rate and ratio 1 1.1rate 1.2ratio 1.3applications of ratios case study chapter summary
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Rate and Ratio1
1.1 Rate
1.2 Ratio
1.3 Applications of Ratios
Case Study
Chapter Summary
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
Country Annual School Fee Exchange Rate
Australia AUD 17 000 AUD 1 HKD 5.85
Great Britain GBP 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.
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.
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) $/year
incomeMonthly (a)
incomeYearly (b)
/month8500$months 6
000 51$
000/year $102/year)128500$(
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
hours 3600
1
minutes60
1s 1
h 3600
1km 0015.0Speed
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
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 9.941
7) (cor. to 2 d. p.) $8450.45 For higher accuracy, we use 9.9417 as the exchange rate in the calculation in part (b).
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). b
a
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 .1
2
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
40
75
A. Basic Concepts of Ratio
15 : 88
15
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
12
6
4
mm
12m 4(m 6) 3m m 6 2m 6 m 3
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
24Number of coins that Stella has 24 2
12
P. 11
1.2 Ratio
mL 13
6650
mL 13
7650
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 1.6TEmily 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
mL 300
Volume of the cup with more juice
mL 350
P. 12
1.2 Ratio
Alternative Solution:
mL 13
6650
mL )300650(
)76(
6
bottle of Volume
juice less with cup theof Volume
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
13
6
Volume of the cup with less juice mL 300
Volume of the cup with more juice mL 350
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.
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
10:9 27 :
30
Solution:
P. 14
1.2 Ratio
7
6
30
35
x
x
∴ 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 manyfemale teachers has the principal hired?
[27 (8 x)] : (30 x) 6 : 7
6(30 x) 7(35 x)
6(30 x) 7(35 x) 180 6x 245 7x 13x 65 x 5
Solution:
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
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.
3
5b
a
5
4c
b∴ 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.
15:12:20:: cba
a : b 5 :3b : c 4 : 5
20 :12
12 : 15
5 4 : 3 4 4 3 : 5 3
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
P. 18
1.2 Ratio
Number of first class seats 27
1540
B. Continued Ratio
Example 1.9TThere 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.
20
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:
I photo ofLength
II photo ofLength
I photo ofHeight
II photo ofHeight
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.
P. 20
In the figure, the 2 parallelograms are similar to each other. Find x and y.
x
5
3
6
6x 15
83
6 y
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.
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.
y
3.1
8.1
2.5
∴ The length of his shadow is 0.45 m.
1.3 Applications of Ratios
A. Similar Figures
Example 1.11T
Solution:
45.0y
34.22.5 y
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.
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 form1 cm : 15 km.
1.3 Applications of Ratios
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
750:1
B. Scaling
Example 1.12T
Solution:
4.8 cm : 3600 cm
3600
8.4
750
1
1.3 Applications of Ratios
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
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) m2 : (2n 1.5n) m2
2.5n mSimilarly, the actual length and the actual width of the kitchen are 2n m and 1.5n m respectively.
6.25 : 3 25 : 12
P. 27
Chapter Summary
1.1 Rate
Rate is the comparison of 2 quantities of different kinds.
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).b
a
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.