f2 1st term maths teaching material (1st edition)
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
Unit 1 Estimation and Approximation
A. IntroductionMeasurements only give approximate values. Instruments with a finer
graduation give a more precise result. Precision have to be specified if we
want to get the same result in measurements.
B. Measurement with different precisionsExample 1.1
Measure the length of the line segments.
A B
X Y
Precisions
Line segments 0.1cm 0.5cm cm
AB 8.8cm 9.0cm 9cm
XY 5.4cm 5.5cm 5cm
C. Numerical Estimation1. Reformulation
Rounding
Round off the numbers to the nearest unit before doing the calculation.
Example 1.2
3.775+2.145-2.603
≈4+2-3
=3
1
Difference between “9.0cm” and “9cm”As an estimated value, the precision of “9.0cm” is less than 1cm and that of “9cm” is or greater than 1cm. In other words, “9.0cm” is more precise.
S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
Compensation
Choose a clustered value to represent all the numbers for estimation.
Example 1.3
61+58+63+59+57+59
≈60+60+60+60+60+60
=60x7
=420
Using Compatible numbers
Change the numbers to make the estimation easier.
Example 1.4
0.3356x181
≈(1 3)x180
=60
2. Compensation
Make adjustments in calculations.
Example 1.5
2217x8.03
=(2000x8.03)+(217x8.03)
≈(2000x8)+(200x8)
=16000+1600
=17600
Note: The remains 217x8.03 is also considered so that the estimated value
will be more precise.
3. Translation
Change the structure of the problem.
Example 1.6
(56.4x3) 22
=56.4x(3 22)
≈56x(3 21)
=56 7
=8
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
4. Rounding up/ Rounding Down
Rounding up
Choose slightly greater values to replace the original values.
Example 1.7a
13+26+18+22+34+48+15
≈20+30+20+30+40+50+20
=210
Rounding down
Choose slightly smaller values to replace the original values.
Example 1.7b
13+26+18+22+34+48+15
≈10+20+10+20+30+40+10
=140
Note: Rounding up is to estimate the greatest possible value of the sum of
numbers and rounding down is to estimate the smallest possible value of it.
Therefore, it is very important to know whether the greatest or smallest
possible value should be estimated before answering some long questions.
Example 1.8
There are 8 people in a lift. The maximum loading of the lift is 600kg and the
people’s weights are 67.3kg, 88.5kg, 53.2kg, 41.1kg, 107.3kg, 56.6kg, 48.9kg
and 50.3kg respectively. Estimate whether the lift will be overload.
Solution:
Rounding up the weights,
The sum of weight
=67.3+88.5+53.2+41.1+107.3+ 56.6+48.9+50.3
≈70+90+60+50+110+60+50+60
=550kg
Since the estimated value is lower than 600kg, the lift will not be overload.
Note: As all the weights of the people have been rounded up, the estimated
sum of weight must be greater than the real value of the sum. Therefore, if the
estimated sum is less than the maximum loading, the lift will not be overload.
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
Example 1.9
Peter, John, Joe and James have $7.1, $6.5, $8.4 and $9.9 respectively. A
football costs $30. Do they have enough money to buy it?
Solution:
Rounding down the values,
The sum of money
=7.1+6.5+8.4+10.9
≈7+6+8+10
=31
Since the estimated sum of money is greater than the price of the football,
they have enough money to buy it.
Note: Rounding down the amount of money each person has, the real sum of
money is greater the estimated one. As the estimated sum is greater than the
price of football, they must have enough money to buy it.
D. Significant numbersOne method of rounding off a number is to take their nearest units. Another
method is to take a few significant numbers (or significant figures) from
the number and delete all the less important digits.
Rules of selecting significant figures from a number
Numbers Rule
For numbers greater than 1 Starting from the first digit, all the digits in the
number are significant.
For numbers between 0 and 1 Digits starting from the first non-zero digit are
significant.
Example 1.10a
Find the first significant figure of the following numbers:
2.737154 1504331 0.093234 1.05113 0.90911
Solution:
2.737154 1504331 0.093234 1.05113 0.90911
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
Example 1.10b
Find the first three significant figures of the following numbers:
2.737154 1504331 0.093234 1.05113 0.90911
Solution:
2.737154 1504331 0.093234 1.05113 0.90911
Rounding off numbers to a certain number of significant figures
To round off a number, we can take the first few significant figures from a
number.
Example 1.11
Correct the following numbers to four significant figures.
637547 1.630299 0.00903993 1.800012 17.7371
Solution:
637500 1.630 0.009040 1.800 17.74
E. ErrorsAbsolute Error
The absolute error is the difference between the actual value and the
measured value. The error is “absolute” because it is always positive.
Example 1.12
Find the absolute errors of the following cases.
The length of a string The weight of a person
Actual value 13.75cm 115.3kg
Measured value 14cm 115kg
Solution:
The length of a string The weight of a person
Actual value 13.75cm 115.3kg
Measured value 14cm 115kg
Absolute error 14cm-13.75cm
=0.25cm
115.3kg-115kg
=0.3kg
5
Upper Limit = Measured value + (Max. absolute error)Lower Limit = Measured Value - (Max. absolute error)
S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
Maximum absolute error, upper and lower limit
However, since the actual figures are usually unknown, we often have to find
out the maximum absolute error so that we can obtain the range of values
of the actual value.
