GCSE Maths Practice: decimals

Question 10 of 10

A challenging multi-step decimals problem: combine multiplication, subtraction and division, then round to the requested accuracy without losing precision.

‹

\( \begin{array}{l} \textbf{Calculate }(0.625\times4-0.95)\div1.5,\\ \textbf{giving your answer to 2 d.p.} \end{array} \)

Choose one option:

Estimate: (≈2.5 − 1) ÷ 1.5 ≈ 1 ÷ 1.5 ≈ 0.67. Closer tracking gives ≈1.03, so a result near 1.0 is sensible. Always round at the final step.

This Higher-tier GCSE task blends several decimal operations in sequence: multiplication, subtraction, division, and final rounding. Questions like this reward careful structure and place-value control more than raw speed.

Key Idea

When multiple decimal operations appear, keep full precision until the end. Rounding early propagates error and can change the final two decimal places.

Method Outline

  1. Multiply exactly: 0.625 × 4 = 2.5 (no rounding).
  2. Subtract carefully: 2.5 − 0.95 = 1.55.
  3. Divide with place-value control: 1.55 ÷ 1.5 = 1.0333…
  4. Round once at the end to the requested accuracy (2 d.p.).

Place-Value Discipline

Write intermediate results vertically when adding or subtracting decimals to preserve tenths/hundredths alignment. For multiplication and division, think in powers of ten: 1.55 ÷ 1.5 is equivalent to 155 ÷ 150 = 31 ÷ 30 = 1.0333…

Common Pitfalls

  • Premature rounding: turning 2.5 − 0.95 into 1.6 by rounding 0.95 to 1, which inflates the result.
  • Decimal drift in division: treating 1.55 ÷ 1.5 as 155 ÷ 15 (missed equal scaling), giving 10.33… instead of 1.033…
  • Dropping trailing zeros: writing 1.5 as 1.50 can clarify the subtraction.

Calculator vs Non-Calculator

Non-calculator: clear the divisor’s decimal by scaling both terms: \(\dfrac{1.55}{1.5}=\dfrac{155}{150}=\dfrac{31}{30}\). Long division gives 1.0333…
Calculator: enter in one line, but still record key steps to show method and to catch entry errors.

Estimation & Reasonableness

Quick check: 0.625 × 4 ≈ 2.5; subtract ~1 gives ~1.5; divide by ~1.5 gives ~1.0. So a result near 1.0 is plausible — 1.03 fits.

Worked Examples (distinct patterns)

Example A: (0.84 + 0.275) ÷ 1.1 → 1.115 ÷ 1.1 = 1.0136… ≈ 1.01 (2 d.p.)

Example B: (1.2 − 0.455) × 0.75 → 0.745 × 0.75 = 0.55875 ≈ 0.56 (2 d.p.)

Example C: (0.39 × 2.4 − 0.18) ÷ 0.6 → (0.936 − 0.18) ÷ 0.6 = 0.756 ÷ 0.6 = 1.26

FAQ

Q1: Why not round 1.55 ÷ 1.5 early?
A1: Because rounding mid-process can change the final two decimal places. Keep exact values until the final step.

Q2: How do I manage repeating decimals when rounding?
A2: Look at the third decimal place. For 1.033…, the third d.p. is 3, so 1.03 is correct to 2 d.p.

Q3: Any quick mental check for dividing by 1.5?
A3: Divide by 3 then multiply by 2: 1.55 ÷ 3 ≈ 0.5166…, ×2 ≈ 1.0333…

Study Tip

Box each intermediate result and label the operation (+, −, ×, ÷). This visual fence stops decimal-place errors and makes exam working crystal clear for method marks.