What a rounded measurement tells you
A length is written as 12 cm, to the nearest cm. The true length was measured and then rounded, so it is not exactly 12 cm. It is some length that rounds to 12.
Every length from 11.5 cm up to just below 12.5 cm rounds to 12 cm, so the true length x satisfies . 11.5 is the lower bound, 12.5 is the upper bound, and the inequality is the error interval.
The filled circle at 11.5 means that 11.5 is included. The hollow circle at 12.5 means that 12.5 is not.
Half a unit each side
The bounds are always half the rounding unit on either side of the rounded value. Rounded to the nearest cm, the unit is 1 cm, so go 0.5 cm each way.
Rounded to the nearest 10, the unit is 10, so go 5 each way: a crowd of 340 to the nearest 10 is between 335 and 345, and . Rounded to one decimal place, the unit is 0.1, so go 0.05 each way: a mass of 3.7 kg to one decimal place means .
48 > 45, above the halfway mark: it is 2 from 50 but 8 from 40, so 48 rounds up to 50
Find the smallest number that rounds up
Slide the number down from 48. Every number from the halfway mark, 45, up to 50 rounds to 50, so to the nearest 10 the lower bound of 50 is 45.
The top of the band stays out
The lower bound is included, because 11.5 rounds up to 12. The upper bound is not: 12.5 is exactly halfway between 12 and 13, and a number exactly halfway rounds up, so 12.5 rounds to 13. That is why the error interval has at the bottom and < at the top.
The true length can be 12.49 cm or 12.499 cm, as close to 12.5 cm as you like, but never 12.5 cm itself. There is no largest length below 12.5, so the upper bound is 12.5, not 12.4 or 12.49.
12.5 rounds to 13, so it lies just outside the band of lengths that round to 12.
Bounds through a sum
Two rods measure 12 cm and 7 cm, each to the nearest cm. The first is between 11.5 cm and 12.5 cm long, and the second is between 6.5 cm and 7.5 cm.
Laid end to end, the total is largest when both rods are at their upper bounds: 12.5 + 7.5 = 20 cm. It is smallest when both are at their lower bounds: 11.5 + 6.5 = 18 cm. So the total length T satisfies .
The rounded values give 12 + 7 = 19 cm, and the true total can be up to 1 cm from 19 either way, because each rod can be up to half a centimeter out.
The total of the two rods lies in a band 2 cm wide, twice as wide as the band for either rod.
Worked example: A Shelf Measured to the Nearest Centimeter
Question A carpenter measures the length of a shelf as 84 cm, correct to the nearest centimeter. (a) Find the lower bound and the upper bound of the length. (b) Write the error interval for the length, L cm, as an inequality, and say whether the shelf could be exactly 84.5 cm long.
1.The length is rounded to the nearest 1 cm, so the true length is at most half of 1 cm away from 84 cm. Half of 1 cm is 0.5 cm.
To the nearest 1 cm, the length is at most 0.5 cm away from 84 cm on either side. 2.The lower bound is 84 − 0.5 = 83.5 cm. A length of exactly 83.5 cm rounds up to 84 cm, so 83.5 cm is a possible length.
The lower bound is 84 − 0.5 = 83.5 cm. It rounds up to 84 cm, so it is a possible length. 3.The upper bound is 84 + 0.5 = 84.5 cm. (a) The lower bound is 83.5 cm and the upper bound is 84.5 cm.
(a) The upper bound is 84 + 0.5 = 84.5 cm. The open circle shows that it is not a possible length. 4.A length of exactly 84.5 cm rounds up to 85 cm, not to 84 cm. Every length below 84.5 cm, such as 84.49 cm or 84.499 cm, rounds to 84 cm. So L can equal the lower bound, but it must be less than the upper bound.
A length of exactly 84.5 cm rounds to 85 cm. Every length below it rounds to 84 cm. 5.(b) The error interval is 83.5 ≤ L < 84.5, and the shelf cannot be exactly 84.5 cm long. Check: the interval is 84.5 − 83.5 = 1 cm wide, which is the unit the length was rounded to.
(b) The error interval is 83.5 ≤ L < 84.5.
Answer: (a) lower bound 83.5 cm, upper bound 84.5 cm; (b) 83.5 ≤ L < 84.5; no, a length of 84.5 cm rounds to 85 cm
Common mistakes
- Giving the upper bound as 84.4 cm or 84.49 cm. A length of 84.495 cm also rounds to 84 cm, so no number below 84.5 is the largest possible length. The upper bound is 84.5 cm, and the sign < shows that it is not reached.
- Writing 83 ≤ L < 85. That goes a whole centimeter either way. A length of 83.2 cm rounds to 83 cm, not to 84 cm, so the bounds are only half a centimeter from 84 cm.
More significant figures and estimation problems, worked step by step →
Worked example: Fencing a Garden Whose Sides Are Given to the Nearest Meter
Question A rectangular garden is 12 m long and 7 m wide, each correct to the nearest meter. A fence is to be put all the way round it. (a) Find the least possible perimeter and the upper bound of the perimeter. (b) The owner works out the perimeter as 2 × (12 + 7) = 38 m. Find the greatest amount by which the true perimeter can differ from 38 m.
1.Each side is rounded to the nearest meter, so it can be 0.5 m either way. The length l m satisfies 11.5 ≤ l < 12.5 and the width w m satisfies 6.5 ≤ w < 7.5.
Each side can be 0.5 m either way: 11.5 ≤ l < 12.5 and 6.5 ≤ w < 7.5. 2.The perimeter is least when both sides are at their lower bounds: 2 × (11.5 + 6.5) = 2 × 18 = 36 m.
The least perimeter uses both lower bounds: 2 × (11.5 + 6.5) = 36 m. 3.The upper bound of the perimeter uses both upper bounds: 2 × (12.5 + 7.5) = 2 × 20 = 40 m. (a) The least possible perimeter is 36 m and the upper bound is 40 m, so the perimeter P m satisfies 36 ≤ P < 40.
(a) The upper bound uses both upper bounds: 2 × (12.5 + 7.5) = 40 m. 4.(b) 38 − 36 = 2 and 40 − 38 = 2, so the true perimeter can differ from 38 m by up to 2 m. Check: the fence has four sides and each can be 0.5 m out, and 4 × 0.5 = 2 m.
(b) 38 − 36 = 2 and 40 − 38 = 2: the true perimeter can differ from 38 m by up to 2 m.
Answer: (a) least 36 m, upper bound 40 m; (b) 2 m
Common mistakes
- Taking 0.5 m off the perimeter and adding 0.5 m to it, to get 37.5 m and 38.5 m. The perimeter was not measured to the nearest meter. The four sides were, and their four errors add up.
- Using the upper bound of the length with the lower bound of the width for the greatest perimeter. In a sum, a larger part always makes a larger total, so the greatest perimeter uses the upper bound of both sides.
More significant figures and estimation problems, worked step by step →