GCSE Maths Practice: decimals

Question 10 of 10

This Higher-tier decimals problem involves three different operations—multiplication, subtraction, and division—then requires rounding to 3 significant figures. It rewards precise step-by-step working and good estimation skills.

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\( \begin{array}{l}\textbf{Calculate } ((0.72\times0.65)-0.017)\div0.934,\\\textbf{giving your answer to 3 significant figures.}\end{array} \)

Choose one option:

Estimate first: (0.7×0.7−0.02)≈0.47; 0.47÷0.93≈0.5. A final answer around 0.48 is therefore realistic.

This Higher-tier GCSE question is designed to test precision with decimals when several operations are combined. It includes multiplication, subtraction, division, and rounding to significant figures, reflecting the kind of multi-step calculation found in challenging exam questions.

Concept Overview

Each operation involving decimals carries the risk of small rounding or alignment errors. The key to full-mark performance is to retain all digits until the end, then round only once according to the question’s instructions. Here, a mixture of place values and significant figures appears, requiring accuracy and conceptual control.

Step-by-Step Breakdown

  1. Multiply accurately: \(0.72 \times 0.65 = 0.468.\) Both factors are less than 1, so the product must be smaller than either number.
  2. Subtract carefully: \(0.468 - 0.017 = 0.451.\) Align decimal points and work from right to left. Even a 0.001 error will change the third significant figure.
  3. Divide: \(0.451 \div 0.934 = 0.48286...\) Because the divisor is just below 1, the quotient should be slightly greater than the dividend, which helps check reasonableness.
  4. Round to 3 significant figures: The first three non-zero digits are 4, 8, and 3, giving \(\mathbf{0.483.}\)

Reasoning and Estimation

Estimate to confirm: 0.7 × 0.7 ≈ 0.49, minus 0.02 gives ≈ 0.47, then ÷ 0.93 ≈ 0.50. The estimate (0.5) is close to the exact answer (0.483), so the final result is consistent and realistic.

Why This Is Higher Level

This problem tests sequencing (following BIDMAS with decimals), rounding to significant figures, and interpreting results. Learners must judge whether the magnitude of each intermediate step is reasonable and keep track of the effect of dividing by a number close to one.

Common Errors

  • Rounding early after multiplication or subtraction, leading to cumulative error.
  • Confusing 3 significant figures with 3 decimal places; they are not the same.
  • Assuming dividing by a number less than 1 always decreases the result — it increases it slightly in this case.
  • Misplacing the decimal point when using a calculator or writing intermediate steps.

Worked Comparisons

Example A: \((0.84\times0.56 - 0.025)\div0.92 = 0.48\) (to 2 d.p.)
Example B: \((1.08\times0.32 - 0.014)\div0.69 = 0.471\) (to 3 s.f.)

Both examples require consistent rounding and alignment of decimals during subtraction and division.

FAQ

Q1: What’s the difference between decimal places and significant figures?
A1: Decimal places count digits after the decimal; significant figures count from the first non-zero digit. For example, 0.483 has three significant figures but only three decimal places by coincidence.

Q2: When dividing by a number close to 1, should the result be larger or smaller?
A2: If the divisor is less than 1, the quotient increases; if it’s greater than 1, it decreases. Always estimate first.

Q3: Can I use a calculator?
A3: Yes, but show structure and final rounding clearly. Examiners award marks for method and accuracy, not just the number displayed.

Study Tip

Carry all digits through to the end and round once. When questions require significant figures, identify the first non-zero digit before rounding. Double-check by estimation to ensure your answer is of the correct order of magnitude.

This style of question reinforces full decimal fluency, precision rounding, and logical estimation — key hallmarks of top-grade performance in GCSE Maths.