Question 1(a): [2 marks]
A wooden toy is 6 cm tall to the nearest cm.
Find the error interval, in which the true height of the toy, h, lies
ANSWER:
5.5 \leq h < 6.55.5 \leq h \leq 6.5 5.5 > h \leq 6.5 5.1 \leq h < 6.9 5 \leq h \leq 7
WORKING:
We need to half the size of the interval, which is 1 (cm).
So 1 \div 2 = 0.5
Lower bound = 6 - 0.5 = 5.5 cm
Upper bound = 6+0.5=6.5 cm
5.5 \leq h < 6.5
Question 1(b): [2 marks]
The mass of the toy is 2.2 kg to the nearest 100 g.
Find the error interval in which the true mass of the toy, m, lies in kg.
Answer type: Multiple answers type 1
ANSWER:
2.15 \leq h \leq 2.25 2.15 \leq h \leq 2.25 2.0\leq h \leq 2.5 1.5 \leq h > 2.5
WORKING:
100 g = \, 0.1 kg
We need to half the size of the interval, which is 0.1 (kg).
So 0.1 \div 2 = 0.05
Lower bound = 2.2 - 0.05 = 2.15 kg
Upper bound = 2.2+0.05=2.25 kg
2.15 \leq h < 2.25
Question 2:
Answer the following upper and lower bounds questions.
Question 2(a): [2 marks]
The mass of a handbag is 3.26 kg correct to 2 decimal places.
Find the upper and lower bounds for the mass of the handbag in kg.
ANSWER:
3.255 \leq h < 3.2653.255 \leq h \leq 3.265 3.255 \leq h < 3.265 3.25 \leq h \leq 3.25 3.2 \leq h < 3.3
WORKING:
We need to half the size of the interval, which is 2 decimal places (0.01)
So 0.01 \div 2 = 0.005
Lower bound = 3.26 - 0.005 = 3.255 kg
Upper bound = 3.26 + 0.005 = 3.265 kg
3.255 \leq h < 3.265
Question 2(b): [2 marks]
The length of the strap on the handbag is 52 cm to the nearest cm.
Find the error interval in which the true length of the strap, l, in cm.
ANSWER:
51.5 \leq h < 52.551.5 \leq h < 52.5 51.5 \leq h \leq 52.5 51.5 < h < 52.5 51.5 < h \leq 52.5
WORKING:
We need to half the size of the interval, which is 1 (cm)
So 1 \div 2 = 0.5
Lower bound = 52 - 0.5 = 51.5 cm
Upper bound = 52 + 0.5 = 52.5 cm
51.5 \leq h < 52.5
Question 3(a): [2 marks]
David measures the amount of hot water into a cup, which is 380 ml to the nearest 10 ml.
Find the upper and lower bounds for amount of water in the cup, in ml.
ANSWER:
375\leq h < 385379.5\leq h < 380.5 370\leq h < 390 377.5\leq h < 382.5
WORKING:
We need to half the size of the interval, which is 10 (ml)
So 10 \div 2 = 5
Lower bound = 380 - 5 = 375 ml
Upper bound = 380 + 5 = 385 ml
375\leq h < 385
Question 3(b): [2 marks]
The temperature of the water is 71.4 \degree\text{C} to 1 decimal place.
Find the upper and lower bounds for the temperature of the water, in \degree \text{C}
ANSWER:
71.35\leq h < 71.4571.3\leq h < 71.5 71.325\leq h < 71.425 71.39\leq h < 71.41
WORKING:
We need to half the size of the interval, which is 1 decimal place (0.1)
So 0.1 \div 2 = 0.05
Lower bound = 71.4 - 0.05 = 71.35 \degree \text{C}
Upper bound = 71.4 + 0.05 = 71.45 \degree \text{C}
71.35\leq h < 71.45
Question 4: [2 marks]
Gwen has £3000 in savings to the nearest £100.
Find the error interval for the amount of money Gwen has in savings, M.
