1. Understand what is meant by UPPER and LOWER BOUNDS
Objectives
Understand what is meant by UPPER and LOWER BOUNDS
Use these to calculate bounded answers
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2. What is the height of each of these herbs to the nearest centimetre?
8 cm
8 cm

3. We have rounded this size up to 8 cm
We have rounded this size down to 8 cm

4. Measuring to the nearest centimetre, we would round this plant height - which looks to be 7.6 cm - up to 8 cm

5. Measuring to the nearest centimetre, we would round this plant height - which looks to be just under 8.4 cm - down to 8 cm

6. We call the numbers 7.5 cm and 8.5 cm the bounds of 8 cm.
To the nearest cm, a length of 8 cm is any length between 7.5 and … (8.5) cm.
8.5 cm is the upper bound
We call the numbers 7.5 cm and 8.5 cm the bounds of 8 cm.
7.5 cm is the lower bound

7. 1 mm is 0.1 cm so it will be plus and minus half of 0.1
To the nearest mm, a length of 2 cm is any length between 1.95 and 2.05 cm.
1 mm is 0.1 cm so it will be plus and minus half of 0.1
± 0.05
2.05 cm is the upper bound
1.95 cm is the lower bound

8. Using the units required for the bounds, what would be the upper and lower bounds of these:
Units for bounds
Size
To nearest
L. Bound
U. Bound
12 cm cm
54 cm mm
27 metres
46 litres
1/10 litres
11.5 cm
12.5 cm
53.95 cm
54.05 cm m m
45.95 litres
46.05 litres

9. What are the upper and lower bounds?
Size
To nearest
Lower B.
Upper B.
19 metres
metre
33 cm cm
7 mm mm
67 cm
63 cm
72 cm
156 metres
81 metre
40 km

10. What are the upper and lower bounds?
Size
To nearest
Lower B.
Upper B.
19 metres
metre
18.5 metres
19.5 metres
33 cm cm
32.5 cm
33.5 cm
7 mm mm
6.5 mm
7.5 mm
67 cm
66.95 cm
67.05 cm
63 cm
62.95 cm
63.05 cm
72 cm
71.95 cm
72.05 cm
156 metres m m
81 metre m m
40 km km km

11. Bounds and Linear Calculations

12. Five cuboids each have a height of 12 cm to the nearest mm
Five cuboids each have a height of 12 cm to the nearest mm. If the cuboids are stacked as shown, what is the minimum and maximum height of the pile?
Height
12 cm

13. To the nearest mm, the bounds of 12 cm are…
Five cuboids each have a height of 12 cm to the nearest mm. If the cuboids are stacked as shown, what is the minimum and maximum height of the pile?
To the nearest mm, the bounds of 12 cm are…
Height
12 cm
Lower Bound is 11.95
Upper Bound is 12.05
Minimum height
11.95 x 5 = cm
Maximum height
12.05 x 5 = cm

14. To the nearest mm, the bounds of 5 cm are…
To the nearest millimetre, the sides of this regular hexagon are 5 cm long. What are the upper and lower bounds of its perimeter?
Lower Bound is 4.95
Upper Bound is 5.05
5 cm
Lower bound of the perimeter is:
4.95 x 8 = 39.6 cm
Upper bound of the perimeter is:
5.05 x 8 = 40.4 cm
To the nearest mm, the bounds of 5 cm are…

15. Here are four for you to try…

16. If the dimensions are to the nearest mm
If the dimensions are to the nearest mm. What are the upper and lower bounds for each shape’s perimeter?
Regular hexagon
7 cm
Isosceles triangle
12 cm
LB cm
UB cm
6.95 x 6
7.05 x 6
LB 41.7 cm
UB 42.3 cm
11.95 x 3
12.05 x 3
Regular pentagon
Rhombus
8.95 x 5
9.05 x 5
LB cm
UB cm
9.95 x 5
10.05 x 5
LB 39.8 cm
UB 40.2 cm
9 cm
10 cm

17. Bounds and Area Calculations

18. Find the bounds of its lengths Use these to find respective areas…
If the lengths are to the nearest mm, what are the upper and lower bounds of this rectangle’s area?
Upper B.
4.55
Find the bounds of its lengths
12.5 cm
4.5 cm
Lower B.
4.45
Upper B
Use these to find respective areas…
Lower B
Lower Bound Area
12.45 x 4.45
= cm2 to 2 dp
Upper Bound Area
12.55 x 4.55
= cm2 to 2 dp

