Year 10 Direct and Inverse Proportion Worksheets & Practice

Worked examples

Year 10 direct and inverse proportion: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

1. Example 1 · Finding the constant of proportionality

   The cost $C$ (in dollars) of buying bags of mulch is directly proportional to the number of bags $n$. When $n = 4$, $C = 30$. Find the equation relating $C$ and $n$, then find the cost of 10 bags.

   Show worked solution

   Key idea: If $C$ is directly proportional to $n$, then $C = kn$ for some constant $k$. Substitute a known pair to find $k$.

   1. Write the proportion equation: $C = kn$
   2. Substitute $n = 4$, $C = 30$: $30 = k \times 4$
   3. Solve for $k$: $k = \frac{30}{4} = 7.5$
   4. The equation is $C = 7.5n$.
   5. Substitute $n = 10$: $C = 7.5 \times 10 = 75$

   Answer: The equation is $C = 7.5n$, and 10 bags cost $75.

2. Example 2 · Solving an inverse proportion problem

   The time $t$ (in hours) taken to complete a delivery route is inversely proportional to the average speed $v$ (in km/h). When $v = 60$, $t = 3$. Find the equation relating $t$ and $v$, then find the time taken when $v = 45$.

   Show worked solution

   Key idea: If $t$ is inversely proportional to $v$, then $t = \dfrac{k}{v}$ for some constant $k$. Substitute a known pair to find $k$.

   1. Write the proportion equation: $t = \dfrac{k}{v}$
   2. Substitute $v = 60$, $t = 3$: $3 = \frac{k}{60}$
   3. Solve for $k$: $k = 3 \times 60 = 180$
   4. The equation is $t = \dfrac{180}{v}$.
   5. Substitute $v = 45$: $t = \frac{180}{45} = 4$

   Answer: The equation is $t = \dfrac{180}{v}$, and the journey takes 4 hours at 45 km/h.

3. Example 3 · Scaling area and volume of a 3D object

   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.

   Show worked solution

   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 $\tfrac{1}{8000}$).

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