5 Approximation and Errors
5 Approximation and Errors
Review Exercise 5 (p. 5.4)
1. (a)
(b)
2. (a)
(b)
3. (a)
(b)
(c)
4. (a)
(b)
(c)
5. (a)
(b)
6. Average price
7. (a) The scale interval of the ruler
(b) (i) Length of the ticket
(ii) Length of the ticket
Activity
Activity 5.1 (p. 5.5)
1. (a) 3 (b) 3, 2
(c) (i)
Digit / 3 / 2 / 4 / 5Place value / thousand / hundred / ten / one
(ii) 3, because it has the greatest place value.
2. (a) 8 (b) 2
(c) 4 (d) 1
Activity 5.2 (p. 5.17)
1. 1 cm
2. A: 7 cm, B: 6 cm, C: 6 cm
3. (a) 6.36 cm, 5.6 cm, 6 cm, 6.42 cm, 5.99 cm
(b) no
(c) Between (6 – 0.5) cm and (6 + 0.5) cm
i.e. between 5.5 cm and 6.5 cm
Quick Practice
Quick Practice 5.1 (p. 5.7)
(a) ∵ The leftmost non-zero digit of 210.0 is 2.
∴ The most significant figure is 2.
(b) ∵ The leftmost non-zero digit of 0.012 is 1.
∴ The most significant figure is 1.
Quick Practice 5.2 (p. 5.7)
(a) The significant figures of 487 are 4, 8 and 7.
∴ 487 has 3 significant figures.
(b) The significant figures of 40 006 are 4, 0, 0, 0 and 6.
∴ 40 006 has 5 significant figures.
(c) The significant figures of 0.0048 are 4 and 8.
∴ 0.0048 has 2 significant figures.
Quick Practice 5.3 (p. 5.9)
(a) 3059 = 3000 (cor. to 1 sig. fig.)
(b) 3059 = 3100 (cor. to 2 sig. fig.)
(c) 3059 = 3060 (cor. to 3 sig. fig.)
Quick Practice 5.4 (p. 5.10)
(a) 0.003 98 = 0.004 (cor. to 1 sig. fig.)
(b) 0.003 98 = 0.0040 (cor. to 2 sig. fig.)
Quick Practice 5.5 (p. 5.10)
(a) 6995 = 7000 (cor. to 1 sig. fig.)
(b) 6995 = 7000 (cor. to 2 sig. fig.)
(c) 6995 = 7000 (cor. to 3 sig. fig.)
Quick Practice 5.6 (p. 5.11)
(a) 180 000 (cor. to the nearest ten thousand) has 2 significant figures.
(b) 180 000 (cor. to the nearest hundred) has 4 significant figures.
(c) 180 000 (cor. to the nearest integer) has 6 significant figures.
Quick Practice 5.7 (p. 5.12)
Average height of the 5 students
Quick Practice 5.8 (p. 5.15)
The required absolute error = 12 756 - 12 000
=
Quick Practice 5.9 (p. 5.16)
(a) (i) Capacity of the container = 403 mL (cor. to the
nearest mL)
∴ The required absolute error
(ii) Capacity of the container = 400 mL (cor. to 2 sig. fig.)
∴ The required absolute error
(b) ∵ 0.5 mL < 2.5 mL
i.e. The absolute error obtained in (a)(i) is smaller than
that in (a)(ii).
∴ The approximation obtained in (a)(i) is closer to the actual capacity.
Quick Practice 5.10 (p. 5.19)
(a) The maximum absolute error
(b) Lower limit of Peter’s actual height = (168 – 0.5) cm
= 167.5 cm
Upper limit of Peter’s actual height = (168 + 0.5) cm
= 168.5 cm
∴ Peter’s actual height lies between 167.5 cm and
168.5 cm.
Quick Practice 5.11 (p. 5.20)
The maximum absolute error
(a) Upper limit of the actual length
Lower limit of the actual length
(b) Upper limit of the actual width
Lower limit of the actual width
(c) Minimum area of the note
Maximum area of the note
∵ 110 < 116.6625 < 118.9625
∴ It is impossible that the actual area of the note is
110 cm2.
Quick Practice 5.12 (p. 5.23)
∴
Quick Practice 5.13 (p. 5.24)
(a) (i) The maximum absolute error of the measured length
of the basketball court
The maximum absolute error of the measured
diameter of the basketball
(ii) The relative error of the measured length of the basketball court
The relative error of the measured diameter of the
basketball
(b) ∵
i.e. The measured length of the basketball court has a smaller relative error.
∴ The measured length of the basketball court is more accurate.
Quick Practice 5.14 (p. 5.25)
(a) The maximum absolute error
(b) The relative error
(c) The percentage error
Further Practice
Further Practice (p. 5.8)
Number / Mostsig. fig. / 2nd
sig. fig. / 3rd
sig. fig. / Number of sig. fig.
