GCSE Maths Practice: decimals

Question 3 of 10

A demanding multi-step decimals question: multiply, subtract, then divide by a small decimal and round the final result to 3 significant figures.

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\( \begin{array}{l} \textbf{Calculate } ((0.625\times0.8)-0.047)\div0.09,\\ \textbf{giving your answer to 3 s.f.} \end{array} \)

Choose one option:

Estimate: (≈0.5−0.05)=0.45 and 0.45÷0.09≈5. So an answer near 5 makes sense. Then calculate exactly and round to 3 significant figures.

This Higher-tier GCSE item is a pure-decimals question that chains three skills: multiplication, subtraction, and division by a small decimal, followed by rounding to significant figures. The difficulty lies in keeping full precision through the intermediate steps and handling the division by 0.09 without losing place value.

Core Technique

When dividing by a decimal, it is often safer to clear the decimals first. For example, \(0.453\div0.09\) can be rewritten by multiplying numerator and denominator by 100: \(\dfrac{45.3}{9}\). Equivalently, convert to fractions: \(\dfrac{453}{1000}\div\dfrac{9}{100}=\dfrac{453}{1000}\times\dfrac{100}{9}=\dfrac{453}{90}=\dfrac{151}{30}\). This shows the exact value is a recurring decimal, \(5.0333\ldots\), which you then round to the required accuracy.

Worked Structure

  1. Multiply: \(0.625\times0.8=0.5\). (You can also see this as fractions: \(\tfrac{5}{8}\times\tfrac{4}{5}=\tfrac{1}{2}\).)
  2. Subtract: \(0.5-0.047=0.453\). Keep all digits; do not round yet.
  3. Divide: \(0.453\div0.09\). Clear decimals → \(\dfrac{45.3}{9}=5.0333\ldots\)
  4. Rounding to 3 s.f.: The first three significant digits are 5, 0, 3; the next digit is 3, so it stays \(\mathbf{5.03}\).

Why This Is Genuinely Higher

The operations are not hard individually, but the sequence demands discipline: exact arithmetic first, rounding last. Division by a small decimal (<1) should increase the size of the number; recognising this helps you sanity-check the result.

Common Mistakes

  • Premature rounding: rounding 0.453 to 0.45 before dividing will shift the final answer.
  • Decimal drift in division: treating 0.453 ÷ 0.09 as 453 ÷ 9 (forgetting to scale both by the same power of ten) gives 50.3… — an order-of-magnitude error.
  • Confusing s.f. with d.p.: 3 significant figures is not the same as 3 decimal places. For 5.033…, 3 s.f. → 5.03; 3 d.p. would be 5.033.
  • Estimation ignored: (≈0.5−0.05)=0.45; 0.45 ÷ 0.09 ≈ 5, so any result near 5 is reasonable. An answer near 50 or 0.5 indicates a place-value mistake.

Parallel Examples (different numbers)

Example A: \((0.84\times0.15-0.012)\div0.03\). 0.84×0.15=0.126; −0.012=0.114; ÷0.03=3.8 (exact).
Example B: \((1.2\times0.4+0.037)\div0.08\). 1.2×0.4=0.48; +0.037=0.517; ÷0.08=6.4625 → 3 s.f. = 6.46.
Example C: \((0.39\times2.4-0.18)\div0.06\). 0.39×2.4=0.936; −0.18=0.756; ÷0.06=12.6.

FAQ

Q1: How do I clear decimals correctly when dividing?
A1: Multiply both numerator and denominator by the same power of ten until the divisor is an integer. This leaves the quotient unchanged.

Q2: When do I round?

A2: Only at the end, after all operations. Intermediate rounding risks moving the final answer outside acceptable tolerance.

Q3: What’s a quick mental check for dividing by 0.09?

A3: Dividing by 0.09 is the same as multiplying by about 11.111… (since \(1/0.09=11\overline{1}\)). If your pre-division value is ~0.45, expect a result around 5.

Study Tip

Box each intermediate value and write a tiny note above it (×, −, ÷). This reduces cognitive load and makes it easy to backtrack if the final magnitude looks off. Always perform a one-line estimation at the end to catch decimal-point slips.