When the precision of a measurement and the measured value are stated, we
can find out the upper limit and lower limit of the actual figure, as well as the
maximum absolute error. The greater the maximum absolute error, the less
precise the figure is.
For example, the length of a line is estimated to be 9cm, correct to the nearest
cm.
In the range of 8.5cm and 9.5cm, the estimated length of the line will be 9cm.
Therefore, the actual length of the line will between 8.5cm and 9.5cm because
only such a range of values can be estimated at 9cm. The upper and lower
limits of the actual value will be 9.5cm and 8.5cm respectively.
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Maximum Absolute Error = ( Upper limit – Lower limit ) 2
S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
Example 1.13
Find the upper limit, the lower limit and the maximum absolute error of the
actual values.
Height of a person Weight of a ball
Measured value 173.0cm 793g
Correct to the nearest… 0.5cm 1g
Solution:
Height of a person Weight of a ball
Measured value 173.0cm 793g
Correct to the nearest… 0.5cm 1g
Upper Limit 173.25cm 793.5g
Lower Limit 172.75cm 792.5g
Maximum absolute error (173.25-172.75) 2
=0.5cm
(793.5-792.5) 2
=1g
Relative error & percentage error
A relative error and percentage error can represent the comparison between
the absolute error and the true value or that between the maximum
absolute error and the measured value.
Note: The value of a relative error and the corresponding percentage error are
the same. However, a relative error is represented by a fraction or a decimal
number and a percentage error is represented by a percentage.
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Relative Error = or
Percentage error = Relative error 100%
S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
Example 1.14a
The weight of a watermelon is measured to be 2.73kg, and the actual weight
of it is 2.7255kg. Find
i) The absolute error
ii) The relative error (corr. to 3 sig. fig.)
iii) The percentage error (corr. to 3 sig. fig.)
Solution:
i) The absolute error
=2.73-2.7255
=0.0045kg
ii) The relative error
=
=0.001651073…
=0.00165 (corr. to 3 sig. fig.)
iii) The percentage error
= x100%
=0.1651073....%
=0.00165% (corr. to 3 sig. fig.)
Example 1.14b
The weight of a watermelon is measured to be 2.73kg, correct to the nearest
0.01kg. Find
i) The upper and lower limit
ii) The maximum absolute error
iii) The relative error (corr. to 3 sig. fig.)
iv) The percentage error (corr. to 3 sig. fig.)
i) The upper limit
=2.735kg
The lower limit
=2.725kg
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
ii) The maximum absolute error
=(2.735-2.725) 2
=0.005kg
iii) The relative error
=
=0.00183150183…
=0.00183 (corr. to 3 sig. fig.)
iv) The percentage error
= x100%
=1.83150183…%
=1.83% (corr. to 3 sig. fig.)
F. AbbreviationsCorrect to 3 significant figures corr. to 3 sig. fig.
Correct to 2 decimal places corr. to 2 d.p.
G. Exercise1. Round off the following numbers.
corr. to 3 d.p. corr. to 3 sig.
fig.
corr. to 4 sig.
fig.
corr. to the
nearest
tenth
1.25703
0.89099
12.7356
0.004504504
corr. to the
nearest 10
corr. to the
nearest 1000
corr. to 3 sig. fig.
49509
290999
195034
13507
2. Round off 2090.999 to
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
the nearest ten
the nearest integer
1 decimal place
2 significant figures
3. Estimate the following.
i) 9.9+10.4+10.6+9.5+10.1+10.2+9.7
ii)
iii)
iv)
v)
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
vi)
vii)
viii)
ix)
x)
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
xi)
4. Complete the following table.
Time for running 100m Length of a chopstick
Absolute error
Measured value 12.93s 9.81cm
Actual value 12.937s 9.83cm
Relative error (in a form
of fraction)
Percentage error (corr.
to 3 sig. fig.)
5. The length of a piece of string is estimated at 10.3cm, correct to the
nearest 0.1cm.
i) Find the range of the actual length.
ii) Find the maximum absolute error.
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
iii) Find the relative error (corr. to 3 sig. fig. when necessary).
iv) Find the percentage error (corr. to 3 sig. fig. when necessary).
6. The length, width and height of a rectangular board are 12.7cm, 15.8cm
& 18.3cm respectively, correct to the nearest 0.1cm.
i) Find the greatest possible volume of the board.
ii) Find the least possible volume of the board.
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
iii) Find the maximum absolute error of the volume of the board.
iv) Find the relative error and the percentage error of the volume of the
board (corr. to 3 sig. fig.).
v) Find the maximum absolute error of the total surface area of the
board.
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
<<END OF UNIT 1>>
Unit 2 Polynomials
A. MonomialsA monomial is a polynomial containing 1 term only and satisfies one of the
following conditions.
Conditions Examples
It contains numbers only.-8, 0, 45, 2.7,
It is a product of number and
variables. The power of the variable
must be positive integers
, -6xy, 3x
Examples of non-monomials
Conditions Examples
The expression contains more than
1 term.
,
The expression is divided by a
variable, i.e. the denominator of the
expression contains a variable.
, ,
The powers are variables. ,
Degree of monomials
Conditions Method to find out the
degree
Example Degree of
monomial
Containing
numbers only
The degree of the
monomial is 0 because
0
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
the power of the
variables is 0.
Product of
numbers and one
variable
The degree of the
monomial equals to the
power of the variable.
3
4 (as
)
Product of
numbers and
more than one
variable
The degree of the
monomial equals to the
sum of power of the
variables.
