Year 9 maths test: free practice questions with answers

Year 9 sample test

10 questions, one from each of 10 topics. Work them out on paper, then open each answer to mark it.

1. Question 1Enlargements and scale factors

   Triangle $ABC$ is enlarged by a scale factor of $3$ to produce triangle $A'B'C'$. Triangle $ABC$ has side lengths $4$ cm, $6$ cm and $9$ cm. Find the side lengths of triangle $A'B'C'$.

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   Answer:

   12 cm, 18 cm and 27 cm

   Key idea: Under an enlargement with scale factor $k$, every side length of the image is $k$ times the corresponding side length of the original shape.

   1. Multiply each side length by the scale factor $k = 3$:
      $\begin{gathered}4 \times 3 = 12 \text{ cm} \\ 6 \times 3 = 18 \text{ cm} \\ 9 \times 3 = 27 \text{ cm}\end{gathered}$
   2. The side lengths of triangle $A'B'C'$ are $12$ cm, $18$ cm and $27$ cm.

   Answer: $12$ cm, $18$ cm and $27$ cm.

   More enlargements and scale factors practice →

2. Question 2Statistics

   The ages (in years) of 8 members of a sports team are listed below.

   $14,\; 15,\; 15,\; 16,\; 17,\; 18,\; 19,\; 21$

   Find the interquartile range (IQR) of these ages.

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   Answer:

   3.5 years

   Key idea: Split the data into a lower half and an upper half, then find the median of each half.

   1. There are $n = 8$ values in order: 14, 15, 15, 16, 17, 18, 19, 21.
   2. Lower half (first 4 values): 14, 15, 15, 16 $\Rightarrow Q_1 = \dfrac{15 + 15}{2} = 15$.
   3. Upper half (last 4 values): 17, 18, 19, 21 $\Rightarrow Q_3 = \dfrac{18 + 19}{2} = 18.5$.
   4. $\text{IQR} = Q_3 - Q_1 = 18.5 - 15 = 3.5$ years.

   Answer: $3.5$ years.

   More statistics practice →

3. Question 3Venn diagrams and two-way tables

   A survey of 80 students found the following results about whether they have a pet dog (D) and a pet cat (C).

   	Has cat	No cat	Total
   Has dog	15	28	43
   No dog	12	25	37
   Total	27	53	80

   Using relative frequencies, estimate $P(\text{dog or cat})$, that is, the probability that a randomly chosen student has a dog, a cat, or both. Give your answer correct to 2 decimal places.

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   Answer:

   $0.69$

   Key idea: $P(D \text{ or } C)$ (inclusive or) counts everyone in the dog region, the cat region, or both, which is every student except those with neither pet.

   1. Read the count with neither dog nor cat from the table: the "No dog, No cat" cell is $25$.
   2. So the number with a dog, a cat, or both is $80 - 25 = 55$.
   3. Alternatively, use the addition rule: $P(D \text{ or } C) = P(D) + P(C) - P(D \text{ and } C)$.
      $P(D) = \frac{43}{80}, P(C) = \frac{27}{80}, P(D \text{ and } C) = \frac{15}{80}$
   4. Applying the rule:
      $P(D \text{ or } C) = \frac{43}{80} + \frac{27}{80} - \frac{15}{80} = \frac{55}{80}$
   5. Convert to a decimal: $\dfrac{55}{80} = 0.6875 \approx 0.69$ (correct to 2 decimal places).

   Answer: $0.69$

   More venn diagrams and two-way tables practice →

4. Question 4Probability

   A bag contains 3 red marbles and 2 blue marbles. Two marbles are drawn without replacement. Using a tree diagram or table to list all outcomes, what is the probability of drawing two red marbles? Give your answer correct to 2 decimal places.

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   Answer:

   $0.30$

   Key idea: Without replacement means the second draw depends on the first. Multiply the probabilities along each branch of the tree diagram.

   1. On the first draw there are 5 marbles: $P(\text{red}_1) = \dfrac{3}{5}$.
   2. After drawing one red marble, 4 marbles remain: 2 red and 2 blue. So $P(\text{red}_2 \mid \text{red}_1) = \dfrac{2}{4} = \dfrac{1}{2}$.
   3. Multiply along the branch: $P(\text{two reds}) = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10}$
   4. Convert to a decimal: $\dfrac{3}{10} = 0.30$.

   Answer: $0.30$

   More probability practice →

5. Question 5Linear equations and graphs

   Solve the equation $3x - 7 = 2x + 5$ for $x$.

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   Answer:

   $x = 12$

   Key idea: To solve a linear equation, collect pronumerals on one side and constants on the other using inverse operations.

   1. Start with the equation: $3x - 7 = 2x + 5$
   2. Subtract $2x$ from both sides: $x - 7 = 5$
   3. Add $7$ to both sides: $x = 12$
   4. Check: $3(12) - 7 = 29$ and $2(12) + 5 = 29$ ✓

   Answer: $x = 12$

   More linear equations and graphs practice →

6. Question 6Measurement error

   A student measures the length of a book as 28 cm. The true length is 27.5 cm. Find the absolute error, the relative error and the percentage error.

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   Answer:

   Absolute error: $0.5$ cm; relative error: $\frac{1}{55}$; percentage error: $\approx 1.82\%$

   Key idea: Absolute error is how far the measurement is from the true value. Relative error compares that gap to the true value. Percentage error converts the relative error to a percentage.

