1. Rate and Ratio 1 1.1Rate 1.2Ratio 1.3Applications of Ratios Case Study Chapter Summary

2. P. 2 Mandy is planning to study abroad next year. She wants to compare the school fees among 3 different countries. The exchange rate is the amount of Hong Kong dollars needed to exchange for one unit of a foreign currency. We can use the exchange rate to convert the foreign school fees into Hong Kong dollars first, and then compare them. Case Study CountryAnnual School FeeExchange Rate AustraliaAUD 17 000 AUD 1  HKD 5.85 Great BritainGBP 8400 GBP 1  HKD 14.80 U.S.A.USD 20 000 USD 1  HKD 7.78 The table below shows the school fees in different currencies and their corresponding exchange rates.

3. P. 3 In the figure, the price of milk per carton is found by comparing 2 different kinds of quantities: ‘the total price’ and ‘the number of cartons’. Rate is the comparison of 2 quantities of different kinds. When expressing the relationship in rate, we express the amount of one quantity as per unit of the other quantity, by the symbol ‘/’. For example, the price of each brand of milk can be expressed as ‘$8/carton’ and ‘$7/carton’ respectively. 1.1 Rate ‘/’ means per.

4. P. 4 Example 1.1T 1.1 Rate Solution: Andy works as a programmer in a computer company. He earns a total of $51 000 in half a year. Find his income in the following units: (a) $/month(b) $/year

5. P. 5 Betty runs at a speed of 1.5 m/s. (a)Express her speed in the unit km/h. (b)How long does she take to run 1350 m? (Give the answer in minutes.) (c)How far can she run in 8 minutes? (Give the answer in m.) (a)1.5 m  (1.5  1000) km  0.0015 km Example 1.2T 1.1 Rate Solution:  5.4 km/h (b) Time required  (1350  1.5) s  900 s  (900  60) minutes  15 minutes (c)Distance run  1.5  (8  60) m  720 m

6. P. 6 Mrs. Wong gets 240 Euros (EU) for HKD 2386. (a)Find the exchange rate in the unit HKD/EU. (b)How much Hong Kong dollars can she get with 850 Euros? (Give the answers correct to 2 decimal places.) (cor. to 2 d. p.) Example 1.3T 1.1 Rate Solution: (a)Exchange rate  HKD 2386  EU 240  9.9417 HKD/EU  9.94 HKD/EU (b) Amount of Hong Kong dollars she can get  $(850  9.9417) (cor. to 2 d. p.)  $8450.45 For higher accuracy, we use 9.9417 as the exchange rate in the calculation in part (b).

7. P. 7 1.2 Ratio In the figure, a fruit punch is mixed by adding a cup of soft drink into 2 cups of orange juice. That means, the volume of orange juice in the fruit punch is always twice that of the soft drink. We compare 2 quantities by division: ‘the volume of orange juice’ and ‘the volume of soft drink’ and these quantities are of the same kind. Ratio is the comparison of quantities of the same kind. The ratio of a to b is usually expressed as a : b or (where a  0 and b  0). A. Basic Concepts of Ratio We say that the ratio of the volume of orange juice to that of the soft drink is 2 : 1. This can be also written in the form.

8. P. 8 1.2 Ratio A ratio is usually expressed in its simplest form, e.g. 75 : 40  15 : 8. A ratio can be written as a fraction and we know that the value of the fraction remains unchanged when we multiply (or divide) both the numerator and the denominator by the same non-zero number. For example,0.75 m : 40 cm  75 cm : 40 cm A. Basic Concepts of Ratio  15 : 8

9. P. 9 1.2 Ratio If m : 4  (m  6) : 12, find the value of m. A. Basic Concepts of Ratio Example 1.4T Solution: m : 4  (m  6) : 12 12m  4(m  6) 3m  m  6 2m  6 m  3

10. P. 10 1.2 Ratio Peter has 18 coins. Nancy has 6 more coins than Peter, and she has twice as many as Stella. Find the ratio of (a)Peter’s coins to Nancy’s coins, (b)Stella’s coins to Peter’s coins. A. Basic Concepts of Ratio Example 1.5T Solution: Note that a : b  b : a. Number of coins that Nancy has  18  6 (a) Required ratio  18 : 24 (b) Required ratio  12 : 18  3 : 4  2 : 3  24 Number of coins that Stella has  24  2  12

