New Progress in Junior Mathematics (Second Edition)
Intensive Practice for EYA Questions (Teacher’s Edition)
Test Bank (Upgraded Edition)
4 Justification by Mathematical Concepts
Intensive Practice for EYA Questions
Chapter 4
Justification by Mathematical Concepts
General Suggestion
In this kind of questions, candidates are asked to justify whether a statement is correct. To solve this
kind of problems, candidates need to understand the definitions or the properties of the
mathematical terms thoroughly and apply these knowledge to justify the statement.
Here are some mathematical concepts appear frequently in this kind of EYA questions in the public
examination:
 Similarity of two solids


Definitions of statistical terms, such as mean, median, mode and quartiles etc.
Use and misuse of statistical graphs
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4 Justification by Mathematical Concepts
Example 4.1
In a laboratory, 3 identical right circular cone-shaped flasks of height 20 cm are filled up with water.
Then the water is filled up 16 smaller identical right circular cone-shaped flasks. It is given that the
base area of a larger flask is 4 times that of a smaller one.
(a) Find the height of a smaller flask.
(b) A scientist claims that a smaller flask and a larger flask are similar. Do you agree? Explain
your answer.
(4 marks)
Solution:
(a) Let h cm and x cm2 be the height and the base area of a
smaller flask respectively.
Then the base area of a larger flask is 4x cm2.
1
1
3   (4x)(20)  16   xh
3
3

h  15
The height of a smaller flask is 15 cm.
1M
1A
(b) Analysis:
The ratio of base areas is given. The height of a smaller
flask is found in (a). Then we can find the ratio of the
heights and the ratio of the base radii. Finally, we can
compare the ratios to check whether the corresponding
sides are proportional.
The height of a smaller flask 15 3


The height of a larger flask
20 4
The base radius of a smaller flask

The base radius of a larger flask
3 1


4 2



heights.
1 1

4 2
The two flasks are not similar.
The claim is disagreed.
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Find the ratio of the
4.2

Find the ratio of the
base radii.
1M

Compare the ratios.
1A

Make a conclusion.
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4 Justification by Mathematical Concepts
Example 4.2
A class of students finished a Physics test. The teacher only inputs the scores of thirty out of
thirty-five students to calculate the mean and the median of the test scores. The mean and the
median of the scores are 70 marks and 72 marks respectively, where only two students got 72 marks.
The mean of the scores of the remaining five students is 84 marks. It is found that the scores of
three of these five students are 90 marks, 92 marks and 93 marks.
(a) Find the mean of the marks of the thirty-five students.
(b) Is it possible that the median of the scores of the thirty-five students is 72 marks? Explain your
answer.
(4 marks)
Solution:
(a) Mean

70(30) + 84(5)
marks
30  5
1M
 72 marks
1A
(b) Analysis:
If the median of the scores of the thirty-five students is
72 marks, then the number of students who got higher
than 72 marks must be 16 or 17. We should check
whether this condition is satisfied.
For the original thirty students, fourteen of them are
higher than 72 marks.
For the five students,
the mean of the scores of the remaining two students


84(5)  90  92  93
marks
2
scores of the remaining
two students.
= 72.5 marks
> 72 marks
 One of the two students got higher than 72 marks
Therefore there are eighteen students got higher than
72 marks in the class.
 It is impossible that the median of the scores of the
thirty-five students is 72 marks.
© Hong Kong Educational Publishing Company
Find the mean of the
4.3
1M

Point out that there are
more than half of the class

1A
got higher than 72 marks.
Make a conclusion.
New Progress in Junior Mathematics (Second Edition)
Intensive Practice for EYA Questions (Teacher’s Edition)
Test Bank (Upgraded Edition)
4 Justification by Mathematical Concepts
Exercise 4
The following table shows the related chapters in New Progress in Junior Mathematics (Second
Edition) for the questions in this exercise.
Question
Book
Chapter
Level 1
Level 2
12
1A
6 Introduction to Statistics
2A
6 More about Statistical Diagrams and
Graphs
3
5
3A
5 Measures of Central Tendency
4
6
3B
7 Area and Volume (III)
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Level 3
7
4.4
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4 Justification by Mathematical Concepts
Level 1
1.
In the figure, the pie chart shows the distribution of the numbers of votes for four candidates P,
1
Q, R and S in an election. It is given that the number of votes for candidate P is 33 % less
3
than that of candidate Q.
Q
R
122
a
P
S
(a) Find a.
(b) Is the number of votes for candidate S more than that for candidate Q? Explain your
answer.
(4 marks)
Answer:
(a)
60
(b)
No
Solution:
(a)
(b)
1 

a = 90 1  33 % 
3 

= 60 
1M
1A
The angle of the sector representing the number of votes for candidate S
= 360  60  90  122
1M
= 88
< 90

