F3 Mock Mid-year Exam (2018-2019)
Name:
(
)
Class:
Time allowed:
1 hour 45 minutes
Full marks:
189
Date:
LQ
Ch1
2,13
Ch2
1,4,15
Ch3
9,10,11
Ch4
3,5,14
Ch5
6,7,17
Ch6
8,12,16
1. Check the answer after you finish
2. Circle the questions you did wrong
3. Check which chapters you should revise more
MC
3,4,12,24
1,7,13,14,15,25
8,18,22,23
2,5,6,16,17
9,10,11,21,27
19,20,26,28,29
Short Questions (57 marks)
1.
Simplify
( x3 y )0
and express the answer with positive indices.
4x2
(3 marks)
2.
Factorize the following polynomials.
(a) 25  d2
(b) 9h2  12hk + 4k2
(4 marks)
3.
Solve the inequality x  4 
7x  8
and represent the solutions graphically.
5
(5 marks)
4. If there are 5.02  1022 atoms in 1 g of carbon, find the number of atoms in
4  105 g of carbon.
(Give the answer in scientific notation and correct to 3 significant figures.)
(4 marks)
5.
If the sum of two consecutive even numbers is not more than 41, find the largest
possible value of the smaller number.
(4 marks)
6.
The numbers of students in six S3 classes are as follows.
25
24
x
28
25
30
It is known that the mean number of students of the six classes is 26.
(a) Find the value of x.
(b) Find the median of the above data.
(c) Find the mode(s) of the above data.
(5 marks)
7. The table below shows the results of Susan and Andy in a Chinese examination.
Part I
Part II
Part III
Susan
72
60
89
Andy
86
44
m
Weight
10
15
30
(a) Find the weighted mean mark of Susan.
(b) If the weighted mean mark of Andy is 2 higher than that of Susan, find the
value of m.
(5 marks)
8. In the figure, AEB and BDC are straight lines. DE is
a perpendicular bisector of △ABC. Prove that AD
A
is an angle bisector of △ABC.
4
8
E
C
D
B
(5 marks)
9. Mr Chan wants to deposit $50 000 in a bank for 2 years. He can choose one of the
following plans offered by the bank.
Plan A: simple interest rate of 7% p.a.
Plan B: interest rate of 6% p.a. compounded half-yearly
Which plan should Mr Chan choose in order to earn more interest? Explain your
answer.
(6 marks)
10. From 2013 to 2015, the value of a house increased at a constant rate of 12% per
year. Then, its value decreased by 24% from 2015 to 2016. It is known that its
value was $5 644 800 in 2015.
(a) Find the value of the house in 2013.
(b) Find the percentage change in the value of the house from 2013 to 2016,
correct to 3 significant figures.
(6 marks)
11. Mr Cheng’s monthly expenditure is composed of 4 parts:
Food $2 400
Clothing $1 200
Transport $1 800
Rent $6 600
It is given that his expenditure on transport decreases by 45%, while the
expenditures on food and clothing increase by 20% and 35% respectively. If his
expenditure on rent remains unchanged, find the percentage change in his
monthly expenditure.
(5 marks)
 12. The lengths of three sides of a triangle are 4 cm, 7 cm and x cm. Is it possible that
the perimeter of the triangle is less than 13 cm? Explain your answer.
(5 marks)
Long Questions (53 marks)
13. (a) Factorize 3  5x  2x2.
(b) Hence, factorize
(i) 3  5(y  1)  2(y  1)2,
(ii) mx + 3m + 3  5x  2x2.
(9 marks)
14. A factory produces dark chocolates and milk chocolates only. A box of dark
chocolates weighs 0.25 kg while a box of milk chocolates weighs 0.65 kg. If 140
boxes of chocolates were produced yesterday and their total weight did not
exceed 50 kg, at least how many boxes of dark chocolates were there?
(7 marks)
 15. (a) Convert FA16 into a denary number.
(b) Does 111110002 represent the same value as FA16? Explain your answer.
(6 marks)
16. In the figure, BDC, AEC and DEF are straight
B
lines. It is given that BC // AF.
(a) Prove that △AFE ~ △CDE.
D
 (b) If E is the mid-point of DF, does the centroid
of △ABC lie on BE? Explain your answer.
A
C
E
F
(9 marks)
17. The stem-and-leaf diagram below shows the prices of 12 toys sold by a shop
yesterday.
Stem ($10) Leaf ($1)
4 0
3
5 2
4
9
6 3
3
7
7 0
5
7
7
(a) Find the mean and the median of the given data.
