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This page sets out the method for error intervals in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Error intervals topic page.
When you see 'rounded to the nearest …' or 'correct to … decimal places / significant figures', find half a unit.
When you see 'truncated to …', the written value is the lower bound.
If a length is 8 cm to the nearest cm, the unit is 1 cm.
The true length can be up to 0.5 cm either side.
The lower bound is 7.5 cm. The upper bound is 8.5 cm.
Write 7.5 ≤ length < 8.5. The upper end is not included, because 8.5 would round up.
To the nearest 10, half the unit is 5.
70 to the nearest 10 sits from 65 up to but not including 75.
To the nearest 0.1, half the unit is 0.05.
The bounds move in by a smaller amount.
A number of people rounded to the nearest 10 is still a whole number.
The continuous rule is for measurements. Counts use the whole numbers that round to the recorded value.
A maximum safe load uses the upper bound of the weight.
A minimum length of cable uses the lower bound.
The rounding unit is the step the question names. Nearest centimetre means a half-unit of 0.5. Nearest 10 means a half-unit of 5. Nearest tenth, which is 1 decimal place, means a half-unit of 0.05. Nearest 100 means a half-unit of 50. Write that half down before you touch the recorded number.
Subtract the half-unit for the lower bound and add it for the upper bound. For 8 cm to the nearest centimetre, 8 - 0.5 = 7.5 and 8 + 0.5 = 8.5. For 70 to the nearest 10, 70 - 5 = 65 and 70 + 5 = 75.
Do not use the whole unit. An error of 1 cm on a reading of 8 cm would run from 7 to 9, twice the true width. Do not use a half-unit of 0.5 when the unit is 10. The number 70 to the nearest 10 is not from 69.5 to 70.5. That thinner interval is for the nearest whole number.
Method: 1. Read the accuracy in the question and call that the rounding unit. 2. Halve the unit to find how far the true value can sit from the recorded value. 3. Use that half-unit on both sides of the recorded number.
The lower bound is part of the interval. 7.5 to the nearest whole number rounds up to 8, so it is a possible true value. The upper bound is not. 8.5 rounds up to 9, and it starts the next interval. Use ≤ on the left and < on the right: 7.5 ≤ x < 8.5.
For 70 to the nearest 10 the interval is 65 ≤ n < 75. A value of 65 rounds to 70, so it stays in. A value of 75 rounds to 80, so it stays out. Two equal signs would keep 75. Two strict signs would drop 65.
Read the inequality before you finish: at least the lower bound, and strictly below the upper bound. Two ≤ signs means the upper end is wrong. Two < signs means the lower end is wrong. For 6.2 to the nearest tenth the half-unit is 0.05, so the interval is 6.15 ≤ x < 6.25.
Method: 1. Include the lower bound, because that value still rounds to the recorded number. 2. Exclude the upper bound, because that value rounds up to the next number. 3. Write the interval as lower bound ≤ value < upper bound.
Truncation cuts off extra digits. It does not look at the next digit and it does not round up. 3.41, 3.47 and 3.49 all truncate to 3.4 at 1 decimal place. Anything below 3.4, such as 3.39, truncates to 3.3 or less. The kept digits are the lower bound, and they are included.
The upper bound is the next number in that place. One more tenth is 0.1, and 3.4 + 0.1 = 3.5. A true value of 3.5 would be written as 3.5, so it is not included. The interval is 3.4 ≤ x < 3.5.
Half of 0.1 is 0.05, and 3.4 ± 0.05 gives 3.35 ≤ x < 3.45. That is the interval for 3.4 rounded to 1 decimal place, not truncated. A truncated reading never sits below the digits that were kept. Rounded uses a half-unit either side. Truncated uses the recorded value as the lower bound and only moves up.
Method: 1. Truncation cuts digits off, so the recorded value is the lower bound. 2. The upper bound is one unit of that accuracy higher. 3. Do not look at a decider digit, because nothing was rounded.
A value is inside when it is greater than or equal to the lower bound and strictly less than the upper bound. For 7.5 ≤ x < 8.5, the value 7.5 is inside, 8.49 is inside, and 8.5 is outside. A value of 7.49 fails the lower test, so it rounds to 7. Both comparisons have to hold.
