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This document delves into the standardized paper sizes, particularly the a series, focusing on the mathematical relationships between different sizes. It explores the ratio of length to width in a4 paper, demonstrates how larger sizes are derived from smaller ones, and applies these concepts to calculate area, perimeter, and mass of different paper sizes. The document also includes practical examples involving carpet calculations and a cable car system, showcasing the real-world applications of these mathematical principles.
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Home Learning Pack:
Problems in Real-world Contexts
The leading Singaporean series adapted for the National Curriculum
Victor Chow
UK Consultant: Robert Wilne
Singapore Consultant: Berinderjeet Kaur
Problems in Real-world Contexts
5
9
7
9
4 0.
9
3
5
waste
C. Brick Wall
65
65
10
215
215
10
D. Magazine Holder
24
10
12
18
30
60
24
8
8
8
30 18 22
Side pieces Back piece Front piece Base
B C
A D
0.94 x
9m^ 0.34 x
70º (^) 70º
x
G. Population Pyramid for the UK
age
Population pyramid for the UK in 2015
Population (thousands) Source: Office for National Statistics
89 85 80 75 70 65 60 55 50 45 40 35 30 25 20 15 10 5 0 500 400 300 200 100 0 0 100 200 300 400 500
males females
Problems in Real-world
Contexts
AO A
A
A
A
1189
A A8A
The diagram shows the dimensions of ‘A series’ paper sizes. The complete sheet is of size A0 with length 1189 mm and area 1000 000 mm^2 (or 1 m^2 ). The dimensions of this series of paper have the special property that one piece of A0 paper can be divided into two pieces of A1 paper, one piece of A1 paper can be divided into two pieces of A2 paper and so on.
1. Find the width of a piece of A0 paper. Give your answer to the nearest mm. 2. (a) Find the area of a piece of A4 paper in m^2 , expressing your answer as a fraction. (b) Find the length and width of a piece of A paper to the nearest mm. 3. The mass of a piece of A0 paper is 80 grams. Find the mass of a piece of A4 paper of the same thickness and quality. 4. Sam puts some A4 posters in an envelope. The envelope has mass 25 grams. To minimise the cost of the postage, the total mass should not be greater than 100 grams. How many A4 posters can be placed in the envelope?
Solution
1. Let x mm be the width of a piece of A0 paper. As the area of a piece of A0 paper is 1000 000 mm^2 , 1189 × x = 1000 000 x = 1000 000 1189 x = 841.04… x = 841 (to the nearest mm) ∴ the width of a piece of A0 paper is 841 mm.
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2. (a) A0 paper = 2 A1 paper = 2 × 2 A2 paper = 2 × 2 × 2 A3 paper = 2 × 2 × 2 × 2 A4 paper = 16 A4 paper You can see how a piece of A0 paper is divided into 16 pieces of A4 paper in the diagram. ∴ area of a piece of A4 paper = 1 16 × area of a piece of A0 paper
= (^1) ×× 1 16
m^2
= 1 16 m^2
1 4 ×^ length of a piece of A0 paper = (^1) ×× 1189 4 = 297. = 297 mm (to the nearest mm)
Width of a piece of A4 paper = 1 4 × width of a piece of A0 paper
= (^1) ×× 841. 4 = 210. = 210 mm (to the nearest mm)
3. Mass of a piece of A4 paper = 1 16
× mass of a piece of A0 paper
= 1 ×× 80 16 = 5 g
4. Let n be the number of A4 posters in the envelope. Consider the total mass of the envelope and the posters: 25 + 5 n = 100 5 n = 75 n = 75 5 n = 15 ∴ 15 A4 posters can be placed in the envelope.
9
4
4 5
3
waste
0.50.
3
Mr Brown can buy 9 + 5 = 14 linear metres of carpet and cut three pieces of dimensions 9 m × 4 m, 5 m × 3 m and 4 m × 0.5 m from it to carpet the room (see diagram above). This way, total cost of carpet = 80 × 14 = £ area of waste = 4 × 0.5 + 1 × 1 = 3 m^2
The standard size of bricks in the UK is 215 mm long, 102.5 mm wide and 65 mm high. When laying bricks, mortar of 10 mm thick is used to hold the bricks together. One-half running bond is the basic pattern when laying a wall, i.e. approximately half of a brick’s length overlaps the brick below.
65
65
10
215
215
10
1. In the diagram, the top and the third rows use three whole bricks, while bricks in the second and bottom rows have to be cut at both ends. Find (a) the length and height of this arrangement in cm, (b) the volume of this arrangement to the nearest 10 cm^3 if the thickness of this arrangement is one brick width. 2. Mr Tamworth is going to build a wall 2 m high, 5 m long and 102.5 mm thick. Estimate the number of bricks required. State any assumptions you make in your calculation.
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3. Mr Tamworth wants to buy 5% more bricks than the number required to allow for breakage and waste. How many bricks should he buy?
