Degree of Accuracy

Decimal places (d.p.). Count digits AFTER the decimal point.

• $3.14159$ to $2$ d.p.: look at 3rd d.p. ($1$) → round down → $3.14$.
• $3.14159$ to $3$ d.p.: look at 4th d.p. ($5$) → round UP → $3.142$.

Significant figures (s.f.). Start counting at the FIRST non-zero digit.

• $0.0067$ has $2$ s.f. ($6, 7$).
• $0.0670$ has $3$ s.f. ($6, 7, 0$ — trailing zero after decimal IS significant).
• $5670$ has $3$ s.f. (trailing zeros in integers are AMBIGUOUS — usually $4$ or $3$).
• $5670.0$ has $5$ s.f. (decimal forces all trailing zeros to count).

Rules:

1. Leading zeros (before first non-zero) are NEVER significant.
2. Zeros between non-zero digits ARE significant. ($304$ has $3$ s.f.)
3. Trailing zeros after a decimal ARE significant. ($30.0$ has $3$ s.f.)
4. Trailing zeros in integers are ambiguous — treat per context.

Round half up. When the digit being rounded is exactly $5$, round UP.

• $0.65$ to $1$ d.p. → $0.7$.
• $1.85$ to $1$ d.p. → $1.9$.

Worked qualitative. Why is $0.0067$ said to have $2$ s.f., not $5$?

• Significant figures convey ACTUAL PRECISION, not how many digits are written.
• The leading zeros just mark the place value (telling us this is a small number) — they don't add precision.
• Two non-zero digits = two significant figures.

Edexcel tip. Always state both numbers when rounding: 'Look at the next digit ($X$); since $X \ge 5$/$X < 5$, round up/down'.

The rule: A value rounded to a unit $u$ lies within $\pm \dfrac{u}{2}$ of the stated value.

Stated as	Rounding unit $u$	Half-unit	Bounds
$7.4$ ($1$ d.p.)	$0.1$	$0.05$	$[7.35, 7.45)$
$7.40$ ($2$ d.p.)	$0.01$	$0.005$	$[7.395, 7.405)$
$250$ (nearest $10$)	$10$	$5$	$[245, 255)$
$250$ (nearest $50$)	$50$	$25$	$[225, 275)$
$5.6$ ($2$ s.f.)	$0.1$	$0.05$	$[5.55, 5.65)$

Notation: square bracket $[$ for inclusive (lower bound), parenthesis $)$ for exclusive (upper bound). Conventionally:

• LOWER bound is INCLUDED.
• UPPER bound is EXCLUDED (because $7.45$ would round UP to $7.5$).

Worked example. A measurement is given as $52$ kg to the nearest kg. Find bounds.

$u = 1$ kg. Half-unit $= 0.5$ kg.

Lower bound = $51.5$ kg. Upper bound = $52.5$ kg.

Worked qualitative. Why is the upper bound 'excluded'?

• $52.5$ would round UP to $53$ (because of round half up).
• So if the original value was exactly $52.5$, it would be reported as $53$, not $52$.
• The actual range of values reported as $52$ is $[51.5, 52.5)$.

Edexcel tip. State the rounding unit and halve it explicitly. 'Rounded to $1$ d.p., so half-unit is $0.05$.' This earns the M1 mark.

Why this matters. When computing $a \times b$ where both are rounded, the answer has its own range. To find the range, combine the bounds correctly.

Addition. Both bounds add naturally:

• $UB(a + b) = UB(a) + UB(b)$.
• $LB(a + b) = LB(a) + LB(b)$.

Subtraction. REVERSE one bound:

• $UB(a - b) = UB(a) - LB(b)$.
• $LB(a - b) = LB(a) - UB(b)$.

(Think: biggest gap when first is largest, second is smallest.)

Multiplication. Both same direction (for positive values):

• $UB(a \times b) = UB(a) \times UB(b)$.
• $LB(a \times b) = LB(a) \times LB(b)$.

Division. Mixed (for positive values):

• $UB(a \div b) = UB(a) \div LB(b)$.
• $LB(a \div b) = LB(a) \div UB(b)$.

Memory aid.

• Want big? Make top big, bottom small.
• Want small? Make top small, bottom big.

Worked example. $a = 8.2$ ($1$ d.p.), $b = 4.5$ ($1$ d.p.). Find UB and LB of $a/b$.

Bounds: $a \in [8.15, 8.25]$, $b \in [4.45, 4.55]$.

• $UB(a/b) = 8.25 / 4.45 = 1.853\ldots$
• $LB(a/b) = 8.15 / 4.55 = 1.791\ldots$

So $a/b \in [1.791, 1.854]$ approximately.

Worked qualitative. Why does division use 'big over small' for the upper bound?

• Bigger numerator → bigger answer.
• Smaller denominator → bigger answer.
• For maximum, want both effects: max numerator, min denominator.

Edexcel tip. Set up bounds explicitly: $a \in [\ldots, \ldots]$, $b \in [\ldots, \ldots]$. Then choose the right combination. Mark schemes credit each bound calculation step.