Bounds and Error Intervals [H]

This lesson covers upper and lower bounds, error intervals, and using bounds in calculations. Bounds is a Higher-tier topic in the Edexcel GCSE Mathematics (1MA1) specification and is commonly worth 3-5 marks on the exam.

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Key Vocabulary

Term	Meaning
Lower bound	The smallest value that rounds (or truncates) to the given measurement
Upper bound	The smallest value that would round UP to the next measurement
Error interval	The range of possible values, written using inequalities
Degree of accuracy	How precisely a measurement has been recorded (e.g. to 1 d.p., to the nearest 10)
Continuous data	Data that can take any value within a range (e.g. length, mass, time)
Discrete data	Data that can only take specific values (e.g. number of people)
Truncation	Cutting off digits without rounding

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Bounds from Rounding

When a value has been rounded, the true value could be anything within half a unit either side.

Formula:

• Lower bound = rounded value - half the degree of accuracy
• Upper bound = rounded value + half the degree of accuracy

Worked Example 1

A length is measured as 8.3 cm, correct to 1 decimal place. Find the error interval.

Degree of accuracy = 0.1 cm, so half = 0.05 cm.

• Lower bound = 8.3 - 0.05 = 8.25 cm
• Upper bound = 8.3 + 0.05 = 8.35 cm

Error interval: $8.25 \leq$8.25≤ length < 8.35

Important: Use $\leq$≤ for the lower bound (it IS included) and < for the upper bound (it is NOT included, because it would round up to 8.4).

Worked Example 2

A mass is 750 g, rounded to the nearest 10 g. Write the error interval.

Degree of accuracy = 10 g, half = 5 g.

• Lower bound = 750 - 5 = 745 g
• Upper bound = 750 + 5 = 755 g

Error interval: $745 \leq mass < 755$745≤mass<755

Worked Example 3

A crowd at a concert is 24,000, rounded to the nearest 1,000. Write the error interval.

Degree of accuracy = 1000, half = 500.

• Lower bound = 24,000 - 500 = 23,500
• Upper bound = 24,000 + 500 = 24,500

Error interval: $23,500 \leq$23,500≤ crowd < 24,500

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Bounds from Truncation

When a value has been truncated (not rounded), the lower bound is the truncated value itself, and the upper bound is one unit of accuracy above.

Worked Example 4

A length is truncated to 4.7 cm (to 1 decimal place). Write the error interval.

• Lower bound = 4.7 (truncation removes digits, so the true value is at least 4.7)
• Upper bound = 4.7 + 0.1 = 4.8

Error interval: $4.7 \leq$4.7≤ length < 4.8

Key Difference from Rounding: When rounding to 1 d.p., 4.7 has bounds 4.65 to 4.75. When truncating to 1 d.p., 4.7 has bounds 4.7 to 4.8. Edexcel specifically tests whether students know this difference.

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Calculations Using Bounds

When performing calculations with rounded values, you need to decide which combination of upper and lower bounds gives the maximum or minimum result.

Rules for Finding Maximum and Minimum Values

Operation	Maximum result	Minimum result
A + B	Upper A + Upper B	Lower A + Lower B
A - B	Upper A - Lower B	Lower A - Upper B
$A \times B$A×B	Upper $A \times$A× Upper B	Lower $A \times$A× Lower B
$A \div B$A÷B	Upper $A \div$A÷ Lower B	Lower $A \div$A÷ Upper B

Memory aid: For subtraction and division, to MAXIMISE the result, make the first number as BIG as possible and the second as SMALL as possible (and vice versa for minimum).

Worked Example 5

p = 6.4 (correct to 1 d.p.) and q = 3.8 (correct to 1 d.p.). Calculate the upper and lower bounds of p + q.

Bounds of p: $6.35 \leq p < 6.45$6.35≤p<6.45 Bounds of q: $3.75 \leq q < 3.85$3.75≤q<3.85

• Maximum of p + q = 6.45 + 3.85 = 10.30
• Minimum of p + q = 6.35 + 3.75 = 10.10

Worked Example 6

a = 12.5 cm (correct to 1 d.p.) and b = 4.3 cm (correct to 1 d.p.). Calculate the upper and lower bounds of a - b.