Solution
1. (a) Length = 3 × length of brick + 2 × thickness of mortar = 3 × 21.5 + 2 × 1 cm = 66.5 cm Height = 4 × height of brick + 3 × thickness of mortar = 4 × 6.5 + 3 × 1 cm = 29 cm (b) Volume = lhw = 66.5 × 29 × 10. = 19 767. = 19 770 cm^3 (to the nearest 10 cm^3 )
65
10
10
Make the assumption that each brick is a ‘nominal brick’ with mortar of thickness 10 mm on the bottom and on the right side as shown above i.e. each nominal brick is 225 mm long and 75 mm high. Area of the front face of a nominal brick = 0.225 × 0.075 m^2 = 0.016 875 m^2 Area of the wall = 5 × 2 m^2 = 10 m^2 Number of nominal bricks required = 10 ÷ 0.016 875 = 593 ( to the nearest integer) ∴ an estimate of the number of bricks required for the wall is 593.
3. Number of bricks that should be bought = 593 + 593 × 5% = 623 (to the nearest integer)
24
30
12
10
18
GCSE Grade 4 –
Problems in Real-world Contexts
Jenny makes a magazine holder using wooden board 1 cm thick. The two side pieces are cut from a board of 60 cm by 24 cm as shown below. A quarter of a circle of radius 12 cm is cut from each side in the middle.
60
24
8
8
8 30 18 22
Back piece
Front piece
Side pieces Base
The back end piece is 30 cm high and 8 cm wide. The front end piece is 18 cm high and 8 cm wide. The base piece is 22 cm long and 8 cm wide. A piece of wood of 90 cm by 24 cm by 1 cm is ordered to make the magazine holder.
1. Find the volume of the piece of wood ordered. 2. Find the perimeter of each side piece to the nearest cm. 3. Find the percentage of wastage from the piece of wood that is ordered to the nearest 1%.
Solution
1. Volume of the piece of wood ordered = 90 × 24 × 1 = 2160 cm^3 2. Perimeter of each side piece = 30 + 24 + (30 − 12) + 2 12 1 4 ××^ ××^ π^ ×× + (24^ −^ 12) = 84 + 6π = 103 cm (to the nearest cm)
8 8
30 Back piece
18 Front piece
22 Base
Side pieces
60
24 8
The diagram shows the way to cut the pieces from the piece of wood bought. The shaded part is the wastage. Total area of the two quarters cut out = 12 1 2 ×× π××^2
= 72 π cm^2 Total area of the shaded rectangles = (30 − 22) × 8 + (30 − 18) × 8 = 160 cm^2
Percentage of wastage = (^72 160) 100% 90 24 ××
++ ××
π
= 18% (to the nearest 1%)
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The above diagram shows a clearway road sign. It is used to indicate no stopping on the main carriageway. The sign is a circle of diameter 60 cm. The red border and the red cross are 6 cm wide. The blue parts are four congruent shapes.
1. (a) State the number of lines of symmetry on the sign. (b) Does the sign possess rotation symmetry? If so, what is the order of the rotation symmetry? 2. Explain why each blue part is not a quarter of the inner circle. 3. Estimate the area of the red region of the sign. State any assumptions you make. Give your answer to the nearest cm^2.
Solution
1. (a)
The sign has 4 lines of symmetry as shown above. (b) The sign has rotation symmetry. It repeats itself 4 times in a full turn about its centre. The order of rotation symmetry is 4.
2.
As shown in the above diagram, the two white diameters of the inner circle divide the inner circle into four quarters. This means the curved line of each blue area is part of the circle with the centre at the centre of the road sign, not at the point of intersection of the straight sides. Therefore, each blue area is only a part of a quarter, not a quarter.
GCSE Grade 2 +
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Problems in Real-world Contexts
A population pyramid shows the age and sex of a population. The diagram shows the population pyramid for the UK in 2015.
age
Population pyramid for the UK in 2015
Population (thousands)
89 85 80 75 70 65 60 55 50 45 40 35 30 25 20 15 10 5 0 500 400 300 200 100 0 0 100 200 300 400 500
males females
Source: Office for National Statistics
1. Assume all Year 7 students are 11 years old. Estimate the total number of Year 7 students from the population pyramid. 2. Among the age groups 5 to 10, 25 to 30 and 45 to 50, which age group has the largest number of people? Explain your answer. 3. Compare the shape of the population by age for males and females. What can you say about the number of males and females under 20, aged 50 to 60 and over 70?
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Solution
1. Reading from the graph, number of boys at age 11 ≈ 370 000 number of girls at age 11 ≈ 370 000 Therefore, the total number of Year 7 students = the total number of boys and girls at age 11 ≈ 370 000 + 370 000 = 740 000 Note: In fact, most students in Year 7 are age 11 or 12, and most students in Year 8 are age 12 or 13, and so on. You can estimate the total number of students in Year 7 by considering all the children who are 11 or all the children who are 12. 2. The bars for the age group 45 to 50 for both male and female are longer than the bars for the other two age groups. Therefore, the age group 45 to 50 has the largest number of people. 3. For people under 20 and aged 50 to 60, the lengths of the bars for males for females are nearly the same for each age. Therefore, the numbers of males and females at these ages are more or less the same. However, the numbers of women over 70 begin to outnumber men. In particular, this is quite obvious for ages over 80. This means, in general, the lifetime of women is longer than the lifetime of men.
