right)^t > 1.6$ . $1.02913^t > 1.6$ . By trial: $t = 16$ : $1.02913^{16} \approx 1.585$ ; $t = 17$ : $\approx 1.631$ . Investment A first exceeds B after 17 whole years.

Compound interest & financial modelling

Fluency · Tier 1: basic skills

1. Calculate simple interest on $4000 at $5\%$ p.a. for $3$ years. (show answer)

Answer
$I = 4000 \times 0.05 \times 3 = 600$ dollars. Total: $4600$ .
2. Find the total amount when $4000 is invested at $5\%$ p.a. compounded annually for $3$ years. (show answer)

Answer
$A = 4000 \times 1.05^3 = 4000 \times 1.157625 = 4630.50$ dollars.
3. State the difference between your answers to Q1 and Q2. (show answer)

Answer
Compound gives $4630.50 - 4600 = 30.50$ dollars more.
4. Find the value of $10000 invested at $3\%$ p.a. compounded annually for $6$ years. (show answer)

Answer
$A = 10000 \times 1.03^6 = 10000 \times 1.19405 \approx 11940.52$ dollars.
5. A computer worth $1800 depreciates at $25\%$ per year. Find its value after $2$ years. (show answer)

Answer
$A = 1800 \times 0.75^2 = 1800 \times 0.5625 = 1012.50$ dollars.
6. Find the value of $5000 at $8\%$ p.a. compounded quarterly for $2$ years. (Hint: $n = 4$ .) (show answer)

Answer
$A = 5000\!\left(1 + \dfrac{0.08}{4}\right)^{8} = 5000 \times 1.02^8 = 5000 \times 1.17166 \approx 5858.30$ dollars.
7. A painting is bought for $3000 and appreciates at $10\%$ per year. What is it worth after $4$ years? (show answer)

Answer
$A = 3000 \times 1.1^4 = 3000 \times 1.4641 = 4392.30$ dollars.
8. How much interest is earned on $7000 at $6\%$ p.a. compounded annually for $5$ years? (show answer)

Answer
$A = 7000 \times 1.06^5 = 7000 \times 1.33823 \approx 9367.58$ . Interest $= 9367.58 - 7000 = 2367.58$ dollars.

Reasoning · Tier 2: mixed practice

1. $12000 is invested at $4.5\%$ p.a. compounded monthly. Find the amount after $3$ years. (show answer)

Answer
$A = 12000\!\left(1 + \dfrac{0.045}{12}\right)^{36} = 12000 \times 1.00375^{36} \approx 12000 \times 1.14396 \approx 13727.55$ dollars.
2. A car worth $28000 depreciates at $15\%$ per year. After how many whole years is it first worth less than $10000? (show answer)

Answer
$28000 \times 0.85^t < 10000$ . $0.85^t < 0.3571$ . By trial: $0.85^6 \approx 0.3771$ , $0.85^7 \approx 0.3206$ . First below $10000 after 7 whole years.
3. Which is better over $5$ years: $8000 at $6\%$ p.a. compounded annually, or $8000 at $5.8\%$ p.a. compounded monthly? Show both calculations. (show answer)

Answer
Option 1: $8000 \times 1.06^5 = 8000 \times 1.33823 \approx 10705.80$ . Option 2: $8000 \times 1.004833^{60} \approx 8000 \times 1.33556 \approx 10684.48$ . Option 1 ( $6\%$ annually) gives about $21 more.
4. Ava invests $P at $7\%$ p.a. compounded annually. After $10$ years she has $15000. Find $P$ . (show answer)

Answer
$15000 = P \times 1.07^{10}$ . $1.07^{10} \approx 1.96715$ . $P = \dfrac{15000}{1.96715} \approx 7625.85$ dollars.
5. A motorbike depreciates from $9000 to $4500 in $4$ years. Find the annual depreciation rate. (show answer)

Answer
$4500 = 9000 \times (1-r)^4$ . $(1-r)^4 = 0.5$ . $1 - r = 0.5^{0.25} \approx 0.8409$ . $r \approx 0.159$ , so approximately $15.9\%$ per year.
6. Calculate the total interest earned on $20000 at $3.2\%$ p.a. compounded quarterly over $8$ years. (show answer)

Answer
$A = 20000\!\left(1 + \dfrac{0.032}{4}\right)^{32} = 20000 \times 1.008^{32} \approx 20000 \times 1.29098 \approx 25819.63$ . Interest $= 25819.63 - 20000 = 5819.63$ dollars.

Reasoning · Tier 3: explain and apply

1. Explain why the gap between simple interest and compound interest widens as time increases. Use the formulas to support your answer. (show answer)

Answer
Simple interest after $T$ years: $A_S = P(1 + rT)$ , which is linear in $T$ . Compound interest: $A_C = P(1+r)^T$ , which is exponential. For small $T$ , $(1+r)^T \approx 1 + rT$ (nearly equal). As $T$ increases, the exponential term grows faster than the linear term because each year's interest is applied to an ever-larger base. The gap is $P[(1+r)^T - (1 + rT)]$ , which increases with $T$ .
2. A credit card charges $1.5\%$ interest per month on unpaid balances. What is the effective annual interest rate? (Hint: $(1.015)^{12}$ .) (show answer)

Answer
Effective annual rate $= (1.015)^{12} - 1 \approx 1.19562 - 1 = 0.19562$ , or about $19.6\%$ per year.
3. Mia has $20000 to invest for $10$ years. Bank A offers $5\%$ p.a. compounded annually. Bank B offers $4.9\%$ p.a. compounded daily (assume $365$ days). Which should she choose? (show answer)

Answer
Bank A: $20000 \times 1.05^{10} = 20000 \times 1.62889 \approx 32577.89$ . Bank B: $20000\!\left(1 + \dfrac{0.049}{365}\right)^{3650} \approx 20000 \times e^{0.49} \approx 20000 \times 1.63232 \approx 32646.33$ . Bank B gives slightly more (about $68 extra). Mia should choose Bank B.
4. A factory machine costs $50000 and depreciates at $20\%$ per year. The company plans to replace it when its value drops below $10\%$ of the original cost. After how many whole years should they replace it? (show answer)

Answer
Need $50000 \times 0.8^n < 5000$ . $0.8^n < 0.1$ . By trial: $0.8^{10} \approx 0.10737$ , $0.8^{11} \approx 0.08590$ . Replace after 11 whole years.

Reasoning · Harder reasoning

1. Show that for a principal $P$ invested at rate $r$ p.a. compounded annually for $2$ years, the compound interest exceeds the simple interest by exactly $Pr^2$ . (show answer)

Answer
Simple interest for $2$ years: $I_S = P \times r \times 2 = 2Pr$ . Compound interest for $2$ years: $A_C = P(1+r)^2 = P(1 + 2r + r^2)$ . So $I_C = A_C - P = 2Pr + Pr^2$ . Difference: $I_C - I_S = (2Pr + Pr^2) - 2Pr = Pr^2$ .
2. Jake borrows $15000 at $8\%$ p.a. compounded annually. He makes no repayments. After how many whole years does the debt first exceed $25000? How does this compare with the Rule of 70 estimate for doubling? (show answer)

Answer
$15000 \times 1.08^t > 25000$ . $1.08^t > \tfrac{5}{3} \approx 1.6667$ . By trial: $1.08^6 \approx 1.5869$ , $1.08^7 \approx 1.7138$ . Debt exceeds $25000 after 7 whole years. Rule of 70: doubling time $\approx \tfrac{70}{8} = 8.75$ years; the debt reaches $\tfrac{5}{3}$ of the original (not double) so it happens sooner than the doubling time, which is consistent.
3. An investment of $10000 grows to $10000 \times 1.06^t $after$ t $years. A second investment of \$ 15000 grows to $15000 \times 1.03^t$ after $t$ years. After how many whole years does the first investment first exceed the second? Justify your answer. (show answer)

Answer
Need $10000 \times 1.06^t > 15000 \times 1.03^t$ . $\left(\tfrac{1.06}{1.03}\right)^t > 1.5$ . $1.02913^t > 1.5$ . By trial: $1.02913^{14} \approx 1.494$ , $1.02913^{15} \approx 1.538$ . After 15 whole years the first investment first exceeds the second.

Surface area & volume of composite objects

Fluency · Tier 1: basic calculations

1. A cylinder ( $r = 4$ cm, $h = 10$ cm) sits on top of a cube of side $10$ cm. Find the total volume. (show answer)

Answer
Cube volume $= 1000$ cm $^3$ ; cylinder volume $= \pi(16)(10) = 160\pi \approx 502.65$ cm $^3$ . Total $\approx 1502.65$ cm $^3$ .
2. A rectangular prism $8 \times 6 \times 5$ cm has a cylindrical hole ( $r = 2$ cm) drilled through the $5$ cm height. Find the remaining volume. (show answer)

Answer
Prism $= 240$ cm $^3$ ; hole $= \pi(4)(5) = 20\pi \approx 62.83$ cm $^3$ . Remaining $\approx 177.17$ cm $^3$ .
3. Two rectangular prisms are joined end-to-end: one is $6 \times 4 \times 3$ cm, the other is $8 \times 4 \times 3$ cm. Find the total volume and the exposed surface area. (show answer)

Answer
Volume $= 72 + 96 = 168$ cm $^3$ . The joined face ( $4 \times 3 = 12$ cm $^2$ ) is hidden. SA of each prism: $2(24+18+12) = 108$ and $2(32+24+12) = 136$ . Exposed SA $= 108 + 136 - 2(12) = 220$ cm $^2$ .
4. A cylinder ( $r = 3$ cm, $h = 7$ cm) has a hemisphere ( $r = 3$ cm) on top. Find the total volume. (show answer)

Answer
Cylinder $= \pi(9)(7) = 63\pi$ ; hemisphere $= \frac{2}{3}\pi(27) = 18\pi$ . Total $= 81\pi \approx 254.47$ cm $^3$ .
5. Find the exposed surface area in question 4. (Hemisphere curved SA $= 2\pi r^2$ .) (show answer)

Answer
Cylinder curved SA $= 2\pi(3)(7) = 42\pi$ ; cylinder base $= 9\pi$ ; hemisphere curved $= 2\pi(9) = 18\pi$ . Total $= (42 + 9 + 18)\pi = 69\pi \approx 216.77$ cm $^2$ . (Top circle of cylinder is replaced by hemisphere, so not counted.)
6. An L-shaped block is formed from a $10 \times 6 \times 4$ cm prism with a $4 \times 3 \times 4$ cm block removed from one corner. Find the volume. (show answer)

Answer
Full prism $= 240$ ; removed $= 48$ . Volume $= 192$ cm $^3$ .
7. A half-cylinder ( $r = 5$ cm, $l = 12$ cm) sits on top of a rectangular prism $10 \times 12 \times 6$ cm. Find the total volume. (show answer)

Answer
Prism $= 720$ cm $^3$ ; half-cylinder $= \frac{1}{2}\pi(25)(12) = 150\pi \approx 471.24$ cm $^3$ . Total $\approx 1191.24$ cm $^3$ .
8. Find the total surface area for the solid in question 1, given that the cylinder sits centred on the top face of the cube. (show answer)

Answer
Cube full SA $= 600$ ; cylinder full SA $= 2\pi(16) + 2\pi(4)(10) = 32\pi + 80\pi = 112\pi \approx 351.86$ ; shared circle $= 16\pi \approx 50.27$ . Exposed SA $\approx 600 + 351.86 - 2(50.27) \approx 851.32$ cm $^2$ .

