Year 10 Core — Sem 1 Exam 1

Tech-free · 60 min · 40 marks · 12 questions

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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

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

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

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

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

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

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

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

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

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

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

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