Bounds of a: $12.45 \leq a < 12.55$12.45≤a<12.55 Bounds of b: $4.25 \leq b < 4.35$4.25≤b<4.35

• Maximum of a - b = 12.55 - 4.25 = 8.30
• Minimum of a - b = 12.45 - 4.35 = 8.10

Worked Example 7

The base of a rectangle is 8.4 cm and the height is 3.6 cm, both correct to 1 decimal place. Calculate the upper and lower bounds of the area.

Bounds of base: $8.35 \leq base < 8.45$8.35≤base<8.45 Bounds of height: $3.55 \leq$3.55≤ height < 3.65

• Maximum area $= 8.45 \times 3.65 =$=8.45×3.65= $30.8425 cm^{2}$30.8425cm2
• Minimum area $= 8.35 \times 3.55 =$=8.35×3.55= $29.6425 cm^{2}$29.6425cm2

Worked Example 8

A car travels 148 km (to the nearest km) in 2.4 hours (to 1 d.p.). Calculate the upper and lower bounds of the average speed.

Bounds of distance: $147.5 \leq d < 148.5$147.5≤d<148.5 Bounds of time: $2.35 \leq t < 2.45$2.35≤t<2.45

Speed = distance $\div$÷ time

• Maximum speed $= 148.5 \div 2.35 =$=148.5÷2.35= 63.191... $\approx 63.2 km/h$≈63.2km/h (to 1 d.p.)
• Minimum speed $= 147.5 \div 2.45 =$=147.5÷2.45= 60.204... $\approx 60.2 km/h$≈60.2km/h (to 1 d.p.)

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Suitable Degree of Accuracy

A common exam question asks: "To what degree of accuracy can you give the answer?"

Method

Calculate the upper and lower bounds of the answer, then find the level of accuracy to which they both round to the same value.

Worked Example 9

A square has side length 8.3 cm, correct to 1 d.p. Calculate the perimeter and give your answer to a suitable degree of accuracy.

Bounds of side length: $8.25 \leq s < 8.35$8.25≤s<8.35

Perimeter = 4s.

• Lower bound of perimeter $= 4 \times 8.25 = 33.0 cm$=4×8.25=33.0cm
• Upper bound of perimeter $= 4 \times 8.35 = 33.4 cm$=4×8.35=33.4cm

To 2 s.f.: 33.0 rounds to 33, and 33.4 rounds to 33. Both agree.

Answer: The perimeter is 33 cm (to 2 significant figures), since both bounds round to 33 at this level of accuracy.

Edexcel Exam Tip: In exam questions asking for a suitable degree of accuracy, the bounds will always agree at some level. Check 1 s.f. first, then 2 s.f., etc. State the degree of accuracy and justify it.

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Bounds with Compound Measures

Worked Example 10

A runner completes a 100 m track (measured to the nearest metre) in 12.3 seconds (measured to 1 d.p.). Calculate the lower bound of the runner's speed.

Bounds of distance: $99.5 \leq d < 100.5$99.5≤d<100.5 Bounds of time: $12.25 \leq t < 12.35$12.25≤t<12.35

For LOWER bound of speed (speed = distance $\div$÷ time):

• Use LOWER distance and UPPER time
• Lower speed $= 99.5 \div 12.35 =$=99.5÷12.35= 8.056... m/s

Answer: 8.06 m/s (to 3 s.f.)

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Bounds and Inequality Notation

Error intervals are always written using inequality notation:

• Rounded: LB $\leq x <$≤x< UB
• Truncated: LB $\leq x <$≤x< LB + one unit of accuracy

Worked Example 11

h = 12 cm, correct to the nearest centimetre. Write the error interval for h.

$11.5 \leq h < 12.5$11.5≤h<12.5

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Common Mistakes and Misconceptions

Mistake	Correction
Using $\leq$≤ for both bounds	Upper bound uses < (strictly less than)
For maximum of A - B, using lower A - upper B	To MAXIMISE a subtraction, use UPPER A - LOWER B
Confusing truncation bounds with rounding bounds	Truncation: lower bound = given value; Rounding: lower bound = given value - half precision
Forgetting to state the degree of accuracy	Always give the appropriate degree of accuracy when asked
Using rounded values instead of bounds in calculations	Use the actual upper/lower bounds, not the rounded values

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Practice Problems