The number of votes for candidate S is not more than that for
candidate Q.
2.
1A
The numbers of members in teams P, Q, R and S are 8, 15, 10 and 7 respectively. Suppose that
the distribution of the numbers of members in the four teams is represented by a pie chart.
(a) Find the angle of the sector representing the number of members in team R.
(b) Some members leave team R. Will the angle of the sector representing the number of
members in team R be halved? Explain your answer.
(4 marks)
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Answer:
(a)
90
(b)
No
Solution:
(a)
(b)
The required angle
10
 360 
=
8  15  10  7
= 90
1M
1A
Let x be the number of members leaving team R.
Assume that the angle of the sector representing the number of members
in team R will be halved.
1
90  
10  x
2
=
40  x
360 
1
10  x
=
40  x
8
1M
80  8x = 40  x
40
x=
, which is not an integer.
7

The angle of the sector representing the number of members in
team R will not be halved.
3.
1A
The following bar charts show the percentage change (in %) of annual revenues in companies
A and B for the years 2014 to 2018.
The percentage change of annual
revenues in company A
The percentage change of annual
revenues in company B
10
Percentage change (%)
Percentage change (%)
10
8
6
4
2
0
8
6
4
2
0
2014 2015 2016 2017 2018
Year
2014 2015 2016 2017 2018
Year
(a) Write down the years that company A and company B has the greatest percentage change
of annual revenue respectively.
(b) Someone claims that the increase in annual revenue in company B must be more than that
in company A in 2017. Do you agree? Explain your answer.
(4 marks)
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4 Justification by Mathematical Concepts
Answer:
(a)
Company A: 2014, company B: 2018
(b)
No
Solution:
(a)
Company A has the greatest percentage change of annual revenue in
2014.
1A
Company B has the greatest percentage change of annual revenue in
2018.
(b)
1A
Although the percentage change of annual revenue in company B is more
than that in company A in 2017, the annual revenues of both companies
are not given. Thus, we cannot calculate the actual increases in annual
4.
revenues of both companies.
1M

1A
The claim is disagreed.
The data below show the percentages of citizens aged over 70 suffering from heart disease in
five cities including city A:
12%
10%
a%
7%
b%
(a) Write down the greatest possible value of the median of the above data.
(b) It is known that the median of the above data is the same as that found in (a). Suppose the
percentage in city A is the median of the above data. The mayor of city A claims that the
number of citizens aged over 70 suffering from heart disease in city A must also be the
median of the numbers of that among the five cities. Do you agree? Explain your answer.
(3 marks)
Answer:
(a)
12%
(b)
No
Solution:
(a)
The greatest possible value of the median of the above data is 12%.
(b)
Since the numbers of citizens aged over 70 of the five cities are not
1A
given, we cannot compare the numbers of citizens aged over 70 suffering
from heart disease in the five cities.
1M

1A
The claim is disagreed.
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4 Justification by Mathematical Concepts
Level 2
5.
The following table shows the distribution of the results (in marks) of a group of students in an
examination.
Result (x marks) Number of boys Number of girls
36
27
x  90
45
29
70  x < 90
17
20
50  x < 70
x < 50
14
4
It is given that the passing score of the examination is 50 marks.
(a) Find the percentage of boys passing the examination.
(b) Since more boys than girls passed the examination, John claims that boys generally
performed better than girls. Do you agree? Explain your answer.
(5 marks)
Answer:
(a)
87.5%
(b)
No
Solution:
(a)
(b)
The required percentage
36  45  17

 100 %
36  45  17  14
 87.5%
1M
1A
The percentage of girls passing the examination
27  29  20

 100 %
27  29  20  4
1M
 95%
> 87.5%
1M
The percentage of girls passing the examination is more than that of boys.
6.

Generally, girls perform better than boys.