Answer type: Multiple choice type 1
A: £2950 \leq M < £3050
B: £2900 leq M < £3100
C: £2995 \leq M < £3005
D: £2500 \leq M < £3500
ANSWER: A: £2950 \leq M < £3050
WORKING:
We need to half the size of the interval, which is £100
So 100 \div 2 = 50
Lower bound = 3000 - 50 = £2950
Upper bound = 3000 + 50 = £3050
Error interval for M is therefore
£2950 \leq M < £3050
Question 5(a): [2 marks]
Diana rounds a number, x, to the nearest whole number.
The result is 1.
Find the error interval for x.
Answer type: Multiple choice type 1
A: 0.5 \leq x < 1.5
B: 0 \leq x < 2
C: 0.75 \leq x < 1.25
D: 0.95 \leq x < 1.05
ANSWER: A: 0.5 \leq x < 1.5
WORKING:
We need to half the size of the interval, which is a whole number (1)
So 1 \div 2 = 0.5
Lower bound = 1 - 0.5 = 0.5
Upper bound = 1 + 1.5 = 1.5
Error interval for x is therefore
0.5 \leq x < 1.5
Question 5(b): [2 marks]
Diana rounds a different number, y, to 2 significant figures.
The result is 0.024
Find the error interval for y.
Answer type: Multiple choice type 1
A: 0.0235 \leq y < 0.0245
B: 0.023 \leq y < 0.025
C: 0.02 \leq y < 0.03
D: 0.235 \leq y < 0.245
ANSWER: A: 0.0235 \leq y < 0.0245
WORKING:
We need to half the size of the interval, which is 2 significant figures.
Here the first significant figure is the 2, since the digits before then are 0
So our interval that we need to half is 0.001
So 0.001 \div 2 = 0.0005
Lower bound = 0.024 - 0.0005 = 0.0235
Upper bound = 0.024 + 0.0005 = 0.0245
Error interval for y is therefore
0.0235 \leq y < 0.0245
Question 6(a): [2 marks]
The population of a small town is 33500 rounded to the nearest 500.
Find the error interval for the population of the town, P.
Answer type: Multiple choice type 1
A: 33250 \leq P < 33750
B: 33000 \leq P < 34000
C: 32500 \leq P < 34500
D: 33450 \leq P < 33550
ANSWER: A: 33250 \leq P < 33750
WORKING:
We need to half the size of the interval, which is 500
So 500 \div 2 = 250
Lower bound = 33500 - 250 = 33250
Upper bound = 33500 + 250 = 33750
Error interval for P is therefore
33250 \leq P < 33750
Question 6(b): [2 marks]
The population of a neighbouring town is 134700 rounded to 4 significant figures.
Find the upper and lower bounds for the population of this town.
Answer type: Multiple answer type 1
ANSWER:
134650\leq P < 134750134650\leq P \leq 134750 134650< P \leq 134750 134650< P < 134750
WORKING:
We need to half the size of the interval, which is 4 significant figures.
Here the first significant figure is the 1
So our interval that we need to half is 100
So 100 \div 2 = 50
Lower bound = 134700 - 50 = 134650
Upper bound = 134700 + 50 = 134750
Question 7:
Ranjit is taking part in his school sports day.
Question 7(a): [1 mark]
Ranjit runs a race in 25.28 seconds to 2 decimal places.
What is his race time truncated to 1 decimal place?
Answer type: Simple text answer
ANSWER: 25.2 seconds
WORKING:
We chop off all digits after the first decimal place.
So we ignore the 8 and get 25.2 seconds
Question 7(b): [2 marks]
Ranjit throws a ball 32.4 feet truncated to 1 decimal place.
What is the interval within which d, the distance he throws the ball, lies?
Answer type: Multiple choice type 1
A: 32.4 \leq d < 32.5
B: 32.4 \leq d < 32.45
C: 32.35 \leq d < 32.45
D: 32.4 \leq d \leq 32.5
ANSWER: A: 32.4 \leq d < 32.5
WORKING:
When this number was truncated, all of the digits after the first decimal place, 4, will have been chopped off.
Therefore the value of d is anything greater than or equal to 32.4, but less than 32.5
So, our error interval is 32.4 \leq d < 32.5