1+5=6
1+5+2=8
B. PolynomialsPolynomials are the sum of 1 or more than 1 monomial.
Monomial, binomial and trinomial
Polynomials Definition Examples
Monomial Polynomial containing 1
term only,
Binomial Polynomial containing 2
terms,
Trinomial Polynomial containing 3
terms
,
Terms of polynomials
Polynomials can be separated into different terms. It can be done by finding
all the monomials in the polynomial.
Examples Terms No. of terms
2x, +y, +1 3
, 2
, , -3y, 4
1
Degree of polynomials
The degree of a polynomial equals to the degree of the term with the highest
degree.
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Example 2.1
Find the degree of the polynomial .
Solution:
There are four terms in the polynomial. They are , , and -3.
Terms Degree
1+2=3
1+1=2
5
-3 0
Since has the highest degree and its degree is 5, the degree of the
polynomial is also 5.
Coefficients of a term
The coefficient of a term is the number multiplied to the variables. The term
with no variables is called a constant term.
Example 2.2
Find the coefficients of the terms in the polynomial.
Polynomials Coefficient Constant
term
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
Solution:
Polynomials Coefficient Constant
term
1 0 -4 0 0
0 0 4 0 0 7
C. Addition and subtraction of polynomialsThe addition and subtraction of polynomials can be done by adding up the
coefficients of the like terms together.
Like terms
Like terms are the terms having the same variables. Their coefficients can be
different. Here are some examples of like terms and unlike terms.
Like terms Reasons
, Same variable multiplied.
,
,
Unlike terms Reasons
, Even though the coefficients are the
same, they are not like terms if they
have different variables multiplied.
, Same variables but with different
powers cannot be regarded as like
terms.,
Addition and subtraction of polynomials in a row
This can be done by adding or subtracting the like terms in the polynomials. If
there are brackets, they have to be removed first.
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
Example 2.3
Simplify the following.
Solution:
=
=
Example 2.4
Simplify the following.
Solution:
=
=
Note: Since and are not like terms, they cannot be added together.
Addition and subtraction of polynomials in a column
This can be done by putting the like terms in the same column, and the
addition or subtraction can be done.
Example 2.5
Evaluate the following.
Solution:
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
Example 2.6
Evaluate the following.
Solution:
D. Multiplication of polynomialsDistributive Law
Some multiplication of polynomials can be done with the distributive law.
Formula of multiplication of polynomials
Consider the figure above. Find the total area of the rectangle.
First solution:
Area of the figure
=Area of I + Area of II + Area of III + Area of IV=
Second solution:
Area of the figure
20
The distribution law
I
III
II
IV
a
b
c d
S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
=Width x Length
=
From the solutions above, since the area of the rectangle doesn’t change, we
can conclude that .
Also from the distributive law, considering (a+b) as one term,
=(a+b)c+(a+b)d
=
Conclusion: formulas for multiplying polynomials
They are:
1.
2.
Example 2.7
Expand .
Solution:
=
Example 2.8
Expand .
Solution:
Using the distributive law,
=
=
=
Or using the formula
21
With the distributive law,
3x
2
21x
2
S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
=
=
=
Example 2.9
Expand .
Solution:
=
=
=
=
=
Example 2.10
Expand .
Solution:
=
=
=
=
=
Expand the first two brackets at
the beginning.
Treat the first two terms in each
bracket as one term.
E. Exercise1. Complete the following table.
Polynomials No. of Coefficients Constant
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terms termx xy y
2. Classify the following expressions.
4a 4x-3-ay -9 4ab2+bc-d a3+b 2 m a10+b+c7 b2-4ac
(a) Monomials :
(b) Binomials :
(c) Trinomials :
(d) None of the above:
In question 3 to 5, arrange the terms of the following polynomials in
descending powers of the variables.
3. 3y + 2 – 5y3 + y5 =
4. 7 + 3x3 – x – x5 =
5. 4c2 – 5c3 + 3 + 2c =
In question 6 to 8, arrange the terms of the following polynomials in ascending
powers of the variables.
6. p4 + 3p3 – 5p2 + p =
7. 2x3 – 4x6 + 7 – 9x =
8. –r2 + 8 – 10r5 – 6r3 =
Expand and simplify the following.
9.
10.
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11.
12.
13.
14.
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
15.
16.
17.
18.
19.
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20.Complete the following table. Put a tick under “monomial”, “binomial”,
“trinomial” and write down the degree of polynomial and number of terms if
they are polynomials. If not, put a tick under the column “not a polynomial”.
Algebraic
expressions
Polynomial Not a
polynomial
Degree of
polynomial
Number
of
terms
Monomial Binomial Trinomial
881903
<<END OF UNIT 2>>
Unit 3 Use of Formulae
A. Algebraic fractionAlgebraic fraction is a fraction in which the denominator is not a constant, i.e.
contains variables. Also, it is assumed that the denominator in an algebraic
fraction is not equal to 0.
, and are some examples of algebraic fractions.
B. Addition and subtraction of algebraic fractionsThe addition and subtraction of algebraic fractions can be done by the
following steps.
For example, now we have to simplify the algebraic fraction .
Ste
p
Description Example
1 Find the lowest common multiple of
the denominators. The lowest common
multiple should be obtained by
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
multiplying the fraction with a non-zero
number.=
=
2 Add or subtract the numerators.
=
3 Simplify the fraction.
=
Example 3.1
Simplify .
