A-Level Number and Arithmetic exam-style questions
Use Clevolab for A-Level exam-style practice in number and arithmetic. This page shows what the topic covers, what skills the current set targets, and a few real examples from the reviewed question bank.
About this topic
Clevolab treats number and arithmetic 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, powers, standard form, and core number methods. 75 reviewed questions currently published for this page.
What you can practise
- Fractions, decimals, and percentages
- Ratio and proportion
- Powers, roots, and indices
- Standard form
- Number fluency and arithmetic 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.
Sample question
Evaluate exactly $27^{2/3}\cdot 9^{-3/2}$.
- A$\dfrac{1}{3}$Correct answer
- B$3$Answer option
- C$\dfrac{1}{9}$Answer option
- D$9$Answer option
Why this answer is right
$27^{2/3}=9$, $9^{-3/2}=1/27$ so the product is $1/3$.
Compute $$27^{2/3}=(\sqrt[3]{27})^2=3^2=9,$$
Sample question
Given $x=2^3 3^{-1} 5^{1/2}$ and $y=2^{-1}3^{2}5^{3/2}$, simplify $$\frac{x^2}{\sqrt{15}\,y}$$ to a rationalised surd in simplest form.
- A$\dfrac{128\sqrt{5}}{1215}$Answer option
- B$\dfrac{128\sqrt{3}}{1215}$Correct answer
- C$\dfrac{256\sqrt{3}}{1215}$Answer option
- D$\dfrac{128\sqrt{3}}{405}$Answer option
Why this answer is right
Use index laws to combine powers, then write $\sqrt{15}=\sqrt{3}\sqrt{5}$ and collect exponents. You get $\dfrac{128}{405\sqrt{3}}$, which rationalises to $\dfrac{128\sqrt{3}}{1215}$.
Square $x$ to get $x^2=2^6 3^{-2} 5^1$. Divide by $y=2^{-1}3^2 5^{3/2}$ and by $\sqrt{15}=3^{1/2}5^{1/2}$.
Sample question
Solve for $x>0$ if $\sqrt{x}-\dfrac{1}{\sqrt{x}}=2$.
- A$3-2\sqrt{2}$Answer option
- B$1+\sqrt{2}$Answer option
- C$2+\sqrt{2}$Answer option
- D$3+2\sqrt{2}$Correct answer
Why this answer is right
Let $t=\sqrt{x}$. Then $t-1/t=2$. Multiply by $t$: $t^2-1=2t$, so $t^2-2t-1=0$. The positive root is $t=1+\sqrt{2}$, hence $x=t^2=3+2\sqrt{2}$.
Set $t=\sqrt{x}>0$. Then $$t-\frac{1}{t}=2\Rightarrow t^2-1=2t\Rightarrow t^2-2t-1=0.$$
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.