23 488 / 2 / 3 / 4 / 5
34.0087 / 3 / 4 / 0 / 6
0.090 005 / 9 / 0 / 0 / 5
101 001 / 1 / 0 / 1 / 6
20.59 / 2 / 0 / 5 / 4
0.000 360 / 3 / 6 / 0 / 3
Further Practice (p. 5.12)
1. (a) ∵ The leftmost non-zero digit of 58.30 is 5.
∴ The most significant figure is 5.
The significant figures of 58.30 are 5, 8, 3 and 0.
∴ 58.30 (cor. to 2 decimal places) has 4 significant figures.
(b) ∵ The leftmost non-zero digit of –0.070 is 7.
∴ The most significant figure is 7.
The significant figures of –0.070 are 7 and 0.
∴ –0.070 (cor. to the nearest 0.001) has 2 significant figures.
(c) ∵ The leftmost non-zero digit of 8 690 000 is 8.
∴ The most significant figure is 8.
8 690 000 (cor. to the nearest hundred) has 5 significant figures.
2. (a) –0.064 85 = –0.06 (cor. to 1 sig. fig.)
–0.064 85 = –0.065 (cor. to 2 sig. fig.)
–0.064 85 = –0.0649 (cor. to 3 sig. fig.)
(b) 0.9029 = 0.9 (cor. to 1 sig. fig.)
0.9029 = 0.90 (cor. to 2 sig. fig.)
0.9029 = 0.903 (cor. to 3 sig. fig.)
(c) 175 000 = 200 000 (cor. to 1 sig. fig.)
175 000 = 180 000 (cor. to 2 sig. fig.)
175 000 = 175 000 (cor. to 3 sig. fig.)
3. (a) Annual rainfall = 1706.9 mm
= 1710 mm (cor. to 3 sig. fig.)
(b) Average monthly rainfall
Further Practice (p. 5.16)
1. For actual value = 203 and approximation = 200,
absolute error = 203 – 200 = 3
For actual value = 2965 and approximation = 3000,
absolute error = 3000 – 2965 = 35
For actual value = 917 356 and approximation = 900 000,
absolute error = 917 356 – 900 000 = 17 356
For actual value = –17.5 and approximation = –18,
absolute error = –17.5 – (–18) = 0.5
∴
Actual value / Approximation / Absolute error203 / 200 / 3
2965 / 3000 / 35
917 356 / 900 000 / 17 356
–17.5 / –18 / 0.5
2. The required absolute error
= (4 – 3.5) kg
= 0.5 kg
3. (a) 2046 = 2050 (cor. to 3 sig. fig.)
149.5 = 150 (cor. to 3 sig. fig.)
(b) Value of the expression
= 2050 + 150
= 2200
(c) Absolute error in (b)
= 2200 − (2046 + 149.5)
= 2200 − 2195.5
= 4.5
Further Practice (p. 5.21)
1. For measured value = 160 cm and scale interval of the measuring tool = 10 cm,
maximum absolute error
lower limit = (170 – 5) cm = 155 cm
upper limit = (160 + 5) cm = 165 cm
For measured value = 340 m and maximum absolute error = 0.5 m,
scale interval of the measuring tool = 2 ´ 0.5 m = 1 m
lower limit = (340 – 0.5) m = 339.5 m
upper limit = (340 + 0.5) m = 340.5 m
For measured value = 17.5 g and the scale interval of the measuring tool = 0.5 g,
Maximum absolute error
lower limit = (17.5 – 0.25) g = 17.25 g
upper limit = (17.5 + 0.25) g = 17.75 g
For maximum absolute error = 0.25 kg and upper limit
= 30.75 kg,
measured value = (30.75 – 0.25) kg = 30.5 kg
scale interval of the measuring tool = 2 ´ 0.25 kg = 0.5 kg
lower limit = (30.5 – 0.25) kg = 30.25 kg
∴
Measured value / Scale intervalof the measuring tool / Maximum absolute error / Lower limit
of the actual value / Upper limit
of the actual value
160 cm / 10 cm / 5 cm / 155 cm / 165 cm
340 m / 1 m / 0.5 m / 339.5 m / 340.5 m
17.5 g / 0.5 g / 0.25 g / 17.25 g / 17.75 g
30.5 kg / 0.5 kg / 0.25 kg / 30.25 kg / 30.75 kg
2. For measured value = 27 500 L (cor. to the nearest 10 L),
lower limit
upper limit
For measured value = 27 500 L (cor. to the nearest 100 L),
lower limit
upper limit
For measured value = 12.5 mm (cor. to 3 sig. fig.),
lower limit
upper limit
For measured value = 14 000 km (cor. to 2 sig. fig.),
lower limit
upper limit
∴
Measured value / Lower limitof the actual value / Upper limit
of the actual value
27 500 L (cor. to the nearest 10 L) / 27 495 L / 27 505 L
27 500 L (cor. to the nearest 100 L) / 27 450 L / 27 550 L
12.5 mm (cor. to 3 sig. fig.) / 12.45 mm / 12.55 mm
14 000 km (cor. to 2 sig. fig.) / 13 500 km / 14 500 km
3. The maximum absolute error
(a) Upper limit of the actual length of AB
Lower limit of the actual length of AB
(b) Upper limit of the actual length of DF
Lower limit of the actual length of DF
(c) Maximum area of square ABCE
Minimum area of square ABCE
Maximum area of △CDE
Minimum area of △CDE
Maximum area of ABCDE
= maximum area of square ABCE + maximum area of
△CDE
Minimum area of ABCDE
= minimum area of square ABCE + minimum area of
△CDE
Further Practice (p. 5.26)
1. For actual value = 54 and approximation = 50,
absolute error = 54 – 50 = 4
relative error (cor. to 2 sig. fig.)