Solution:
=
=
=
Example 3.2
Simplify .
Solution:
Simply add the numerators when the
denominators are the same.
Simplify the sum of the fractions.
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
=
=
Example 3.3
Simplify .
Solution:
=
=
=
Example 3.4
Simplify .
Solution:
=
=
=
Example 3.5
Simplify .
Solution:
The lowest common multiple of and
is . Multiply by 2 to
obtain .
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
=
=
=
=
=
If there is not any common factor between two
denominators, the common denominator will be
the product of the two denominators.
C. Basic concept of factorizationThe idea of factorization is to take out the common factor from different terms.
The common factor may be a number or a variable.
Factorization by taking common factors
A number or a monomial is multiplied by some factors. Common factors
indicate the same numbers or variables contained in both monomials at the
same time.
For example, in the algebraic expression ,
The factors of pq are p and q. (1 is also a factor of this monomial, but it is
meaningless to be taken out in factorization.)
The factors of qr are q and r.
Therefore they have a common factor q.
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Taking out the common factor,
=
Therefore the result of factorizing is .
Example 3.6Factorize .
Solution:
=
Example 3.7Factorize .
Solution:
=
=
Example 3.8Factorize .
Solution:
=
=
Example 3.9
Factorize .
Solution:
2 is the common of the factor of 2a and 2b.
2 is a factor of 6 because .
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
=
=
=
=
Example 3.10
Factorize .
Solution:
=
=
=
Example 3.11
Factorize .
Solution:
=
=
=
Treat (n-1) as a term.
In the expression , and
.You may convert the expression into
this form to avoid mistakes.
Take out the common factors of the variables.
Take out the common factors of the numbers.
D. Formulae and substitutionSubstitution is to put the values of the variables into a formula.
Example 3.12
Given that , find the value of S
when a=3, b=7 and n=-2.
Solution:
=
=
Example 3.13
Given that , find the value of b
Simply substitute the values
into the values to find the
unknown.
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
when a=2, c=-7 and d=-1.
Solution:
E. Change the subject of the formula
Do you still remember the formula of distance - , or ?
This formula helps us to find the speed when the distance and time are given.
However, when the value we want to find is the distance but not the speed,
we have to change the subject of the formula to help us find it out more
easily.
Example 3.14
Change the subject of
to a.
Solution:
Example 3.15
Change the subject of the
following formula.
Put the terms containing the subject to the left
side in order to separate them from the other
terms.
Divide the both side by the coefficient (-3) of
the subject (a).
Usually, the denominator of a fraction will not
be a negative number.
The symbol [w] means to change the subject
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Solution:
Example 3.16
Change the subject of the
following formula.
Solution:
of the formula to w.
Take out the terms containing the subject (w).
Put the terms containing the subject to the left.
Divide both sides by the term multiplied by the
subject.
When there are fractions in the formula,
multiply both side by the lowest common
multiple of the denominator.
Put the terms containing the subject to the left.
Take out the subject from the terms.
Divide both sides by the term multiplied by the
subject.
F. Exercise1. Factorize the following expressions.
a. b.
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c. d.
e. f.
g. h.
2. Simplify the following expressions.
a. b.
c. d.
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
e. f.
g. h.
3. Change the subject of the following formulae.
a. b.
c. d.
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
e. f.
g. h.
4. The Body Mass Index (BMI) is used to indicate whether a person is
obese or not. The relationship between the BMI (B), the weight of a
person in kilograms (m) and his height in meters (h) is .
a. Change the subject of the formula to m.
b. Given that the BMI of a healthy person is 21. If his height is 1.7m, using
the formula in (a), find his ideal weight.
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
c. Amy’s height is 160cm, find her weight in terms of B.
d. Given that people with their BMI over 23 is regarded is overweight. If
Mary is 155cm tall and weighs 120 pounds, is she overweight (given that
1kg is equal to 2.2 pounds)?
e. Alex’s height is 210cm, if his weight is 90kg, find the difference between
the BMI of Alex and Mary (corr. to 3 sig. fig.)
f. Hiromi is overweight. Her height is 160cm and her weight is 70kg. She
wants to decrease her BMI to 21.5. Find the weight she needs to lose in
order to achieve this target.
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5. It is known that the volume of a cone can be calculated with the formula
, where V, r and h represents the volume (in ), radius of the
base (in cm) and the height (in cm) respectively.
a. When , and , find V.
b. When the base radius and the volume are 3cm and 66 respectively,
find the height of the cone. (Take )
<<END OF UNIT 3>>
Unit 4 Identities and factorization
A. The concept of identityAn equation is only true for one or a few special values of the unknowns.
However, an identity is true for all values of the unknowns.
How to prove and disprove an identity
An identity can be proved or disproved with the following steps.
Example 4.1
Determine whether the equation is an
identity.
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Solution:
L.H.S. =
=
=
=
R.H.S =
Therefore, L.H.S.=R.H.S.
Therefore the equation is an identity.
Example 4.2
Determine whether the equation is an
identity.
Solution:
L.H.S. =
=
=
R.H.S.=
=
Therefore, L.H.S R.H.S.
Therefore the equation is not an
identity.
Expand both sides to see
whether the coefficient of every
term is the same.
Finding a counter example is also a method to disprove an identity. However,
since an identity satisfies all values of unknowns, giving examples is not
enough to prove an identity. Therefore, giving an example can only disprove
an identity but cannot prove one.