11. P. 11 1.2 Ratio Since the volumes of juice are in the ratio 6 : 7, we can imagine that the bottle of apple juice is divided into (6  7)  13 equal parts. A. Basic Concepts of Ratio Example 1.6T Emily bought a bottle of apple juice of volume 650 mL. She pours the juice into 2 cups such that the volumes of juice in these cups are in the ratio 6 : 7. Find the volume of juice in these 2 cups. Solution: Volume of the cup with less juice Volume of the cup with more juice

12. P. 12 1.2 Ratio Alternative Solution: If we divide the juice into 13 parts, then 6 parts belong to the cup with less juice and the other 7 parts belong to the cup with more juice. A. Basic Concepts of Ratio Example 1.6T Volume of the cup with less juice Volume of the cup with more juice We can compare the ratios directly, without finding the exact value of each small part. Emily bought a bottle of apple juice of volume 650 mL. She pours the juice into 2 cups such that the volumes of juice in these cups are in the ratio 6 : 7. Find the volume of juice in these 2 cups.

13. P. 13 1.2 Ratio (a)Number of male teachers : Number of female teachers  27 : (57  27) Educational Secondary School has a total of 57 teachers, of which 27 of them are male teachers. (a)Find the ratio of the number of male teachers to the number of female teachers. A. Basic Concepts of Ratio Example 1.7T  27 : 30 Solution:

14. P. 14 1.2 Ratio ∴ 5 female teachers has been hired. A. Basic Concepts of Ratio Example 1.7T (b)Let x be the number of female teachers hired. Number of male teachers hired  8  x. Educational Secondary School has a total of 57 teachers, of which 27 of them are male teachers. (b)The principal has just hired 8 new teachers. The ratio of male teachers to female teachers now becomes 6 : 7. How many female teachers has the principal hired? [27  (8  x)] : (30  x)  6 : 7 6(30  x)  7(35  x) 180  6x  245  7x 13x  65 x  5 Solution:

15. P. 15 1.2 Ratio We can also use ratio to compare 3 or more quantities of the same kind. For example, the expression a : b : c  4 : 5 : 9 compares the 3 quantities a, b and c, with a : b  4 : 5, b : c  5 : 9 and a : c  4 : 9. Such an expression is called a continued ratio. For 3 quantities given, if we only know the ratio between individual quantities, we can rewrite the ratios into a continued ratio. Continued ratios can only be expressed in the form a : b : c, but not in a fraction. B. Continued Ratio

16. P. 16 1.2 Ratio If 3a  5b  4c, find the ratio a : b : c. B. Continued Ratio Example 1.8T Solution: Since 3a  5b  4c, we have 3a  5b and 5b  4c. ∴ and ∴ a : b  5 : 3 and b : c  4 : 5 1.First, find the ratios a : b and b : c. 2. Then, make the common terms equal in both ratios. a :b 5 :3b : c 4 : 5a :b 5 :3b : c 4 : 5  20 :12  12 : 15  5  4 :3  4  4  3 : 5  3

17. P. 17 1.2 Ratio There are 540 seats in a plane. The number of economy class seats and business class seats are in the ratio 12 : 1. The number of business class seats and first class seats are in the ratio 2 : 1. (a)Find the ratio of the number of economy class seats : the number of business class seats : the number of first class seats. (b)Find the number of first class seats. B. Continued Ratio Example 1.9T Solution: (a) Economy :Business  12 :1 Business : First  2 : 1  24 :2  2 : 1 Required ratio  24 : 2 : 1

18. P. 18 1.2 Ratio Number of first class seats B. Continued Ratio Example 1.9T There are 540 seats in a plane. The number of economy class seats and business class seats are in the ratio 12 : 1. The number of business class seats and first class seats are in the ratio 2 : 1. (a)Find the ratio of the number of economy class seats : the number of business class seats : the number of first class seats. (b)Find the number of first class seats. Solution: (b)We can imagine that the total number of seats can be divided into (24  2  1)  27 equal parts.

19. P. 19 A. Similar Figures 1.3 Applications of Ratios If we compare the lengths of the corresponding sides in the 2 photos, we will have: In general, similar figures have the following property: If 2 figures have the same shape but their sizes are not the same, then the 2 figures are said to be similar. For 2 similar figures, the ratios of the corresponding sides are always the same.

20. P. 20 In the figure, the 2 parallelograms are similar to each other. Find x and y. 6x  15 1.3 Applications of Ratios A. Similar Figures Example 1.10T 3y  48 x  2.5 (m) y  16 (m) Solution: When finding the side lengths of similar figures, we should identify which of them are the corresponding sides.

