T Madas T Madas The weight of a

The weight of a chocolate bar and its packaging is 72 grams, correct to

The following table shows some items and information about their measurements. Fill in the

If p = 7. 4 and q = 2. 8, both correct to 1

A granite block weighs 700 ± 5 kg and its volume is given as

A right angled triangle has a hypotenuse of 71 cm and one of its

A digestive biscuit has its diameter measured as 70 ± 1 mm. Its thickness

The firing of a projectile can be used to estimate the gravitational acceleration g

A certain type of compass including its package weighs 95 grams, correct to the

The radius of the Earth is 6400 km and the radius of planet Jupiter

A cylinder has a height of 12 cm and the radius of its base

Fifi estimates the value of 20. 7 x 61. 5 187 to be 6.

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The weight of a chocolate bar and its packaging is 72 grams, correct to the nearest gram. What is the maximum and the minimum possible weight of this object according to this information? 68 69 grams 70 71 72 73 Lower Bound 74 75 Upper Bound Minimum Weight: 71. 5 g Maximum Weight: 72. 5 g 71. 5 ≤ W < 72. 5 © T Madas

The following table shows some items and information about their measurements. Fill in the table showing the minimum and maximum possible values for these measurements Minimum Maximum Value Item Measurement Length of a Pen 16 cm nearest whole number 15. 5 cm 16. 5 cm Weight of a Needle 1. 36 g 2 decimal places 1. 355 g 1. 365 g Can of Drink 330 ml 2 significant figures 325 ml 335 ml © T Madas

If p = 7. 4 and q = 2. 8, both correct to 1 decimal place, complete the following table giving your answers correct to 2 decimal places where appropriate. 7. 45 10. 3 7. 4 7. 35 al le Sm p q st pq p – q st p–q 10. 1 7. 45 + 2. 85 = 10. 3 rg e p+q 7. 35 + 2. 75 = 10. 1 Upper Bound La Lower Bound p – q 2. 85 2. 8 2. 75 © T Madas

If p = 7. 4 and q = 2. 8, both correct to 1 decimal place, complete the following table giving your answers correct to 2 decimal places where appropriate. Lower Bound p+q p–q pq p q 7. 35 + 2. 75 = 10. 1 7. 35 – 2. 85 = 7. 45 + 2. 85 = 10. 3 4. 5 7. 35 x 2. 75 = 20. 2125 20. 21 7. 35 ÷ 2. 85 = 2. 5789… 2. 58 Upper Bound 7. 45 – 2. 75 = 4. 7 7. 45 x 2. 85 = 21. 2325 21. 23 7. 45 ÷ 2. 75 = 2. 7091… 2. 71 © T Madas

A granite block weighs 700 ± 5 kg and its volume is given as 0. 13 m 3 correct to 2 significant figures. Using these figures calculate the maximum possible density of granite as kg per m 3. mass density = volume Upper Bound Lower Bound © T Madas

A granite block weighs 700 ± 5 kg and its volume is given as 0. 13 m 3 correct to 2 significant figures. Using these figures calculate the maximum possible density of granite as kg per m 3. 705 mass = = 5640 kg/m 3 density = volume 0. 125 © T Madas

A right angled triangle has a hypotenuse of 71 cm and one of its other sides 30 cm. These measurements are correct to 2 significant figures. Calculate the minimum and maximum possible lengths for the third side, correct to 2 significant figures. 1. 5 – 7 29. 5 – 30. 5 (70. 5)2 – (30. 5)2 4970. 25 – 930. 25 4040 64 cm Maximum Length x 2= x= (71. 5)2 – (29. 5)2 5112. 25 – 870. 25 4242 c c c . 5 70 x x 2= x= c c c Minimum Length 65 cm © T Madas

A digestive biscuit has its diameter measured as 70 ± 1 mm. Its thickness is measured as 7 ± 0. 5 mm. Calculate to 3 significant figures: 1. the maximum possible volume of the biscuit in cm 3. 2. the minimum possible surface area of this biscuit in cm 2. 70 ± 1 mm V = π r 2 h V = π x 35. 52 x 7. 5 V ≈ 29694 mm 3 V ≈ 29. 7 cm 3 [3 s. f. ] 7 ± 0. 5 mm 1 cm 3 = 1000 mm 3 © T Madas

A digestive biscuit has its diameter measured as 70 ± 1 mm. Its thickness is measured as 7 ± 0. 5 mm. Calculate to 3 significant figures: 1. the maximum possible volume of the biscuit in cm 3. 2. the minimum possible surface area of this biscuit in cm 2. 70 ± 1 mm 7 ± 0. 5 mm A = π r 2 + 2π r h A=π x 34. 52 + 2 x π x 34. 5 x 6. 5 A ≈ 8888 mm 2 V ≈ 88. 9 cm 2 [3 s. f. ] 1 cm 2 = 100 mm 2 © T Madas