Example 4.3
Determine whether the following
equation is an identity.
Solution:
When ,
L.H.S. =
Substitute one value of the
unknown into the both sides of the
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=
R.H.S.
=4
Since L.H.S R.H.S,
the equation is not an identity.
equation. If the value of both sides
are the same, use the method in
example 4.1 and 4.2 to prove or
disprove the identity and do not
use this method to prove an
identity. You may continue to try
other values if you think that the
equation is not an identity.
Comparing coefficients of like terms
You may use this method to find some unknowns in the identity.
Example 4.4
If , find the
values of A and B.
L.H.S. =
=
=
=
Therefore, = .
Therefore, A=5 and B=6.
Expand the left hand side so the
coefficients of the terms of both
sides can be compared.
Method of substitution
This method can also find some unknowns in an identity. Since it is an identity,
the coefficients of all terms are always unchanged. Therefore this method can
be used.
Example 4.5
If , find the
values of A and B.
Solution:
When x=0,
L.H.S. =
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=5
R.H.S.
=B
Therefore, B=5.
Therefore the identity is
.
When ,
L.H.S. =
=
R.H.S.
=
Since both L.H.S.=R.H.S.,
Therefore A=6.
A=6, B=5
Find a value so that one of the
unknown will be multiplied by 0 (so
that it will be eliminated).
”1” is used because this makes
the calculation easier. You may also
substitute other values.
B. Some important algebraic identitiesThe following identities can be used to expand and factorize some algebraic
expressions.
1.
[Proof]
L.H.S. =
=
=
=
=
R.H.S.
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
,
is true.
2.
[Proof]
L.H.S. =
=
=
=
=
=
R.H.S.
is true.
3.
[Proof]
L.H.S. =
=
=
=
=
=
R.H.S.
is true.
Expanding expressions using the algebraic identities
Example 4.6
Expand .
Solution:
=
=
Example 4.7
Expand .
Use the identity .
Treat and .
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
Solution:
=
=
Example 4.8
Expand .
Solution:
=
=
Example 4.9
Expand .
Solution:
=
=
Example 4.10
Expand .
Solution:
=
=
=
=
=
Use the identity .
Treat and . Add a bracket
outside 4p and 3q in order to show a
clear work and avoid mistakes.
Use the identity .
Treat and .
Use the identity
. Treat and .
Use the identity .
Treat and .
Use the identity .
Treat and .
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C. FactorizationFactorization, the reverse of expansion, is the process rewriting an algebraic
expression in a form of the product of its factors. An algebraic expression
can be factorized by taking common factors, grouping terms, using
algebraic identities. The method of taking common factors has been
introduced in Unit 3 so only the remaining two will be talked about in this unit.
Factorization by grouping terms
Not every term in an algebraic expression has one common factor. There may
be two groups of terms that have their own common factors.
Example 4.11
Factorize .
Solution I:
=
=
=
Solution II:
=
=
=
Example 4.12
Factorize .
Solution:
=
=
=
Example 4.13
Factorize .
Put pq and qr in one group, ap and ar
into another group.
Taking out the common factors from both
groups.
Taking out the common factor -
from both groups.
You may also put pq and ap into one
group, qr and ar into another group.
Separate the terms into 2 groups.
Take out the common factors from the
groups.
Take out the common factor .
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Solution:
=
=
=
=
Example 4.14
Factorize .
Solution:
=
=
=
=
=
Example 4.15 (Non-essential)
Factorize
.
Solution:
=
=
=
=
Arrange the terms to make the
calculation easier.
Find the common factors of all terms
first, and then separate the terms into
groups.
Take out the common factor among
these six terms.
Group the terms into groups. (Other
groupings are also possible)
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
Example 4.16
Factorize .
Solution:
<Trial I>
=
=
<Trial II>
=
=
Therefore, this expression cannot be
factorized.
Take out the common factor
.
Try to separate the first and last two
terms into two groups first.
However, after taking the common
factors from each bracket, there are no
common terms left.
Try to put the first and third term into one
group and the remaining ones into another.
However, there is still no common factor
left. Therefore the expression cannot be
factorized.
Note: Before using the method of grouping terms, it is necessary to check
whether there are any common factors of all terms.
Factorization by using identities
We have learnt three identities. They are:
1. or
2. or
3. or
We can apply these identities to factorize algebraic expressions apart from
expanding them.
Example 4.17
Factorize .
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
Solution:
=
=
Example 4.18
Factorize .
Solution:
=
=
Example 4.19
Factorize .
Solution:
=
=
=
=
Example 4.20
Factorize .
Solution:
=
=
Example 4.21
Factorize .
Solution:
Change the terms in the form of .
Use the identity . Treat
and .
Change the terms in the form of .
Use the identity . Treat
and .
Take out the common factor of the terms before
using the identities.
Change the terms in the form of .
Use the identity . Treat
and .
Change the expression in the form of .
Use the identity . Treat
and .
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
=
=
=
Example 4.22
Factorize .
Solution:
=
=
Example 4.23
Factorize .
Solution:
=
=
=
=
Example 4.23 (Non-essential)
Factorize .
Solution:
=
=
=
=
=
=
Take out the common factor of all terms. Take -1
out of the expression so that the coefficient of
won’t be a negative number.
Change the expression in the form of .
Use the identity . Treat
and .
Change the expression in the form of .
Use the identity . Treat
and .
