Year 10 Core — Exam 1 + Exam 2

Full Semester 1 review

Use ◀ ▶ arrow keys, Space to advance, F for fullscreen. Click "Jump" top-right.

Exa Q 1 2 marks Simplify expressions

(a) $4f^{2}g \times 5fg$ (1 mark)
(b) $\dfrac{x}{3} + \dfrac{2x}{7}$ (1 mark)

Approach

For part (a), multiply the numbers together first, then handle each letter. Same letters get added powers (so $f \times f^2 = f^3$).
For part (b), the bottoms are different, so we can't add yet. Find a common bottom — the smallest number both 3 and 7 divide into is 21.

Key step

(a) $4 \times 5 = 20$, $f^{2} \times f = f^{3}$, $g \times g = g^{2}$.
(b) Rewrite as $\dfrac{7x}{21}+\dfrac{6x}{21}$, then add the tops.

Answer

(a) $20f^{3}g^{2}$ (b) $\dfrac{13x}{21}$

Watch out

Lots of students write $\dfrac{x+2x}{21}$ — that's wrong. You must multiply the top by whatever you multiplied the bottom by.

$Solution page 2$

Handwritten solution

Exa Q 2 2 marks Expand and simplify

(a) $(3x+2)(3x-2)$ (1 mark)
(b) $(5-3x)(12-5x)$ (1 mark)

Approach

Part (a) is a special pattern called 'difference of two squares': $(A+B)(A-B) = A^{2} - B^{2}$. The middle terms always cancel.
Part (b) is regular expanding. Multiply every term in the first bracket by every term in the second bracket. Four terms in total, then collect any like terms.

Key step

(a) $(3x)^{2} - 2^{2} = 9x^{2} - 4$.
(b) $60 - 25x - 36x + 15x^{2}$, then add the two $x$ terms together.

Answer

(a) $9x^{2}-4$ (b) $15x^{2}-61x+60$

Watch out

Signs in (b) catch people out. Write out all four terms before combining $-25x$ and $-36x$.

$Solution page 2$

Handwritten solution

Exa Q 3 6 marks Line on a graph — equation, distance, midpoint

Use the linear graph in the question booklet (two marked points on a straight line).
(a) Find the equation of the line. (2)
(b) Find the exact distance between the two points. (2)
(c) Find the midpoint and draw it on the graph. (2)

Approach

For (a) read the slope (rise over run) and the y-intercept off the graph and plug into $y = mx + c$.
For (b) use the distance formula $d = \sqrt{(x_2-x_1)^{2} + (y_2-y_1)^{2}}$. 'Exact' means keep the square root — don't turn it into a decimal.
For (c) the midpoint is just the average of the two x-coordinates and the average of the two y-coordinates.

Key step

Distance: $d = \sqrt{(x_2-x_1)^{2} + (y_2-y_1)^{2}}$.
Midpoint: $M = \left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$.

Answer

See your handwritten working — answers depend on the points marked on the graph.

Watch out

'Exact' = leave the square root. Turning $\sqrt{20}$ into 4.47 loses you a mark.

Handwritten solution

Exa Q 4 2 marks Simultaneous equations

Solve $3x+2y=19$ and $y=2x-8$.

Approach

The second equation already tells us what $y$ is, so we can swap it straight into the first equation. This is substitution.

Key step

Replace $y$ with $2x-8$: $3x + 2(2x-8) = 19$.
That gives $3x + 4x - 16 = 19$, so $7x = 35$ and $x = 5$.
Then $y = 2(5) - 8 = 2$.

Answer

$x = 5,\ y = 2$

Watch out

Don't forget to find $y$ once you have $x$. One mark for each.

Handwritten solution

Exa Q 5 2 marks Perpendicular line through a point

Find the equation of the line perpendicular to $y = -3x + 5$ that passes through $(3, 2)$.

Approach

Perpendicular means the gradients multiply to $-1$. So if one gradient is $-3$, flip it and change the sign — the new gradient is $\dfrac{1}{3}$. Then use the point-gradient formula $y - y_1 = m(x - x_1)$ with the new gradient and the given point.

