Transcript Ch1 Rate and Ratio - Wah Yan College, Kowloon

1
Rate and Ratio
Case Study
1.1 Rate
1.2 Ratio
1.3 Applications of Ratios
Chapter Summary
Case Study
Mandy is planning to study abroad next
year. She wants to compare the school
fees among 3 different countries.
The table below shows the school fees
in different currencies and their
corresponding exchange rates.
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 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.
P. 2
1.1 Rate
Rate is the comparison of 2 quantities of different kinds.
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’.
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.
P. 3
‘/’ means per.
1.1 Rate
Example 1.1T
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
Solution:
$51 000
6 months
 $8500/month
(a) Monthly income 
(b) Yearly income  $(8500 12) /year
 $102 000/year
P. 4
1.1 Rate
Example 1.2T
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.)
Solution:
(a) 1.5 m  (1.5  1000) km
 0.0015 km
1
1 s  minutes
60
1

hours
3600
1
Speed  0.0015 km 
h
3600
 5.4 km/h
P. 5
(b)
Time required
 (1350  1.5) s
 900 s
 (900  60) minutes
 15 minutes
(c) Distance run  1.5  (8  60) m
 720 m
1.1 Rate
Example 1.3T
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.)
Solution:
(a) Exchange rate  HKD 2386  EU 240
 9.9417 HKD/EU
 9.94 HKD/EU (cor. to 2 d. p.)
(b)
Amount of Hong Kong dollars she can get
 $(850  9.9417)
 $8450.45 (cor. to 2 d. p.)
P. 6
For higher accuracy,
we use 9.9417 as the
exchange rate in the
calculation in part (b).
1.2 Ratio
A. Basic Concepts of Ratio
Ratio is the comparison of quantities of the same kind. The ratio of
a
a to b is usually expressed as a : b or (where a  0 and b  0).
b
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.
We say that the ratio of the volume of orange juice to that of the soft
2
drink is 2 : 1. This can be also written in the form .
1
P. 7
1.2 Ratio
A. Basic Concepts of 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
75

40
15

8
 15 : 8
P. 8
1.2 Ratio
A. Basic Concepts of Ratio
Example 1.4T
If m : 4  (m  6) : 12, find the value of m.
Solution:
m : 4  (m  6) : 12
m m6

4
12
12m  4(m  6)
3m  m  6
2m  6
m3
P. 9
1.2 Ratio
A. Basic Concepts of Ratio
Example 1.5T
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.
Solution:
Number of coins that Nancy has  18  6
 24
Number of coins that Stella has  24  2
 12
(a) Required ratio
(b) Required ratio
 18 : 24
 12 : 18
3:4
2:3
P. 10
Note that a : b  b : a.
1.2 Ratio
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:
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.
6
Volume of the cup with less juice  650 mL
13
 300 mL
7
Volume of the cup with more juice  650 mL
13
 350 mL
P. 11
1.2 Ratio
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.
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.
Volume of the cup with less juice
6

Volume of bottle
(6  7 )
6

13
We can compare the
ratios directly, without
finding the exact value
of each small part.
6
mL  300 mL
13
Volume of the cup with more juice  (650  300) mL  350 mL
Volume of the cup with less juice  650 
P. 12
1.2 Ratio
A. Basic Concepts of Ratio
Example 1.7T
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.
Solution:
(a)
Number of male teachers : Number of female teachers
 27 : (57  27)
 27 : 30
 9 : 10
P. 13
1.2 Ratio
A. Basic Concepts of Ratio
Example 1.7T
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?
Solution:
(b) Let x be the number of female teachers hired.
Number of male teachers hired  8  x.
[27  (8  x)] : (30  x)  6 : 7
6(30  x)  7(35  x)
180  6x  245  7x
35  x 6

13x  65
30  x 7
x 5
6(30  x)  7(35  x)
∴ 5 female teachers has been hired.
P. 14
1.2 Ratio
B. Continued 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
Continued ratios can
only be expressed in the
form a : b : c, but not in
a fraction.
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.
P. 15
1.2 Ratio
B. Continued Ratio
Example 1.8T
If 3a  5b  4c, find the ratio a : b : c.
Solution:
Since 3a  5b  4c, we have 3a  5b and 5b  4c.
a 5
b 4


∴
and
b 3
c 5
∴ a : b  5 : 3 and b : c  4 : 5
a :b  5: 3  5  4 : 3  4
 20 : 12
b : c  4: 5 
43:53
12 : 15
a : b : c  20 : 12 : 15
P. 16
1. First, find the ratios
a : b and b : c.
2. Then, make the
common terms equal
in both ratios.
1.2 Ratio
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:
(a) Economy : Business
 12 : 1  24 : 2
Business : First 
2:1
2:1
Required ratio  24 : 2 : 1
P. 17
1.2 Ratio
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.
1
Number of first class seats  540
27
 20
P. 18
1.3 Applications of Ratios
A. Similar Figures
If 2 figures have the same shape but their sizes are not the same,
then the 2 figures are said to be similar.
If we compare the lengths of the corresponding sides in the 2 photos,
we will have:
Height of photo II Length of photo II

Height of photo I Length of photo I
In general, similar figures have the following property:
For 2 similar figures, the ratios of the
corresponding sides are always the same.
P. 19
1.3 Applications of Ratios
A. Similar Figures
Example 1.10T
In the figure, the 2 parallelograms are similar to each
other. Find x and y.
Solution:
6 5

3 x
6 y

3 8
6x  15
3y  48
x  2.5 (m)
y  16 (m)
P. 20
When finding the side
lengths of similar
figures, we should
identify which of them
are the corresponding
sides.
1.3 Applications of Ratios
A. Similar Figures
Example 1.11T
In the figure, a boy with a height of 1.8 m stands
in front of the tree. Assume that DABC and DDEF
are similar triangles. What is the length of his
shadow?
Solution:
Let y m be the length of his shadow, i.e., EF  y m.
5.2 1.3

1.8 y
5.2 y  2.34
y  0.45
∴ The length of his shadow is 0.45 m.
P. 21
1.3 Applications of Ratios
B. Scaling
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.
Note that 1 : n  n : 1.
P. 22
1.3 Applications of Ratios
B. Scaling
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.
We can also express
the scale in the form
1 cm : 15 km.
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.
P. 23
1.3 Applications of Ratios
B. Scaling
Example 1.12T
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.
Solution:
Scale of the picture  4.8 cm : 36 m
 4.8 cm : 3600 cm
4 .8

3600
1

750
 1 : 750
P. 24
1.3 Applications of Ratios
B. Scaling
Example 1.13T
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.
Solution:
Actual distance  (3.2  20 000) cm
 64 000 cm
 640 m
 0.64 km
P. 25
1.3 Applications of Ratios
B. Scaling
Example 1.14T
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.)
Solution:
Actual side length of the master bedroom  (2.5  n) m
 2.5n m
Similarly, the actual length and the actual width of the kitchen are
2n m and 1.5n m respectively.
∴ The required ratio  (2.5n  2.5n) m2 : (2n  1.5n) m2
 6.25 : 3
 25 : 12
P. 26
Chapter Summary
1.1 Rate
Rate is the comparison of 2 quantities of different kinds.
P. 27
Chapter Summary
1.2 Ratio
1. Ratio is the comparison of quantities of the same kind. The ratio of
a
a to b is usually expressed as a : b or (where a  0 and b  0).
b
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.
P. 28
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.
P. 29