Change the expression in the form of .
Use the identity . Treat
and .
Change the expression in the form of .
Use the identity . Treat
and .
Factorize the first three terms first.
Change first three terms in the form of
.
Use the identity . Treat
and .
Change expression in the form of .
Use the identity . Treat
and .
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D. Simplification, multiplication and division of algebraic fractionsThe simple addition and subtraction of algebraic expression has been
introduced in the previous unit, and we will go further in this unit.
Simplification of algebraic fractions
The simplification of algebraic fractions is to cancel the common factors
between the numerator and the denominator.
Example 4.24
Simplify .
Solution:
=
=
=
Example 4.25
Simply .
Solution:
=
=
Take out the common factor mn from the
numerator and the denominator.
Cancel the common factor.
Take out the common of the terms of the
numerator.
Cancel the common factors between the
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
=
=
Example 4.26
Simplify .
Solution:
=
=
=
=
Example 4.27
Simplify .
Solution:
=
=
numerator and the denominator.
Factorize the numerator and the
denominator.
Cancel the common factor from
the numerator and the denominator.
Cancel the common factor b from the
numerator and the denominator.
Factorize the numerator and the
denominator by taking out the common
factors in all terms.
Factorize the numerator and the
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
=
=
=
denominator using identities.
Cancel the common factors.
Multiplication and division of algebraic fractions
With factorization, we can simplify the product and quotient of algebraic
fractions and make the calculation easier.
Example 4.28
Simplify .
Solution:
=
=
=
Example 4.29
Simplify .
Solution:
=
=
Cancel the common factors between
the numerator and denominator of the
fractions.
Change all the division signs to times
signs first.
Factorize the numerators and
denominators of all fractions.
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=
=
=
Example 4.30
Simplify .
Solution:
=
=
=
=
=
=
Cancel the common factors.
Change the division sign to a times
sign first.
Factorize the numerators and
denominators of the fractions.
Cancel all the common factors.
E. Exercise1. Expand the following.
a. b.
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c. d.
e. f.
g. h.
i. j.
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2. Determine whether the following equations are identities.
a.
b.
c.
3. Factorize the following.
a. b.
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c. d.
e. f.
g. h.
i. j.
k. l.
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m. n.
o. p.
q. r.
s. t.
4. If , find A and B.
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5. If , find A and B.
6. Answer the following questions.
a. Factorize .
b. Factorize .
c. Using the result of (a) and (b), factorize .
7. Simplify the following.
a. b.
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c. d.
e. f.
g. h.
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i. j.
k. l.
<<END OF UNIT 4>>
Unit 5 Pythagoras’ Theorem
A. Square and square rootsThere are two rules about square roots:
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1. If , then x is the square root of y.
2. The square roots of are and . is the positive root and is
the negative root.
Example 5.1
Find the square roots of 16, 0.64,
and .
Solution:
and , the
square roots of 16 are 4 and -4.
and ,
the square roots of 0.64 are 0.8
and -0.8.
,
,
The square roots of are
and .
is a negative number,
there is no real square roots of
.
There are no real square roots for a
negative number.
B. Values of square rootsSome square roots cannot be expressed as a fraction or an integer when its
square is not a perfect square number. They can only be expressed as surd
forms. The numbers that can only be expressed as surd forms are surds.
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Examples of surd forms: , and .
Estimation of square roots
We can use two consecutive square numbers to estimate the value of a
square root. If , then .
Example 5.2
Find two consecutive numbers
between which lies.
,
Therefore lies between 1 and 2.
Example 5.3
Find two consecutive numbers
between which lies.
,
Therefore lies between 14 and
15.
Find two consecutive perfect square
numbers between which 3 lies.
Operations of square roots
There are two rules about the operations of square roots
1. , where and .
2. , where and .
Example 5.4
Simplify .
Solution:
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S.2 Mathematics Study GroupDesigned by Yu Yat Tin Sunny
=
=
=
Example 5.5
Simplify .
Solution:
=
=
=
Example 5.6
Simplify .
Solution:
=
=
=
=
Example 5.7
Simplify .
Solution:
Express in the form of
product of prime factors.
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=
=
=
=
=2
C. Pythagoras’ TheoremPythagoras’ Theorem
In , if , then . (Pyth. Theorem)
Converse of Pythagoras’ Theorem
In , if , then . (Converse of Pyth. Theorem)
Example 5.8
Determine whether the following
triangles are right-angled triangles.
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Solution:
In ,
=
=
=
=
=
,
is a right-angled triangle.
(Converse of Pyth. Theorem)
In ,
=
=
=
=
=144
, is not a
right-angled triangle.
Example 5.9
Only check whether the sum of
squares of the shorter sides
is equal to the square of
the longest side . Other
groupings such as and
are not needed.
As , only check
whether is correct
or not.
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Find AC.
Solution:
In ,
(Pyth. Theorem)
Example 5.10
Find the unknown.
Solution:
In ,
(Pyth. Theorem)
Example 5.11
Even though is also
correct, as AC must be positive, only
is possible.
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A ladder is leaning against the wall.
The distance between the foot of
the ladder and the top of the wall is
4m and the height of the wall is
2.4m. Find the distance between
the foot of the ladder and the wall.
Solution:
Let the distance be m.
(Pyth. Theorem)
Therefore the distance is 3.2m.
Remember to add back the unit
after solving the equation.