(b) Today, the shop sells 4 toys only, where two of them are sold for $96 and
$97. The mean price of these 4 toys is $80.
(i) Find the mean price of the 16 toys sold.
(ii) Is it possible that the median price of the 16 toys sold is the same as the
median found in (a)? Explain your answer.
 (iii) Later, an error is found in the above stem-and-leaf diagram because the
shopkeeper forgot to include a toy of price $y in the diagram. If the
lowest price of the toys sold today is also $y, is the actual mean price of
all the toys sold in the two days less than the mean found in (b)(i)?
Explain your answer without calculating the mean.
(12 marks)
Multiple-choice Questions (90 marks)
Choose the best answer for each question.
1. Express 0.000 024 5 in scientific notation.
A.
B.
C.
D.
2.
245  107
2.45  106
2.45  105
0.0245  103
Which of the following represents the solutions of x < 5?
A.
5
0
5
0
5
0
5
0
B.
C.
D.
3.
Which of the following has a + 1 as a factor?
A. a2 + 5a + 6
B. a2  5a + 6
C. a2 + 5a  6
D. a2  5a  6
4.
Which of the following is/are factor(s) of 12x2  43x  20?
I.
II.
III.
A.
B.
C.
D.
5.
x5
6x + 5
12x + 5
I only
III only
I and III only
II and III only
Find the least integer x such that 9(x  2)  2x + 15.
A. 2
B. 3
C. 4
D. 5
6. If p  q and p, q  0, which of the following must be true?
I. p – 2  q – 2
p
q
II.   
5
5
2 2
III.

p q
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
7.
If x, y  0, then (x2y)1(x2) =
A. x4y.
1
B.
.
x4 y
1
C.  .
y
1
D.  4 .
x y
8.
The value of a machine depreciates by 25% every 2 years. Its value was $84 375
in 2016. Find the value of the machine in 2010.
A. $43 200
B. $164 795
C. $200 000
D. $474 074
9.
The following table shows the number of siblings of each student in a class.
Number of siblings
0
1
2
3
Frequency
54
32
16
7
Find the median of the data.
A. 0
B. 0.5
C. 1
D. 2
10. The following table shows the time taken by some students to run 200 m.
Time (s)
Frequency
2023
2427
2831
5
12
8
Find the mean of the data.
A. 22.9 s
B. 24.48 s
C. 25.5 s
D. 25.98 s
11. The following cumulative frequency polygon shows the ages of a group of
students.
Ages of a group of students
Cumulative frequency
50
40
30
20
10
0
12
Find the first quartile of the data.
A. 14
B. 15
C. 16
D. 20
12. a2  b2 + 2b  1 =
A. (a  b  1)(a + b  1)
B. (a  b  1)(a + b + 1)
C. (a  b + 1)(a + b  1)
D. (a  b + 1)(a  b  1)
13. If a  0, then (a0 + a0)0 + a0 =
A. a.
B. 2.
C. 1.
D. 0.
14
16
Age
18
20
 x 1 y 
14. If x, y then  2 
 2y 
A. 2x3y3.
3
=
B. 8x3y3.
x3
.
2 y3
y3
D.
.
8x3
C.
15. If n is an integer and a  0, then
an  2  an  1
=
an  2
A. an  1.
B. 1 + an  1.
1
C. 1 + .
a
D. 1 + a.
16. Solve the inequality 1 +
1 x
< x.
3
A. x > 1
B. x < 1
C. x > 1
D. x 
17. The adult fare for a bus ride is $x, while the child fare is half of the adult fare. If
the total fare for 3 adults and 4 children is less than $50, which of the following
inequalities can be used to find the range of values of x?
A. 3(2x) + 4x  50
B. 3(2x) + 4x < 50
 x
C. 3 x  4   50
2
 x
D. 3x  4   50
2
18. If the height of a cylinder increases by 15% and its base radius decreases by
7.5%, find the percentage change in its volume, correct to 1 decimal place.
A.
B.
C.
D.
22.3%
1.6%
+1.6%
+6.4%
19. In △ABC, M is the mid-point of BC. Which of the following must pass
through M?
I. The altitude of BC
II. The median of BC
III. The perpendicular bisector of BC
A. I only
B. II only
C. I and III only
D. II and III only
20. In the figure, ADC is a straight line and BAC = DBC.
B
A
D
Which of the following must be true?
I. △ABC ~ △BDC
II. △ADB ~ △BDC
III. BC2 = AC  DC
A. I only
B. II only
C. I and III only
D. II and III only
C
21. The median of the five numbers 5, 11, 13, y, y is 8. Find the mean of the five
numbers.
