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Cambridge IGCSE0580

Limits of accuracy

Mathematics 0580 Chapter Notes

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Limits of accuracy
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In this chapter

  1. 1Bounds of a rounded value
  2. 2Decimal places and significant figures
  3. 3Mixed units
  4. 4Which bound to use in a calculation
  5. 5Perimeter and area
  6. 6Working back to a missing length

1. Bounds of a rounded value

A value that has been rounded is not exact. The lower bound is the smallest value that rounds to it. The upper bound is the limit it cannot reach.

Take the degree of accuracy, for example the nearest 10 or the nearest 50. Half of it goes below the value and half goes above.

Write the bounds as an inequality: lower bound leqslant x < upper bound.

lower bound = x - 1/2 × accuracy

upper bound = x + 1/2 × accuracy

Examiner insight

“A common error showing a lack of understanding of the different inequality symbols was the upper bound of 635.4.” “Other errors included adding and subtracting 5, 0.5, 0.1 or 1.”

Worked example 12 marks

The mass, m grams, of some butter is 650 grams, correct to the nearest 50 grams. Complete this statement about the value of m. … leqslant m < …

  1. 1

    half of the accuracy: 50 ÷ 2 = 25 [1]

  2. 2

    650 - 25 = 625 and 650 + 25 = 675, so 625 leqslant m < 675 [1]

2. Decimal places and significant figures

For decimal places, the accuracy is the place value of the last digit. A value given to 1 decimal place is accurate to the nearest 0.1. So the bounds are 0.05 either side.

For significant figures, find the place value of the last significant figure. Then take half of it.

3.6 (1 d.p.): 3.55 leqslant x < 3.65

12.7 (3 s.f.): 12.65 leqslant x < 12.75

Examiner insight

“The most common error was giving the answer 13.7 from 14.2 – 0.5 rather than 14.2 – 0.05.”

Worked example 12 marks

Do not use a calculator. The time, t seconds, for the winner of a race is 12.7 seconds, correct to 3 significant figures. Complete this statement about the value of t. … leqslant t < …

  1. 1

    the last significant figure is in the tenths, so the accuracy is 0.1 and half of it is 0.05 [1]

  2. 2

    12.7 - 0.05 = 12.65 and 12.7 + 0.05 = 12.75, so 12.65 leqslant t < 12.75 [1]

3. Mixed units

The accuracy may be in different units from the value. A height in metres can be correct to the nearest 10 centimetres.

Change the accuracy into the units of the value first. Then take half. Give the bounds in the units the question uses.

10 cm = 0.1 m, half = 0.05 m

Examiner insight

“Some candidates combined decimals and mixture of units in their response.” “Working in centimetres, but not reverting back to metres for the answer did gain 1 mark, as did the often seen correct values but the wrong way round.”

Worked example 12 marks

Do not use a calculator. The height, h metres, of a tree is 37.4 m, correct to the nearest 10 centimetres. Complete this statement about the value of h. … leqslant h < …

  1. 1

    10 cm = 0.1 m, so half the accuracy is 0.05 m [1]

  2. 2

    37.4 - 0.05 = 37.35 and 37.4 + 0.05 = 37.45, so 37.35 leqslant h < 37.45 [1]

4. Which bound to use in a calculation

Find the bounds of each rounded value before the calculation. Never calculate with the given figures and adjust after.

For a sum or product, use the same type of bound for each value. For a difference or quotient, choose the pair that makes the answer as large or as small as possible.

To make a quotient small, divide a small number by a large one.

upper bound of a - b = a_U - b_L

lower bound of a - b = a_L - b_U

upper bound of a/b = a_U/b_L

lower bound of a/b = a_L/b_U

Examiner insight

“The key to success in this question was realising that the lower bound of the mass and the upper bound of the volume were required.” “Some gave the correct bounds but selected the wrong one(s).”

Worked example 13 marks

The volume of a solid metal cuboid is 22 cm³, correct to the nearest cm³. The mass of the metal cuboid is 140 grams, correct to the nearest 10 grams. Calculate the lower bound of the density of the metal. [Density = mass ÷ volume]

  1. 1

    lower bound of mass = 135 g [1]

  2. 2

    upper bound of volume = 22.5 cm³ [1]

  3. 3

    lower bound of density = 135 ÷ 22.5 = 6 g/cm³ [1]

5. Perimeter and area

Find the bounds of each side first. Then work out the perimeter or area from those bounds.