Key step

New gradient $=\dfrac{1}{3}$. Substitute: $y - 2 = \dfrac{1}{3}(x - 3)$. Expand and tidy.

Answer

$y = \dfrac{1}{3}x + 1$

Watch out

Flip AND change the sign. A common mistake is keeping the negative.

Handwritten solution

Exa Q 6 2 marks Rectangular prism — height and surface area

A rectangular prism has length $5\,\text{cm}$, width $3\,\text{cm}$, and volume $60\,\text{cm}^{3}$.
(a) Find the height. (1)
(b) Find the surface area. (1)

Approach

Volume = length × width × height. Rearrange to find the height.
Surface area has three pairs of matching faces: top/bottom, front/back, left/right. Add up all six faces.

Key step

(a) $h = \dfrac{V}{l \cdot w} = \dfrac{60}{5 \cdot 3} = 4\,\text{cm}$.
(b) $SA = 2(lw + lh + wh) = 2(15 + 20 + 12)$.

Answer

(a) $h = 4\,\text{cm}$ (b) $SA = 94\,\text{cm}^{2}$

Watch out

Six faces in total — three pairs. Easy to count only three.

Handwritten solution

Exa Q 7 4 marks Composite solid — surface area and volume

Use the solid in the question booklet.
(a) Find the surface area in $\text{cm}^{2}$. (2)
(b) Find the volume in $\text{mm}^{3}$. (2)

Approach

Break the solid into faces and count each one. For the volume part, you need it in $\text{mm}^{3}$. The cleanest way: convert all the measurements to mm first, then calculate. Otherwise you'll multiply by 1000 at the end (since $1\,\text{cm}^{3} = 1000\,\text{mm}^{3}$).

Key step

Convert before calculating. Don't mix units mid-problem.

Answer

See your handwritten working.

Watch out

$1\,\text{cm} = 10\,\text{mm}$, but $1\,\text{cm}^{3} = 1000\,\text{mm}^{3}$. Cubing the conversion factor catches people out.

Handwritten solution

Exa Q 8 1 mark Python: does '=' mean 'is equal to'?

In Python, does '=' mean 'is equal to'? Explain briefly.

Approach

In Python, a single equals sign assigns a value to a variable. Comparing two things uses a double equals sign.

Key step

Assignment vs comparison are two different operators in code.

Answer

No. A single '=' is the assignment operator (it puts a value into a variable). To check if two things are equal you use '==' (double equals).

Watch out

The mark is for the explanation, not just 'no'. Say what '=' actually does.

Handwritten solution

Exa Q 9 4 marks Composite shape — find x and perimeter

Use the composite shape in the question booklet.
(a) Find $x$. (2)
(b) Find the perimeter. (2)

Approach

Look for a right-angled triangle hidden inside the shape — that's where Pythagoras gives you the missing length.
For the perimeter, only add the lengths that are on the OUTSIDE of the shape. Internal lines don't count.

Key step

Pythagoras: $a^{2} + b^{2} = c^{2}$. Then trace the outline edge-by-edge.

Answer

See your handwritten working.

Watch out

Perimeter is the outside boundary only.

Handwritten solution

Exa Q 10 2 marks Factorise without expanding first

Show how $(x+3)^{2}-(x-4)^{2}$ can be factorised and simplified WITHOUT expanding the brackets first.

Approach

This is the 'difference of two squares' pattern in disguise. Treat $A = x+3$ and $B = x-4$. Then $A^{2} - B^{2} = (A-B)(A+B)$.

Key step

$A - B = (x+3) - (x-4) = 7$.
$A + B = (x+3) + (x-4) = 2x - 1$.
So the expression equals $7(2x-1)$.

Answer

$7(2x-1)$, which simplifies to $14x - 7$.

Watch out

Be careful with the brackets when computing $A - B$. The minus flips every sign inside: $-(x-4) = -x + 4$.

