This set has ten original questions that mix biology, chemistry and physics contexts. All data is invented for practice and is not from any exam paper. Attempt each question on paper, write the unit on every line, then open the answer.
The questions move from easier to harder. Take the whole set in one sitting without a calculator for the first six, then use a calculator for the rest if you need one. The module overview is three-science numerical fluency.
Questions
1. A cyclist travels at 54 km/h. Give the speed in m/s.
Show answer
Divide by 3.6: 54 ÷ 3.6 = 15 m/s. Check: 15 × 3.6 = 54. ✓
2. A student uses 400 cm³ of a 0.25 mol/dm³ solution. How many moles are in that volume?
Show answer
400 cm³ = 0.400 dm³. Moles = 0.25 × 0.400 = 0.100 mol.
3. A cell drawing is 42 mm long. The real cell is 0.35 mm long. Find the magnification.
Show answer
Both lengths are already in mm. Magnification = 42 ÷ 0.35 = 120. Check: 0.35 × 120 = 42. ✓
4. A reaction gives off 150 cm³ of gas in 2.5 minutes at a steady rate. Give the rate in cm³/s.
Show answer
2.5 min = 150 s. Rate = 150 ÷ 150 = 1.0 cm³/s.
5. A kettle has a power of 2000 W and runs for 3 minutes. How much energy does it transfer, in kJ?
Show answer
3 min = 180 s. Energy = 2000 × 180 = 360 000 J = 360 kJ.
6. At rest, a student’s heart beats 75 times per minute and pumps 64 cm³ of blood per beat. Find the volume pumped per minute and the total volume in 5 minutes.
Show answer
Per minute: 75 × 64 = 4800 cm³/min. In 5 minutes: 4800 × 5 = 24 000 cm³. The first is a rate and the second is a total.
7. A distance-time graph is a straight line from (0 s, 0 m) to (12 s, 30 m). Find the gradient and say what it represents.
Show answer
Gradient = 30 ÷ 12 = 2.5 m/s. It represents the speed of the object.
8. A velocity-time graph passes through (0 s, 3 m/s) and (6 s, 21 m/s) in a straight line. Find the acceleration.
Show answer
Change in velocity = 21 − 3 = 18 m/s. Change in time = 6 s. Acceleration = 18 ÷ 6 = 3 m/s².
9. A stopwatch has a limit of ±0.2 s. Give the percentage uncertainty for a time of 8.0 s and for a time of 2.0 s. What does the comparison show?
Show answer
For 8.0 s: 0.2 ÷ 8.0 × 100 = 2.5%. For 2.0 s: 0.2 ÷ 2.0 × 100 = 10%. The same limit matters four times more on the shorter time, so short timings are less reliable.
10. A student obtains 7.2 g of product when the theoretical maximum is 6.0 g. Calculate the percentage yield and comment.
Show answer
7.2 ÷ 6.0 × 100 = 120%. A yield above 100% is not possible in a correct calculation, so recheck the data and working. In an invented lab result, the sample may still be wet or impure.
If you got these wrong
| Type of error | Where it shows | Go back to |
|---|---|---|
| Mixed or unconverted units | Questions 1, 2, 3, 4 | Carry units consistently across mixed tasks |
| Rate given as total, or the reverse | Questions 4, 5, 6 | Distinguish a rate from a total amount |
| Gradient wrong or unexplained | Questions 7, 8 | Read a graph gradient and explain its meaning |
| Limits and percentages | Question 9 | Propagate a stated measurement limitation conceptually |
| No sense-check | Question 10 | Check a numerical conclusion against the physical situation |
Record each slip in the mistake log and retest queue and test your reasoning on an invented investigation with the scientific investigation critic.
What next?
When the set feels comfortable, take the next module on cross-topic experimental evaluation, where these number habits support longer evaluation answers.
If one error type keeps returning even after you review the lesson, online one-to-one Co-ordinated Sciences tuition lets a teacher watch you work through new questions and fix the habit at its source.