A. 13
B. 11
C. 9
D. 8
22. If the height of a triangle decreases by 8% and its area remains unchanged, find
the percentage change in its base, correct to 3 significant figures.
A. +7.41%
B. +8.00%
C. +8.70%
D. +12.5%
23. The simple interest on a sum of money at r% p.a. for 3 years is equal to the
interest on the same principal at 8% p.a. for 3 years compounded quarterly. Find
the value of r, correct to 3 significant figures.
A. 2.04
B. 8.66
C. 8.84
D. 8.94
 24. Which of the following is not a factor of 432x3  128y3?
A. 16
B. 3x  2y
C. 9x2  12xy + 4y2
D. 9x2 + 6xy + 4y2
 25. C00002D0000016 =
A. 12  1611 + 45  165
B. 12  1612 + 45  166
C. 13  1611 + 46  165
D. 13  1612 + 46  166
 26. Each of the following options lists the lengths of three line segments. Which of
them cannot form a triangle?
A. 4 cm, 6 cm, 8 cm
B. 2 cm, 4 cm, 5 cm
C. 4 cm, 6 cm, 9 cm
D. 3 cm, 5 cm, 8 cm
 27. The mode of a set of data consisting of 5 negative numbers is m. If each datum is
multiplied by m, which of the following must be true?
I. The mode becomes a positive number.
II. The mode becomes larger.
III. The mode becomes m2.
A. I only
B. III only
C. I and II only
D. II and III only
 28. If △ABC is an acute-angled triangle, which of the following must lie inside
△ABC?
I. The circumcentre of △ABC
II. The orthocentre of △ABC
III. The in-centre of △ABC
A.
B.
C.
D.
I and II only
I and III only
II and III only
I, II and III
 29. In the figure, BDC is a straight line. It is given that AB = AC, AD = BD and
ABC = 30.
A
B
30
D
Which of the following must be true?
I. △ABC ~ △DAB
II. △DAC is a right-angled triangle.
III. A is the orthocentre of △ADC.
A. I only
B. II only
C. I and II only
D. I, II and III
— End —
C
Multiple-choice Questions Answer Sheet
No.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
Answer
F3 Mock Mid-year Exam (2018-2019)
Solutions
Short Questions
1.
2.
3.
1
( x3 y )0
=
2
4x2
4x
1
=
1
4 2
x
2
x
=
4
1M
1M
1A
(a) 25  d2 = 52  d2
= (5 + d)(5  d)
1M
1A
(b) 9h2  12hk + 4k2 = (3h)2  2(3h)(2k) + (2k)2
= (3h  2k)2
1M
1A
7x  8
5
5(x  4)  7x  8
5x  20  7x  8
5x  7x  8 + 20
2x  12
x
12

2
2
x  6
x4
1M
1M
1A
Graphical representation:
6
0
1A+1A
4.
Number of atoms in 4  105 g of carbon
= (4  105)  (5.02  1022)
= 4  5.02  105 + 22
= 20.08  1017
= 2.008  1018
= 2.01  1018, cor. to 3 sig. fig.
5.
Let x be the smaller even number.
Then, the larger even number is x + 2.
x + (x + 2)  41
2x + 2  41
6.
1M
1A
1M
1M
2x  39
x  19.5
∵ x is an even number.
1A
∴ The largest possible value of the smaller number is 18.
1A
(a)
Mean = 26
25  24  x  28  25  30
= 26
6
132 + x = 156
x = 24
(b) Arranging the data in ascending order:
24, 24, 25, 25, 28, 30
the 3rd datum  the 4th datum
Median =
2
25  25
=
2
= 25
7.
1M
1M
1M
1A
1M
1A
(c) The modes are 24 and 25.
1A
(a) Weighted mean mark of Susan
72  10  60  15  89  30
=
10  15  30
= 78
1M
1A
(b) Weighted mean mark of Andy = weighted mean mark of Susan + 2
86  10  44  15  m  30
= 78 + 2
10  15  30
1 520 + 30m = 4 400
30m = 2 880
m = 96
8.
1A
In △AED and △ACD,
∵ DE is a perpendicular bisector of △ABC.
∴ AED = 90 and
1
1
AE = AB =  8 = 4.
2
2
AED = ACD = 90
AD = AD
AE = AC = 4
∴ △AED  △ACD
EAD = CAD
∴ AD is an angle bisector of △ABC.
9.