Handwritten solution

Exa Q 11 7 marks Sets and probability — basketball and netball

A class of 15 students were surveyed. Let $B$ = basketball, $N$ = netball.
9 play basketball, 5 play netball, 3 play both, the rest play neither.
(a) Complete the 2-way table. (1)
(b) Find $n(B')$ and $n(B \cup N)$. (2)
(c) Find $\Pr(B)$, $\Pr(B \cap N')$, $\Pr(B|N)$. (3)
(d) Are $B$ and $N$ independent? Show working. (1)

Approach

Start by filling in the 'play both' cell (3 students). Then work out the rest: basketball only = 9 − 3 = 6; netball only = 5 − 3 = 2; neither = 15 − (6+3+2) = 4.
$n(B')$ means 'not basketball'. $n(B \cup N)$ means 'play at least one of the two'.
$\Pr(B|N)$ is the conditional 'given they play netball, what's the chance they play basketball?'
Independent means $\Pr(B|N) = \Pr(B)$ — knowing one event doesn't change the other.

Key step

Table fills: B∩N=3, B-only=6, N-only=2, neither=4. So $n(B)=9$, $n(N)=5$, $n(B')=6$, $n(N')=10$.
$n(B \cup N) = 9 + 5 - 3 = 11$.
$\Pr(B) = \dfrac{9}{15} = \dfrac{3}{5}$. $\Pr(B \cap N') = \dfrac{6}{15} = \dfrac{2}{5}$. $\Pr(B|N) = \dfrac{3}{5}$.
Since $\Pr(B|N) = \Pr(B) = \dfrac{3}{5}$, the events are independent.

Answer

$B$ and $N$ are independent because $\Pr(B|N) = \Pr(B) = \dfrac{3}{5}$.

Watch out

For $\Pr(B|N)$, divide by 5 (the netball players), not 15. Conditional probability narrows the sample.

Handwritten solution

Exa Q 12 6 marks Factorise quadratics

(a) $x^{2}-7x-30$ (1)
(b) $x^{2}-32$ (1)
(c) $x^{2}+8x+4$ (2)
(d) $2x^{2}-10x+4$ (2)

Approach

(a) Two numbers that multiply to $-30$ and add to $-7$: try $-10$ and $3$.
(b) Difference of two squares — $32$ isn't a perfect square but we can still write $x^{2} - 32 = x^{2} - (\sqrt{32})^{2}$. Then simplify the surd.
(c) and (d) don't factorise as nice integers. Use 'completing the square' instead: take half the $x$ coefficient, square it, add and subtract.
(d) Pull out the 2 first so the $x^{2}$ term is on its own.

Key step

(a) $(x-10)(x+3)$.
(b) $x^{2} - 32 = (x - 4\sqrt{2})(x + 4\sqrt{2})$ since $\sqrt{32} = 4\sqrt{2}$.
(c) $x^{2}+8x+4 = (x+4)^{2} - 16 + 4 = (x+4)^{2} - 12 = (x+4-2\sqrt{3})(x+4+2\sqrt{3})$.
(d) $2x^{2}-10x+4 = 2(x^{2}-5x+2) = 2\!\left[\left(x-\tfrac{5}{2}\right)^{2} - \tfrac{17}{4}\right]$.

Answer

See above for each part.

Watch out

Always pull out the highest common factor FIRST (part d). And simplify surds: $\sqrt{32}$ becomes $4\sqrt{2}$.

Handwritten solution

Exa MC 1 1 mark Triangular prism — surface area

A wood prism has a right-angled triangular base with sides $3$, $4$ and $5\,\text{cm}$. The volume is $60\,\text{cm}^{3}$. Find the surface area.

Approach

First find how long the prism is (the depth). Volume = base area × length. The base is a right triangle with legs 3 and 4, so its area is $\tfrac{1}{2} \cdot 3 \cdot 4 = 6$. Length is therefore $60 \div 6 = 10\,\text{cm}$.
Then add up the faces: two triangle ends, plus three rectangular sides (one for each edge of the triangle).

