1. Rounding to a place value
The rule
- Locate the digit sitting in the place value being rounded to
- Look at the single digit immediately to its right, and ignore everything beyond it
- That digit decides the direction: 5 or more rounds up, 4 or less leaves the digit unchanged
- Fill every place between the rounded digit and the decimal point with zeros, since dropping them would change the size of the number
- 4826 to the nearest hundred is 4800
- 4872 to the nearest hundred is 4900
2. Rounding to decimal places
Counting the places
- Decimal places are counted rightwards from the decimal point
- Apply the same rule: inspect the first digit beyond the place being kept
- 6.7183 to two decimal places is 6.72
Keeping the trailing zero
- An answer must show exactly as many decimal places as were asked for, including a final zero
- 0.4961 to two decimal places is 0.50, not 0.5
- This differs from place-value rounding, where the zeros before the decimal point keep the size of the number, while a final zero after the point shows the accuracy
- Where rounding up cascades through a 9, carry it as with ordinary addition
- 3.897 to two decimal places is 3.90, because 89 becomes 90
Worked example
Rounding where the digits carry
Round 0.4961 to two decimal places.
Solution:
- The second decimal place holds 9, and the digit to its right is 6
- Since 6 is at least 5, the 9 rounds up, which carries into the tenths column
- 49 hundredths becomes 50 hundredths
- The answer is 0.50, written with the final zero because two decimal places were requested
3. Rounding to significant figures
Finding the first significant figure
- Reading left to right, the first significant figure is the first non-zero digit
- In 5081 it is the 5
- In 0.00407 it is the 4
- Zeros in front of that digit only position the number and are never significant
- Zeros sitting between non-zero digits are significant, because they carry information
- The second significant figure of 5081 is the 0
Rounding and padding
- Count rightwards to the required figure, then apply the usual rounding rule
- Large numbers are padded with zeros out to the decimal point
- 5081 to two significant figures is 5100
- For a decimal, zeros fill the gap from the point up to where the significant figures begin
- 0.00407 to two significant figures is 0.0041
Exam tip
Three significant figures unless told otherwise
On the calculator paper, an answer that is not exact goes to 3 significant figures, and an angle in degrees to 1 decimal place, whenever the question names no accuracy of its own; an exact answer stays in full. The schemes print such answers to 3 significant figures or longer, as in 4.27 or 4.268…, so a shorter 4.3 misses the final mark. The non-calculator paper sets no such default.
Worked example
Rounding to significant figures
Write 0.004 078 2 to 3 significant figures, and 46 851 to 2 significant figures.
Solution:
- The first significant figure is the first non-zero digit, so leading zeros never count
- In 0.004 078 2 the first significant figure is the 4, the second is the 0 after it, and the third is the 7
- The next digit is 8, which is 5 or more, so the 7 rounds up to 8
- 0.004 078 2 = 0.004 08 to 3 significant figures
- In 46 851 the first two significant figures are 4 and 6
- The next digit is 8, so the 6 rounds up to 7
- The place holders must stay, so the answer is 47 000, not 47
4. Estimation
Making the estimate
- Rounding every value to one significant figure turns an awkward calculation into one that can be done mentally
- 19.4 becomes 20 and 4.7 becomes 5
- When the question says 1 significant figure, round to exactly that, even where another rounding would make the arithmetic simpler
- Never round a value to zero: a zero denominator makes the calculation undefined, and a zero numerator destroys the estimate
Deciding whether the estimate is too big or too small
- For a sum or a product, rounding both values the same way settles it
| Operation | Both rounded up | Both rounded down |
|---|---|---|
| a + b | Overestimate | Underestimate |
| a × b | Overestimate | Underestimate |
- For a difference or a quotient the second value works in the opposite direction
| Operation | a up, b down | a down, b up |
|---|---|---|
| a − b | Overestimate | Underestimate |
| a ÷ b | Overestimate | Underestimate |
- Where the two roundings pull opposite ways, the direction cannot be decided without more care
Worked example
Estimating and judging the direction
Estimate the value of 19.4 × 4.7, and state whether the estimate is too large or too small.