1M+1A
given
1M
common side
RHS
corr. s,  △s
1M
1A
1M
1A
For plan A,
interest = $50 000  7%  2
= $7 000
For plan B,
interest rate every 6 months =
1M
1A
6%
= 3%
2
Taking 6 months as a period,
number of periods in 2 years = 2 
12
= 4
6
interest = $50 000[(1 + 3%)4  1]
= $6 275, cor. to the nearest dollar
< $7 000
∴ He should choose plan A.
1M
1A
1M
1A
10. (a) Let $P be the value of the house in 2013.
P(1 + 12%)2 = 5 644 800
P(1.12)2 = 5 644 800
5 644 800
P=
1.122
= 4 500 000
∴ The value of the house was $4 500 000 in 2013.
1M
1A
(b) Value of the house in 2016
= $5 644 800  (1  24%)
= $4 290 048
Percentage change in the value of the house
4 290 048  4 500 000
=
 100%
4 500 000
= 4.67%, cor. to 3 sig. fig.
11. Original monthly expenditure
= $(2 400 + 1 200 + 1 800 + 6 600)
= $12 000
New monthly expenditure
1M
1A
1M
1A
1M
= $[1 800  (1  45%) + 2 400  (1 + 20%) + 1 200  (1 + 35%) + 6 600]
1M
= $(990 + 2 880 + 1 620 + 6 600)
= $12 090
∴ Percentage change in his monthly expenditure
12 090  12 000
=
 100%
1M+1A
12 000
= +0.75%
1A
12. If the perimeter of the triangle is less than 13 cm, then
4 + 7 + x < 13
x<2
∵ Sum of the lengths of two shorter line segments
= (x + 4) cm
< (2 + 4) cm
= 6 cm
< 7 cm
∴ The three line segment of lengths 4 cm, 7 cm and x cm (x < 2)
cannot form a triangle.
i.e. It is impossible that the perimeter of the triangle is less than 13 cm.
1M
1A
1M
1A
1A
Long Questions
13. (a)
3  5x  2x2
= 2x2  5x + 3
= (2x2 + 5x  3)
= (x + 3)(2x  1)
(b) (i) Let x = y  1.
3  5(y  1)  2(y  1)2
= 3  5x  2x2
= (x + 3)(2x  1)
= [(y  1) + 3][2(y  1)  1]
= (y  1 + 3)(2y  2  1)
= (y + 2)(2y  3)
(ii)
mx + 3m + 3  5x  2x2
= m(x + 3) + [(x + 3)(2x  1)]
= m(x + 3)  (x + 3)(2x  1)
= (x + 3)[m  (2x  1)]
= (x + 3)(m  2x + 1)
1M
1A
1M
1M
1A
1M+1M
1M
1A
14. Let x be the number of boxes of dark chocolates produced.
Then, the number of boxes of milk chocolates produced is 140  x.
0.25x + 0.65(140  x)  50
0.25x + 91  0.65x  50
0.4x  41
x  102.5
∵ x is an integer.
∴ The minimum value of x is 103.
∴ There were at least 103 boxes of dark chocolates.
1M
1M+1A
1A
1M
1A
1A
15. (a) FA16 = 15  16 + 10  1
= 240 + 10
= 250
1M
1A
(b) 111110002 = 1  27 + 1  26 + 1  25 + 1  24 + 1  23 +
0  22 + 0  2 + 0  1
= 128 + 64 + 32 + 16 + 8
= 248
 250
∴ 111110002 does not represent the same value as FA16.
1M
1A
1M
1A
16. (a) In △AFE and △CDE,
AFE = CDE
FAE = DCE
AEF = CED
∴ △AFE ~ △CDE
alt. s, AF // BC
alt. s, AF // BC
vert. opp. s
AAA
(b) ∵ E is the mid-point of DF.
∴ FE = DE
∵ △AFE ~ △CDE
AE
FE
∴
=
CE
DE
AE
=1
CE
AE = CE
i.e. E is the mid-point of AC.
∴ BE is a median of △ABC.
∴ The centroid of △ABC lies on BE.
1M
1M
1M
1A
given
1M
proved in (a)
corr. sides, ~△s
1M
1A
1M
1A
40  43  52  54  59  63  2  67  3  70  75
12
= $60
the 6th datum  the 7th datum
Median =
2
63  63
=$
2
= $63
17. (a) Mean = $
(b) (i) Mean = $
1M
1A
1M
1A
60  12  80  4
16
= $65
(ii) Let $a and $b be the prices of the other two toys sold today.