Key step

$SA = 2(\text{triangle}) + \text{length} \times (\text{triangle perimeter}) = 2(6) + 10(3+4+5) = 12 + 120$.

Answer

D — $132\,\text{cm}^{2}$

Watch out

The slanted (hypotenuse) face counts. That's three rectangles, not two.

Handwritten solution

Exa MC 2 1 mark Cube capacity

A cube has side length $200\,\text{mm}$. What is its capacity when full of water?

Approach

Volume of the cube is $200^{3} = 8\,000\,000\,\text{mm}^{3}$. Now convert to litres. The trick: $1\,\text{L} = 1000\,\text{cm}^{3} = 1\,000\,000\,\text{mm}^{3}$.

Key step

$8\,000\,000\,\text{mm}^{3} \div 1\,000\,000 = 8\,\text{L}$.

Answer

A — $8\,\text{litres}$

Watch out

Option (d) says $8\,000\,000\,\text{mL}$ which would be 8000 litres — clearly wrong, but easy to pick if you forget the conversion.

Handwritten solution

Exa MC 3 1 mark Simultaneous equations

Solve $3x - y = 18$ and $4x + y = 10$.

Approach

Add the two equations together — the $y$ terms cancel out straight away. Then solve for $x$, then plug back in to find $y$.

Key step

$7x = 28 \Rightarrow x = 4$. Substitute into the first equation: $3(4) - y = 18 \Rightarrow y = -6$.

Answer

A — $(4, -6)$

Watch out

Always check your answer in BOTH original equations. Easy to read the $-6$ as $6$ in a rush.

Handwritten solution

Exa MC 4 1 mark Sector perimeter

Find the perimeter of the sector (pie-slice shape) in the question booklet.

Approach

The perimeter of a sector is the curved arc PLUS the two straight radii. Arc length = $\dfrac{\theta}{360} \times 2\pi r$ where $\theta$ is the angle in degrees.

Key step

$P = \text{arc} + 2r$. Plug in the angle and radius from the diagram.

Answer

See your handwritten working.

Watch out

Don't forget to add the two straight edges. Many students compute the arc only.

Handwritten solution

Exa MC 5 1 mark Right triangle area

Find the area of the given right-angled triangle.

Approach

Area = $\tfrac{1}{2} \times \text{base} \times \text{height}$, using the two sides that meet at the right angle. NOT the hypotenuse.

Key step

Identify the two perpendicular sides, multiply them, then halve.

Answer

See your handwritten working.

Watch out

If only one leg and the hypotenuse are given, use Pythagoras first to find the missing leg.

Handwritten solution

Exa MC 6 1 mark Completed square form

Write $2x^{2} + 8x + 3$ in completed-square form.

Approach

Pull out the 2 from the $x^{2}$ and $x$ terms first — $2(x^{2} + 4x) + 3$. Then complete the square inside the bracket: take half of 4, square it (=4), add and subtract.
Be careful: when you bring the $-4$ outside the brackets, it gets multiplied by the 2.

Key step

$2(x^{2}+4x) + 3 = 2((x+2)^{2} - 4) + 3 = 2(x+2)^{2} - 8 + 3 = 2(x+2)^{2} - 5$.

Answer

A — $2(x+2)^{2} - 5$

Watch out

The $-4$ inside the bracket becomes $-8$ when you multiply by the 2 outside. That's where the answer differs from the bad options.

Handwritten solution

Exa MC 7 1 mark Silo volume — cylinder + cone

A silo is a cylinder with a cone on top. Find the volume in $\text{m}^{3}$ (closest answer).

Approach

Total volume = $V_{\text{cyl}} + V_{\text{cone}}$, where $V_{\text{cyl}} = \pi r^{2} h$ and $V_{\text{cone}} = \tfrac{1}{3}\pi r^{2} h_{\text{cone}}$. Use CAS to calculate. Read the radius and BOTH heights carefully from the diagram.

Key step

Cylinder height and cone height are usually different. Don't mistake the total height for one of them.

Answer

See your handwritten circled option.

Watch out

The cone needs the perpendicular height, not the slant. The diagram should label which is which.

