When practice shows the same wrong answer in different topics, look for a small skill underneath. In Additional Science, the most common one is rearranging a formula.
This lesson belongs to SPM subjects and learning resources. After it, return to the original mixed practice and retry the calculation questions.
How do I trace a wrong answer back to its cause?
Put the wrong answer beside the working and ask three questions.
- Was the science idea correct? If so, the gap is not the topic.
- Were the numbers read correctly from the question, with units?
- Did the formula get turned around correctly?
If the answer to the third question is no, the problem is algebra and not science. That is good news, because it is quick to fix.
Worked example: density, step by step
Question (original): A piece of aluminium has mass 540 g. The density of aluminium is 2.7 g/cm³. Find its volume.
The formula is density = mass ÷ volume. The unknown is volume, so it has to be isolated.
- Write the formula: 2.7 = 540 ÷ V.
- Multiply both sides by V: 2.7 × V = 540.
- Divide both sides by 2.7: V = 540 ÷ 2.7.
- Calculate: V = 200 cm³.
Check by putting the answer back: 540 ÷ 200 = 2.7. It matches the given density.
The mistake that costs marks
The common slip is to multiply the two numbers given: 540 × 2.7 = 1458. The working looks like a formula, but it answers a different question.
| Step | Wrong | Right |
|---|---|---|
| Formula used | volume = mass × density | volume = mass ÷ density |
| Sense check | 1458 cm³ for a 540 g block | 200 cm³ for a 540 g block |
| Units | Nothing to check against | g ÷ (g/cm³) = cm³ |
A quick sense check catches this. A metal heavier than water, about 1 g/cm³, should take up less volume than its mass in grams. So 540 g must be less than 540 cm³, which rules out 1458.
Where the same skill returns
The pattern of three linked quantities appears across the syllabus.
| Topic | Formula | Rearranged for the unknown |
|---|---|---|
| Speed | speed = distance ÷ time | time = distance ÷ speed |
| Concentration | amount = concentration × volume | volume = amount ÷ concentration |
| Electricity | V = I × R | R = V ÷ I |
Doing the same operation to both sides works for each. Once that habit is firm, the three formulas feel like one idea.
Check yourself
A solution has concentration 0.50 mol/dm³. How many dm³ of it contain 0.20 mol of solute?
Answer
The formula is amount = concentration × volume, so 0.20 = 0.50 × V. Divide both sides by 0.50: V = 0.20 ÷ 0.50 = 0.40 dm³.
Check: 0.50 × 0.40 = 0.20. Also, 0.20 mol is less than one 0.50 mol portion, so the volume must be less than 1 dm³. A result of 0.10 or 2.5 would fail this test.
What to study next
Return to the mixed practice set and retry Questions 2, 4 and 6. Use the structured answer self-review tool to check that your working shows the formula, the rearranged form and the units.
For a teacher to go through your own reviewed practice and find the first gap, see online one-to-one Additional Science tuition.