The leading Singaporean series adapted for the National Curriculum
Victor Chow
UK Consultant: Robert Wilne
Singapore Consultant: Berinderjeet Kaur
Problems in Real-world Contexts
C. Prepaid Mobile Phone Plans
D. Paper Clip
E. Medals Won in 2016 Summer Olympics
Rank Country Gold Silver Bronze
F. Stretching a Metal Spring
(^1) (^2) (^3) (^4) (^5) (^6) (^7) (^8)
(^9) 10 11 12 13 14 15 16 17
19 20 21 22 23
25
24
18
clamp stand
helical spring
pointer
mass hanger
vertical ruler
cm
3. A pie chart would not be suitable as there are too many cities.
The charges for prepaid mobile phone plans offered by two communications companies are as follows.
Company A National calls will be charged at 16 pence per minute during day time and 8 pence per minute during night time.
Company B National calls will be charged at 22 pence for the first minute and 8 pence per minute thereafter for the whole day.
Your friend from the United States is visiting the United Kingdom for seven days. He would like you to suggest a phone plan to purchase during his visit.
1. What would you need to consider when helping your friend decide on which phone plan to purchase? 2. Assume that he makes 10 phone calls in the day and 5 phone calls at night for each day, and each call lasts about 3 minutes. Calculate his daily call charges for each plan. 3. Suppose you wish to propose another plan that charges the same rate per minute for each phone call throughout the day. What would be your plan? How would your plan be more economical? Justify your explanation.
Solution
1. You need to consider the preference of the caller to make his calls during the day or night time and how long each phone call lasts. 2. (^) Company A Company B
Daily call charge = £0.16 × 10 × 3 + £0.08 × 5 × 3 = £4.80 + £1. = £6.
Daily call charge = £0.22 × 15 + £0.08 × 15 × 2 = £3.30 + £2. = £5.
3. Answer depends on plan chosen.
The diagram shows a paper clip which is made from a circular wire of diameter 1 mm. The dimensions shown are in centimetres. The clip consists of three semicircular arcs and four straight edges.
Suppose you are the manufacturer of the paper clip and you would like to know more about the amount of material required to make the clip.
1. Calculate the total length of the wire needed to make a clip. State the assumption you have to make in your calculation. 2. What is the volume (in cm^3 ) of the material used for making each clip? 3. The material of the clips has a density of 7.6 grams/cm^3. The clips are packed in a box of 200 pieces. Find the total mass of the clips in a box.
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Problems in Real-world Contexts
4. The wire of the clip is opened out and then bent into these shapes: (i) a square, (ii) a rectangle, (iii) a circle. (a) Find a possible set of dimension(s) for each shape. (b) Find the area of each shape. (c) Given the same perimeter (in this case, the length of wire is fixed), which shape gives the largest area?
Solution
1. Assume that all 3 semicircles are of diameter 1 cm. Total length of wire = 3 × π × 0.5 + 4 + 3 + 3 + 2 = 16.7123... = 16.7 cm (to 3 sf ) 2. Volume = π × 0.05^2 × 16.7123... = 0.1312... = 0.131 cm^3 (to 3 sf ) 3. Total mass = 7.6 × 0.1312... × 200 = 199.512... = 200 g (to 3 sf ) 4. (^) (i) Square (ii) Rectangle (iii) Circle
(a) Dimension(s)
Length = 16.7123... ÷ 4 = 4.1780... = 4. ∴ 4.18 cm by 4.18 cm
Length + Width = 16.7123... ÷ 2 = 8.3561... ∴ 3.36 cm by 5 cm 4.36 cm by 4 cm 5.36 cm by 3 cm etc.
2 π r = 16.7123... r = 2.66 cm
(b) Area
= 17.5 cm^2
3.36 × 5 = 16.8 cm^2 4.36 × 4 = 17.4 cm^2 5.36 × 3 = 16.1 cm^2
π r^2 = π(2.6598...)^2 = 22.2 cm^2
(c) From the table above, the circle gives the largest area.
The table shows the number of medals won by the top 10 countries in the 2016 Summer Olympics in Rio de Janeiro.
Rank Country Gold Silver Bronze 1 United States 46 37 38 2 Great Britain 27 23 17 3 China 26 18 26 4 Russia 19 17 20 5 Germany 17 10 15 6 Japan 12 8 21 7 France 10 18 14 8 South Korea 9 3 9 9 Italy 8 12 8 10 Australia 8 11 10
1. Represent the data using a graph, showing the gold, silver and bronze medals won by the countries. Hint: You may use a spreadsheet program to help you draw the appropriate graph. 2. Besides ranking by gold medals, what other ways would you suggest ranking the countries? Will the ranking be different? Which way of ranking do you think is most reasonable? Explain your answer.
GCSE Grade 5
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