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Rounding, Estimation & BoundsEdexcel IGCSE Maths: Revision notes

Section 1

How do you round to decimal places and significant figures?

Decimal places (d.p.) count digits after the decimal point. Significant figures (s.f.) count from the first non-zero digit, reading left to right.

  • To round, look at the digit after the one you're keeping. If it is 5 or more, round up; otherwise round down.
  • Leading zeros (e.g. in 0.004560.00456) are never significant. Trailing zeros in a whole number (e.g. 24002400) may or may not be significant — context/rounding tells you.
Number2 d.p.3 s.f.
3.141593.141593.143.143.143.14
0.0205810.0205810.020.020.02060.0206
128456128456128456.00128456.00128000128000
Key termsdecimal placessignificant figures
Common mistake

Don't confuse d.p. and s.f. — 0.00480.0048 to 2 d.p. is 0.000.00, but to 2 s.f. it is 0.00480.0048 itself (already only 2 s.f.).

Section 2

Why do we estimate calculations?

Estimation rounds each number to 1 significant figure before calculating, giving a quick check that an exact answer is sensible.

Steps:

  1. Round every value in the calculation to 1 s.f.
  2. Carry out the simplified calculation.
  3. Compare the estimate to your exact answer — if they're wildly different, you likely made an arithmetic slip.

Example: estimate 58.7×3.920.51\dfrac{58.7 \times 3.92}{0.51}. Round: 60×40.5=2400.5=480\dfrac{60 \times 4}{0.5} = \dfrac{240}{0.5} = 480.

Key termsestimation
Example

Estimate 212×4.989.87\dfrac{212 \times 4.98}{9.87}: round to 200×510=100\dfrac{200 \times 5}{10} = 100.

Exam tip

Always show the rounded values used in an estimation question — marks are given for the method, not just the final number.

Section 3

What are upper and lower bounds?

When a measurement is rounded, the true value lies within a range. The upper bound (UB) and lower bound (LB) define that range.

For a value rounded to the nearest unit uu: add/subtract half of uu.

LB=x−u2,UB=x+u2\text{LB} = x - \dfrac{u}{2}, \qquad \text{UB} = x + \dfrac{u}{2}

Example: a length of 7.4 cm7.4\text{ cm} measured to the nearest 0.1 cm0.1\text{ cm} has u=0.1u = 0.1, so: LB=7.4−0.05=7.35 cm,UB=7.4+0.05=7.45 cm\text{LB} = 7.4 - 0.05 = 7.35\text{ cm}, \qquad \text{UB} = 7.4 + 0.05 = 7.45\text{ cm}

This is written as 7.35≤x<7.457.35 \leq x < 7.45. Note the upper bound uses a strict inequality (<<) because 7.457.45 would itself round up to 7.57.5.

Key termsupper boundlower bound
Think of it like this

Think of bounds like a safety net either side of a rounded number — the true value is guaranteed to be caught somewhere inside that range.

Section 4

How do you combine bounds in calculations?

When combining measurements, work out which combination of bounds gives the extreme result you want.

  • Maximum of a+ba + b: use UBa+UBb\text{UB}_a + \text{UB}_b
  • Minimum of a+ba + b: use LBa+LBb\text{LB}_a + \text{LB}_b
  • Maximum of a−ba - b: use UBa−LBb\text{UB}_a - \text{LB}_b
  • Minimum of a−ba - b: use LBa−UBb\text{LB}_a - \text{UB}_b
  • Maximum of a÷ba \div b: use UBa÷LBb\text{UB}_a \div \text{LB}_b
  • Minimum of a÷ba \div b: use LBa÷UBb\text{LB}_a \div \text{UB}_b

Multiplication follows the same pattern as division: maximum uses UBa×UBb\text{UB}_a \times \text{UB}_b, minimum uses LBa×LBb\text{LB}_a \times \text{LB}_b.

Common mistake

For subtraction and division, mixing up which bound goes where gives the wrong extreme — always ask which combination makes the result as large or as small as possible.

Section 5

How do bounds apply to area and speed problems?

Bounds questions often disguise themselves as area, volume or speed calculations.

  • Area of a rectangle =length×width= \text{length} \times \text{width}: maximum area uses UBlength×UBwidth\text{UB}_{\text{length}} \times \text{UB}_{\text{width}}.
  • Speed =distancetime= \dfrac{\text{distance}}{\text{time}}: maximum speed uses UBdistance÷LBtime\text{UB}_{\text{distance}} \div \text{LB}_{\text{time}} — travelling the furthest possible distance in the shortest possible time.

Always identify the formula first, then decide whether each variable needs its upper or lower bound for the extreme you're asked for.

Example

A car travels 150 km150\text{ km} (to nearest 10 km10\text{ km}) in 2 hours2\text{ hours} (to nearest hour). Maximum speed =1551.5=103.3 km/h= \dfrac{155}{1.5} = 103.3\text{ km/h} (1 d.p.).

Must Know

  • Decimal places count digits after the decimal point; significant figures count from the first non-zero digit.
  • To estimate, round every value to 1 s.f. before calculating.
  • For a value rounded to the nearest uu: LB=x−u2\text{LB} = x - \frac{u}{2} and UB=x+u2\text{UB} = x + \frac{u}{2}, written as LB≤x<UB\text{LB} \leq x < \text{UB}.
  • Addition/multiplication maximum uses both upper bounds; minimum uses both lower bounds.
  • Subtraction/division maximum uses (UB of first value) combined with (LB of second value); minimum uses the reverse.
  • Always state which bounds you used — method marks depend on it.

That's the notes covered.

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