The claim is disagreed.
1A
The following are the ages of 20 interviewees in a street survey:
7 8 18 20 20 22 23 24 26 27
29 29 30 51 63 63 63 80 87 90
(a) Find the mean, the mode and the median of the above data.
(b) 4 more people are interviewed. It is given that the mean of the ages of these four
interviewees is 30 and the ages of two of these four interviewees are 33 and 34. Is it
possible that the median of the ages of the 24 interviewees is the same as the median
found in (a)? Explain your answer.
(6 marks)
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Answer:
(a)
Mean = 39, mode = 63, median = 28
(b)
Yes
Solution:
(a)
Mean
 7  8  18  20  20  22  23  24  26  27




29

29

30

51

63

63

63

80

87

90


=
2
780
=
20
= 39
1A
Mode = 63
1A
Median =
(b)
27  29
= 28
2
1A
Let x and y be the ages of the remaining two interviewees.
x  y  33  34
 30 .
Note that
4

x + y = 53
1M
If the two medians are the same, then we have x  27 and y  27.
1M
Hence, we have x + y  54.
Note that x + y = 53, which is less than 54.

It is possible that the two medians are the same.
1A
7. In the figure, ABCDEFGH is a prism with square base EFGH. It is given that AB = 12 cm and
AF = 50 cm. APQRSFTU is a prism with square base SFTU cut from ABCDEFGH where P, R,
S and T are the mid-points of AB, AD, EF and FG respectively.
A
B
D
A
G
Q
F
E
F
R
P
C
T
H
S
U
(a) Find the volumes of ABCDEFGH and APQRSFTU.
(b) Are ABCDEFGH and APQRSFTU similar? Explain your answer.
(6 marks)
Answer:
(a)
Volume of ABCDEFGH = 7200 cm3,
volume of APQRSFTU = 1800 cm3
(b)
No
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4 Justification by Mathematical Concepts
Solution:
(a)
Volume of ABCDEFGH
 (12)2(50) cm3
 7200 cm 3
1M
1A
Volume of APQRSFTU
 (6)2(50) cm3
3
 1800 cm
(b)
Volume of APQRSFTU
Volume of ABCDEFGH
1A
3

3
1800 1
1
 AP 
 6 
 and 
   
7200 4
AB
12
8


 
1M

The two ratios are not equal.
1M

ABCDEFGH and APQRSFTU are not similar.
1A
Level 3
8.
In the figure, an hourglass consisting of two similar right circular cones is held vertically on a
table. The base radii of the larger cone and the smaller cone are 9 cm and 6 cm respectively.
The height of the hourglass is 20 cm. Initially, the smaller cone is full of liquid.
9 cm
20 cm
6 cm
(a) Find the volume of the liquid in terms of .
(b) Find the area of the curved surface of the smaller cone in contact with the liquid in terms
of .
(c) The hourglass is flipped over such that the base of the larger cone lies on the table. The
liquid starts falling to the larger cone and forms a frustum. Peter claims that the final area
of the curved surface of the larger cone in contact with the liquid is at least 88 cm2. Do
you agree? Explain your answer.
(8 marks)
Answer:
(a)
96 cm3
(b)
60 cm2
(c)
Yes
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Solution:
(a)
Height of the smaller cone
6 

=  20 
 cm
69

= 8 cm
Volume of the liquid
1
2
 π (6 )(8) cm3
3
 96π cm 3
(b)
(c)
1M
1A
The required area of the curved surface
 π (6) 6 2  8 2 cm2
1M
 60π cm
1A
2
Let r cm be the radius of the upper base of the frustum.
Let h cm be the distance between the apex and the upper base of the
frustum.
Height of the larger cone = (20  8) cm = 12 cm
Refer to the following figure.
h cm
12 cm
r cm
9 cm
h
r

12 9
4r
h
3
1M
1 2  4r 
1
πr    96 π = π (9 2 )(12 )
3
3
 3 
1M
4 3
πr = 228
9
r = 3 513
Final area of the curved surface in contact with the liquid

2

  π (9) 9 2  12 2  π (3 513 ) (3 513 ) 2   4 3 513   cm2


3
 

1M
 88.5756 cm2
 88 cm2

The claim is agreed.
1A
END OF TEST
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New Progress in Junior Mathematics (Second Edition)
Intensive Practice for EYA Questions (Teacher’s Edition)
Answers
Exercise 4
1.
2.
3.
4.
5.
6.
7.
(a)
60
(b)
No
(a)
90
(b)
No
(a)
Company A: 2014, company B: 2018
(b)
No
(a)
12%
(b)
No
(a)
87.5%
(b)
No
(a)
Mean = 39, mode = 63, median = 28
(b)
Yes
(a)
Volume of ABCDEFGH = 7200 cm3,
volume of APQRSFTU = 1800 cm3
8.
(b)
No
(a)
96 cm3
(b)
60 cm2
(c)
Yes
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4.12
Test Bank (Upgraded Edition)
4 Justification by Mathematical Concepts