A-Level Fractions, ratios and proportions exam-style questions

Use Clevolab for A-Level exam-style practice in fractions, ratios and proportions. This page shows what the topic covers, what skills the current set targets, and a few real examples from the reviewed question bank.

Maths A-Level Fractions, ratios and proportions

About this topic

Clevolab treats fractions, ratios and proportions as repeated practice with explanation, not just answer checking. This page is designed to make the topic legible before you open the app.

The current A-Level set covers fractions, percentages, ratio, direct proportion, inverse proportion, and proportional reasoning. 27 reviewed questions currently published for this page.

What you can practise

  • Fractions, decimals, and percentages
  • Ratio simplification and sharing
  • Percentage change and reverse percentages
  • Direct and inverse proportion
  • Multi-step proportional reasoning

Real sample questions from the current set

These examples come from the reviewed questions currently stored for this topic. They are here so the page shows the actual flavour of Clevolab, not just a summary.

A-Level

Sample question

Two similar triangles have side ratio $3:7$. If the smaller has area $54\,\mathrm{cm^2}$, what is the larger’s area?

  • A294 cm$^2$Correct answer
  • B126 cm$^2$Answer option
  • C882 cm$^2$Answer option
  • D98 cm$^2$Answer option

Why this answer is right

Area scales with the square of the linear factor. Factor $\left(\tfrac{7}{3}\right)^2=\tfrac{49}{9}$. So $54\times \tfrac{49}{9}=6\times 49=294$.

For similar figures, $A\propto L^2$. Here the scale factor from smaller to larger is $\tfrac{7}{3}$, so $$A_\text{large}=54\times\left(\frac{7}{3}\right)^2=54\times\frac{49}{9}=6\times 49=294.$$ Using $\tfrac{7}{3}$ linearly gives $126$, which is too small.

A-Level

Sample question

A population doubles every $4$ hours. What is the growth factor per hour (to three d.p.)?

  • A1.250Answer option
  • B1.189Correct answer
  • C2.000Answer option
  • D1.150Answer option

Why this answer is right

If $r$ is the hourly factor, then $r^4=2$. So $r=2^{1/4}\approx 1.189$.

Let hourly factor be $r$. Four hours multiply by $r^4$. Since doubling occurs in $4$ hours: $$r^4=2\quad\Rightarrow\quad r=2^{1/4}\approx 1.189.$$ Arithmetic means like $1.25$ are not correct for multiplicative growth.

A-Level

Sample question

Solutions of $30\%$ acid and $10\%$ acid are mixed in the ratio $2:3$ by volume. What is the overall concentration?

  • A20%Answer option
  • B18%Correct answer
  • C16%Answer option
  • D24%Answer option

Why this answer is right

Use a weighted average: $\frac{2\cdot 0.30+3\cdot 0.10}{2+3}=\frac{0.60+0.30}{5}=0.18=18\%$.

Concentration combines linearly by volume. Compute total acid over total volume: $$\text{acid} = 2\cdot 0.30 + 3\cdot 0.10 = 0.60+0.30 = 0.90,$$ $$\text{volume} = 2+3=5.$$ Overall concentration: $$\frac{0.90}{5}=0.18=18\%.$$ The simple mean $20\%$ ignores the $2:3$ weighting.

How this page fits into Clevolab

Clevolab is broader than any one exam mode. GCSE and A-level pages are useful entry points, while the wider project is about sharpening understanding through repeated topic practice.

Related topics