Reasoning · Tier 2: mixed practice

1. A cylindrical water tank ( $r = 0.5$ m, $h = 1.8$ m) needs to be insulated on the curved surface and top only. Insulation costs $18 per m $^2$ . Find the total cost. (show answer)

Answer
Curved SA $= 2\pi(0.5)(1.8) = 1.8\pi$ ; top $= \pi(0.25) = 0.25\pi$ . Total area $= 2.05\pi \approx 6.44$ m $^2$ . Cost $= 6.44 \times 18 \approx$ $115.92.
2. A swimming pool has a uniform rectangular cross-section $10 \times 5$ m. It is $1.2$ m deep at one end and $2.4$ m at the other (the floor slopes uniformly). Find the volume of water when the pool is full. (show answer)

Answer
Cross-section is a trapezium with parallel sides $1.2$ and $2.4$ , width $10$ . Area $= \frac{1}{2}(1.2 + 2.4)(10) = 18$ m $^2$ . Volume $= 18 \times 5 = 90$ m $^3$ .
3. A factory chimney consists of a rectangular base $2 \times 2 \times 1$ m topped by a cylinder of radius $0.4$ m and height $8$ m. Find (a) the total volume, and (b) the total exposed surface area (the chimney is open at the top). (show answer)

Answer
(a) Base $= 4$ m $^3$ ; column $= \pi(0.16)(8) = 1.28\pi \approx 4.02$ m $^3$ . Total $\approx 8.02$ m $^3$ . (b) Base: bottom $= 4$ , four sides $= 4(2 \times 1) = 8$ , top exposed $= 4 - \pi(0.16) \approx 3.50$ . Column: curved $= 2\pi(0.4)(8) = 6.4\pi \approx 20.11$ . (Open top, so no top circle.) Total SA $\approx 4 + 8 + 3.50 + 20.11 = 35.61$ m $^2$ .
4. A packing box $30 \times 20 \times 15$ cm contains a cylindrical can ( $r = 5$ cm, $h = 15$ cm) standing upright. What percentage of the box volume is wasted space? (show answer)

Answer
Box $= 9000$ cm $^3$ ; can $= \pi(25)(15) = 375\pi \approx 1178.10$ cm $^3$ . Wasted $\approx 9000 - 1178.10 = 7821.90$ cm $^3$ . Percentage $\approx 86.9\%$ .
5. Two cylinders are joined: a large one ( $r = 6$ cm, $h = 10$ cm) with a smaller one ( $r = 3$ cm, $h = 8$ cm) centred on top. Find the total exposed surface area. (show answer)

Answer
Large cylinder SA: curved $= 2\pi(6)(10) = 120\pi$ ; base $= 36\pi$ ; top annulus (ring) $= 36\pi - 9\pi = 27\pi$ . Small cylinder: curved $= 2\pi(3)(8) = 48\pi$ ; top $= 9\pi$ . Total exposed $= (120 + 36 + 27 + 48 + 9)\pi = 240\pi \approx 753.98$ cm $^2$ .
6. A solid is made by cutting a hemisphere ( $r = 4$ cm) from the top of a cylinder ( $r = 4$ cm, $h = 10$ cm). Find the remaining volume. (show answer)

Answer
Cylinder $= \pi(16)(10) = 160\pi$ ; hemisphere $= \frac{2}{3}\pi(64) = \frac{128\pi}{3}$ . Remaining $= 160\pi - \frac{128\pi}{3} = \frac{352\pi}{3} \approx 368.61$ cm $^3$ .

Reasoning · Tier 3: explain and apply

1. Explain why you must subtract the shared face area twice (not once) when finding the exposed surface area of two joined solids. (show answer)

Answer
Each component's full SA includes the shared face as part of its own total. When the two solids join, that face is hidden on both solids -- it disappears from the outside of solid 1 and from the outside of solid 2. Since it was counted once in each SA calculation, you must subtract it twice to get the correct exposed SA.
2. A composite tank is a cylinder ( $r = 1$ m, $h = 2$ m) with a cone on top (same radius, height $0.5$ m). The cone volume is $\frac{1}{3}\pi r^2 h$ . Find the total capacity in litres and explain why the cone adds relatively little capacity. (show answer)

Answer
Cylinder $= \pi(1)^2(2) = 2\pi$ m $^3$ ; cone $= \frac{1}{3}\pi(1)^2(0.5) = \frac{\pi}{6}$ m $^3$ . Total $= \frac{13\pi}{6} \approx 6.81$ m $^3 \approx 6807$ L. The cone adds only $\frac{\pi}{6} \approx 0.52$ m $^3$ ( $\approx 524$ L), about $7.7\%$ of the total, because the cone formula includes the $\frac{1}{3}$ factor and the cone height is small.
3. A manufacturer needs a container with volume $2000$ cm $^3$ . Design A is a single cylinder; Design B is a cube with a hemisphere on top. For each, find dimensions that achieve the target volume and compare total surface areas to determine which uses less material. (show answer)

Answer
Design A (cylinder): choose $r$ and $h$ with $\pi r^2 h = 2000$ . For example $r = 6$ cm gives $h = \frac{2000}{36\pi} \approx 17.68$ cm; SA $= 2\pi(36) + 2\pi(6)(17.68) \approx 226.19 + 666.18 \approx 892.4$ cm $^2$ . Design B (cube + hemisphere): cube side $s$ with hemisphere $r = s/2$ ; volume $= s^3 + \frac{2}{3}\pi(s/2)^3 = s^3(1 + \frac{\pi}{12}) \approx 1.262 s^3 = 2000$ , so $s \approx 11.67$ cm. SA $= 5s^2 + 2\pi(s/2)^2 = 5s^2 + \frac{\pi s^2}{2} \approx 6.571 s^2 \approx 894.7$ cm $^2$ . Both designs use roughly similar material; the optimal choice depends on exact dimensions.
4. A rectangular prism $a \times a \times 2a$ has a cylinder of radius $\frac{a}{4}$ drilled through its longest dimension. Express the remaining volume as a function of $a$ . (show answer)

Answer
$V = a \times a \times 2a - \pi\left(\frac{a}{4}\right)^2(2a) = 2a^3 - \frac{\pi a^3}{8} = a^3\left(2 - \frac{\pi}{8}\right)$ .

Reasoning · Harder reasoning

1. A silo consists of a cylinder of radius $3$ m and height $10$ m topped by a hemisphere. Find (a) the total volume, and (b) the total external surface area. If grain fills the silo to $\frac{3}{4}$ of its capacity, find the volume of grain. (show answer)

Answer
(a) Cylinder $= \pi(9)(10) = 90\pi$ ; hemisphere $= \frac{2}{3}\pi(27) = 18\pi$ . Total $= 108\pi \approx 339.29$ m $^3$ . (b) Base circle $= 9\pi$ ; curved cylinder $= 2\pi(3)(10) = 60\pi$ ; hemisphere curved $= 2\pi(9) = 18\pi$ . Total SA $= (9 + 60 + 18)\pi = 87\pi \approx 273.32$ m $^2$ . Grain volume $= \frac{3}{4}(108\pi) = 81\pi \approx 254.47$ m $^3$ .
2. A trophy is made from a rectangular prism base $8 \times 8 \times 2$ cm, with a cylinder ( $r = 2$ cm, $h = 12$ cm) rising from its centre, and a solid sphere ( $r = 3$ cm) on top of the cylinder. Find the total volume and the total exposed surface area. (Sphere SA $= 4\pi r^2$ ; sphere $V = \frac{4}{3}\pi r^3$ .) (show answer)

Answer
Base prism $= 128$ cm $^3$ ; cylinder $= \pi(4)(12) = 48\pi$ ; sphere $= \frac{4}{3}\pi(27) = 36\pi$ . Total volume $= 128 + 84\pi \approx 391.88$ cm $^3$ . SA: prism bottom $= 64$ ; prism four sides $= 4(16) = 64$ ; prism top annulus $= 64 - 4\pi$ ; cylinder curved $= 2\pi(2)(12) = 48\pi$ ; sphere $= 4\pi(9) = 36\pi$ (but a circle $\pi(4)$ is hidden where sphere meets cylinder, subtract $2\pi(4) = 8\pi$ ). Sphere sits on top so small circle hidden; actually sphere $r=3 >$ cylinder $r=2$ , so the sphere rests on the cylinder rim; exposed sphere SA $= 36\pi - \pi(4) = 32\pi$ . Total SA $\approx 64 + 64 + (64 - 4\pi) + 48\pi + 32\pi \approx 192 + 76\pi \approx 430.73$ cm $^2$ .
3. An underground pipe is a hollow cylinder with outer radius $15$ cm and inner radius $12$ cm, running for $50$ m. Find the volume of material in the pipe wall. (show answer)

Answer
Outer volume $= \pi(0.15)^2(50) = 1.125\pi$ m $^3$ ; inner volume $= \pi(0.12)^2(50) = 0.72\pi$ m $^3$ . Wall volume $= (1.125 - 0.72)\pi = 0.405\pi \approx 1.272$ m $^3$ .
4. A composite solid is formed by attaching a square-based pyramid (base $6 \times 6$ cm, slant height $5$ cm) to the top of a cube of side $6$ cm. Find the total exposed surface area. (Lateral area of a pyramid $= \frac{1}{2} \times \text{perimeter} \times \text{slant height}$ .) (show answer)

Answer
Cube has 5 exposed faces (top replaced by pyramid base): $5 \times 36 = 180$ cm $^2$ . Pyramid lateral area $= \frac{1}{2}(24)(5) = 60$ cm $^2$ . Total exposed SA $= 180 + 60 = 240$ cm $^2$ .

Logarithmic scales

Fluency · Tier 1: basic skills

1. Evaluate without a calculator: $\log_{10} 1000$ . (show answer)

Answer
$\log_{10} 1000 = 3$ , since $10^3 = 1000$ .
2. Evaluate: $\log_{10} 0.01$ . (show answer)

Answer
$\log_{10} 0.01 = -2$ , since $10^{-2} = 0.01$ .
3. If $\log_{10} x = 4$ , find $x$ . (show answer)

Answer
$x = 10^4 = 10\,000$ .
4. How many orders of magnitude separate $500$ from $5\,000\,000$ ? (show answer)

Answer
$\dfrac{5\,000\,000}{500} = 10\,000 = 10^4$ . They are $4$ orders of magnitude apart.
5. A sound of $30$ dB is how many times more intense than the threshold of hearing ( $0$ dB)? (show answer)

Answer
$10^{30/10} = 10^3 = 1000$ times more intense.
6. An earthquake of magnitude $6.0$ has wave amplitudes how many times larger than one of magnitude $4.0$ ? (show answer)

Answer
Amplitude ratio: $10^{6.0 - 4.0} = 10^2 = 100$ times larger.
7. A solution has $[\text{H}^+] = 10^{-3}$ mol/L. Find its pH. (show answer)

Answer
$\text{pH} = -\log_{10}(10^{-3}) = 3$ .
8. If the pH drops from $7$ to $5$ , by what factor has the hydrogen-ion concentration increased? (show answer)

Answer
$10^{7-5} = 10^2 = 100$ times greater.

Reasoning · Tier 2: mixed practice

1. Two earthquakes measure $4.5$ and $6.5$ on the Richter scale. (a) Compare their wave amplitudes. (b) Estimate the energy ratio. (show answer)

Answer
(a) Amplitude ratio: $10^{6.5 - 4.5} = 10^2 = 100$ times. (b) Energy ratio: $31.6^2 \approx 1000$ times.
2. A vacuum cleaner produces $75$ dB and a whisper is $20$ dB. How many times more intense is the vacuum cleaner? (show answer)

Answer
Difference: $75 - 20 = 55$ dB. Intensity ratio: $10^{55/10} = 10^{5.5} \approx 316\,228$ times more intense.
3. A scientist records data points $2, 20, 200, 2000, 20\,000$ . Explain why a log scale is more suitable for graphing this data. (show answer)

Answer
The values span $4$ orders of magnitude ( $2$ to $20\,000$ ). On a linear scale, $2$ and $20$ would be indistinguishable near the axis while $20\,000$ dominates. A log scale spaces all five points evenly, revealing the constant factor-of- $10$ pattern.
4. Coffee has pH $5$ and household ammonia has pH $11.5$ . Which is more acidic, and by what factor of hydrogen-ion concentration? (show answer)

Answer
Coffee is more acidic (lower pH). Concentration ratio: $10^{11.5 - 5} = 10^{6.5} \approx 3\,162\,278$ times more hydrogen ions in the coffee.
5. The population of a town doubles every $10$ years. If the current population is $5000$ , calculate the population after $50$ years and explain why a log-scale graph of this growth would appear as a straight line. (show answer)

Answer
After $50$ years (5 doublings): $5000 \times 2^5 = 160\,000$ . On a log scale, exponential growth (constant doubling time) appears as a straight line because $\log(P) = \log(5000) + t \times \frac{\log 2}{10}$ , which is linear in $t$ .
6. On a log-scaled graph, two data points appear $3$ cm apart and each centimetre represents one order of magnitude. What is the ratio of the larger value to the smaller? (show answer)

Answer
Ratio $= 10^3 = 1000$ .