D. Rationalization of denominators
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We can rationalize the denominator of a fraction so that its denominator will
not be an irrational number.
Example 5.12
Rationalize .
Solution:
=
=
Example 5.13
Rationalize .
Solution:
=
=
=
=
=
Multiply the denominator by .
Check if the denominator is a
rational number after the fraction is
rationalized.
Multiply the denominator by .
E. Exercise
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1. Simplify the following. Rationalize the denominator if necessary. Your
answer should be in a surd form.
a. b.
c. d.
e. f.
g. h.
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2. Find two consecutive numbers between which the following numbers.
a.
b.
3. Rationalize the denominator of the following fractions.
a. b.
c. d.
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e. f.
4. Determine whether the following triangles are right-angled triangles.
a.
b.
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c.
5. Find the unknowns in the figures.
a.
b.
c.
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d.
6. ABCDEFGH is a cuboid.
a. Find BD.
b. Find BE.
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7. There is a ladder lying against the wall. The length of the ladder is 4.2m
and its foot is 2.5cm apart from the wall. If it moves 0.5m farther from the
wall, how far will the top of the ladder slide down? (correct your answer to
3 significant figures)
8. Mr. Wong bought a TV. Its length is 65cm and its width is 58cm, correct
to the nearest cm. Find the upper and lower limit of the length of its
diagonal. (correct your answer to 3 significant figures)
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<<END OF UNIT 5>>
Unit 6 Analysis of Statistical Graphs
A. Grouping of dataFrequency distribution
Suppose the scores of a group of 50 students has been recorded and shown
in the following table.
25 71 44 58 61 73 90 81 61 57
40 45 42 62 74 68 77 70 49 51
53 60 77 48 33 39 88 22 49 47
56 55 60 62 83 79 78 54 51 49
48 51 53 90 82 60 52 48 30 42
A few steps should be done in order to see the trend of the data.
1. Find the largest and smallest values of the data.
From the data above, the largest value and the smallest value are 90 and 22
respectively.
2. Decide the number of classes and the range of each class. The
classes should cover all data.
Suppose the range of each class is 10. They are 21-30, 31-40, 41-50, 51-60,
61-70, 71-80 and 81-90. You may also use other ranges but the classes
should cover all data.
3. Use tallies to record the number of data in each class.
The following table shows the data separated into different classes.
Score Tally Frequency
21-30 3
31-40 3
41-50 11
51-60 14
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61-70 6
71-80 7
81-90 6
B. Terminologies about classesThere are some terminologies about classes.
(1) Range – The difference between the maximum and the minimum value of
the data. The largest value and the smallest value are 90 and 22
respectively. Therefore the range in this example is (90-22)=68.
(2) Class interval – Intervals with equal length cover the range of data
between the maximum and the minimum without overlapping. For
example, the class interval of the first class is 21-30.
(3) Class limit – Two ends of each class interval. The smaller value is the
lower class limit and the greater value is the upper class limit. For
example, the lower class limit of the second class is 31 and the upper
class limit is 40.
(4) Class mark – The average of the upper class limit and the lower class
limit. For example, the class limit of the second class is .
(5) Lower class boundary – The minimum possible value in the class. For
example, the lower class boundary of the second class is 30.5.
(6) Upper class boundary – The maximum possible value in the class. For
example, the upper class boundary of the second class is 40.5.
(7) Class width – The difference between the upper class boundary and the
lower class boundary. For example, the class width of the second class is
.
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C. HistogramThe data in a histogram is presented by rectangular bars with no gaps and
each bar is divided by its class boundaries. The following figure is the
histogram showing the score of the 50 students referring to the table in part A.
Before drawing the histogram, we use this frequency distribution table to
present the data.
Score Class mark Frequency
21-30 25.5 3
31-40 35.5 3
41-50 45.5 11
51-60 55.5 14
61-70 65.5 6
71-80 75.5 7
81-90 85.5 6
Something that you should be careful with:
(1) A title must be given.
(2) The vertical axis (y-axis) represents frequency.
(3) The horizontal (x-axis) and vertical axes (y-axis) should be labelled clearly
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with suitable units.
(4) There should be no gaps between 2 bars. Each bar is perpendicular to
the horizontal axis (x-axis).
(5) Only class mark is marked for each class.
(6) The area of each bar is proportional to the frequency of the
corresponding class. If the class width of all classed are equal, the height
of the bar is proportional to the frequency of the corresponding class.
D. Frequency polygons and frequency curvesWe can also use a frequency polygon to present the grouped data. Two class
intervals with zero frequency have to be added to both ends of the data when
constructing the frequency polygon. Therefore, the frequency distribution
polygon based on the scores of the 50 students will be as follows:
Process of constructing a frequency polygon:
1. Construct a frequency distribution table with the class mark of each class.
2. Mark the class mark of each class on the x-axis. Mark the frequency on
the y-axis.
3. Give the title to the frequency polygon.
4. Mark the frequency corresponding to its class mark on the graph.
5. Join the points to form a frequency polygon.
A frequency polygon can also be drawn by smoothing the frequency polygon.
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Something you should be careful with:
1. If a frequency polygon is marked with the crosses (X), the frequencies can
be read from them and the total number of data can be calculated. If not,
we can only know the trend of the data but cannot know the frequencies.
Therefore, in examinations, when constructing a frequency polygon, the
crosses should be marked unless special instructions are given.
2. We can construct two or a few frequency polygons in one graph in order to
compare the sets of data but we cannot use a histogram to do so.
