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Year 10 · Measurement and scale · 18 min

The effect of rounding and repeated approximations

Retain exact quantities and guard digits until the final calculation.

01 / LEARN

The idea, step by step

Round once, not at every step

Rounded intermediate values feed later operations. Their errors can grow when multiplied or squared, or dominate when close quantities are subtracted. Retain exact fractions, roots and pi until the end where possible.

On a calculator use the stored result rather than retyping a short display value. This prevents avoidable numerical error, although it does not improve an imprecise original measurement.

Retaining the exact intermediate quantity
CalculationExact resultEarly rounded result
3 × (10/3)103 × 3.33 = 9.99
6 × (1/3)26 × 0.33 = 1.98
(√50)²507.07² = 49.9849

Match precision to the task

An exact fraction such as one third is not the same as its two-decimal approximation. A long final decimal cannot restore digits lost in an earlier step.

The final answer still needs an appropriate unit and sensible precision. Rough measured dimensions do not justify claiming a capacity correct to many decimal places.

A common trap

Assuming later decimal places recover digits discarded earlier.

Instead: Keep exact values or guard digits through the chain.

02 / WORKED EXAMPLES

See the thinking

Example 1

Cancellation first

Compare 3×(10/3) exactly with 3×3.33.

  1. Exact cancellation gives 10.
  2. Early rounding gives 9.99.
  3. The rounded result is 0.01 too small.
10 versus 9.99
Example 2

Retaining pi

A circle has exact radius 10 cm. Compare circumference using pi and using 3.14.

  1. C=20⁢π≈62.831853 cm.
  2. Using 3.14 gives 62.8 cm.
  3. The difference is about 0.03185 cm.
20⁢π cm; about 0.03185 cm difference

Check your understanding

Compare 6×(1/3) and 6×0.33.

Show answer

2 and 1.98

One third is exact; 0.33 is approximate.

03 / WORKSHEET

Your turn

6 questions · 13 marks · Work at your own pace.

Read the questions online
  1. Compare 7×(1/7) exactly with 7×0.14.
    2 marks
  2. Compare(√50)² exactly with 7.07².
    2 marks
  3. Calculate 4×(2/3) exactly and compare with 4×0.67.
    2 marks
  4. Find the circumference of an exact-radius 4 cm circle to 2 decimal places, rounding at the end.
    2 marks
  5. Compare 100√5 with 100×2.24, both to 2 decimal places.
    2 marks
  6. A ruler gives 4.2 cm to the nearest 0.1 cm. Does reporting a derived area to 8 decimal places make that measurement more precise?
    3 marks
Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

  1. Compare 7×(1/7) exactly with 7×0.14.
    2 marks

    Exact factors cancel.

    7×0.14=0.98.

    1 and 0.98
  2. Compare(√50)² exactly with 7.07².
    2 marks

    The exact square root squares to 50.

    7.07²=49.9849.

    50 and 49.9849
  3. Calculate 4×(2/3) exactly and compare with 4×0.67.
    2 marks

    The exact result is 8⁢/3.

    4×0.67=2.68.

    83 and 2.68
  4. Find the circumference of an exact-radius 4 cm circle to 2 decimal places, rounding at the end.
    2 marks

    8⁢π≈25.132741.

    Round once to 25.13.

    25.13 cm
  5. Compare 100√5 with 100×2.24, both to 2 decimal places.
    2 marks

    100√5≈223.60679.

    2.24×100=224.00.

    223.61 and 224.00
  6. A ruler gives 4.2 cm to the nearest 0.1 cm. Does reporting a derived area to 8 decimal places make that measurement more precise?
    3 marks

    The length remains uncertain by up to 0.05 cm.

    Extra arithmetic digits cannot remove instrument uncertainty.

    No.
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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum

Surface area of composite solids; volume of composite solids; proportion/scaling of objects; related length/area/volume factors; measurement error; the effect of rounding and repeated approximations; practical design models; evaluating assumptions and suitable precision.

Official curriculum source ↗
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