For the upper bound of a perimeter or an area, use all the upper bounds. For the lower bound, use all the lower bounds.

Do not find the perimeter from the given sides and then add or subtract a half. A bound from a calculation is exact, so do not round it.

P = 2(l + w)

A = l × w

Examiner insight

“A common error was for candidates to calculate the perimeter using the original figures 16 and 14 to reach 60, then try to deal with the bounds, for example by giving 59.5 as a final answer.”

Worked example 12 marks

A rectangle measures 4.2 cm by 6.8 cm. Each measurement is given correct to the nearest millimetre. Calculate the lower bound of the area of the rectangle.

  1. 1

    nearest millimetre is 0.1 cm, so the lower bounds are 4.15 cm and 6.75 cm [1]

  2. 2

    lower bound of area = 4.15 × 6.75 = 28.0125 cm² [1]

6. Working back to a missing length

Sometimes the result is given and a side is missing. Rearrange first. Then choose the bounds that make the missing side as small or as large as possible.

For the lower bound of the height, divide the lower bound of the volume by the upper bounds of the other sides.

h = V/(l × w), h_L = V_L/(l_U × w_U)

Examiner insight

“Some candidates used the lower bound for volume and mistakenly used the lower bounds for the length and width.” “Some calculated the height using the given figures and attempted to give the bounds for their answer.”

Worked example 13 marks

The length of a cuboid is 2 cm, correct to the nearest integer. The width of the cuboid is 4 cm, correct to the nearest integer. The volume of the cuboid is 123.8 cm³, correct to 1 decimal place. Find the lower bound of the height of the cuboid.

  1. 1

    lower bound of volume = 123.75 cm³ [1]

  2. 2

    upper bounds of length and width are 2.5 cm and 4.5 cm, so l × w = 11.25 [1]

  3. 3

    h = 123.75 ÷ 11.25 = 11 cm [1]

Recap

  • Bounds are half the degree of accuracy below and above the value.
  • Write them as lower bound leqslant x < upper bound.
  • Decimal places and significant figures: half of the last digit's place value.
  • Change the accuracy into the units of the value, then back again for the answer.
  • Extended: find every bound before the calculation, never after it.
  • Extended: for a quotient, lower bound = a_L ÷ b_U and upper bound = a_U ÷ b_L.
  • Extended: upper bound of a perimeter or area uses all upper bounds.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Do not use a calculator. The height of a wall is 3 m, correct to the nearest metre. Write down the upper bound of the height.1 mark
  2. A length is 7.4 m, correct to 1 decimal place. Write down the lower bound and the upper bound.2 marks
  3. Do not use a calculator. The population of a town is 4300, correct to 2 significant figures. Write down the lower bound and the upper bound.2 marks
  4. Do not use a calculator. A pole is 12.5 m long, correct to the nearest 0.5 m. Write down the lower bound and the upper bound.2 marks
  5. Do not use a calculator. A table is 120 cm long, correct to the nearest 10 cm. Write down the lower bound and the upper bound of its length in metres.2 marks
  6. A rope is 2.4 m long, correct to the nearest centimetre. Write down the lower bound and the upper bound of its length in metres.2 marks
  7. Do not use a calculator. A rectangle is 8 cm long and 5 cm wide. Each is correct to the nearest centimetre. Calculate the upper bound of the perimeter.3 marks
  8. Do not use a calculator. A cyclist rides 50 m, correct to the nearest 10 m, in 4 s, correct to the nearest second. Calculate the lower bound of the speed.3 marks
  9. A sand pit is a cuboid. Its length is 130 cm, correct to the nearest 10 cm. Its width is 0.4 m, correct to 1 decimal place. It holds 0.87 m³ of sand, correct to 2 decimal places, and the sand is level. Calculate the upper bound of the depth of the sand, in metres.4 marks
  10. A car starts a journey with 60 litres of fuel, correct to the nearest litre. At the end it has 43 litres, correct to the nearest litre. The journey is 230 km, correct to the nearest 10 km. Calculate the upper bound of the distance travelled per litre of fuel used.5 marks

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