Reasoning · Tier 3: explain and apply

1. Explain in your own words why $\log_{10} 0$ is undefined and what this means for graphing on a log scale. (show answer)

Answer
$\log_{10} 0$ is undefined because there is no power $n$ such that $10^n = 0$ (powers of $10$ are always positive). On a log-scale graph, zero cannot be plotted -- the axis extends toward $-\infty$ in log-space as values approach zero. This means log scales can only represent strictly positive data.
2. The apparent magnitude scale for stars decreases by $1$ for each factor of $\approx 2.512$ increase in brightness. A star of magnitude $1$ is how many times brighter than a star of magnitude $6$ ? Show your working. (show answer)

Answer
Each magnitude step is a factor of $2.512$ . Over $5$ steps: $2.512^5 \approx 100$ . A magnitude- $1$ star is about $100$ times brighter than a magnitude- $6$ star.
3. A student says "an earthquake of magnitude $8$ is twice as strong as one of magnitude $4$ ." Explain why this statement is incorrect and calculate the actual amplitude ratio. (show answer)

Answer
The Richter scale is logarithmic, not linear. A magnitude $8$ quake has $10^{8-4} = 10^4 = 10\,000$ times the wave amplitude of a magnitude $4$ quake -- not $2$ times. The student confused additive and multiplicative differences.
4. Create a table listing five quantities from everyday life that span at least $8$ orders of magnitude (e.g. mass, distance, or time). Explain why a log scale is useful for displaying them together. (show answer)

Answer
Example table (masses): electron $\approx 10^{-30}$ kg, grain of sand $\approx 10^{-6}$ kg, human $\approx 10^2$ kg, Earth $\approx 10^{24}$ kg, Sun $\approx 10^{30}$ kg. These span about $60$ orders of magnitude. A log scale is essential because a linear axis from $10^{-30}$ to $10^{30}$ would make all but the largest value invisible.

Reasoning · Harder reasoning

1. The energy $E$ (in joules) released by an earthquake of Richter magnitude $M$ is approximately $\log_{10} E = 1.5M + 4.8$ . Find the energy released by earthquakes of magnitude $5.0$ and $8.0$ , and verify that the ratio is approximately $31.6^3$ . (show answer)

Answer
For $M = 5.0$ : $\log_{10} E = 1.5(5) + 4.8 = 12.3$ , so $E \approx 2.0 \times 10^{12}$ J. For $M = 8.0$ : $\log_{10} E = 1.5(8) + 4.8 = 16.8$ , so $E \approx 6.3 \times 10^{16}$ J. Ratio: $\dfrac{6.3 \times 10^{16}}{2.0 \times 10^{12}} \approx 31\,500 \approx 31.6^3$ (since $31.6^3 \approx 31\,623$ ). Confirmed.
2. Two sound sources produce $80$ dB and $80$ dB respectively. When played simultaneously, the total intensity doubles but the combined level is not $160$ dB. Find the actual combined decibel level using the formula $L = 10\log_{10}(I_1/I_0 + I_2/I_0)$ . (show answer)

Answer
Each source has intensity $I = I_0 \times 10^{80/10} = 10^8 I_0$ . Combined intensity $= 2 \times 10^8 I_0$ . Combined level $= 10\log_{10}(2 \times 10^8) = 10(\log_{10} 2 + 8) \approx 10(0.301 + 8) = 83.01$ dB. Doubling intensity adds about $3$ dB, not $80$ dB.
3. A culture of bacteria grows from $100$ to $100\,000\,000$ in $24$ hours at a constant rate. (a) How many orders of magnitude of growth is this? (b) If you plot the count on a log scale against time, what shape will the graph be? (c) Find the hourly growth factor. (show answer)

Answer
(a) $\log_{10}\!\left(\frac{10^8}{10^2}\right) = \log_{10}(10^6) = 6$ orders of magnitude. (b) A straight line, because $\log(\text{count})$ increases linearly with time for exponential growth. (c) Total growth factor $= 10^6$ over $24$ hours. Hourly factor $= (10^6)^{1/24} = 10^{0.25} \approx 1.778$ .
4. The Moment Magnitude Scale (used for large earthquakes) is defined by $M_w = \frac{2}{3}\log_{10}(M_0) - 6.07$ , where $M_0$ is the seismic moment in Nm. If $M_0$ increases by a factor of $1000$ , by how much does $M_w$ increase? (show answer)

Answer
If $M_0$ increases by a factor of $1000 = 10^3$ , then $\log_{10}(M_0)$ increases by $3$ . So $M_w$ increases by $\frac{2}{3} \times 3 = 2$ units.

Pythagoras & trigonometry in 3D

Fluency · Tier 1: basic skills

1. Find the space diagonal of a box $6 \times 8 \times 10$ cm. (show answer)

Answer
$d = \sqrt{36 + 64 + 100} = \sqrt{200} \approx 14.14$ cm.
2. A cube has side $5$ cm. Find its space diagonal. (show answer)

Answer
$d = \sqrt{25 + 25 + 25} = \sqrt{75} = 5\sqrt{3} \approx 8.66$ cm.
3. From a point on level ground, the angle of elevation to the top of a $25$ m tree is $40°$ . Find the horizontal distance to the tree. (show answer)

Answer
$\tan 40° = \frac{25}{d}$ , so $d = \frac{25}{\tan 40°} \approx \frac{25}{0.8391} \approx 29.79$ m.
4. A drone at height $80$ m observes a point on the ground at an angle of depression of $50°$ . Find the horizontal distance. (show answer)

Answer
$\tan 50° = \frac{80}{d}$ , so $d = \frac{80}{\tan 50°} \approx \frac{80}{1.1918} \approx 67.13$ m.
5. A ship sails $10$ km on a bearing of $045°$ . How far north and how far east has it travelled? (show answer)

Answer
North $= 10\cos 45° \approx 7.07$ km; East $= 10\sin 45° \approx 7.07$ km.
6. A measurement of $8.0 \pm 0.2$ cm is squared. Find the percentage error in the original measurement and the approximate percentage error in the squared value. (show answer)

Answer
Percentage error $= \frac{0.2}{8.0} \times 100 = 2.5\%$ . Squared value error $\approx 2 \times 2.5\% = 5\%$ .
7. Find the length of the longest rod that fits inside a rectangular box $12 \times 5 \times 4$ cm. (show answer)

Answer
$d = \sqrt{144 + 25 + 16} = \sqrt{185} \approx 13.60$ cm.
8. A pyramid has a square base of side $10$ cm and a vertical height of $12$ cm. Find the slant edge length (from a base corner to the apex). (show answer)

Answer
Half-diagonal of base $= \frac{10\sqrt{2}}{2} = 5\sqrt{2} \approx 7.071$ cm. Slant edge $= \sqrt{12^2 + (5\sqrt{2})^2} = \sqrt{144 + 50} = \sqrt{194} \approx 13.93$ cm.

Reasoning · Tier 2: mixed practice

1. A rectangular room is $6 \times 4 \times 3$ m. A spider at one top corner wants to reach the opposite bottom corner by walking along walls. Find the shortest path. (Hint: unfold two walls into a flat net.) (show answer)

Answer
Unfold the $6 \times 3$ wall and the $4 \times 3$ wall into a single rectangle $10 \times 3$ . The spider walks diagonally: $\sqrt{10^2 + 3^2} = \sqrt{109} \approx 10.44$ m. (Other unfoldings give longer paths; the shortest is $\sqrt{(6+3)^2 + 4^2} = \sqrt{97} \approx 9.85$ m by unfolding the $6 \times 3$ wall with the floor. Check all configurations; the minimum is approximately $9.85$ m.)
2. A hiker walks $5$ km on bearing $030°$ then $8$ km on bearing $120°$ . Find the distance and bearing from start to finish. (show answer)

Answer
Leg 1: N $= 5\cos 30° \approx 4.33$ km, E $= 5\sin 30° = 2.5$ km. Leg 2: N $= -8\cos 60° = -4$ km (south), E $= 8\sin 60° \approx 6.93$ km. Total: N $\approx 0.33$ km, E $\approx 9.43$ km. Distance $= \sqrt{0.33^2 + 9.43^2} \approx 9.44$ km. Bearing $= \tan^{-1}(9.43/0.33) \approx 88°$ .
3. From the top of a $60$ m cliff, the angles of depression to two boats in a line directly out to sea are $45°$ and $30°$ . Find the distance between the boats. (show answer)

Answer
Boat 1: $d_1 = \frac{60}{\tan 45°} = 60$ m from cliff base. Boat 2: $d_2 = \frac{60}{\tan 30°} \approx 103.92$ m. Distance between boats $\approx 103.92 - 60 = 43.92$ m.
4. A surveyor measures the angle of elevation to a mountain peak as $22°$ from point A and $15°$ from point B, which is $500$ m further away on level ground in a direct line from the peak. Find the height of the mountain. (show answer)

Answer
Let height $= h$ and distance from A to the base $= x$ . From A: $\tan 22° = h/x$ . From B: $\tan 15° = h/(x + 500)$ . So $x = h/\tan 22°$ and $x + 500 = h/\tan 15°$ . Subtracting: $500 = h(1/\tan 15° - 1/\tan 22°) = h(3.732 - 2.475) = 1.257h$ . So $h \approx 397.8$ m.
5. A cylinder has $r = 10.0 \pm 0.3$ cm and $h = 20.0 \pm 0.5$ cm. Calculate the volume and estimate the maximum percentage error. (show answer)

Answer
$V = \pi(100)(20) = 2000\pi \approx 6283.19$ cm $^3$ . Error in $r$ : $\frac{0.3}{10} = 3\%$ , doubled for $r^2$ : $6\%$ . Error in $h$ : $\frac{0.5}{20} = 2.5\%$ . Total $\approx 8.5\%$ .
6. A tent pole $2.5$ m tall is supported by guy ropes pegged $1.5$ m from the base. Find (a) the length of each rope, and (b) the angle each rope makes with the ground. (show answer)

Answer
(a) Rope $= \sqrt{2.5^2 + 1.5^2} = \sqrt{8.5} \approx 2.92$ m. (b) $\tan\theta = \frac{2.5}{1.5}$ , so $\theta = \tan^{-1}(1.667) \approx 59.0°$ .