Handwritten solution

Exa MC 8 1 mark Shortest rod that won't fit in the locker

Locker is $1.2 \times 0.8 \times 0.5\,\text{m}$. Find the shortest rod that will NOT fit inside.

Approach

The longest rod that DOES fit is the space diagonal: $d = \sqrt{l^{2} + w^{2} + h^{2}}$. Anything longer than that won't fit. So we want the smallest option that's bigger than this diagonal.

Key step

$d = \sqrt{1.2^{2} + 0.8^{2} + 0.5^{2}} = \sqrt{1.44 + 0.64 + 0.25} = \sqrt{2.33} \approx 1.526\,\text{m}$.

Answer

C — $1.55\,\text{m}$ (smallest option strictly greater than $1.526\,\text{m}$).

Watch out

The wording is negative ('will NOT fit'). Pick the smallest option BIGGER than the diagonal.

Handwritten solution

Exa MC 9 1 mark Table of values for y = 6x − 11

Which table of values matches the rule $y = 6x - 11$?

Approach

Test ONE entry from each table. The fastest test: at $x = 0$, the rule gives $y = -11$. Eliminate any tables where that pair isn't there. If two tables both pass, test a second point.

Key step

At $x = 0$: $y = -11$. At $x = 1$: $y = -5$.

Answer

B — the table where $x$ runs $-3$ to $2$ and includes $(0, -11)$ and $(1, -5)$.

Watch out

Two of the tables include $(0, -11)$. Use a second point to break the tie.

Handwritten solution

Exa MC 10 1 mark Two marbles without replacement

A bag has 5 red and 3 blue marbles. Two are drawn without replacement. What is the probability that BOTH are not red?

Approach

'Both not red' means both are blue. After the first blue, only 7 marbles are left and only 2 are blue. So multiply the two probabilities.

Key step

$\Pr(BB) = \dfrac{3}{8} \times \dfrac{2}{7} = \dfrac{6}{56} = \dfrac{3}{28}$.

Answer

D — $\dfrac{3}{28}$

Watch out

Without replacement means the bottom number (and the top) drops by 1 the second time. Don't just square the first probability.

Handwritten solution

Exa MC 11 1 mark Perpendicular line through (-2, 5)

Line: $3x - 2y + 6 = 0$. Find the equation of the perpendicular line through $(-2, 5)$.

Approach

Rearrange the line into $y = mx + c$ form first to read off the gradient. Then flip and change the sign for the perpendicular gradient. Finally use $y - y_1 = m(x - x_1)$.

Key step

Rearrange: $y = \dfrac{3}{2}x + 3$. So $m = \dfrac{3}{2}$ and perpendicular gradient $= -\dfrac{2}{3}$.
Substitute: $y - 5 = -\dfrac{2}{3}(x + 2)$, expand to get the answer.

Answer

A — $y = -\dfrac{2}{3}x + \dfrac{11}{3}$

Watch out

You MUST rearrange before reading the gradient. Reading 3 and $-2$ straight off gives the wrong number.

Handwritten solution

Exa MC 12 1 mark Valid first step in solving

$4 - 3(x-2) = 5$. Which of the following could be a correct FIRST step?

Approach

Test each option. Either an expansion is done correctly, or the same operation is done to both sides. Anything else is wrong.

Key step

Expanding the bracket: $-3(x-2) = -3x + 6$. So the equation becomes $4 - 3x + 6 = 5$. That matches option (a).

Answer

A.

Watch out

Option (c) drops the bracket without distributing the $-3$ properly. (b) and (d) do bogus operations.

Handwritten solution

Exa MC 13 1 mark Find P(A ∩ B) from the union

$\Pr(A)=0.6,\ \Pr(B)=0.35,\ \Pr(A\cup B)=0.8$. Find $\Pr(A\cap B)$.

Approach

Use the addition rule: $\Pr(A\cup B) = \Pr(A) + \Pr(B) - \Pr(A\cap B)$. Rearrange for the unknown.