Mean price of the toys sold today = $80
a  b  96  97
= 80
4
a + b + 193 = 320
a + b = 127
If the median price of the 16 toys sold is the same as the median
found in (a), then a  63 and b  63.
∴ a + b  63 + 63
a + b  126
Since a + b = 127, it is impossible that a  63 and b  63.
∴ It is impossible that the median price of the 16 toys sold
is the same as the median found in (a).
1A
1M
1A



 1M



1A
(iii) Suppose the prices of the 4 toys sold today are $y, $96, $97 and $x.
By (b)(ii), x + y = 127.
From the question, we have y  x.
∴ y + y  127
2y  127
y  63.5
∵ 63.5 is smaller than the mean 65 found in (b)(i).
∴ The actual mean price of all the toys sold in the two days
is less than the mean found in (b)(i).
1A
1M
1A
Multiple-choice Questions
Answers
No.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
Answer
C
C
D
B
B
A
D
C
C
D
A
C
B
B
D
A
D
B
D
C
C
C
D
C
A
D
B
D
D
Solutions
1.
The answer is C.
2.
The answer is C.
For A:
It represents the solutions of x > 5.
For B:
It represents the solutions of x  5.
For D:
It represents the solutions of x  5.
3.
The answer is D.
a2  5a  6 = (a + 1)(a  6)
∴ a2  5a  6 has a + 1 as a factor.
4.
The answer is B.
12x2  43x  20
= (12x + 5)(x  4)
∴ Only III is the factor of 12x2  43x  20.
5.
The answer is B.
9(x  2)  2x + 15
9x  18  2x + 15
9x + 2x  15 + 18
11x  33
x3
∴ The least integer x is 3.
6.
The answer is A.
For I:
pq
p + (2)  q + (2)
p2q2
∴ I must be true.
For II:
pq
p  q
p
q
  
5
5
∴ II must be true.
For III:
If p > 0 and q < 0, then
1
1
> 0 and
< 0.
p
q
1
1
>
p q
1
1
2>
2
p
q
2
2
>
p q
∴ III may not be true.
∴
∴ Only I and II must be true.
7.
The answer is D.
1
1
 2
2
x y x
1
=  22
x y
1
=  4
x y
(x2y)1(x2) =
8.
The answer is C.
2016  2010
2
=3
Number of periods =
Let $P be the value of the machine in 2010.
P(1  25%)3 = 84 375
P(0.75)3 = 84 375
84 375
P=
0.753
= 200 000
∴ The value of the machine in 2010 was $200 000.
9.
The answer is C.
Total frequency = 54 + 32 + 16 + 7 = 109
The middle datum is the 55th datum.
∴ Median = the 55th datum
=1
10. The answer is D.
Time (s)
2023
2427
2831
Class mark
21.5
25.5
29.5
Frequency
5
12
8
Mean =
21.5  5  25.5  12  29.5  8
s
5  12  8
= 25.98 s
11. The answer is A.
Ages of a group of students
Cumulative frequency
50
40
30
20
10
Q1
0
12
14
16
Age
Q1 = the 10th datum
= 14
12. The answer is C.
a2  b2 + 2b  1
= a2  (b2  2b + 1)
= a2  (b  1)2
= [a + (b  1)][a  (b  1)]
= (a + b  1)(a  b + 1)
= (a  b + 1)(a + b  1)
13. The answer is B.
(a0 + a0)0 + a0
=1+1
=2
14. The answer is B.
 x 1 y 
 2 
 2y 
3
3
 x 1 

= 
2
y


( 1)  ( 3)
x
= 3 3
2 y
= 23x3y3
= 8x3y3
18
20
15. The answer is D.
an  2  an  1
an  2 an 1
=

an  2 an  2
an  2
= 1 + a(n  1)  (n  2)
= 1 + an  1  n + 2
=1+a
16. The answer is A.
1 x
1+
<x
3
 1 x 
3 1 
 > 3x
3 

3 + 1  x > 3x
x + 3x > 3  1
2x > 2
x>1
17. The answer is D.
18. The answer is B.
Let h and r be the height and the base radius of the cylinder respectively.
Original volume = r2h
New height = h  (1 + 15%)
= 1.15h
New base radius = r  (1  7.5%)
= 0.925r
New volume = (0.925r)2(1.15h)
= (0.9252)(1.15)r2h
Percentage change in its volume
(0.9252 )(1.15) πr 2 h  πr 2 h
 100%
πr 2 h
[(0.9252 )(1.15)  1]πr 2 h
=
 100%
πr 2 h
= 1.6%, cor. to 1 d.p.