Reasoning · Tier 3: explain and apply

1. Explain why the space diagonal formula $d = \sqrt{l^2 + w^2 + h^2}$ works by describing the two right-angled triangles used in its derivation. (show answer)

Answer
First, in the base rectangle, we form a right triangle with legs $l$ and $w$ to get the base diagonal $d_b = \sqrt{l^2 + w^2}$ . Second, this base diagonal becomes one leg of a new right triangle whose other leg is the height $h$ and whose hypotenuse is the space diagonal. Applying Pythagoras again: $d = \sqrt{d_b^2 + h^2} = \sqrt{l^2 + w^2 + h^2}$ .
2. A pilot flies from airport A on bearing $070°$ for $200$ km to point B, then on bearing $160°$ for $150$ km to airport C. Find the bearing and distance for the direct return flight from C to A. (show answer)

Answer
Leg 1 (A to B): N $= 200\cos 70° \approx 68.40$ km, E $= 200\sin 70° \approx 187.94$ km. Leg 2 (B to C): N $= -150\cos 20° \approx -140.95$ km, E $= 150\sin 20° \approx 51.30$ km. C relative to A: N $\approx -72.55$ km, E $\approx 239.24$ km. Return C to A: N $\approx 72.55$ , W $\approx 239.24$ . Distance $= \sqrt{72.55^2 + 239.24^2} \approx 250.0$ km. Bearing from C to A: $\theta = \tan^{-1}(239.24/72.55) \approx 73.1°$ west of north $= 360° - 73.1° = 286.9°$ .
3. A building casts a shadow $30$ m long when the sun's angle of elevation is $55°$ . An hour later the shadow is $45$ m long. Find the new angle of elevation and determine whether the sun rose or fell during this hour. (show answer)

Answer
Building height $h$ . When shadow $= 30$ : $\tan 55° = h/30$ , so $h = 30\tan 55° \approx 42.84$ m. When shadow $= 45$ : $\tan\alpha = 42.84/45$ , so $\alpha = \tan^{-1}(0.952) \approx 43.6°$ . The angle decreased from $55°$ to $43.6°$ , so the sun fell (moved lower in the sky).
4. Discuss why percentage error in a calculated volume based on measured radius is approximately double the percentage error of the radius itself, using the formula $V = \frac{4}{3}\pi r^3$ as an example. (What multiplier applies here instead of double?) (show answer)

Answer
For $V = \frac{4}{3}\pi r^3$ , the radius is cubed, so the percentage error in $V$ is approximately $3$ times the percentage error in $r$ (not double). For example, a $1\%$ error in $r$ gives roughly a $3\%$ error in volume. The multiplier equals the exponent of the variable in the formula.

Reasoning · Harder reasoning

1. A right square pyramid has base side $12$ cm and slant height $10$ cm. Find (a) the vertical height, (b) the angle between a slant face and the base, and (c) the angle between a slant edge and the base. (show answer)

Answer
Base half-diagonal $= \frac{12\sqrt{2}}{2} = 6\sqrt{2}$ . Slant height $= 10$ (distance from midpoint of base edge to apex). Half base edge $= 6$ . Vertical height from base edge midpoint: $h_{\text{face}} = \sqrt{10^2 - 6^2} = 8$ cm. (a) Vertical height of pyramid: the midpoint of a base edge is $6$ cm from the centre (for a $12 \times 12$ base). So $H = \sqrt{10^2 - 6^2} = 8$ cm. (Alternatively, using the half-diagonal: $H = \sqrt{10^2 - (6\sqrt{2})^2}$ ... but slant height goes from base edge midpoint, not corner. Let's recalculate. Slant height $l = 10$ is from midpoint of a base edge to apex. Half base $= 6$ . Distance from centre to midpoint of edge $= 6$ . $H = \sqrt{l^2 - 6^2} = \sqrt{100 - 36} = 8$ cm.) (b) Angle between slant face and base: $\tan\alpha = H/6 = 8/6$ , so $\alpha = \tan^{-1}(4/3) \approx 53.1°$ . (c) Slant edge (corner to apex): distance from centre to corner $= 6\sqrt{2}$ . Slant edge $= \sqrt{8^2 + (6\sqrt{2})^2} = \sqrt{64 + 72} = \sqrt{136} \approx 11.66$ cm. Angle with base: $\tan\beta = 8/(6\sqrt{2}) = 8/8.485$ , so $\beta \approx 43.3°$ .
2. Two observers A and B are $300$ m apart on level ground. Both observe the same drone. Observer A measures the angle of elevation as $40°$ and the bearing to the drone as $060°$ . Observer B (due east of A) measures the angle of elevation as $55°$ . Find the height of the drone. (show answer)

Answer
Let A be at the origin, B at $(300, 0)$ . Drone bearing $060°$ from A means the drone's horizontal position is along direction $060°$ from A. Let horizontal distance from A to point below drone $= d_A$ . Then $d_A \tan 40°...$ -- actually the drone is at some point $(x, y, h)$ . From A: bearing $060°$ means $x = d_A \sin 60°$ , $y = d_A \cos 60°$ and $\tan 40° = h/d_A$ . From B at $(300, 0)$ : $\tan 55° = h/d_B$ where $d_B$ is horizontal distance from B to drone. Using $d_A = h/\tan 40°$ and $d_B = h/\tan 55°$ , and the geometry: $x = d_A \sin 60° = h\sin 60°/\tan 40°$ , $y = d_A \cos 60° = h\cos 60°/\tan 40°$ . Also $d_B^2 = (x - 300)^2 + y^2 = h^2/\tan^2 55°$ . Substituting and solving numerically gives $h \approx 218$ m. (Accept reasonable numerical solutions with clear working.)
3. A sphere of radius $r$ fits exactly inside a cube. Show that the ratio of the space diagonal of the cube to the diameter of the sphere is $\sqrt{3}:1$ . (show answer)

Answer
Cube side $= 2r$ (the sphere diameter equals the cube side). Space diagonal of cube $= 2r\sqrt{3}$ . Diameter of sphere $= 2r$ . Ratio $= \frac{2r\sqrt{3}}{2r} = \sqrt{3}:1$ .
4. A surveyor measures two sides of a triangle as $a = 50.0 \pm 0.5$ m and $b = 80.0 \pm 0.5$ m, with the included angle $C = 60° \pm 1°$ . Using the area formula $A = \frac{1}{2}ab\sin C$ , estimate the area and discuss how the angle error and the length errors each contribute to the total error in the area. (show answer)

Answer
Nominal area $= \frac{1}{2}(50)(80)\sin 60° = 2000 \times 0.8660 \approx 1732$ m $^2$ . Length errors: $\frac{0.5}{50} = 1\%$ and $\frac{0.5}{80} = 0.625\%$ . Since each length appears to the first power, their contributions are $1\%$ and $0.625\%$ . Angle error: $1°$ in $60°$ . The sensitivity factor is $\frac{d}{dC}(\sin C) \cdot \frac{C}{\sin C}$ . At $C = 60°$ : $\frac{\cos 60°}{\sin 60°} \times 1° \times \frac{\pi}{180} \approx 0.577 \times 0.01745 \approx 1.007\%$ . Total error $\approx 1 + 0.625 + 1.007 \approx 2.6\%$ , i.e. $\pm 45$ m $^2$ . The angle error contributes roughly as much as the length errors combined.

Geometric proofs & deductive reasoning

Fluency · Tier 1: basic skills

1. Name the four congruence tests and for each, state what must be equal. (show answer)

Answer
SSS: three pairs of corresponding sides equal. SAS: two sides and the included angle equal. AAS: two angles and a corresponding side equal. RHS: right angle, hypotenuse, and one other side equal.
2. In $\triangle ABC$ and $\triangle DEF$ : $AB = DE$ , $BC = EF$ , $\angle B = \angle E$ . Which congruence test applies? (show answer)

Answer
SAS (two sides and the included angle between them).
3. Triangle $PQR$ has $PQ = PR$ and $\angle QPR = 70°$ . Find $\angle PQR$ and $\angle PRQ$ . (show answer)

Answer
$\angle PQR = \angle PRQ = \frac{180° - 70°}{2} = 55°$ .
4. Lines $l_1 \parallel l_2$ are cut by a transversal. One alternate angle is $55°$ . Find the other alternate angle and the co-interior angle on the same side. (show answer)

Answer
The other alternate angle is $55°$ . The co-interior angle on the same side is $180° - 55° = 125°$ .
5. In $\triangle XYZ$ , the exterior angle at $Z$ is $130°$ . If $\angle X = 80°$ , find $\angle Y$ . (show answer)

Answer
Exterior angle $= \angle X + \angle Y$ , so $130° = 80° + \angle Y$ , giving $\angle Y = 50°$ .
6. Explain the difference between a demonstration and a deductive proof using an example. (show answer)

Answer
A demonstration checks specific cases (e.g. measuring angles in three triangles and finding they sum to $180°$ ). A deductive proof uses logical steps from axioms to show the result holds for all cases. The demonstration could fail for an untested case; the proof cannot.
7. $ABCD$ is a parallelogram. State two properties of its diagonals that can be proven using congruent triangles. (show answer)

Answer
The diagonals of a parallelogram bisect each other (proven by showing two pairs of congruent triangles using SAS with vertically opposite angles).
8. Triangle $ABC$ has $\angle A = 90°$ , $AB = 6$ , $AC = 8$ , $BC = 10$ . Triangle $DEF$ has $\angle D = 90°$ , $DE = 6$ , $DF = 8$ . Are the triangles congruent? State the test. (show answer)

Answer
Yes, by RHS: right angle, hypotenuse $BC = \sqrt{36 + 64} = 10 =$ hypotenuse of $\triangle DEF$ , and $AB = DE = 6$ .

Reasoning · Tier 2: structured proofs

1. In $\triangle ABC$ , $M$ is the midpoint of $BC$ . $AM$ is drawn and $AM = \frac{1}{2}BC$ . Prove that $\angle BAC = 90°$ . (Hint: show $\triangle ABM$ and $\triangle ACM$ are both isosceles.) (show answer)

Answer
Since $AM = \frac{1}{2}BC$ and $M$ is the midpoint, $BM = MC = AM$ . So $\triangle ABM$ is isosceles ( $AM = BM$ ) and $\triangle ACM$ is isosceles ( $AM = CM$ ). Let $\angle ABM = \alpha$ and $\angle ACM = \beta$ . In $\triangle ABM$ : $\angle BAM = 180° - 2\alpha$ . In $\triangle ACM$ : $\angle CAM = 180° - 2\beta$ . Also $\alpha + \beta = \angle AMB$ or more directly: $\angle BAC = \angle BAM + \angle CAM = (180° - 2\alpha) + (180° - 2\beta) = 360° - 2(\alpha + \beta)$ . But $\alpha + \beta = \angle ABC + \angle ACB = 180° - \angle BAC$ . So $\angle BAC = 360° - 2(180° - \angle BAC) = 360° - 360° + 2\angle BAC$ , giving $\angle BAC = 2\angle BAC - 0$ ... Let us redo cleanly. In $\triangle ABC$ : $\angle A + \alpha + \beta = 180°$ , so $\alpha + \beta = 180° - \angle A$ . In $\triangle ABM$ (isosceles with $AM = BM$ ): $\angle BAM = \alpha$ , so $\angle AMB = 180° - 2\alpha$ . In $\triangle ACM$ (isosceles with $AM = CM$ ): $\angle CAM = \beta$ , so $\angle AMC = 180° - 2\beta$ . But $\angle AMB + \angle AMC = 180°$ (straight line). So $(180° - 2\alpha) + (180° - 2\beta) = 180°$ , giving $360° - 2(\alpha + \beta) = 180°$ , so $\alpha + \beta = 90°$ . Therefore $\angle A = 180° - 90° = 90°$ .
2. $ABCD$ is a kite with $AB = AD$ and $CB = CD$ . Prove that diagonal $AC$ bisects diagonal $BD$ at right angles. (show answer)

Answer
In kite $ABCD$ with $AB = AD$ and $CB = CD$ , let diagonals meet at $O$ . In $\triangle ABC$ and $\triangle ADC$ : $AB = AD$ , $CB = CD$ , $AC$ common. By SSS, $\triangle ABC \cong \triangle ADC$ . So $\angle BAC = \angle DAC$ . Now in $\triangle ABO$ and $\triangle ADO$ : $AB = AD$ , $\angle BAO = \angle DAO$ , $AO$ common. By SAS, $\triangle ABO \cong \triangle ADO$ . So $BO = DO$ (diagonal bisected) and $\angle AOB = \angle AOD = 90°$ (supplementary and equal).
3. Two circles of equal radius intersect at points $P$ and $Q$ . Prove that the line joining the centres is the perpendicular bisector of $PQ$ . (show answer)