The firing of a projectile can be used to estimate the gravitational acceleration g (in ms-2). The formula can be used for this estimate. The time t = 4. 22 s correct to 2 d. p. The projection speed u = 41. 41 ms-1 correct to 2 d. p. The projection angle θ = 30. 1° correct to 1 d. p. Calculate the upper and lower bound for g, giving your answers to 4 decimal places Write down an estimate for g to an appropriate degree of accuracy 4. 215 ≤ t ≤ 4. 225 41. 405 ≤ u ≤ 41. 415 30. 05 ≤ θ ≤ 30. 15 Upper bound for g g = 2 x 41. 415 x sin 30. 15°= 9. 8701 4. 215 © T Madas

The firing of a projectile can be used to estimate the gravitational acceleration g (in ms-2). The formula can be used for this estimate. The time t = 4. 22 s correct to 2 d. p. The projection speed u = 41. 41 ms-1 correct to 2 d. p. The projection angle θ = 30. 1° correct to 1 d. p. Calculate the upper and lower bound for g, giving your answers to 4 decimal places Write down an estimate for g to an appropriate degree of accuracy The gut feeling is to take the mean of these measurements as an estimate for g however the real question is to how many digits must the answer be quoted? The appropriate degree of accuracy is to give the answer to the number of common digits starting from the 1 st significant figure. = 9. 8701 = 9. 8148 © T Madas

A certain type of compass including its package weighs 95 grams, correct to the nearest 5 grams. 5000 of these compasses are loaded on a lorry, and the invoice states the weight of the load is 470 kg. 1. Calculate the minimum and the maximum possible weight of the load. 2. Calculate the maximum possible error in the stated weight. Min weight: 92. 5 Max weight: 97. 5 450 Lower Bound 460 470 5000 = 462. 5 462500 kgg x 5000 = 487500 487. 5 kgg x 480 Max error 17. 5 kg 490 500 Kg Upper Bound Stated weight © T Madas

The radius of the Earth is 6400 km and the radius of planet Jupiter is 11 times larger, both numbers given to 2 significant figures. 1. Calculate the minimum and the maximum possible value for the radius of Jupiter. 2. Write down a figure for the radius of Jupiter to an appropriate degree of accuracy. The gut feeling is to take a halfway figure as Jupiter’s radius however the real question is to how many digits must the answer be quoted? Jupiter radius The appropriate degree of accuracy is to give the answer to the number of common digits, starting from the 1 st significant figure. 74175 66675 Our estimates do not agree even in the 1 st significant figure, thus we can only quote the answer as 70000 km, correct to 1 significant figure. © T Madas

A cylinder has a height of 12 cm and the radius of its base is 4. 4 cm. These measurements are correct to the nearest mm. Calculate: 1. the volume of the cone to the nearest cm 3. 2. the upper and lower bound for this volume to the nearest cm 3. 3. the percentage error if the actual volume is 715 cm 3 and we use the figure we found in part (1) instead. V = π x 4. 4 x 12 V ≈ 730 cm 3 If the measurements are to the nearest mm, then: 4. 35 ≤ r < 4. 45 11. 95 ≤ h < 12. 05 V = π r 2 h V = π x 4. 352 x 11. 95 c c 2 lower bound of the volume V ≈ 710 cm 3 upper bound of the volume V = π r 2 h V = π x 4. 452 x 12. 05 c c V = πr h 2 c c volume of the cylinder V ≈ 750 cm 3 © T Madas

A cylinder has a height of 12 cm and the radius of its base is 4. 4 cm. These measurements are correct to the nearest mm. Calculate: 1. the volume of the cone to the nearest cm 3. 2. the upper and lower bound for this volume to the nearest cm 3. 3. the percentage error if the actual volume is 715 cm 3 and we use the figure we found in part (1) instead. V = π r 2 h V = π x 4. 42 x 12 c c volume of the cylinder V ≈ 730 cm 3 If the measurements are to the nearest mm, then: If we use the figure of 730 cm 3 instead of 715 cm 3 … … we have an error of 15 cm 3 … … calculated over the actual volume of 715 cm 3 … 15 x 100 ≈ 2. 1% 715 4. 35 ≤ r < 4. 45 11. 95 ≤ h < 12. 05 © T Madas

Fifi estimates the value of 20. 7 x 61. 5 187 to be 6. 1. Write down the 3 numbers Fifi used in this estimate. 2. Without performing any calculations explain why the value must be more than 6. 20. 7 x 61. 5 20 x 60 ≈ 187 200 The value of 6 obtained, is an underestimate because: • both numbers on the numerator have been rounded down • the number in the denominator has been rounded up © T Madas