19. Here are four for you to try…

20. If the lengths are to the nearest mm, what are the upper and lower bounds of the shape’s areas?
15.2 cm
5.6 cm
12.7 cm
8.9 cm
LB  cm2 to 2 dp
UB  cm2 to 2 dp
LB  cm2 to 2 dp
UB  cm2 to 2 dp
10.3 cm
6.2 cm
2.7 cm
Assume
∏ = 3.14
10.2 cm
diameter
LB  cm2 to 2 dp
UB  cm2 to 2 dp
LB  cm2 to 2 dp
UB  cm2 to 2 dp

21. Bounds and Volume Calculations

22. Find the bounds of its lengths Use these to find respective volumes…
If the lengths are to the nearest mm, what are the upper and lower bounds of this cuboid’s volume?
Find the bounds of its lengths
Use these to find respective volumes…
UB 10.05
LB 9.95
UB 17.55
LB 17.45
10 cm
17.5 cm
UB 9.25
LB 9.15
9.2 cm
Lower Bound Volume
9.15 x 9.95 x 17.45
= cm3 to 2 dp
Upper Bound Volume
9.25 x x 17.55
= cm3 to 2 dp

23. Here are two for you to try…

24. Find the bounds of its lengths
If the lengths are to the nearest mm, what are the upper and lower bounds of this cuboid’s volume?
Find the bounds of its lengths
Assume
∏ = 3.14
8.3 cm radius
UB 8.35
LB 8.25
15.2 cm
UB 15.25
LB 15.15
Lower Bound Volume
3.14 x x 15.15
= cm3 to 2 dp
Upper Bound Volume
3.14 x x 15.25
= cm3 to 2 dp

25. Find the bounds of its lengths Use these to find respective volumes…
If the lengths are to the nearest cm, what are the upper and lower bounds of this triangular prism’s volume?
Find the bounds of its lengths
1.5 metres
1.7 metres
2.5 metres
Use these to find respective volumes…
UB 1.505
LB 1.495
UB 2.505
LB 2.495
UB 1.705
LB 1.695
Lower Bound Volume
1.695 x ÷ 2 x 2.495
= 3.16 metres3 to 2 dp
Upper Bound Volume
1.705 x ÷ 2 x 2.505
= 3.21 metres3 to 2 dp

26. Bounds and Calculations

27. A joiner has lengths of timber that are 2
A joiner has lengths of timber that are 2.4 metre long to the nearest cm. It is possible to set a joiner’s circular saw so that it will consistently cut wood accurately to the nearest mm of the setting size. The saw’s blade is exactly 2 mm wide. If the joiner cuts 5 pieces on a setting of 36 cm long, what will be the lower bound length of the unused timber?
36 cm
Unused length
Do a sketch of the problem
5 pieces 36 cm to nearest mm from a length 2.4 metres long to the nearest cm
We can assume blade cut is exactly 2 mm wide
We want the upper bound of the unused length

28. A joiner has lengths of timber that are 2
A joiner has lengths of timber that are 2.4 metre long to the nearest cm. It is possible to set a joiner’s circular saw so that it will consistently cut wood accurately to the nearest mm of the setting size. The saw’s blade is exactly 2 mm wide. If the joiner cuts 5 pieces on a setting of 36 cm long, what will be the lower bound length of the unused timber?
36 cm
Unused length
Work in cm
Find the bounds
UB 36.05
LB 35.95
UB 240.5
LB 239.5
36 cm
240 cm
Unused LB = – 5 x – 1  cm
Blade 5 x 2 mm = 1 cm

29. Here is one for you to try…

30. Upper Bound Petrol consumption =
Notice that to find upper bound consumption, the upper bound is divided by the lower bound
Kathy drove 238 miles correct to the nearest mile. She used 23.7 litres of petrol correct to the nearest tenth of a litre. Workout the upper bounds for the petrol consumption for Kathy’s journey. Give your answer to 2 d.p.
Miles travelled
Litres of petrol used
Petrol consumption =
Find the bounds
238 miles
UB 238.5
LB 237.5
23.7 litres
UB 23.75
LB 23.65
Upper Bound Petrol consumption =
10.08 miles/litre