Answer
Let the centres be $O_1$ and $O_2$ with equal radius $r$ . Then $O_1P = O_1Q = r$ and $O_2P = O_2Q = r$ . So $O_1$ and $O_2$ are each equidistant from $P$ and $Q$ , meaning both lie on the perpendicular bisector of $PQ$ . The line $O_1O_2$ is therefore the perpendicular bisector of $PQ$ .
4. In $\triangle ABC$ , $D$ is on $BC$ such that $AD \perp BC$ . If $AB = 13$ cm, $AD = 12$ cm, and $DC = 4$ cm, find $BD$ and $AC$ . (show answer)

Answer
$BD = \sqrt{AB^2 - AD^2} = \sqrt{169 - 144} = 5$ cm. $AC = \sqrt{AD^2 + DC^2} = \sqrt{144 + 16} = \sqrt{160} = 4\sqrt{10} \approx 12.65$ cm.
5. Prove that the diagonals of a rhombus bisect each other at right angles. (show answer)

Answer
In rhombus $ABCD$ , all sides equal. Diagonals $AC$ and $BD$ meet at $O$ . In $\triangle AOB$ and $\triangle COB$ : $AB = CB$ (rhombus), $OB$ common, $AO = OC$ (to prove). Instead: in $\triangle ABD$ and $\triangle CBD$ : $AB = CB$ , $AD = CD$ , $BD$ common. By SSS, $\triangle ABD \cong \triangle CBD$ . So $\angle ABD = \angle CBD$ . Now in $\triangle ABO$ and $\triangle CBO$ : $AB = CB$ , $\angle ABO = \angle CBO$ , $BO$ common. By SAS, congruent. So $AO = CO$ and $\angle AOB = \angle COB = 90°$ .
6. $\triangle ABC$ has $AB = AC$ . $D$ and $E$ are points on $AB$ and $AC$ respectively such that $BD = CE$ . Prove that $DE \parallel BC$ . (show answer)

Answer
Let $AD = AE$ (since $BD = CE$ and $AB = AC$ , we have $AD = AB - BD = AC - CE = AE$ ). In $\triangle ADE$ : $AD = AE$ , so $\triangle ADE$ is isosceles with $\angle ADE = \angle AED$ . Also $\angle ADE + \angle AED + \angle A = 180°$ . In $\triangle ABC$ : $\angle ABC = \angle ACB$ (isosceles) and $\angle ABC + \angle ACB + \angle A = 180°$ . So $\angle ADE = \angle ABC$ . Since these are corresponding angles with transversal $AB$ cutting $DE$ and $BC$ , we get $DE \parallel BC$ .

Reasoning · Tier 3: explain and apply

1. A student claims that if two triangles have three pairs of equal angles, they must be congruent. Is this correct? Explain your reasoning with an example. (show answer)

Answer
Incorrect. Three equal angle pairs (AAA) guarantee similarity, not congruence. For example, an equilateral triangle with side $3$ cm and an equilateral triangle with side $6$ cm both have all angles $= 60°$ but are clearly not congruent. AAA does not fix the size of the triangle.
2. Prove that the angle in a semicircle is $90°$ . (Hint: let $O$ be the centre, and use the isosceles triangle property for $\triangle OAB$ and $\triangle OAC$ where $BC$ is the diameter.) (show answer)

Answer
Let $BC$ be a diameter with centre $O$ , and $A$ a point on the circle. Then $OA = OB = OC = r$ . In $\triangle OAB$ : $OA = OB$ , so $\angle OAB = \angle OBA = \alpha$ . In $\triangle OAC$ : $OA = OC$ , so $\angle OAC = \angle OCA = \beta$ . In $\triangle ABC$ : $\angle BAC = \alpha + \beta$ and $\angle ABC + \angle BCA + \angle BAC = 180°$ , i.e. $\alpha + \beta + (\alpha + \beta) = 180°$ , so $2(\alpha + \beta) = 180°$ , giving $\angle BAC = 90°$ .
3. In quadrilateral $ABCD$ , $AB = CD$ and $AB \parallel CD$ . Prove that $ABCD$ is a parallelogram. (show answer)

Answer
In $\triangle ABC$ and $\triangle CDA$ : $AB = CD$ (given), $AC$ common, $\angle BAC = \angle DCA$ (alternate angles since $AB \parallel CD$ ). By SAS, $\triangle ABC \cong \triangle CDA$ . Therefore $BC = DA$ and $\angle BCA = \angle DAC$ (alternate angles), so $BC \parallel AD$ . Both pairs of opposite sides are parallel, hence $ABCD$ is a parallelogram.
4. Explain why SSA (two sides and a non-included angle) is not a valid congruence test. Illustrate with two non-congruent triangles that satisfy SSA. (show answer)

Answer
SSA is ambiguous because the given angle is not between the two known sides. For example: $\triangle_1$ has $a = 5$ , $b = 7$ , $\angle A = 30°$ ; $\triangle_2$ has $a = 5$ , $b = 7$ , $\angle A = 30°$ but $\angle B$ can be either acute or obtuse (since $\sin B = \frac{7\sin 30°}{5} = 0.7$ gives $B \approx 44.4°$ or $B \approx 135.6°$ ). Two different triangles satisfy the same SSA conditions.

Reasoning · Harder reasoning

1. In $\triangle ABC$ , $\angle A = 60°$ , $AB = AC = 10$ cm. Point $P$ lies inside the triangle such that $PA = PB = PC$ . Find $PA$ . (Hint: $P$ is the circumcentre; use the circumradius formula for a triangle.) (show answer)

Answer
In $\triangle ABC$ with $\angle A = 60°$ and $AB = AC = 10$ , the triangle is isosceles. By the cosine rule: $BC^2 = 100 + 100 - 2(100)\cos 60° = 200 - 100 = 100$ , so $BC = 10$ . The triangle is equilateral. The circumradius $R = \frac{a}{2\sin A} = \frac{10}{2\sin 60°} = \frac{10}{\sqrt{3}} = \frac{10\sqrt{3}}{3} \approx 5.77$ cm. So $PA = \frac{10\sqrt{3}}{3}$ cm.
2. Prove that the medians of a triangle are concurrent. (Hint: let two medians meet at $G$ and show $G$ divides each in the ratio $2:1$ , then prove the third median also passes through $G$ .) (show answer)

Answer
Let medians from $A$ and $B$ meet at $G$ . The median from $A$ to midpoint $M$ of $BC$ : by the vector approach, $G = A + \frac{2}{3}(M - A) = \frac{A + 2M}{3} = \frac{A + B + C}{3}$ . Similarly, the median from $B$ to midpoint $N$ of $AC$ gives the same point $G = \frac{A + B + C}{3}$ . The third median from $C$ to midpoint $P$ of $AB$ also passes through $\frac{A + B + C}{3}$ . Since all three medians yield the same intersection point, they are concurrent, and $G$ divides each median in the ratio $2:1$ from vertex.
3. In a cyclic quadrilateral $ABCD$ , prove that opposite angles sum to $180°$ . Use the fact that the angle at the centre is twice the angle at the circumference. (show answer)

Answer
Let $\angle AOB = 2\alpha$ and $\angle AOD = 2\beta$ where $O$ is the centre (central angles subtended by arcs $AB$ and $AD$ ). Then $\angle ACB = \alpha$ and $\angle ACD = \beta$ (angle at circumference = half central angle). Opposite angle $\angle B =$ angle subtended by arc $ADC$ at circumference. Arc $ADC$ has central angle $2\beta + 2\gamma$ (where $2\gamma$ is the central angle for arc $DC$ ). The key result: $\angle B + \angle D$ corresponds to arcs that together make a full circle ( $360°$ ), so the sum of the half-angles is $180°$ .
4. Two triangles $\triangle ABC$ and $\triangle A'B'C'$ have $AB = A'B'$ , $AC = A'C'$ , and $\angle A > \angle A'$ . Prove that $BC > B'C'$ (the hinge theorem). You may use the cosine rule. (show answer)

Answer
By the cosine rule in $\triangle ABC$ : $BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cos A$ . Similarly $B'C'^2 = A'B'^2 + A'C'^2 - 2 \cdot A'B' \cdot A'C' \cos A'$ . Since $AB = A'B'$ and $AC = A'C'$ , and $\angle A > \angle A'$ , we have $\cos A < \cos A'$ (cosine is decreasing on $[0°, 180°]$ ). Therefore $-2 \cdot AB \cdot AC \cos A > -2 \cdot A'B' \cdot A'C' \cos A'$ , so $BC^2 > B'C'^2$ , hence $BC > B'C'$ .

Networks & network diagrams

Fluency · Tier 1: basic skills

1. A network has $4$ vertices with degrees $2, 3, 3, 4$ . How many edges does it have? (show answer)

Answer
Sum of degrees $= 2 + 3 + 3 + 4 = 12$ . Number of edges $= \dfrac{12}{2} = 6$ .
2. Draw a network with $4$ vertices where every vertex has degree $2$ . (show answer)

Answer
A square (cycle graph $C_4$ ): four vertices arranged in a loop, each connected to its two neighbours.
3. A connected planar graph has $4$ vertices and $6$ edges. How many faces does it have? (show answer)

Answer
$F + V = E + 2$ , so $F + 4 = 6 + 2 = 8$ , giving $F = 4$ .
4. A network has vertices with degrees $2, 2, 4, 4$ . Does an Eulerian circuit exist? Explain. (show answer)

Answer
Yes. All vertex degrees ( $2, 2, 4, 4$ ) are even, so an Eulerian circuit exists.
5. A network has vertices with degrees $1, 2, 2, 3$ . Does an Eulerian path exist? Explain. (show answer)

Answer
Vertices with odd degree: the vertex with degree $1$ and the vertex with degree $3$ (two odd-degree vertices). So an Eulerian path exists (but not a circuit). It must start at one odd-degree vertex and end at the other.
6. True or false: a connected graph with $5$ vertices must have at least $4$ edges. (show answer)

Answer
True. A connected graph with $n$ vertices needs at least $n - 1$ edges (a spanning tree). For $n = 5$ , at least $4$ edges.
7. State Euler's formula and define each variable. (show answer)

Answer
$F + V = E + 2$ , where $F$ = number of faces (regions), $V$ = number of vertices, $E$ = number of edges. Applies to connected planar graphs.
8. A network has $8$ edges. What is the sum of all vertex degrees? (show answer)

Answer
Sum of degrees $= 2 \times 8 = 16$ .

Reasoning · Tier 2: mixed practice

1. A connected planar graph has $5$ faces and $8$ edges. How many vertices does it have? (show answer)

Answer
$F + V = E + 2$ , so $5 + V = 8 + 2 = 10$ , giving $V = 5$ .
2. Draw a network representing the direct flights between four cities: Melbourne, Sydney, Brisbane, and Perth. Melbourne has direct flights to all three cities; Sydney and Brisbane are connected; Perth has no direct flight to Brisbane or Sydney. How many edges does your network have? (show answer)

Answer
Edges: Melbourne--Sydney, Melbourne--Brisbane, Melbourne--Perth, Sydney--Brisbane. Total $= 4$ edges.
3. A postman must walk along every street in a neighbourhood. The network has vertex degrees: $2, 2, 2, 4, 4, 4$ . Can the postman complete an Eulerian circuit? Justify your answer. (show answer)

Answer
Yes. All degrees ( $2, 2, 2, 4, 4, 4$ ) are even, so an Eulerian circuit exists. The postman can walk every street exactly once and return to the starting point.
4. A network has $6$ vertices and $10$ edges. Verify that it cannot be planar using the inequality $E \leq 3V - 6$ for simple planar graphs. (show answer)

Answer
For a simple planar graph: $E \leq 3V - 6$ . Here $3 \times 6 - 6 = 12$ and $E = 10 \leq 12$ , so this test alone does not prove non-planarity. In fact, a graph with $6$ vertices and $10$ edges could be planar. (The complete graph $K_4$ has $6$ edges; $K_5$ has $10$ edges and is not planar.) Since $K_5$ is the unique simple graph on $5$ vertices with $10$ edges, but we have $6$ vertices, a graph with $6$ vertices and $10$ edges may or may not be planar depending on its structure. The inequality is necessary but not sufficient.
5. Three houses and three utilities (water, gas, electricity) must each be connected to every utility. Draw this as a bipartite network. How many edges are there? Is this graph planar? (show answer)

Answer
Three houses and three utilities: each house connects to each utility, giving $3 \times 3 = 9$ edges. This is the complete bipartite graph $K_{3,3}$ , which is not planar (by Kuratowski's theorem).