percentage error (cor. to 2 sig. fig.)
For actual value = 204 and approximation = 200,
absolute error = 204 – 200 = 4
relative error (cor. to 2 sig. fig.)
percentage error (cor. to 2 sig. fig.)
For actual value = 106.3 and approximation = 110,
absolute error = 110 – 106.3 = 3.7
relative error (cor. to 2 sig. fig.)
percentage error (cor. to 2 sig. fig.)
For actual value = 6368 and approximation = 8000,
absolute error = 8000 – 6368 = 1632
relative error (cor. to 2 sig. fig.)
percentage error (cor. to 2 sig. fig.)
∴
Actual value / Approximation / Absolute error / Relative error / Percentage error54 / 50 / 4 / 0.074 / 7.4%
204 / 200 / 4 / 0.020 / 2.0%
106.3 / 110 / 3.7 / 0.035 / 3.5%
6368 / 8000 / 1632 / 0.26 / 26%
2. For actual value = 60°C (cor. to the nearest 10°C),
maximum absolute error
relative error(cor. to 2 sig. fig.)
percentage error(cor. to 2 sig. fig.)
For measured value = 1700 kg (cor. to the nearest 100 kg),
maximum absolute error
relative error(cor. to 2 sig. fig.)
percentage error(cor. to 2 sig. fig.)
For measured value = 3.8 mL (cor. to 2 fig. fig.),
maximum absolute error
relative error(cor. to 2 sig. fig.)
percentage error (cor. to 2 sig. fig.)
For measured value = 95 000 cm2 (cor. to 3 fig. fig.),
maximum absolute error
relative error (cor. to 2 sig. fig.)
percentage error (cor. to 2 sig. fig.)
∴
Measured value / Maximum absolute error / Relative error / Percentage error60°C (cor. to the nearest 10°C) / 5°C / 0.083 / 8.3%
1700 kg (cor. to the nearest
100 kg) / 50 kg / 0.029 / 2.9%
3.8 mL (cor. to 2 sig. fig.) / 0.05 mL / 0.013 / 1.3%
95 000 cm2 (cor. to 3 sig. fig.) / 50 cm2 / 0.000 53 / 0.053%
3. (a) ∵
∴ Maximum absolute error
(b) Lower limit of the actual volume of the pack of drink
Upper limit of the actual volume of the pack of drink
∴ The actual volume of the pack of drink falls between 242.5 mL and 257.5 mL.
(c) ∵ 242.5 mL < 257 mL <257.5 mL
∴ The actual volume of the pack of drink could
be 257 mL.
Exercise
Exercise 5A (p. 5.12)
Level 1
1. (a) ∵ The leftmost non-zero digit of 2173 is 2.
∴ The most significant figure is 2.
(b) ∵ The leftmost non-zero digit of 18 900 is 1.
∴ The most significant figure is 1.
(c) ∵ The leftmost non-zero digit of 10.250 is 1.
∴ The most significant figure is 1.
(d) ∵ The leftmost non-zero digit of 0.64 is 6.
∴ The most significant figure is 6.
(e) ∵ The leftmost non-zero digit of 0.043 09 is 4.
∴ The most significant figure is 4.
(f) ∵ The leftmost non-zero digit of 0.005 07 is 5.
∴ The most significant figure is 5.
2. (a) The significant figures of 3026 are 3, 0, 2 and 6.
∴ 3026 has 4 significant figures.
(b) The significant figure of 0.005 is 5.
∴ 0.005 has 1 significant figure.
(c) The significant figures of 6408 are 6, 4, 0 and 8.
∴ 6408 has 4 significant figures.
(d) The significant figures of 0.705 are 7, 0 and 5.