Solution:
- Round each value to one significant figure: 19.4 becomes 20 and 4.7 becomes 5
- Multiply the rounded values: 20 × 5 = 100
- The estimate is 100
- Both values were rounded up, and this is a product, so the estimate is an overestimate
- The exact value is 91.18, which confirms the direction
5. Upper and lower bounds
What the bounds are
- A rounded value stands for a range of true values, and the bounds are the ends of that range
- The lower bound is the smallest value that would still round to the figure given
- The upper bound is the value the true figure must stay below
- Both lie half a unit of accuracy away from the rounded value, so they are the midpoints between it and its neighbours
- A true value may sit exactly on the lower bound, but reaching the upper bound would round it the other way, so only one end of the range is inclusive
Finding them
- Halve the degree of accuracy the value was rounded to
- Subtract that half-unit for the lower bound and add it for the upper bound
- A mass of 6.4 kg given to one decimal place has a half-unit of 0.05
- The bounds are 6.35 ⩽ m < 6.45
Worked example
Writing down a pair of bounds
A length is 8.6 cm, correct to 1 decimal place. Write down its lower and upper bounds.
Solution:
- 1 decimal place means the value was rounded to the nearest 0.1
- Half of 0.1 is 0.05, so the true length is within 0.05 cm of 8.6
- Lower bound = 8.6 − 0.05 = 8.55 cm
- Upper bound = 8.6 + 0.05 = 8.65 cm
- The lower bound is included and the upper bound is not, so 8.55 ⩽ length < 8.65
- 8.65 itself would round up to 8.7, which is why it is excluded even though it is written as the upper bound
6. Calculating with bounds
Choosing which bound to use
- A calculation using rounded data has bounds of its own, and each one comes from a particular combination of the input bounds
- Adding or multiplying: the largest result uses both upper bounds, the smallest uses both lower bounds
- Subtracting or dividing: the largest result uses the upper bound of the first value with the lower bound of the second, and the smallest reverses that
- Dividing by a smaller number gives a larger answer, which is the step most easily taken the wrong way round
Worked example
Bounds of an area
A rectangle measures 8 cm by 5 cm, each correct to the nearest centimetre. Find the upper and lower bounds of its area.
Solution:
- The half-unit for the nearest centimetre is 0.5 cm
- The length lies in 7.5 ⩽ l < 8.5 and the width in 4.5 ⩽ w < 5.5
- Area is a product, so the upper bound uses both upper bounds: 8.5 × 5.5 = 46.75 cm²
- The lower bound uses both lower bounds: 7.5 × 4.5 = 33.75 cm²
Exam tip
Write each measurement's bounds down first
Questions on the bounds of a calculation were asked in 10 of the 21 papers, and every one of those schemes gave a mark for a correct bound of a measurement seen in the working, such as 7.5 for 8 cm to the nearest centimetre. Write each bound down before you combine them: if you then pair them the wrong way or slip in the arithmetic, that mark is still yours.
Worked example
Lower bound of a density
A metal block has a mass of 90 g, correct to the nearest 5 g, and a volume of 12 cm³, correct to the nearest cm³. Calculate the lower bound of its density. [density = mass ÷ volume]
Solution:
- Half of 5 g is 2.5 g, so the mass lies in 87.5 ⩽ m < 92.5
- Half of 1 cm³ is 0.5 cm³, so the volume lies in 11.5 ⩽ V < 12.5
- Density is a division, so the smallest density takes the smallest mass over the largest volume
- Lower bound = 87.5 ÷ 12.5 = 7 g/cm³
- Using both lower bounds instead gives 87.5 ÷ 11.5 = 7.6…, which is bigger than 7 and so cannot be the lower bound