These ten questions mix the skills from cross-science data interpretation: converting units, choosing a graph, handling anomalies, testing proportionality and checking answers. All data is invented for practice.
Work on paper, show every step, then open the answer. Order runs from easier to harder. Log slips in the mistake log and retest queue, and use the scientific investigation critic to test your reasoning.
Questions
Q1 (units). One beaker holds 0.65 L of solution and another holds 600 mL. Which holds more, and by how much?
Show answer
Convert 0.65 L to 650 mL. The first beaker holds more, by 650 − 600 = 50 mL.
Q2 (percentage change). Dry mass of seedlings rose from 5.0 g to 6.2 g. Resting pulse in a student rose from 80 to 100 beats per minute. Which rose by the larger percentage?
Show answer
Mass: 1.2 ÷ 5.0 × 100 = 24%. Pulse: 20 ÷ 80 × 100 = 25%. The pulse rose by the larger percentage, though only by a small margin.
Q3 (graph choice). Name the suitable display for each: (a) number of seeds germinating in four soil types, (b) water temperature against time as it cools, (c) current against voltage for a resistor.
Show answer
(a) Bar chart, because soil type is a category. (b) Line graph with a smooth curve, because both variables are continuous. (c) Scatter graph with a straight line of best fit, because both variables are numbers.
Q4 (axis scale). Your temperatures run from 52 °C to 79 °C. Suggest a sensible y-axis range and step.
Show answer
Use 50 °C to 80 °C in steps of 5 °C. All points then fill most of the grid and the pattern is visible.
Q5 (anomaly). Titre values in cm³ are 23.10, 23.20, 24.60 and 23.15. Identify the anomaly and give the mean with and without it.
Show answer
24.60 is the anomaly, since the other three agree within 0.10 cm³. Without it: (23.10 + 23.20 + 23.15) ÷ 3 = 69.45 ÷ 3 = 23.15 cm³. With it: 94.05 ÷ 4 = 23.5125, which is 23.51 cm³. Report the mean of the three and state which value was left out and why.
Q6 (proportionality). A resistor gives these readings: voltage 2.0, 4.0, 6.0 V and current 0.5, 1.0, 1.5 A. Is current directly proportional to voltage? Find V ÷ I.
Show answer
V ÷ I: 2.0 ÷ 0.5 = 4.0, 4.0 ÷ 1.0 = 4.0, 6.0 ÷ 1.5 = 4.0. The ratio is constant, so yes, proportional, and the resistance is 4.0 Ω.
Q7 (trend only). Concentration of a reactant (mol/dm³) is 0.5, 1.0, 1.5, 2.0 and the rate (arbitrary units) is 1.0, 1.9, 2.6, 3.0. Is rate proportional to concentration?
Show answer
Rate ÷ concentration: 2.0, 1.9, 1.73, 1.5. The ratio falls steadily, so the rate increases but is not proportional. The increase levels off at higher concentration.
Q8 (independent check). Calcium carbonate has relative formula mass 100. Find the mass of 0.50 mol, then check by estimating.
Show answer
Mass = 0.50 × 100 = 50 g. Estimate: half of 100 is 50. Reverse: 50 ÷ 100 = 0.50 mol.
Q9 (energy with units). Water of mass 0.50 kg is heated by 10 °C. Specific heat capacity is 4200 J/(kg °C). Find the energy, then name one check.
Show answer
E = 0.50 × 4200 × 10 = 21 000 J. Check by estimate: 0.5 × 4000 × 10 = 20 000 J, which is close. Reverse: 21 000 ÷ (0.50 × 4200) = 10 °C.
Q10 (mixed conversion). A metal block has density 2.7 g/cm³ and volume 50 cm³. Find its mass in kg. A car travels at 90 km/h: give this in m/s.
Show answer
Mass = 2.7 × 50 = 135 g = 0.135 kg. Speed = 90 × 1000 ÷ 3600 = 25 m/s. Check: 25 × 3.6 = 90.
If you got these wrong
| Error type | Questions | Go to |
|---|---|---|
| Mixed units or raw numbers compared | Q1, Q2, Q10 | Comparing datasets with different units |
| Wrong graph, poor axis scale | Q3, Q4 | Selecting a graph type |
| Anomaly handled badly | Q5 | Explaining an anomaly |
| Trend confused with proportion | Q6, Q7 | Proportionality and trend |
| Unchecked numerical slips | Q8, Q9, Q10 | Independent answer checks |
What next?
Retry the questions you missed after a day, using changed numbers. If the same error type returns, that is a sign of a gap rather than a slip.
If your working is fine but your explanations in words lose marks, online one-to-one Combined Science tuition lets a teacher work from your own answers.