   1. Find the absolute error: $\text{absolute error} = \lvert \text{measured} - \text{true} \rvert = \lvert 28 - 27.5 \rvert = 0.5 \text{ cm}$
   2. Find the relative error: $\text{relative error} = \frac{\text{absolute error}}{\text{true value}} = \frac{0.5}{27.5} = \frac{1}{55}$
   3. Convert to a percentage error: $\text{percentage error} = \frac{1}{55} \times 100 \approx 1.82\%$
   4. Answer: The absolute error is $0.5$ cm, the relative error is $\dfrac{1}{55}$, and the percentage error is approximately $1.82\%$.

   More measurement error practice →

7. Question 7Exponent laws

   Simplify each expression.

   (a) $(m^3)^4$

   (b) $(2p^2)^3$

   (c) $5x^0 + (3y)^0$

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   Answer:

   (a) $m^{12}$ (b) $8p^6$ (c) $6$

   Key idea: For a power of a power, multiply the exponents. Any non-zero base raised to the power of zero equals 1.

   1. (a) $(m^3)^4 = m^{3 \times 4} = m^{12}$
   2. (b) Apply the power to both the coefficient and the variable: $(2p^2)^3 = 2^3 \times p^{2 \times 3} = 8p^6$
   3. (c) $x^0 = 1$ and $(3y)^0 = 1$, so $5x^0 + (3y)^0 = 5(1) + 1 = 6$

   Answer: (a) $m^{12}$ (b) $8p^6$ (c) $6$

   More exponent laws practice →

8. Question 8Pythagoras (applied)

   A $6.5$ m ladder leans against a vertical wall. The base of the ladder is on horizontal ground and sits $2.5$ m from the base of the wall. A safety rule says the angle the ladder makes with the ground must be between $70°$ and $80°$. Use trigonometry to find the angle the ladder makes with the ground, correct to one decimal place, and state whether the ladder is positioned safely.

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   Answer:

   $\theta \approx 67.4°$, not safe

   Key idea: The ladder, wall and ground form a right-angled triangle. Use the cosine ratio, since the adjacent side (ground distance) and hypotenuse (ladder) are known, to find the angle at the base.

   1. Identify the sides relative to the angle at the ground: adjacent $= 2.5$ m, hypotenuse $= 6.5$ m.
   2. Write the cosine ratio:
      $\cos(\theta) = \frac{2.5}{6.5}$
   3. Solve for $\theta$:
      $\theta = \cos^{-1}\!\left(\frac{2.5}{6.5}\right) \approx \cos^{-1}(0.3846) \approx 67.4°$
   4. Compare with the safe range of $70°$ to $80°$: $67.4° < 70°$, so the ladder is outside the safe range.

   Answer: The ladder makes an angle of approximately $67.4°$ with the ground, which is not safe.

   More pythagoras (applied) practice →

9. Question 9Coordinate geometry

   A park has two water fountains located at $F(1, 3)$ and $G(7, 11)$ on a map, where each unit represents 10 metres. A garden bed is to be planted exactly halfway between the two fountains. A path also runs through the points $H(0, 6)$ and $K(3, 0)$. (a) Find the coordinates of the garden bed. (b) Find the actual straight-line distance between the two fountains in metres. (c) Determine the gradient of the path $HK$ and explain what it means in context.

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   Answer:

   Garden bed at $(4, 7)$; distance $= 100$ metres; gradient of $HK = -2$

   Key idea: Apply the midpoint, distance and gradient formulas, then interpret the answers using the map scale.

   1. Midpoint (garden bed): $M = \left(\frac{1+7}{2},\, \frac{3+11}{2}\right) = (4,\, 7)$
   2. Distance $FG$ in map units: $d = \sqrt{(7-1)^2 + (11-3)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ units}$
   3. Actual distance: $10 \times 10 = 100$ metres.
   4. Gradient of path $HK$: $m = \frac{0 - 6}{3 - 0} = \frac{-6}{3} = -2$
   5. Interpretation: A gradient of $-2$ means the path falls 2 units on the map for every 1 unit across, so it runs steeply downward from left to right.

   Answer: Garden bed at $(4, 7)$; distance between fountains is 100 metres; gradient of path $HK$ is $-2$ (steeply downward from left to right).

   More coordinate geometry practice →

10. Question 10Surface area and volume

    A cylindrical tin has a radius of 4 cm and a height of 10 cm. What is the total surface area of the tin, correct to 2 decimal places, in cm²?

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    Answer:

    $351.86$

    Key idea: The total surface area of a cylinder is $SA = 2\pi r^2 + 2\pi r h$, where $r$ is the radius and $h$ is the height.

    1. Identify the values: $r = 4$ cm, $h = 10$ cm.
    2. Calculate the area of the two circular ends: $2\pi r^2 = 2 \times \pi \times 4^2 = 2 \times \pi \times 16 = 32\pi$.
    3. Calculate the curved surface area: $2\pi r h = 2 \times \pi \times 4 \times 10 = 80\pi$.
    4. Add the parts: $SA = 32\pi + 80\pi = 112\pi$.
    5. Evaluate: $SA = 112\pi \approx 351.858\ldots$ cm².

    Answer: $351.86$ cm² (correct to 2 decimal places)

    More surface area and volume practice →