A subtraction shows the gap. 8.5 - 8.49 = 0.01, so 8.49 is still below the upper bound. There is no positive gap under 8.5 itself, so 8.5 is outside. For a truncated reading of 3.4, 3.49 - 3.4 = 0.09 and 3.5 - 3.49 = 0.01, so 3.49 is inside. 3.5 matches the upper bound, so it is outside.
If the question asks for a limit, pick the bound it needs. A shortest possible length uses the lower bound. A greatest possible mass uses the upper bound. The upper bound itself is not a possible true value, but every possible true value sits just below it, so it is the limit you use in a calculation.
Method: 1. A value is possible if it is at least the lower bound and below the upper bound. 2. Test the lower bound itself: it is included. 3. Test the upper bound itself: it is not included.
The rule is the same whatever the rounding is: find the size of one unit in the last place that was kept, halve it, then go that far below and above the rounded value. To 1 decimal place the last place is a tenth, so go 0.05 either way. To 2 decimal places go 0.005. A value of 4.8 correct to 1 decimal place lies in 4.75 ≤ x < 4.85.
Significant figures need one extra thought. For 380 to 2 significant figures the last kept digit is the 8, which is worth ten, so the half-unit is 5. The error interval is 375 ≤ y < 385. The zero in 380 is only a place holder. Students often write 379.5 ≤ y < 380.5 because they treat 380 as rounded to the nearest whole number. Work out what the last kept digit is worth first.
Method: 1. Identify the last digit that was kept and what it is worth, such as tenths, tens or hundreds. 2. Halve that value to get the half-unit. 3. Subtract the half-unit for the lower bound and add it for the upper bound. 4. Write the lower bound with ≤ and the upper bound with <.
Tip: For significant figures, ask what the last non-placeholder digit is worth. In 380 to 2 significant figures it is worth 10.
Real questions ask for a "minimum possible" value or "which of these could be the true value". A crowd of 6 300 to the nearest 100 has a minimum possible size of 6 250, which is the lower bound. That value does count, because the lower bound is included. For the maximum, you can write the upper bound 6 350 as a limit, but it is not itself possible, so say "less than 6 350".
If the quantity can only be a whole number, such as a number of people, the largest actual value is one less than the upper bound. The number k is a whole number and 70 to the nearest 10 means 65 up to 74. To test a possible value, check it against both bounds. A value of 75 fails because it equals the upper bound, and 64 fails because it is below the lower bound.
Method: 1. Write the error interval first. 2. For a minimum, use the lower bound, which is included. 3. For a maximum of a measurement, use the upper bound as a limit and say "less than". 4. For whole numbers, the largest value is one less than the upper bound.
When several equal sides are added, the error grows with the number of sides. A regular hexagon with side 5.2 cm to 1 decimal place has a side in 5.15 ≤ s < 5.25. The perimeter is six sides, so multiply both bounds by 6 to get 30.9 ≤ P < 31.5. The perimeter is not 31.2 plus or minus 0.05.
A common slip is to give the perimeter the same half-unit as one side. Another is to find the perimeter first and then round. Always find the bounds of one side, then multiply. Check by asking if the interval for the perimeter is wider than the interval for a side. It must be, because six small errors add up.
Method: 1. Write the error interval for one side. 2. Count how many equal sides the shape has. 3. Multiply both bounds by that number. 4. Check that the new interval is wider than the one for a single side.
Higher questions ask for the largest or smallest result of a calculation with rounded measurements. For an addition, the largest total uses both upper bounds and the smallest total uses both lower bounds. For a subtraction, the largest result uses the upper bound of the first number and the lower bound of the number being taken away.
This catches a lot of students out in "length left over" problems. If a reel of rope is 50 m to the nearest metre and a piece 3.4 m to the nearest 10 cm is cut off, the most rope that can be left is 50.5 - 3.35 = 47.15 m. For a rectangle, the greatest area multiplies the greatest length by the greatest width. Write which bound you used for each number.
Method: 1. Write the error interval for each rounded value. 2. Decide which bound makes the answer as large or as small as the question asks. 3. For a subtraction, pair the upper bound of the first number with the lower bound of the second. 4. Do the calculation and state whether the answer is a bound or a possible value.
Tip: Label each number as UB or LB beside the working. It stops the wrong pairing in subtractions.
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