Year 10 maths test: free practice questions with answers

Year 10 sample test

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

1. Question 1Exponential equations

   Solve $3^{2x-1} = 81$. What is the value of $x$?

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   Answer:

   $x = 2.5$

   Key idea: If both sides have the same base, the exponents must be equal.

   1. Write 81 as a power of 3: $81 = 3^4$.
   2. The equation becomes $3^{2x-1} = 3^4$.
   3. Equate the exponents: $2x - 1 = 4$.
   4. Solve: $2x = 5$, so $x = 2.5$.
   5. Check: $3^{2(2.5)-1} = 3^4 = 81$ ✓

   Answer: $x = 2.5$

   More exponential equations practice →

2. Question 2Statistics

   A dataset of ten reaction times (in milliseconds) is: 210, 230, 245, 250, 260, 275, 280, 290, 310, 480. For this dataset, $Q1 = 245$, $Q3 = 290$, and $\text{IQR} = 45$. Use the $1.5 \times \text{IQR}$ rule to determine whether any values are outliers.

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   Answer:

   $480$ ms is an outlier; no lower outliers.

   Key idea: A value is an outlier if it lies below $Q1 - 1.5 \times \text{IQR}$ or above $Q3 + 1.5 \times \text{IQR}$.

   1. Calculate the lower fence: $Q1 - 1.5 \times \text{IQR} = 245 - 1.5 \times 45 = 245 - 67.5 = 177.5.$
   2. Calculate the upper fence: $Q3 + 1.5 \times \text{IQR} = 290 + 1.5 \times 45 = 290 + 67.5 = 357.5.$
   3. Any value below $177.5$ or above $357.5$ is an outlier. Check the extremes: the minimum is $210 > 177.5$ (not an outlier), and the maximum is $480 > 357.5$ (outlier).

   Answer: $480$ ms is an outlier. No values fall below the lower fence.

   More statistics practice →

3. Question 3Compound interest

   Mia borrows $12 000 to pay for a car. The loan charges 8% per annum compound interest, compounded annually. Mia makes no repayments. After 5 years, how much more does she owe compared with what she would owe under a simple interest loan at the same rate over the same period? Give your answer to the nearest dollar.

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   Answer:

   $832 more

   Key idea: Calculate each type of interest separately and find the difference.

   1. Compound interest final amount:
      $A = 12000 \times (1.08)^5 = 12000 \times 1.46932807 \approx 17631.94$
   2. Simple interest final amount:
      $A = P(1 + rn) = 12000 \times (1 + 0.08 \times 5) = 12000 \times 1.4 = 16800.00$
   3. Find the difference:
      $17631.94 - 16800.00 = 831.94$

   Answer: Mia owes approximately $832 more under compound interest than under simple interest.

   More compound interest practice →

4. Question 4Linear inequalities

   Anika has already saved $85. She earns $12 per hour working at a cafe and wants to save enough to buy a jacket that costs at least $181. Write and solve an inequality to find the minimum number of whole hours she must work.

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   Answer:

   At least 8 hours

   Key idea: Translate the real-life situation into an inequality, solve it, then interpret the answer in context.

   1. Let $h$ be the number of hours Anika works. Her total savings will be $85 + 12h$ dollars. She needs this to be at least $181: $85 + 12h \geq 181$
   2. Subtract 85 from both sides: $12h \geq 96$
   3. Divide both sides by 12 (positive, sign stays): $h \geq 8$
   4. Since $h$ must be a whole number and $h \geq 8$, the minimum number of hours is 8.

   Answer: Anika must work at least 8 hours.

   More linear inequalities practice →

5. Question 5Linear equations

   Solve for $x$: $\dfrac{2x - 3}{5} = 7$

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   Answer:

   $x = 19$

   Key idea: Multiply both sides by the denominator to eliminate the fraction, then isolate $x$.

   1. Start with $\dfrac{2x - 3}{5} = 7$.
   2. Multiply both sides by $5$: $2x - 3 = 35$.
   3. Add $3$ to both sides: $2x = 38$.
   4. Divide both sides by $2$: $x = 19$.
   5. Check: $\dfrac{2(19) - 3}{5} = \dfrac{35}{5} = 7$ ✓

   Answer: $x = 19$

   More linear equations practice →

6. Question 6Algebraic fractions

   Simplify $\frac{4x}{9} \times \frac{3}{2x}.$

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   Answer:

   $\dfrac{2}{3}$

   Key idea: Multiply numerators together and denominators together, then cancel common factors before or after multiplying.