Reasoning · Tier 3: explain and apply

1. A council wants to inspect every road in a town. The town's network has $6$ intersections. Four intersections have degree $3$ and two have degree $4$ . Explain why an Eulerian path does not exist, and describe a strategy the council could use to inspect all roads. (show answer)

Answer
Four vertices have odd degree ( $3$ ). Since more than $2$ vertices have odd degree, no Eulerian path (or circuit) exists. Strategy: the council could identify pairs of odd-degree vertices and add duplicate edges (roads walked twice) to make all degrees even, minimising the total extra distance. This is the Chinese Postman approach.
2. Prove that in any network, the number of vertices with odd degree is always even. (Hint: use the handshaking lemma.) (show answer)

Answer
The handshaking lemma states $\sum \text{degrees} = 2E$ . The left side is the sum of all degrees. Suppose $k$ vertices have odd degree. Each odd degree contributes an odd number to the sum, and each even degree contributes an even number. The total must be $2E$ (even). The sum of the even degrees is even. So the sum of the odd degrees must also be even, which requires $k$ to be even. Therefore the number of odd-degree vertices is always even.
3. A connected planar graph has $10$ vertices, each of degree $3$ . Find the number of edges and faces. (show answer)

Answer
Edges: sum of degrees $= 10 \times 3 = 30$ , so $E = \dfrac{30}{2} = 15$ . Faces: $F + V = E + 2$ , so $F + 10 = 15 + 2 = 17$ , giving $F = 7$ .
4. Explain the difference between a Hamiltonian path and an Eulerian path. Give an example of a graph that has an Eulerian circuit but no Hamiltonian circuit. (show answer)

Answer
An Eulerian path visits every edge exactly once. A Hamiltonian path visits every vertex exactly once. Example: consider a graph shaped like a "bowtie" (two triangles sharing a single vertex). It has $5$ vertices and $6$ edges. The central vertex has degree $4$ and all others have degree $2$ -- all even, so an Eulerian circuit exists. However, no Hamiltonian circuit exists because the graph is not $2$ -connected (removing the central vertex disconnects it).

Reasoning · Harder reasoning

1. The complete graph $K_5$ has $5$ vertices, each connected to every other vertex. Calculate the number of edges, the degree of each vertex, and determine whether an Eulerian circuit exists. (show answer)

Answer
$K_5$ : each vertex connects to $4$ others, so each has degree $4$ . Number of edges $= \dbinom{5}{2} = 10$ . All degrees are even ( $4$ ), so an Eulerian circuit exists.
2. A network has $V$ vertices and is a tree (connected with no cycles). Prove that it has exactly $V - 1$ edges. (show answer)

Answer
Base case: $V = 1$ , edges $= 0 = 1 - 1$ . Inductive step: a tree with $V > 1$ vertices has at least one leaf (vertex of degree $1$ ). Remove the leaf and its edge: the remaining graph is a connected tree with $V - 1$ vertices and (by hypothesis) $V - 2$ edges. Adding the leaf back gives $V - 1$ edges. So a tree on $V$ vertices has $V - 1$ edges.
3. Seven bridges connect four land masses. The vertex degrees are $3, 3, 3, 5$ . Show that it is impossible to walk a route crossing each bridge exactly once. (This is the Konigsberg bridge problem.) (show answer)

Answer
The vertex degrees are $3, 3, 3, 5$ -- all four are odd. Since there are $4$ odd-degree vertices (more than $2$ ), no Eulerian path or circuit exists. It is impossible to cross each bridge exactly once.
4. A delivery driver must visit $6$ locations. The distances (km) between each pair are given. The driver starts and ends at the depot (location A). Outline how you would use a network to find an efficient route, and explain why finding the absolute shortest route is computationally difficult for large networks. (show answer)

Answer
Model locations as vertices and distances as weighted edges. An efficient route visiting all locations and returning to the start is a Hamiltonian circuit. You can try nearest-neighbour or other heuristics. Finding the absolute shortest Hamiltonian circuit (the Travelling Salesman Problem) is computationally difficult because the number of possible routes grows factorially: for $n$ locations there are $\dfrac{(n-1)!}{2}$ distinct routes. For large $n$ , checking all routes is impractical, so heuristic or approximation methods are used.

Boxplots & comparing distributions

Fluency · Tier 1: basic skills

1. Find the five-number summary for: $5, 8, 12, 15, 18, 20, 22, 25, 30$ . (show answer)

Answer
Min $= 5$ , $Q_1 = 10$ (median of $5, 8, 12, 15$ : average of $8$ and $12$ ), median $= 18$ (5th value), $Q_3 = 23.5$ (median of $20, 22, 25, 30$ : average of $22$ and $25$ ), max $= 30$ .
2. Calculate the IQR for the data in Q1. (show answer)

Answer
IQR $= Q_3 - Q_1 = 23.5 - 10 = 13.5$ .
3. A data set has $Q_1 = 10$ , $Q_3 = 30$ . Find the upper and lower fences for outlier detection. (show answer)

Answer
IQR $= 30 - 10 = 20$ . Lower fence $= 10 - 1.5 \times 20 = 10 - 30 = -20$ . Upper fence $= 30 + 1.5 \times 20 = 30 + 30 = 60$ .
4. The five-number summary for a data set is: $2, 8, 14, 20, 28$ . Sketch a boxplot. (show answer)

Answer
Boxplot with whisker at $2$ , box from $8$ to $20$ , median line at $14$ , whisker to $28$ .
5. A boxplot has its median closer to $Q_1$ than to $Q_3$ . Is the distribution positively or negatively skewed? (show answer)

Answer
Positively skewed (the data is more spread out above the median than below).
6. In a two-way table, $40$ out of $100$ people surveyed are left-handed. What proportion is left-handed? (show answer)

Answer
$\dfrac{40}{100} = 0.4$ or $40\%$ .
7. A data set has $Q_1 = 25$ and IQR $= 12$ . What is $Q_3$ ? (show answer)

Answer
$Q_3 = Q_1 + \text{IQR} = 25 + 12 = 37$ .
8. True or false: the median always lies exactly in the centre of the box in a boxplot. (show answer)

Answer
False. The median is only centred if the distribution is symmetric. In a skewed distribution, the median is closer to one quartile.

Reasoning · Tier 2: mixed practice

1. The heights (cm) of $15$ students are: $152, 155, 158, 160, 162, 164, 165, 167, 170, 172, 175, 178, 180, 195, 198$ . Find the five-number summary, identify any outliers, and sketch a boxplot. (show answer)

Answer
Five-number summary: min $= 152$ , $Q_1 = 160$ (median of positions 1--7), median $= 167$ (8th value), $Q_3 = 178$ (median of positions 9--15), max $= 198$ . IQR $= 178 - 160 = 18$ . Upper fence $= 178 + 1.5 \times 18 = 178 + 27 = 205$ . Lower fence $= 160 - 27 = 133$ . Both $195$ and $198$ are below $205$ , so there are no outliers.
2. Two classes have the following five-number summaries for a maths test (out of $50$ ): - Class X: $15, 28, 35, 40, 48$ . - Class Y: $20, 25, 30, 42, 50$ . Draw parallel boxplots and write two comparison statements. (show answer)

Answer
Class X has a higher median ( $35$ vs $30$ ) and a smaller IQR ( $40 - 28 = 12$ vs $42 - 25 = 17$ ). Class X performed better overall and more consistently. Class Y has a higher maximum ( $50$ ) but also a lower minimum ( $20$ vs $15$ -- actually Class Y min is higher). Both classes have similar ranges.
3. A two-way table shows transport mode and year level:

| | Bus | Car | Walk | Total | |--|-----|-----|------|-------| | Year 9 | 30 | 20 | 10 | 60 | | Year 10 | 15 | 35 | 10 | 60 | | Total | 45 | 55 | 20 | 120 |

Find $P(\text{bus} \mid \text{Year 9})$ and $P(\text{bus} \mid \text{Year 10})$ . What do you notice? (show answer)

Answer
$P(\text{bus} \mid \text{Year 9}) = \dfrac{30}{60} = 0.5$ . $P(\text{bus} \mid \text{Year 10}) = \dfrac{15}{60} = 0.25$ . Year 9 students are twice as likely to catch the bus as Year 10 students.
4. A newspaper reports "Average house prices rose by $20\%$ ." Explain why the median might be a better measure than the mean for house prices, and how a few expensive sales could distort the mean. (show answer)

Answer
House prices are often positively skewed: most houses cluster around a typical value, but a few very expensive properties pull the mean upward. The median is resistant to extreme values and better represents the "typical" house price. A few multi-million-dollar sales can raise the mean significantly without affecting most buyers' experience.
5. A data set has values: $3, 5, 7, 8, 10, 12, 14, 15, 50$ . Show that $50$ is an outlier using the $1.5 \times \text{IQR}$ rule. (show answer)

Answer
Ordered: $3, 5, 7, 8, 10, 12, 14, 15, 50$ . $Q_1 = \dfrac{5 + 7}{2} = 6$ . $Q_3 = \dfrac{14 + 15}{2} = 14.5$ . IQR $= 14.5 - 6 = 8.5$ . Upper fence $= 14.5 + 1.5 \times 8.5 = 14.5 + 12.75 = 27.25$ . Since $50 > 27.25$ , the value $50$ is an outlier.

Reasoning · Tier 3: explain and apply

1. A study claims students who eat breakfast score higher on tests. The data shows a correlation. Explain why this does not prove causation and suggest a confounding variable. (show answer)

Answer
Correlation does not prove causation because a third variable could explain both. For example, students from families with higher socioeconomic status may be more likely to eat breakfast and have access to tutoring, quiet study spaces, and parental support. The breakfast itself may not cause higher scores; the underlying variable (family resources) may drive both outcomes.
2. Two factories produce bolts. Factory A: median length $50.2$ mm, IQR $= 0.8$ mm. Factory B: median length $50.0$ mm, IQR $= 2.5$ mm. Which factory produces more consistent bolts? Which is closer to the target of $50.0$ mm? Discuss trade-offs. (show answer)

Answer
Factory A is more consistent (IQR $= 0.8$ mm vs $2.5$ mm). Factory B has a median closer to the target of $50.0$ mm. Trade-off: Factory A produces bolts of very uniform length but slightly above target; Factory B hits the target on average but with much greater variability. If precision matters (e.g. safety-critical components), Factory A is preferable despite the slight offset, which could be corrected by recalibrating.
3. A survey of $200$ people finds that $60\%$ support a new policy. The survey was conducted online and only advertised on one social media platform. Identify two sources of potential bias and explain how each could affect the results. (show answer)

Answer
Sources of bias: (i) Self-selection bias -- only people who chose to respond are counted; those with strong opinions may be overrepresented. (ii) Platform bias -- users of that particular social media platform may not be representative of the general population (e.g. younger demographic, specific political leanings). Both could overestimate or underestimate true support depending on the platform's user base.
4. Explain the difference between the range and the IQR as measures of spread. Give an example where the range is misleading but the IQR is not. (show answer)

Answer
Range uses only the two most extreme values, so a single outlier can make the range very large. IQR uses the middle $50\%$ and is resistant to outliers. Example: $\{10, 12, 14, 15, 16, 18, 100\}$ . Range $= 100 - 10 = 90$ (misleadingly large). IQR $= 18 - 12 = 6$ (reflects the actual spread of most data).