21. P. 21 In the figure, a boy with a height of 1.8 m stands in front of the tree. Assume that  ABC and  DEF are similar triangles. What is the length of his shadow? Let y m be the length of his shadow, i.e., EF  y m. ∴ The length of his shadow is 0.45 m. 1.3 Applications of Ratios A. Similar Figures Example 1.11T Solution:

22. P. 22 If we want to draw something which is very large or small in size, such as a country or an insect, we need to reduce or enlarge it according to a specified ratio in a diagram. This kind of drawing is called scale drawing. When using scale drawing, we need to specify the ratio in which the object is enlarged or reduced in the picture. This ratio is called the scale of the drawing, and is usually represented in the form 1 : n or n : 1. 1.3 Applications of Ratios B. Scaling Note that 1 : n  n : 1.

23. P. 23 For example, the map of Hong Kong Island shown has a scale of 1 : 1 500 000. This means that a length of 1 cm on the map represents an actual length of 1 500 000 cm. In the figure, a length of 1 cm on the figure represents an actual length of 0.2 cm. Thus the scale is 1 : 0.2, i.e., 5 : 1. B. Scaling We can also express the scale in the form 1 cm : 15 km. 1.3 Applications of Ratios

24. P. 24 The picture on the right shows the top view of a tennis court of actual length 36 m. If the length of the picture is 4.8 cm, find the scale of the picture. Scale of the picture  4.8 cm : 36 m B. Scaling Example 1.12T Solution:  4.8 cm : 3600 cm 1.3 Applications of Ratios

25. P. 25 Consider a map of a city with a scale of 1 : 20 000. If the distance between 2 buildings is 3.2 cm, find the actual distance between them. Give the answer in the unit of km. B. Scaling Example 1.13T Solution: Actual distance  (3.2  20 000) cm  64 000 cm  640 m  0.64 km 1.3 Applications of Ratios

26. P. 26 According to the floor plan, find the ratio of the actual area of the master bedroom to the actual area of the kitchen. (Hint: Assume the scale of the floor plan to be 1 cm : n m.) 1.3 Applications of Ratios B. Scaling Example 1.14T Solution: Actual side length of the master bedroom  (2.5  n) m ∴ The required ratio  (2.5n  2.5n) m 2 : (2n  1.5n) m 2  2.5n m Similarly, the actual length and the actual width of the kitchen are 2n m and 1.5n m respectively.  6.25 : 3  25 : 12

27. P. 27 Chapter Summary 1.1 Rate Rate is the comparison of 2 quantities of different kinds.

28. P. 28 Chapter Summary 1.2 Ratio 2.If the ratios a : b and b : c are given, we can find the continued ratio a : b : c by finding the L.C.M. of the values corresponding to the common term b. 1.Ratio is the comparison of quantities of the same kind. The ratio of a to b is usually expressed as a : b or (where a  0 and b  0).

29. P. 29 Chapter Summary 1.3 Applications of Ratios 1. Similar figures For 2 similar figures, the ratios of the corresponding sides are always the same. 2. Scale drawing If we reduce or enlarge the drawing of the real object by a certain scale, the drawing is similar to the original object.

30. In a supermarket, 4 bottles of orange juice are sold at $18. Express the cost of the orange juice in the following units: (a)$/bottle(b)$/dozen (b)1 dozen  12 bottles 1.1 Rate Follow-up 1.1 Solution: (a)Rate Rate

31. Alan cycles at a speed of 5 m/s. (a)Express his speed in km/h. (b)How far can he travel in 15 minutes? (c)If Tim takes 4.5 hours to finish 72 km, who is faster? 1.1 Rate Follow-up 1.2 Solution: (a)5 m  (5  1000) km  0.005 km  18 km/h (b) Distance travelled  18 km/h  15 minutes  18 km/h  0.25 h  4.5 km (c)Tim’s speed ∴ Alan is faster.

32. Suppose that we can exchange HKD 9000 for GBP 740 (British pound, the currency in Great Britain). (a)Find the exchange rate in the unit HKD/GBP. (b)How much HKD can be exchanged for GBP 110? (Give the answers correct to 2 decimal places.) 1.1 Rate Follow-up 1.3 Solution: (cor. to 2 d. p.) (a)Exchange rate  HKD 9000  GBP 740  12.1622 HKD/GBP  12.16 HKD/GBP (b) Amount of HKD we can get  $(110  12.1622) (cor. to 2 d. p.)  $1337.84

33. 1.2 Ratio A. Basic Concepts of Ratio If 6 : (p  2)  5 : 15, find the value of p. Solution: 6 : (p  2)  5 : 15 Follow-up 1.4 6  15  5(p  2) 90  5(p  2) 18  p  2 p  20

