Progressive Number and Arithmetic practice questions
Use Clevolab for progressive 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 Progressive set covers fractions, percentages, ratio, powers, standard form, and core number methods. 239 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
Maria has $8$ stickers and gets $3$ more. She has
- A11Correct answer
- B12Answer option
- C10Answer option
- D13Answer option
Why this answer is right
Add to find the new total: $8+3=11$.
Total after gaining items is initial plus added. Compute $8+3=11$. So she has $11$ stickers.
Sample question
Find the highest common factor of $56$ and $98$.
- A$14$Correct answer
- B$7$Answer option
- C$28$Answer option
- D$21$Answer option
Why this answer is right
$56=2^3\cdot7$, $98=2\cdot7^2$. Common lowest powers: $2^1\cdot7^1=14$.
HCF uses the smallest powers of common primes. $56=2^3\cdot7$ and $98=2\cdot7^2$. Common primes: $2$ and $7$ with least powers $2^1$ and $7^1$. So $$\text{HCF}=2\cdot7=14.$$
Sample question
Which statement is always true for all integers $n$?
- A$n^2-1$ is divisible by $6$Answer option
- B$n^3+n$ is divisible by $6$Answer option
- C$n^3-n$ is divisible by $6$Correct answer
- D$n^4-1$ is divisible by $8$Answer option
Why this answer is right
$n^3-n=n(n-1)(n+1)$ is the product of three consecutive integers, so it is even and divisible by $3$, thus divisible by $6$.
Factor $$n^3-n=n(n^2-1)=n(n-1)(n+1).$$ Among three consecutive integers, one is even and one is a multiple of $3$. Therefore the product is a multiple of $2\cdot 3=6$. Other claims fail for some $n$ (e.g., $n=2$ makes $n^4-1=15$, not a multiple of $8$).
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.