Reasoning · Harder reasoning

1. A data set of $20$ values has $Q_1 = 15$ , median $= 22$ , $Q_3 = 30$ . If the value $60$ is added to the data set, explain qualitatively how each part of the five-number summary might change and whether $60$ would be classified as an outlier. (show answer)

Answer
Adding $60$ : the minimum stays at $15$ (or whatever it was), the maximum becomes $60$ . IQR $= 30 - 15 = 15$ . Upper fence $= 30 + 1.5 \times 15 = 52.5$ . Since $60 > 52.5$ , yes, $60$ is an outlier. The median may shift slightly upward (from the average of the 10th and 11th values to the 11th value of the new 21-value set). $Q_1$ and $Q_3$ may shift slightly but the effect is small.
2. Two data sets both have median $= 50$ and IQR $= 10$ , but one is symmetric and the other is positively skewed. Sketch boxplots for both and explain how the whisker lengths differ. (show answer)

Answer
Symmetric: both whiskers are approximately equal length, extending evenly from the box. Positively skewed: the right whisker is much longer than the left; data extends further above $Q_3$ than below $Q_1$ . Both have the same box size (IQR $= 10$ ) and median ( $50$ ), but the skewed version has the median closer to $Q_1$ .
3. A researcher collects data from $500$ people and presents a boxplot showing no outliers. A critic argues that with $500$ data points, some outliers are expected. Evaluate this argument. (show answer)

Answer
The argument has some merit: in a normal distribution, about $0.7\%$ of values lie beyond $Q_1 - 1.5 \times \text{IQR}$ or $Q_3 + 1.5 \times \text{IQR}$ , so we might expect roughly $0.007 \times 500 \approx 3$ -- $4$ outliers. However, if the data is truly free of measurement errors and follows a tight distribution, it is possible (though unlikely) to have no outliers. The researcher should report the distribution shape and explain why outliers are absent.
4. Design a two-way table for $80$ students that shows an association between "plays sport" and "gets more than $8$ hours of sleep." Then modify it so there is no association. Explain the difference. (show answer)

Answer
With association: Sport-yes/Sleep-yes $= 30$ , Sport-yes/Sleep-no $= 10$ , Sport-no/Sleep-yes $= 15$ , Sport-no/Sleep-no $= 25$ . $P(\text{sleep} \mid \text{sport}) = \dfrac{30}{40} = 0.75$ , $P(\text{sleep} \mid \text{no sport}) = \dfrac{15}{40} = 0.375$ . These differ, showing an association. No association: Sport-yes/Sleep-yes $= 22.5$ , Sport-yes/Sleep-no $= 17.5$ , Sport-no/Sleep-yes $= 22.5$ , Sport-no/Sleep-no $= 17.5$ (using whole numbers: $23, 17, 22, 18$ ). Now $P(\text{sleep} \mid \text{sport}) \approx P(\text{sleep} \mid \text{no sport}) \approx 0.5625$ , so the variables are approximately independent.

Scatterplots & bivariate data

Fluency · Tier 1: basic skills

1. Define bivariate data and give an example of two variables you might investigate. (show answer)

Answer
Bivariate data consists of pairs of measurements on two variables for each individual. Example: height (cm) and weight (kg) for each student in a class.
2. Which variable goes on the horizontal axis: the explanatory variable or the response variable? (show answer)

Answer
The explanatory (independent) variable goes on the horizontal axis.
3. A scatterplot shows points rising steeply from left to right with little scatter. Describe the association. (show answer)

Answer
Strong, positive, linear association.
4. A scatterplot shows points scattered randomly with no pattern. Describe the association. (show answer)

Answer
No association (no pattern).
5. A line of good fit passes through $(2, 10)$ and $(8, 28)$ . Find the gradient. (show answer)

Answer
Gradient $= \dfrac{28 - 10}{8 - 2} = \dfrac{18}{6} = 3$ .
6. Using the line from Q5, find the equation and predict $y$ when $x = 5$ . (show answer)

Answer
Using $(2, 10)$ : $10 = 3 \times 2 + c$ , so $c = 4$ . Equation: $y = 3x + 4$ . When $x = 5$ : $y = 3 \times 5 + 4 = 19$ .
7. Is predicting $y$ for $x = 5$ (data range $2$ -- $8$ ) interpolation or extrapolation? (show answer)

Answer
Interpolation (5 is within the range 2--8).
8. Is predicting $y$ for $x = 12$ interpolation or extrapolation? (show answer)

Answer
Extrapolation (12 is outside the range 2--8).
9. State whether each is positive or negative association: (a) height and shoe size, (b) altitude and temperature, (c) practice hours and error count. (show answer)

Answer
(a) Positive. (b) Negative. (c) Negative.
10. True or false: a strong correlation between two variables proves that one causes the other. (show answer)

Answer
False. Correlation does not prove causation.

Reasoning · Tier 2: mixed practice

1. Plot the following data on a scatterplot and describe the association:

| $x$ | 2 | 4 | 6 | 8 | 10 | 12 | |-----|---|---|---|---|----|----| | $y$ | 35 | 30 | 24 | 20 | 14 | 10 | (show answer)

Answer
The scatterplot shows a strong, negative, linear association. As $x$ increases, $y$ decreases steadily.
2. Draw a line of good fit for the data in Q1, find its equation, and predict $y$ when $x = 7$ . (show answer)

Answer
A line of good fit through approximately $(2, 35)$ and $(12, 10)$ gives gradient $= \dfrac{10 - 35}{12 - 2} = \dfrac{-25}{10} = -2.5$ . Using $(2, 35)$ : $35 = -2.5 \times 2 + c$ , so $c = 40$ . Equation: $y = -2.5x + 40$ . When $x = 7$ : $y = -2.5 \times 7 + 40 = -17.5 + 40 = 22.5$ .
3. A researcher finds a strong positive correlation between the number of firefighters at a fire and the amount of damage caused. Does this mean firefighters cause damage? Explain. (show answer)

Answer
No, firefighters do not cause damage. The confounding variable is the size of the fire. Larger fires cause more damage and also require more firefighters. The number of firefighters and the damage are both consequences of the fire's severity.
4. Data on advertising spend ($'000) and sales ($'000) for $6$ months is:

| Advertising | 5 | 10 | 15 | 20 | 25 | 30 | |------------|---|----|----|----|----|-----| | Sales | 40 | 55 | 65 | 80 | 90 | 100 |

Find the equation of the line of good fit and predict sales for an advertising spend of $18,000. (show answer)

Answer
Line through $(5, 40)$ and $(30, 100)$ : gradient $= \dfrac{100 - 40}{30 - 5} = \dfrac{60}{25} = 2.4$ . Using $(5, 40)$ : $40 = 2.4 \times 5 + c$ , so $c = 28$ . Equation: $y = 2.4x + 28$ . For $x = 18$ : $y = 2.4 \times 18 + 28 = 43.2 + 28 = 71.2$ . Predicted sales: $71,200.
5. Explain why extrapolating the line from Q4 to predict sales for $100,000 in advertising is unreliable. (show answer)

Answer
At $x = 100$ : $y = 2.4 \times 100 + 28 = 268$ , predicting $268,000 in sales. This is extrapolation far beyond the data range ( $5$ -- $30$ ). The linear trend may not continue: there could be diminishing returns on advertising, market saturation, or budget constraints. The prediction is unreliable.

Reasoning · Tier 3: explain and apply

1. A scatterplot of study hours vs exam mark shows a strong positive linear association for $0$ -- $6$ hours, but the points flatten out beyond $6$ hours. Explain this pattern and discuss the limitations of using a single straight line for the entire data set. (show answer)

Answer
The flattening suggests diminishing returns: beyond a certain point, additional study hours produce smaller improvements (perhaps due to fatigue or already knowing the material). A single straight line would overestimate marks at high hours and underestimate them in the middle range. A curve or two separate line segments would better fit the data.
2. Two scatterplots are shown: (A) shows a strong linear pattern; (B) shows a moderate curved pattern. A student claims that (A) always provides better predictions. Evaluate this claim. (show answer)

Answer
The claim is not always correct. If the true relationship is curved, a straight line in (A) may give poor predictions at the extremes despite appearing strong. Plot (B), if fitted with an appropriate curve, could give better predictions than a straight line forced onto (A). The best model matches the form of the data, not just the apparent strength.
3. Explain the difference between an observed association, a confounding variable, and a causal relationship. Use a real-world example to illustrate all three concepts. (show answer)

Answer
Observed association: data shows that students who eat breakfast tend to score higher on tests. Confounding variable: family income -- wealthier families may provide both regular meals and better educational resources. Causal relationship: to establish that breakfast causes higher scores, you would need a controlled experiment where students are randomly assigned to eat or skip breakfast, with other factors held constant. Without this, the association may be driven by the confounding variable.
4. A line of good fit has the equation $y = -2.5x + 80$ . The data ranges from $x = 5$ to $x = 25$ . For what values of $x$ does the line predict negative $y$ values? Explain why these predictions are meaningless. (show answer)

Answer
$y = 0$ when $-2.5x + 80 = 0$ , so $x = 32$ . For $x > 32$ , the line predicts negative $y$ values. Since $x = 32$ is outside the data range ( $5$ to $25$ ), these predictions are extrapolations. Negative values may be physically meaningless (e.g. you cannot have negative sales, negative height, etc.), confirming that extrapolation beyond the data range is unreliable.

Reasoning · Harder reasoning

1. Two students draw different lines of good fit for the same scatterplot. Student A's line passes through $(5, 20)$ and $(15, 50)$ . Student B's line passes through $(3, 14)$ and $(17, 56)$ . Show that both lines have the same gradient but different $y$ -intercepts. Which line would you trust more and why? (show answer)

Answer
Student A: gradient $= \dfrac{50 - 20}{15 - 5} = \dfrac{30}{10} = 3$ . Equation: $y = 3x + 5$ . Student B: gradient $= \dfrac{56 - 14}{17 - 3} = \dfrac{42}{14} = 3$ . Equation: $y = 3x + 5$ . Both lines have gradient $3$ and $y$ -intercept $5$ -- they are actually the same line. If the $y$ -intercepts differed, you would trust the line whose reference points are closer to the centre of the data cloud, as it is less influenced by extreme points.
2. A data set of $12$ points has a strong positive linear association. One additional point is added far from the trend (an outlier). Describe how this outlier could affect (a) the position of the line of good fit, (b) the strength of the association, and (c) predictions made using the line. (show answer)

Answer
(a) The outlier can "pull" the line of good fit toward it, tilting or shifting the line. (b) The strength of association decreases because the outlier increases the scatter around the line. (c) Predictions near the outlier become less reliable, and the line may give poorer predictions for the rest of the data if it has been pulled off course.
3. A study finds that countries with more mobile phones per person also have higher life expectancy. A journalist writes "Mobile phones increase life expectancy." Write a critique of this claim, identifying at least two confounding variables and explaining why a controlled experiment would be needed. (show answer)

Answer
The claim confuses correlation with causation. Confounding variables include: (i) GDP per capita -- wealthier countries can afford both more mobile phones and better healthcare, nutrition, and sanitation. (ii) Education levels -- higher education leads to both greater technology adoption and healthier lifestyles. A controlled experiment (randomly assigning mobile phones and measuring life expectancy) is impractical and ethically complex. Without controlling for confounders, we cannot conclude that mobile phones increase life expectancy.
4. The residual for a data point is defined as $\text{residual} = \text{observed } y - \text{predicted } y$ . For the data $(3, 22), (5, 30), (7, 35), (9, 44)$ and the line $y = 3x + 12$ , calculate each residual. What does the pattern of residuals tell you about the fit? (show answer)

Answer
Predicted values: $y(3) = 3 \times 3 + 12 = 21$ ; $y(5) = 27$ ; $y(7) = 33$ ; $y(9) = 39$ . Residuals: $(22 - 21) = 1$ ; $(30 - 27) = 3$ ; $(35 - 33) = 2$ ; $(44 - 39) = 5$ . All residuals are positive and increasing, suggesting the line slightly underestimates $y$ values, and the underestimation grows for larger $x$ . This may indicate a slight curve (non-linearity) in the data, or that the gradient of the line is slightly too small.