   1. Write as a single fraction: $\frac{4x \times 3}{9 \times 2x}$
   2. Simplify numerator and denominator: $\frac{12x}{18x}$
   3. Cancel $x$ (assuming $x \neq 0$): $\frac{12}{18}$
   4. Simplify the numerical fraction by dividing by 6: $\frac{2}{3}$

   Answer: $\dfrac{2}{3}$

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7. Question 7Direct and inverse proportion

   A garden shed is modelled by a rectangular prism with length 3 m, width 2 m and height 2.5 m. A scale model of the shed is built using a scale factor of $\dfrac{1}{20}$ (that is, every length on the model is $\dfrac{1}{20}$ of the real length). Find the surface area and volume of the model, and compare them with the real shed's surface area and volume by stating the scale factors for area and volume.

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   Answer:

   Model surface area $= 0.0925 \text{ m}^2$; model volume $= 0.001875 \text{ m}^3$; area scales by $\tfrac{1}{400}$, volume scales by $\tfrac{1}{8000}$

   Key idea: When lengths scale by a factor of $s$, areas scale by $s^2$ and volumes scale by $s^3$.

   1. Find the real shed's surface area. The six faces of the rectangular prism have areas: two $3 \times 2.5 = 7.5 \text{ m}^2$ faces, two $2 \times 2.5 = 5 \text{ m}^2$ faces, and two $3 \times 2 = 6 \text{ m}^2$ faces.
      $SA_{\text{real}} = 2(7.5) + 2(5) + 2(6) = 15 + 10 + 12 = 37 \text{ m}^2$
   2. Find the real shed's volume:
      $V_{\text{real}} = 3 \times 2 \times 2.5 = 15 \text{ m}^3$
   3. The length scale factor is $s = \dfrac{1}{20}$, so the area scale factor is $s^2 = \dfrac{1}{400}$ and the volume scale factor is $s^3 = \dfrac{1}{8000}$.
   4. Model surface area:
      $SA_{\text{model}} = \frac{37}{400} = 0.0925 \text{ m}^2$
   5. Model volume:
      $V_{\text{model}} = \frac{15}{8000} = 0.001875 \text{ m}^3$

   Answer: The model's surface area is $0.0925 \text{ m}^2$ (area scale factor $\tfrac{1}{400}$) and its volume is $0.001875 \text{ m}^3$ (volume scale factor $\tfrac{1}{8000}$).

   More direct and inverse proportion practice →

8. Question 8Index laws with algebra

   Simplify $3a^4 \times 5a^3$.

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   Answer:

   $15a^7$

   Key idea: When multiplying terms with the same base, add the exponents: $a^m \times a^n = a^{m+n}$.

   1. Multiply the coefficients: $3 \times 5 = 15$.
   2. Apply the product law to the powers of $a$: $a^4 \times a^3 = a^{4+3} = a^7$.
   3. Write the result: $15a^7$.

   Answer: $15a^7$

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9. Question 9Geometric reasoning and proofs

   Triangle $ABC$ has $AB = 5$ cm, $BC = 7$ cm and $\angle ABC = 48°$. Triangle $DEF$ has $DE = 5$ cm, $EF = 7$ cm and $\angle DEF = 48°$. Prove that $\triangle ABC \cong \triangle DEF$.

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   Answer:

   $\triangle ABC \cong \triangle DEF$ by SAS.

   Key idea: Two triangles are congruent by SAS if two pairs of corresponding sides and the included angle are equal.

   1. $AB = DE = 5$ cm (given)
   2. $BC = EF = 7$ cm (given)
   3. $\angle ABC = \angle DEF = 48°$ (given), and this angle is included between the two known sides in each triangle.
   4. Therefore $\triangle ABC \cong \triangle DEF$ (SAS).

   Answer: $\triangle ABC \cong \triangle DEF$ by SAS.

   More geometric reasoning and proofs practice →

10. Question 10Linear equations with algebraic fractions

    Solve the equation $\frac{x+1}{3} = \frac{2x-1}{5}$

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    Answer:

    $x = 8$

    Key idea: Multiply both sides by the LCD to clear all fractions, then solve the resulting linear equation.

    1. Find the LCD of 3 and 5: $\text{LCD} = 15$.
    2. Multiply every term on both sides by 15:
       $15 \times \frac{x+1}{3} = 15 \times \frac{2x-1}{5}$
    3. Simplify each side:
       $5(x+1) = 3(2x-1)$
    4. Expand the brackets:
       $5x + 5 = 6x - 3$
    5. Solve for $x$:
       $\begin{aligned} 5 + 3 &= 6x - 5x \\ 8 &= x \end{aligned}$
    6. Check by substituting $x = 8$ back into the original equation: $\dfrac{8+1}{3} = \dfrac{9}{3} = 3$ and $\dfrac{2(8)-1}{5} = \dfrac{15}{5} = 3$. Both sides are equal.

    Answer: $x = 8$

    More linear equations with algebraic fractions practice →