Key step

$0.8 = 0.6 + 0.35 - \Pr(A\cap B) \Rightarrow \Pr(A\cap B) = 0.15$.

Answer

A — $0.15$.

Watch out

The intersection is SUBTRACTED in the formula. If you add it instead, you get a wrong-sign answer.

Handwritten solution

Exa MC 14 1 mark Simplify rational expression

Simplify $\dfrac{x-4}{8x} \times \dfrac{2x+8}{x^{2}-16}$.

Approach

Factorise every part first, THEN cancel. $x^{2}-16 = (x-4)(x+4)$ (difference of two squares). $2x+8 = 2(x+4)$.

Key step

$\dfrac{(x-4)}{8x} \times \dfrac{2(x+4)}{(x-4)(x+4)} = \dfrac{2}{8x} = \dfrac{1}{4x}$.

Answer

B — $\dfrac{1}{4x}$

Watch out

Always factorise first. If you multiply out before cancelling, you'll be stuck with a quartic.

Handwritten solution

Exa MC 15 1 mark Expand and simplify

Fully expand and simplify $(2x-3)^{2} - (2x+3)(2x-3)$.

Approach

First piece is a perfect square: $(2x-3)^{2} = 4x^{2} - 12x + 9$. Second piece is difference of two squares: $(2x+3)(2x-3) = 4x^{2} - 9$. Subtract — bracket the second piece carefully.

Key step

$(4x^{2} - 12x + 9) - (4x^{2} - 9) = -12x + 18$.

Answer

C — $-12x + 18$

Watch out

The minus sign flips EVERY term in the second bracket. $-(-9) = +9$.

Handwritten solution

Exa MC 16 1 mark Identify line 3y = 1 − 2x

Which line in the diagram represents $3y = 1 - 2x$?

Approach

Rearrange to $y = mx + c$ form: divide by 3 → $y = \dfrac{1}{3} - \dfrac{2}{3}x$. So gradient is $-\dfrac{2}{3}$ (negative, gentle) and y-intercept is $\dfrac{1}{3}$.

Key step

$y = \dfrac{1}{3} - \dfrac{2}{3}x$. Negative gradient (line goes down left-to-right), y-intercept just above the origin.

Answer

C — Line C (the negative-slope line crossing y-axis above the origin).

Watch out

Always rearrange to $y = mx + c$ form first. The original $3y = 1 - 2x$ hides both the gradient and the intercept.

Handwritten solution

Exa MC 17 1 mark Mutually exclusive — true statement

Sets $A$ and $B$ are mutually exclusive. Which statement must be true?

Approach

Mutually exclusive means the two sets can't both happen — they don't overlap. So their intersection is empty, which means $n(A\cap B) = 0$.

Key step

Mutually exclusive ⇔ $A \cap B = \emptyset$ ⇔ $n(A \cap B) = 0$.

Answer

B — $n(A \cap B) = 0$.

Watch out

Mutually exclusive is NOT the same as independent, and doesn't mean either set is empty.

Handwritten solution

Exa MC 18 1 mark Conditional probability — tea and coffee

50 teachers surveyed: 28 like tea, 20 like coffee, 8 like both. A teacher is picked and we know they like tea. What is the probability they also like coffee?

Approach

This is conditional. The sample is restricted to tea drinkers (28 of them). Of those 28, how many also like coffee? That's the 8 who like both.

Key step

$\Pr(C \mid T) = \dfrac{8}{28} = \dfrac{2}{7}$.

Answer

B — $\dfrac{2}{7}$

Watch out

Denominator is 28 (not 50). Conditional probability narrows the sample.

Handwritten solution

Exa MC 19 1 mark Fully factorise 12x² − 12x − 45

When fully factorised, $12x^{2} - 12x - 45$ is:

Approach

Always pull out the common factor first. All three coefficients (12, 12, 45) divide by 3, giving $3(4x^{2} - 4x - 15)$. Then factorise the quadratic inside — CAS gives it instantly.

Key step

$3(4x^{2} - 4x - 15) = 3(2x+3)(2x-5)$.