34. 1.2 Ratio Eric, Frank and Gary won prizes that are worth a total of $6000 in a lucky draw. Among them, Eric’s prize worth $1500, which is $500 less than Frank’s prize. Find the ratio of (a)the value of Frank’s prize to the value of Gary’s prize, (b)the value of Eric’s prize to the value of Gary’s prize. A. Basic Concepts of Ratio Follow-up 1.5 Solution: Value of Frank’s prize  $(1500  500) (a) Required ratio  $2000 : $2500 (b) Required ratio  $1500 : $2500  4 : 5  3 : 5  $2000 Value of Gary’s prize  $(6000  1500  2000)  $2500

35. 1.2 Ratio There is a total of 72 novels and textbooks in a bookshelf. The ratio of the number of novels to the number of textbooks is 5 : 4. Find the number of each kind of book. Since the total number of books are in the ratio 5 : 4, we can imagine that the total number of books is divided into (5  4)  9 equal portions. Number of novels  5  8 books Number of each part A. Basic Concepts of Ratio Follow-up 1.6 Solution: Number of textbooks  4  8 books  40 books  32 books

36. 1.2 Ratio Educational Secondary School has a total of 1200 students, of which 480 of them are boys. (a)Find the ratio of the number of boys to the number of girls. A. Basic Concepts of Ratio Follow-up 1.7 (a)Number of boys : Number of girls  480 : (1200  480)  480 : 720  2 : 3 Solution:

37. 1.2 Ratio A. Basic Concepts of Ratio Educational Secondary School has a total of 1200 students, of which 480 of them are boys. Follow-up 1.7 Solution: (b)Since the ratio of the numbers of boys and girls are in the ratio 2 : 3, we can imagine that the total number of students is divided into (2  3)  5 equal parts. ∴ Number of students in each part  1400  5  280 ∴ Number of girls  3  280  840 (b)The ratio of the number of boys to the number of girls in the Commercial School is the same as in the Educational Secondary School. If there is a total of 1400 students in the Commercial School, how many girls are there?

38. 1.2 Ratio If 2a  3b and b : c  8 : 7, find the ratio a : b : c. B. Continued Ratio Follow-up 1.8 Solution: a :b 3 :2b : c 8 : 7a :b 3 :2b : c 8 : 7  3  4 :2  4  8  1 : 7  1  12 :8  8 : 7

39. 1.2 Ratio Martin wants to mix sugar, salt and pepper in the ratio 3 : 8 : 5 by weight. If the total weight of the seasonings used is 24 g, find the weight of each seasoning used. We can imagine that the seasonings used can be divided into (3  8  5)  16 equal parts. Weight of saltWeight of pepper Weight of sugar B. Continued Ratio Follow-up 1.9 Solution:

40. A. Similar Figures If the 2 triangles shown in the figure are similar triangles, find the values of x and y. 4x  303y  18 x  7.5 y  6 Solution: Follow-up 1.10 1.3 Applications of Ratios

41. Let x m be the length of the shadow of the tree, i.e., BC  x m. ∴ The length of the shadow of the tree is 1.95 m. Eric is 1.6 m tall. He is standing in front of a tree which is 5.2 m tall. He casts a shadow of length 0.6 m. Assume that  ABC and  DEF are similar triangles. Find the length of the shadow of the tree. A. Similar Figures Follow-up 1.11 Solution: 1.3 Applications of Ratios

42. The picture shows the top view of a basketball court of actual length 39 m. If the length of the picture is 3.9 cm, find the scale of the picture. Scale of the picture B. Scaling Follow-up 1.12 Solution: 1.3 Applications of Ratios

43. In a map of scale 1 : 41 000, the length of the Eastern Harbour Tunnel on the map is 4.9 cm. Find the actual length of the Eastern Harbour Tunnel in m. Actual length of the Eastern Harbour Tunnel  (4.9  41 000) cm B. Scaling Follow-up 1.13 Solution:  200 900 cm  2009 m 1.3 Applications of Ratios

44. The figure shows the floor plan of a flat with a scale of 1 : 200. (a)Find the actual area of the bathroom in m 2. (b)Find the total actual area of the 2 bedrooms in m 2. B. Scaling Follow-up 1.14 Solution: (a)Actual length  (1.5  200) cm  3 m Actual width  (1  200) cm  2 m ∴ Actual area of the bathroom  (3  2) m 2  6 m 2 (b)Actual length  (4  200) cm  8 m Actual width  (2.5  200) cm  5 m ∴ Total actual area  (8  5) m 2  40 m 2 1.3 Applications of Ratios