Number and ratio · GCSE Maths
Bounds and error intervals
GCSE Maths error intervals: write upper and lower bounds for rounded or truncated values, then find the maximum or minimum of a calculation from those bounds.
A value rounded to the nearest unit can be up to half a unit either way. Truncation only goes down. For a calculation, max / min comes from combining bounds, not from rounding the answer twice.
The important bits
What you need to know
- 1
If x is rounded to the nearest 10, the error interval is 5 units either side: 75 to the nearest 10 means 70 ≤ x < 80, or 70 ≤ x ≤ 80 depending on how the paper writes the upper bound.
- 2
Lower bound is the smallest value that still rounds to the given number. Upper bound is halfway to the next rounding step. To the nearest 0.1, 3.6 means 3.55 ≤ x < 3.65.
- 3
Truncation is not rounding. 3.69 truncated to 1 d.p. is 3.6, and the true value satisfies 3.6 ≤ x < 3.7. There is no “half a unit down” on the lower side.
- 4
Write the inequality with the degree of accuracy in the question. “To the nearest centimetre” on a length of 12 cm is 11.5 ≤ L < 12.5, often written 11.5 ≤ L ≤ 12.5 in GCSE work.
- 5
Maximum of a sum uses both upper bounds. Minimum of a sum uses both lower bounds. Subtraction is the opposite pairing: max of a − b is upper a minus lower b.
- 6
For a product, maximum is upper × upper when both quantities are positive. For a quotient a ÷ b with positive values, maximum is upper a ÷ lower b, because dividing by a smaller number makes a larger result.
- 7
A discrete count (number of people, number of sweets) cannot be 12.5 people. Bounds on integers to the nearest 10 still use 5 either way, but interpret the context at the end.
- 8
Show the bounds before you calculate. The method mark sits on “UB = 12.5, LB = 11.5”, not on a lone maximum written from memory.
Quotations worth analysing
Short evidence. Real method.
“11.5 ≤ L < 12.5”
Half a centimetre either way. Some papers use ≤ on both ends. Use the inequality the question’s answer line expects, and keep both bounds visible.
“Maximum = upper ÷ lower”
To make a ÷ b as large as possible, make a as large as possible and b as small as possible. The opposite pairing gives the minimum.
“Truncated to 1 decimal place”
Truncation chops extra digits. 3.69 truncated to 1 d.p. is 3.6, so 3.6 ≤ x < 3.7, not 3.55 ≤ x < 3.65.
Go deeper
Half a unit, not a whole unit
The degree of accuracy is the size of the rounding step. Nearest 10 means the step is 10, so half a step is 5. A mass of 80 kg to the nearest 10 kg satisfies 75 ≤ m < 85. Nearest 0.1 means half a step is 0.05: 6.4 to 1 d.p. is 6.35 ≤ x < 6.45. Students who write 6.3 ≤ x ≤ 6.5 have used a whole tenth either way, which is the interval for truncation to 1 d.p., or for rounding to the nearest integer on a different number. Always write “half of the accuracy” as a tiny working line: nearest 100, half is 50; nearest 0.01, half is 0.005. Discrete rounding still uses this arithmetic; you then decide whether 12.5 people is allowed in the inequality the paper wants, or whether you jump to 13 in a later interpretation part.
Go deeper
Maximum and minimum of a calculation
A rectangle is measured as 12 cm by 8 cm, each to the nearest centimetre. Lower bounds 11.5 and 7.5, upper bounds 12.5 and 8.5. Maximum perimeter uses both uppers: 2 × (12.5 + 8.5) = 42 cm. Minimum area uses both lowers: 11.5 × 7.5 = 86.25 cm². Maximum area is 12.5 × 8.5 = 106.25 cm². For a speed calculated as distance ÷ time, with distance 150 m to the nearest 10 m and time 12 s to the nearest second, maximum speed is 155 ÷ 11.5, because the largest distance over the smallest time is the fastest. Minimum speed is 145 ÷ 12.5. Write the pairing in words: “max speed = UB distance ÷ LB time.” That sentence is the method. Computing 150 ÷ 12 and then rounding the answer is not bounds; it is a different question.
Go deeper
Rounded versus truncated
Rounding looks at the next digit and may go up. Truncation never goes up; it simply stops. 9.849 rounded to 2 d.p. is 9.85; truncated to 2 d.p. is 9.84. If a calculator display of 3.6 is known to be truncated to 1 d.p., the true value is in [3.6, 3.7). If 3.6 is rounded to 1 d.p., the true value is in [3.55, 3.65). Mixing these intervals is a Higher-tier favourite. When a question says “correct to 3 significant figures”, the same half-unit rule applies to the place of the last significant figure: 2.36 to 3 s.f. has bounds 2.355 and 2.365. Write both numbers before you start a max/min product so the pairing stays honest.
See the idea in action
A length is 12 cm to the nearest centimetre. A time is 5.0 s to 1 decimal place. Find the error interval for the length, and the maximum possible speed in cm/s if speed = length ÷ time. Step 1: Length L is to the nearest 1 cm, so half a unit is 0.5 cm. Step 2: 11.5 ≤ L < 12.5. Upper bound 12.5 cm, lower bound 11.5 cm. Step 3: Time T is to 1 d.p., so half a unit is 0.05 s. Step 4: 4.95 ≤ T < 5.05. Lower bound 4.95 s, upper bound 5.05 s. Step 5: Maximum speed = upper length ÷ lower time = 12.5 ÷ 4.95 = 2.52525… = 2.53 cm/s to 3 s.f. Check pairing: bigger distance, smaller time, faster speed. Minimum would be 11.5 ÷ 5.05 ≈ 2.28 cm/s.
Exam technique
Turn knowledge into marks
Write both bounds, then write which pairing you need in words (upper ÷ lower, and so on) before you calculate. The pairing is the method mark; the decimal is the accuracy mark.
Common mistakes
Do not give these marks away
- 01
Using a whole unit either side, such as 11 ≤ L ≤ 13 for 12 cm to the nearest centimetre, instead of half a unit.
- 02
Using both upper bounds for a quotient, which does not give the maximum when the denominator is positive.
- 03
Treating a truncated value as if it had been rounded, so the lower bound is wrongly taken half a unit below the displayed number.
A length of 12 cm is measured to the nearest centimetre. The lower bound is
A11.5 cm
B11 cm
C12.5 cm
D10 cm
Show the answer
11.5 cm. Half of 1 cm is 0.5 cm, so the lower bound is 12 − 0.5 = 11.5 cm. 11 cm is a whole centimetre down. 12.5 cm is the upper bound. 10 cm is the nearest 10, the wrong degree of accuracy.
Quick questions
If this is the bit you searched
How do you find upper and lower bounds GCSE?
Take half of the degree of accuracy. Subtract it for the lower bound and add it for the upper bound. For 8.4 to 1 d.p., half is 0.05, so 8.35 and 8.45.
What is the difference between rounded and truncated?
Rounding may increase the last kept digit. Truncation never does; extra digits are chopped. Truncated 3.6 to 1 d.p. means 3.6 ≤ x < 3.7, not 3.55 ≤ x < 3.65.
How do you find the maximum of a ÷ b using bounds?
If a and b are positive, maximum is (upper bound of a) ÷ (lower bound of b). Minimum is lower a ÷ upper b. Write the pairing before you divide.
Why do some mark schemes use ≤ on the upper bound and some use <?
Convention on the exact halfway case. Follow the inequality the question prints. The numerical bounds 11.5 and 12.5 are what you calculate with either way.