E. Cumulative frequency polygons and cumulative frequency curvesA cumulative frequency polygon or cumulative frequency curve is used to
present cumulative quantities. The following table shows the height of 50
students.
Height (cm) Class boundaries (cm) Frequency
141-145 140.5-145.5 3
146-150 145.5-150.5 4
151-155 150.5-155.5 4
156-160 155.5-160.5 6
161-165 160.5-165.5 12
166-170 165.5-170.5 11
171-175 170.5-175.5 7
176-180 175.5-180.5 3
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The frequency distribution table above can be changed to the following
cumulative frequency table:
Height less than (cm) Cumulative frequency
140.5 0
145.5 3
150.5 7
155.5 11
160.5 17
165.5 29
170.5 40
175.5 47
180.5 50
From the table, we can know the number of students who get less than a
certain score. Note that the phrase “cumulative frequency” is used to
describe the number of data with their values lower than a certain level.
Using the data above, we can construct a cumulative frequency polygon.
(Note that the unit should be added to the frequency distribution table and the
label of the x-axis.)
Also, a cumulative frequency curve can be obtained.
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F. PercentilesThe percentile indicates the percentage of data with their values lower than
that datum. The following graph is the cumulative frequency polygon of the
height of 100 students.
For example, when we want to find the weight for which 40% of the students’
weights are under it (i.e. to find the 40th percentile, , we have to do the
following steps:
1. Find its corresponding cumulative frequency.
The cumulative frequency
=
=40
2. Find the weight corresponding to the cumulative frequency.
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From the yellow line, the weight is 55.5kg. Therefore . Do you
know how to find out with the graph above?
Also, some percentiles can be represented by quartiles. They are
a. The lower quartile, , where .
b. The median, , where .
c. The upper quartile, , where .
Example 6.1
The cumulative frequency polygon shows the weight of 50 students. Find
(a) ;
(b) the median;
(c) .
Solution:
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(a)
The cumulative frequency
=
=12.5
From the graph above,
.
(b)
The cumulative frequency
=
=25
From the graph above, the
median is 58kg.
(c)
The cumulative frequency
=
=45
From the graph above,
.
G. Abuse of statisticsThese are the common abuses of statistics:
1. The scales of the axes are adjusted to exaggerate the relative change.
Description Example
The values of
the markings
on the y-axis
do not start
from 0.
2. The sizes of the figures are overcastted to exaggerate the differences.
Description Example
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A bar in the
chart is wider
than the others
so that the area
of the bar will
be much larger
than the others.
3. There may be hidden information in the figures which would lead the
readers to make wrong conclusions.
Description Example
Only a part of
the values are
compared. (The
sales of
magazines with
higher rankings
are not shown.)
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Regard some
values as
“others”. (In this
pie chart, some
of the “other”
candidates may
have more
supporters than
Candidate B
does.)
Only present the
”percentage
increase” but not
the actual value.
(The actual
sales of the soft
drinks are not
shown.)
Example 6.2
The following chart shows the percentage increase of sales of different brands
of video games.
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a. List two misleading points of the bar chart.
b. State the objective of the bar chart.
a. The vertical axis does not start from zero. Therefore it overstates the
percentage increase of sales Ouendan. Also, only the percentage
increase, but not the actual sales, is shown, we cannot conclude which
video game sells better.
b. To let the readers think that Ouendan sells better.
H. Exercise1. Here are the weights, in kg, of 50 newborn babies in King’s Hospital.
2.3 1.9 3.4 3.5 3.2 2.9 3.6 2.8 1.8 3.1
3.0 3.1 3.3 2.4 2.7 1.7 1.8 2.4 0.9 1.6
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2.3 2.2 3.0 3.2 3.1 3.0 4.1 3.7 3.3 1.4
3.5 3.6 3.7 2.7 2.9 2.5 2.5 2.3 2.0 2.1
2.4 2.9 3.4 4.2 3.3 2.8 2.6 1.7 4.0 2.2
a. Use the above data to complete the following table.
Weight
(kg)
Class boundaries
(kg)
Class mark
(kg)
Tally Frequency
0.6-1.0
1.1-1.5
1.6-2.0
2.1-2.5
2.6-3.0
3.1-3.5
3.6-4.0
4.1-4.5
b. Construct a frequency polygon using the above data.
c. Construct a histogram using the above data.
d. Construct a cumulative frequency polygon using the above data.
e. Find the first quartile, median and using the cumulative frequency
polygon.
2. The following frequency polygon shows the monthly income, in $1000, of
100 people. Given that the class interval of the first class is $0-$5000.
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a. Which class interval has the highest frequency?
b. How many people have the income between $20000.5 and $24999.5?
c. Find the number of people who has the income below $14999.5.
d. Use the above data to complete the following table.
Income lower than ($) Cumulative frequency
5000
10000
15000
20000
25000
30000
35000
e. Use the above data to construct a cumulative frequency polygon and find
the median.
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3. The following graph shows the sales of different magazines.
a. From the figure above, what can you say about the sales between 88
Magazine and the others?
b. State two misleading points of the graph.
4. Here is the cumulative frequency polygon showing the scores of F.2A and
F.2B students in a mathematics test.
a. Find the lower quartile, median and upper quartile of the score of F.2A
and F.2B students respectively.
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b. What conclusion can you make between the result of F.2A and F.2B
students?
<<END OF UNIT 6>>
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