Answer

D — $3(2x+3)(2x-5)$

Watch out

'Fully' factorised means don't stop early. Option (b) leaves an unfactorised quadratic — wrong.

Handwritten solution

Exa MC 20 1 mark Python dice simulation

The Python program simulates rolling two dice and finding the chance of a sum of 8. Which statement about the code is correct?

Approach

The loop runs 1000 trials. As the number of trials grows, the experimental probability gets closer to the theoretical one.

Key step

Theoretical $\Pr(\text{sum}=8) = \dfrac{5}{36}$ (combinations: (2,6),(3,5),(4,4),(5,3),(6,2)).

Answer

B — code runs the experiment 1000 times so the answer will be close to the theoretical probability.

Watch out

Option (d) says $\dfrac{6}{36}$ — that's wrong. The correct theoretical chance is $\dfrac{5}{36}$.

Handwritten solution

Exa ER 1 10 marks Play area + water tank

Three regions A, B, C make up a play area (see diagram).
(a) Show length $l = 17\,\text{m}$. (2)
(b) Exact area of region C. (2)
(c) Who is correct, Bob or Barbara, about working out the perimeter? (2)
(d) Volume of water tank — a cylinder with a hemisphere on top, $r=2\,\text{m}$, cylinder height $2\,\text{m}$. Answer to 2 d.p. (2)
(e) Write the equation for the surface area to paint. (1)
(f) Exact total external surface area. (1)

Approach

For 'show that' parts you must write the rule, sub in the numbers, and arrive at the required value — not just write the value at the end.
For the dispute in (c), Barbara is right. Pythagoras needs a right triangle, and we don't have enough info about the bottom edge of region C to form one.
For the tank: it's a cylinder with a HEMISPHERE on top. Volume = $\pi r^{2} h + \dfrac{1}{2}\cdot\dfrac{4}{3}\pi r^{3}$. The tank sits on concrete, so the surface you paint is the curved side of the cylinder plus the top half-sphere — NO base, NO top of cylinder (the hemisphere covers it).

Key step

Volume $= \pi(2)^{2}(2) + \dfrac{2}{3}\pi(2)^{3} = 8\pi + \dfrac{16\pi}{3} \approx 41.89\,\text{m}^{3}$.
Painted SA $= 2\pi r h + 2\pi r^{2} = 2\pi(2)(2) + 2\pi(2)^{2} = 8\pi + 8\pi = 16\pi\,\text{m}^{2}$.

Answer

(c) Barbara is correct. (d) Volume $\approx 41.89\,\text{m}^{3}$. (f) Painted SA $= 16\pi\,\text{m}^{2}$.

Watch out

'Show that' marks require explicit working. Just writing 17 m at the end loses you a mark. The tank top is a HEMISPHERE (half-sphere) — half of $\dfrac{4}{3}\pi r^{3}$, and curved surface is half of $4\pi r^{2}$.

Handwritten solutions

Exa ER 2 10 marks Taxi linear models + inequation

Awesome Cabs: \$8 flag-fall + \$1.80 per km. Better Cabs: from two receipts (10 km = \$26.50, 5 km = \$15.50).
(a) Cost equation for Awesome. (1)
(b) Flag-fall, gradient and rule for Better Cabs. (3)
(c) Verify intersection $(12, 62)$ for $y = 3.5x + 20$ and $y = 5x + 2$. (2)
(d) Graph both lines, label intercepts and intersection. (3)
(e) Solve $\dfrac{5 - 3(2x+1)}{2} > x - 7$. (1)

Approach

Cost = (per km) × distance + (flag-fall). Gradient is the change in cost over the change in distance. For (c) substitute $x = 12$ into both equations and show both give $y = 62$. For the inequation, multiply both sides by 2, expand, collect $x$ terms — and only flip the inequality if you divide by a negative.