Conditional probability & independence

Fluency · Tier 1: basic skills

1. In a class of $30$ students, $12$ play basketball. What is $P(\text{basketball})$ ? (show answer)

Answer
$P(\text{basketball}) = \dfrac{12}{30} = \dfrac{2}{5} = 0.4$ .
2. Of the $12$ basketball players, $5$ are also in the swim team. What is $P(\text{swim} \mid \text{basketball})$ ? (show answer)

Answer
$P(\text{swim} \mid \text{basketball}) = \dfrac{5}{12} \approx 0.417$ .
3. A bag has $6$ red and $4$ blue marbles. One marble is drawn and not replaced. If the first marble was red, what is $P(\text{red on 2nd draw})$ ? (show answer)

Answer
After removing a red marble: $5$ red and $4$ blue remain out of $9$ . $P(\text{red on 2nd draw}) = \dfrac{5}{9}$ .
4. Events $A$ and $B$ satisfy $P(A) = 0.3$ , $P(B) = 0.5$ , $P(A \text{ and } B) = 0.15$ . Find $P(A \mid B)$ . (show answer)

Answer
$P(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)} = \dfrac{0.15}{0.5} = 0.3$ .
5. Are $A$ and $B$ in Q4 independent? Justify. (show answer)

Answer
Yes, $A$ and $B$ are independent because $P(A \mid B) = 0.3 = P(A)$ . Knowing $B$ occurred does not change the probability of $A$ .
6. A two-way table shows: $20$ males own a pet, $15$ males do not, $25$ females own a pet, $10$ females do not. Find $P(\text{pet} \mid \text{male})$ . (show answer)

Answer
$P(\text{pet} \mid \text{male}) = \dfrac{20}{20 + 15} = \dfrac{20}{35} = \dfrac{4}{7} \approx 0.571$ .
7. Using the same table, find $P(\text{male} \mid \text{pet})$ . (show answer)

Answer
$P(\text{male} \mid \text{pet}) = \dfrac{20}{20 + 25} = \dfrac{20}{45} = \dfrac{4}{9} \approx 0.444$ .
8. A coin is tossed and a die is rolled. Are the events "heads" and "rolling a 6" independent? Explain. (show answer)

Answer
Yes, they are independent. The outcome of the coin does not affect the die, and vice versa. $P(\text{heads and 6}) = \dfrac{1}{2} \times \dfrac{1}{6} = \dfrac{1}{12} = P(\text{heads}) \times P(\text{6})$ .
9. Two cards are drawn without replacement from a deck of $52$ . Find $P(\text{2nd card is heart} \mid \text{1st card is heart})$ . (show answer)

Answer
After removing one heart, $12$ hearts remain out of $51$ cards. $P(\text{2nd heart} \mid \text{1st heart}) = \dfrac{12}{51} = \dfrac{4}{17}$ .
10. State the formula for $P(A \mid B)$ . (show answer)

Answer
$P(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)}$ , where $P(B) > 0$ .

Reasoning · Tier 2: mixed practice

1. A box contains $5$ green and $3$ yellow balls. Two balls are drawn without replacement. Draw a tree diagram with conditional probabilities on each branch, and find $P(\text{both green})$ . (show answer)

Answer
First draw: $P(G) = \dfrac{5}{8}$ , $P(Y) = \dfrac{3}{8}$ . If 1st is green: $P(G) = \dfrac{4}{7}$ , $P(Y) = \dfrac{3}{7}$ . If 1st is yellow: $P(G) = \dfrac{5}{7}$ , $P(Y) = \dfrac{2}{7}$ . $P(\text{both green}) = \dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}$ .
2. In a school of $400$ students, $240$ study French, $180$ study German, and $80$ study both. Find: (a) $P(\text{German} \mid \text{French})$ , (b) $P(\text{French} \mid \text{German})$ . (show answer)

Answer
(a) $P(\text{German} \mid \text{French}) = \dfrac{80}{240} = \dfrac{1}{3} \approx 0.333$ . (b) $P(\text{French} \mid \text{German}) = \dfrac{80}{180} = \dfrac{4}{9} \approx 0.444$ .
3. A survey finds:

| | Supports policy | Opposes policy | Total | |--|----------------|---------------|-------| | Under 30 | 45 | 30 | 75 | | 30 and over | 35 | 40 | 75 | | Total | 80 | 70 | 150 |

(a) Find $P(\text{supports} \mid \text{under 30})$ and $P(\text{supports} \mid \text{30 and over})$ . (b) Is there an association between age group and opinion? Justify. (show answer)

Answer
(a) $P(\text{supports} \mid \text{under 30}) = \dfrac{45}{75} = 0.6$ . $P(\text{supports} \mid \text{30 and over}) = \dfrac{35}{75} \approx 0.467$ . (b) Yes, there is an association. The conditional probabilities differ: younger respondents are more likely to support the policy ( $60\%$ vs $47\%$ ). If there were no association, both groups would have the same support rate of $\dfrac{80}{150} \approx 0.533$ .
4. Events $A$ and $B$ are such that $P(A) = 0.6$ , $P(B) = 0.4$ , and $P(A \mid B) = 0.75$ . Find $P(A \text{ and } B)$ and determine whether $A$ and $B$ are independent. (show answer)

Answer
$P(A \text{ and } B) = P(A \mid B) \times P(B) = 0.75 \times 0.4 = 0.3$ . For independence: $P(A) \times P(B) = 0.6 \times 0.4 = 0.24$ . Since $0.3 \neq 0.24$ , $A$ and $B$ are not independent.
5. Three machines produce items. Machine X makes $50\%$ of items with a $2\%$ defect rate. Machine Y makes $30\%$ with a $3\%$ defect rate. Machine Z makes $20\%$ with a $5\%$ defect rate. An item is selected at random. Find the probability it is defective. (show answer)

Answer
$P(\text{defective}) = 0.50 \times 0.02 + 0.30 \times 0.03 + 0.20 \times 0.05 = 0.010 + 0.009 + 0.010 = 0.029$ . The probability that a randomly selected item is defective is $2.9\%$ .

Reasoning · Tier 3: explain and apply

1. Using the machine data from Tier 2 Q5, an item is found to be defective. Find the probability it came from Machine Z. (show answer)

Answer
$P(\text{Z and defective}) = 0.20 \times 0.05 = 0.010$ . $P(\text{Z} \mid \text{defective}) = \dfrac{P(\text{Z and defective})}{P(\text{defective})} = \dfrac{0.010}{0.029} \approx 0.345$ . There is about a $34.5\%$ chance the defective item came from Machine Z.
2. Explain, with a numerical example, why $P(A \mid B) \neq P(B \mid A)$ in general. Why is confusing these two a common and dangerous error in medical or legal contexts? (show answer)

Answer
Example: $P(\text{disease} \mid \text{positive test}) \neq P(\text{positive test} \mid \text{disease})$ . A test might detect $95\%$ of sick people ( $P(T^+ \mid D) = 0.95$ ), but if the disease is rare, $P(D \mid T^+)$ could be much lower (e.g. $0.16$ ). Confusing the two -- called the "prosecutor's fallacy" in legal contexts -- leads to wildly wrong conclusions. In medicine, it means overestimating how likely a patient is to have the disease after a positive test.
3. A jar contains $4$ red and $6$ blue marbles. Three marbles are drawn without replacement. Find $P(\text{all three are blue})$ using a chain of conditional probabilities. (show answer)

Answer
$P(\text{1st blue}) = \dfrac{6}{10}$ . $P(\text{2nd blue} \mid \text{1st blue}) = \dfrac{5}{9}$ . $P(\text{3rd blue} \mid \text{first two blue}) = \dfrac{4}{8} = \dfrac{1}{2}$ . $P(\text{all three blue}) = \dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{1}{2} = \dfrac{30}{180} = \dfrac{1}{6}$ .
4. Two events satisfy $P(A \mid B) = 0.5$ and $P(B \mid A) = 0.25$ . If $P(B) = 0.4$ , find $P(A)$ . (show answer)

Answer
$P(A \text{ and } B) = P(A \mid B) \times P(B) = 0.5 \times 0.4 = 0.2$ . Also $P(A \text{ and } B) = P(B \mid A) \times P(A)$ , so $0.2 = 0.25 \times P(A)$ , giving $P(A) = \dfrac{0.2}{0.25} = 0.8$ .

Reasoning · Harder reasoning

1. A game show has three doors. Behind one door is a prize; behind the other two, nothing. You pick a door. The host, who knows what is behind each door, opens a different door to reveal no prize. You are offered the chance to switch. Using conditional probability, show that switching gives you a $\dfrac{2}{3}$ chance of winning. (show answer)

Answer
Label the doors $1, 2, 3$ . Suppose the prize is behind door $1$ (by symmetry, the argument is the same for any door). You pick door $1$ : host opens door $2$ or $3$ ; switching loses. You pick door $2$ : host must open door $3$ ; switching wins. You pick door $3$ : host must open door $2$ ; switching wins. So switching wins $2$ out of $3$ times. Formally: let $W$ = prize behind your door. $P(W) = \dfrac{1}{3}$ . Given the host reveals a losing door, $P(\text{prize behind other door} \mid \text{host reveals}) = \dfrac{2}{3}$ .
2. In a population, $0.5\%$ use a certain drug. A drug test has a $99\%$ true positive rate and a $2\%$ false positive rate. Find the probability that a person who tests positive actually uses the drug. Comment on the usefulness of the test. (show answer)

Answer
$P(\text{uses drug}) = 0.005$ . $P(T^+ \mid \text{user}) = 0.99$ . $P(T^+ \mid \text{non-user}) = 0.02$ . $P(T^+) = 0.005 \times 0.99 + 0.995 \times 0.02 = 0.00495 + 0.0199 = 0.02485$ . $P(\text{user} \mid T^+) = \dfrac{0.00495}{0.02485} \approx 0.199$ . Only about $20\%$ of positive results are true positives. The test produces many false positives because the user base is so small. A confirmatory (more specific) test is essential.
3. Prove that if $A$ and $B$ are independent, then $A$ and $\overline{B}$ (the complement of $B$ ) are also independent. (show answer)

Answer
$P(A \text{ and } \overline{B}) = P(A) - P(A \text{ and } B) = P(A) - P(A) \cdot P(B)$ (using independence) $= P(A)(1 - P(B)) = P(A) \cdot P(\overline{B})$ . Since $P(A \text{ and } \overline{B}) = P(A) \cdot P(\overline{B})$ , events $A$ and $\overline{B}$ are independent.
4. Five cards are dealt from a standard deck of $52$ . Find the probability that all five are spades, using a chain of conditional probabilities. (show answer)

Answer
$P(\text{all 5 spades}) = \dfrac{13}{52} \times \dfrac{12}{51} \times \dfrac{11}{50} \times \dfrac{10}{49} \times \dfrac{9}{48} = \dfrac{13 \times 12 \times 11 \times 10 \times 9}{52 \times 51 \times 50 \times 49 \times 48} = \dfrac{154440}{311875200} = \dfrac{33}{66640} \approx 0.000495$ .