Rounding, Estimation & BoundsCambridge IGCSE Maths: Revision notes

Decimal places (d.p.): count digits after the decimal point. Look at the next digit — round up if it is 5 or more, round down otherwise.

Significant figures (s.f.): the first significant figure is the first non-zero digit. Count from there, keeping placeholder zeros to preserve place value.

Example: 3.14159 to 2 d.p. = 3.14; to 3 s.f. = 3.14 Example: 0.002378 to 2 s.f. = 0.0024

To estimate, round every value in a calculation to 1 significant figure first, then calculate.

Example: Estimate $\frac{58.7 \times 21.3}{9.8}$ Round: $\frac{60 \times 20}{10} = \frac{1200}{10} = 120$

Estimation checks whether a calculator answer is sensible and is often required to show working without a calculator.

The context of a question determines a sensible degree of accuracy.

• Money: round to 2 decimal places (nearest cent/fils)
• Number of people, buses, boxes: round to a whole number, often rounding UP even if the decimal is below 0.5 (you can't have half a bus)
• Measurements: match the precision given in the question

When a measurement is rounded, the true value lies within a range called its bounds.

For a value rounded to the nearest unit $u$:

• Lower bound = rounded value − $u/2$
• Upper bound = rounded value + $u/2$

Example: A length is 12 cm to the nearest cm. Lower bound = 11.5 cm, Upper bound = 12.5 cm

The upper bound is not included in the range (it would round to the next value up), so bounds are often written as $11.5 \leq l < 12.5$

When combining rounded measurements, choose bounds carefully to get the extreme result:

• Maximum (upper bound) of a sum or product: use the upper bounds of all values
• Minimum (lower bound) of a sum or product: use the lower bounds of all values
• Maximum of a division/subtraction: use the upper bound of the numerator/first term and the lower bound of the denominator/second term
• Minimum of a division/subtraction: use the lower bound of the numerator/first term and the upper bound of the denominator/second term

Example: distance = 100 m (nearest m), time = 20 s (nearest s). Maximum speed = upper bound of distance ÷ lower bound of time = 100.5 ÷ 19.5 = 5.15 m/s (3 s.f.)