Key step

Better Cabs gradient = $\dfrac{26.50 - 15.50}{10 - 5} = \dfrac{11}{5} = 2.20$. Flag-fall = $15.50 - 5(2.20) = 4.50$. Rule: $y = 2.2x + 4.5$.
Inequation: $5 - 6x - 3 > 2x - 14 \Rightarrow 2 - 6x > 2x - 14 \Rightarrow 16 > 8x \Rightarrow x < 2$.

Answer

Better Cabs: $y = 2.2x + 4.5$. Intersection verified at $(12, 62)$. $x < 2$.

Watch out

Only flip the inequality when you divide by a negative. We divided by 8 (positive) here, so no flip.

Handwritten solutions

Exa ER 3 10 marks Two-way table + spinner tree

(a) Fill in missing values in the table (Left/Right handed × Male/Female). (2)
(b) Probability a left-handed female is picked. (1)
(c) Tree diagram for two spins of a 1–7 spinner with outcomes O (odd) or E (even). (3)
(d) Probability of at least one even. (1)
(e) Two example spins where the product is even. (1)
(f) Probability that Student 2 wins (product is odd). (2)

Approach

For the table, use row and column totals to back-fill. For the spinner: 4 odd numbers (1, 3, 5, 7), 3 even (2, 4, 6), so $\Pr(O) = \dfrac{4}{7}$, $\Pr(E) = \dfrac{3}{7}$.
'Product is odd' ONLY happens when BOTH spins are odd. So 'product is even' is everything else — use the complement.

Key step

$\Pr(\text{at least one E}) = 1 - \Pr(OO) = 1 - \dfrac{4}{7} \cdot \dfrac{4}{7} = 1 - \dfrac{16}{49} = \dfrac{33}{49}$.
$\Pr(\text{product odd}) = \Pr(OO) = \dfrac{16}{49}$.

Answer

$\Pr(\text{product odd}) = \dfrac{16}{49}$; Student 2 has 16 out of 49 chance.

Watch out

'Product is odd' requires BOTH to be odd. Product is even if EITHER is even — easiest via the complement.

Handwritten solutions

Exa ER 4 10 marks Expansion, completing square, simplifying, factorising

(a) Complete Mrs Cardno's two-step expansion of $3(x-4)^{2} - 2(x+1)$. (2)
(b) Fill in the missing parts of her completed-square version of $x^{2} - 8x + 11$. (3)
(c) Show that $\dfrac{t^{2}-49}{5(t-8)} \div \dfrac{2t^{2}-8t-42}{(t-8)(t+3)}$ simplifies to $\dfrac{t+7}{10}$. (3)
(d) Profit after 5 years. (1)
(e) Fully factorise $3x(2x+4) - 6(2x+4)$. (1)

Approach

(a) Expand $(x-4)^{2} = x^{2} - 8x + 16$ first. Multiply through by 3, then distribute the $-2$.
(b) Add and subtract 16 inside, so you can write it as $(x - 4)^{2} - 5$. Then it's difference of two squares with $\sqrt{5}$.
(c) Dividing fractions = flip the second one and multiply. Factor every quadratic; $t^{2}-49$ is difference of squares, $2t^{2}-8t-42 = 2(t^{2}-4t-21) = 2(t-7)(t+3)$. Things cancel.
(d) Just sub $t = 5$ into the simplified $\dfrac{t+7}{10}$.
(e) Both terms share $(2x+4)$. Factor it out and simplify.

Key step

(b) $(x-4-\sqrt{5})(x-4+\sqrt{5})$.
(c) $\dfrac{(t-7)(t+7)}{5(t-8)} \times \dfrac{(t-8)(t+3)}{2(t-7)(t+3)} = \dfrac{t+7}{10}$.
(d) Profit $= \dfrac{5+7}{10} = 1.2$ million dollars = \$1,200,000.
(e) $(2x+4)(3x - 6) = 2(x+2) \cdot 3(x-2) = 6(x+2)(x-2)$.

Answer

Profit at $t=5$: \$1,200,000. Final factorisation: $6(x+2)(x-2)$.

Watch out

In (e) most students stop at $(2x+4)(3x-6)$ — that's not FULLY factorised. Both brackets share another common factor.

Handwritten solutions