=
19. The answer is D.
For I:
There is no sufficient information to show that the altitude of BC passes
through M.
For II:
∵ M is the mid-point of BC.
∴ AM is the median of BC.
i.e. The median of BC passes through M.
For III:
∵ M is the mid-point of BC.
∴ MB = MC
∴ The perpendicular bisector of BC passes through M.
∴ Only II and III must pass through M.
20. The answer is C.
For I and III:
In △ABC and △BDC,
BAC = DBC
ACB = BCD
ABC = 180  BAC  ACB
BDC = 180  DBC  BCD
∴ ABC = BDC
∴ △ABC ~ △BDC
AC
BC
∴
=
BC
DC
AC  DC = BC  BC
i.e.
BC2 = AC  DC
∴ I and III must be true.
given
common angle
 sum of △
 sum of △
AAA
corr. sides, ~△s
For II:
There is no sufficient information to show that △ADB ~ △BDC.
∴ II may not be true.
∴ Only I and III must be true.
21. The answer is C.
When arranging the data in ascending order, there are 4 possible cases:
case 1: y, y, 5, 11, 13
case 2: 5, y, y, 11, 13
case 3: 5, 11, y, y, 13
case 4: 5, 11, 13, y, y
∵ Median (the 3rd datum) = 8
∴ Case 1, case 3 and case 4 are impossible.
i.e. Only case 2 is possible.
∴ y=8
5  11  13  y  y
Mean =
5
29  2 y
=
5
29  2(8)
=
5
=9
22. The answer is C.
Let h, b and B be the height, the base and the new base of the triangle respectively.
New height = h  (1  8%)
= 0.92h
∵ Its area remains unchanged.
1
1
∴
hb = (0.92h)B
2
2
b
B=
0.92
B b
 100%
b
b
b
= 0.92
 100%
b
1
1
0
.
92
=
 100%
1
= +8.70%, cor. to 3 sig. fig.
Percentage change in its base =
23. The answer is D.
Let $P be the sum of the money.
Simple interest = P  r%  3
r
=P
3
100
3Pr
=
100
For the compound interest,
8%
interest rate per quarter =
= 2%
4
Take a quarter as a period,
number of periods in 3 years = 3  4 = 12
compound interest = P[(1 + 2%)12  1]
Simple interest = compound interest
3Pr
= P[(1 + 2%)12  1]
100
3Pr
= P(1.0212  1)
100
(1.0212  1)  100
r=
3
= 8.94, cor. to 3 sig. fig.
24. The answer is C.
432x3  128y3
= 16(27x3  8y3)
= 16[(3x)3  (2y)3]
= 16(3x  2y)[(3x)2 + (3x)(2y) + (2y)2]
= 16(3x  2y)(9x2 + 6xy + 4y2)
∴ 9x2  12xy + 4y2 is not a factor of 432x3  128y3.
25. The answer is A.
C00002D0000016 = 12  1611 + 2  166 + 13  165
= 12  1611 + 2  (165  16) + 13  165
= 12  1611 + (2  16)  165 + 13  165
= 12  1611 + (32 + 13)  165
= 12  1611 + 45  165
26. The answer is D.
For D:
∵ (3 + 5) cm = 8 cm
∴ The three line segments of lengths 3 cm, 5 cm and 8 cm cannot form
a triangle.
27. The answer is B.
For I and III:
New mode = m  (m)
= m2
<0
∴ I is not true and III must be true.
For II:
Since all data are negative and the mode must be one of the data, the mode is
negative.
Take m = 1.
New mode = m2
= (1)2
= 1
i.e. The mode remains unchanged.
∴ II may not be true.
∴ Only III must be true.
28. The answer is D.
For I:
The circumcentre of an acute-angled triangle must lie inside the triangle.
For II:
The orthocentre of an acute-angled triangle must lie inside the triangle.
For III:
The in-centre of a triangle always lies inside the triangle.
∴ I, II and III must lie inside △ABC.
29. The answer is D.
For I:
In △ABC and △DAB,
∵ AB = AC and DA = DB.
AB
AC
∴
=
DA
DB
DAB = DBA = 30
ACB = ABC = 30
BAC = 180  30  30 = 120
ADB = 180  30  30 = 120
∴ BAC = ADB
∴ △ABC ~ △DAB
∴ I must be true.
For II and III:



DAB + DAC = BAC
DAC = 120
DAC = 90
∴ △DAC is a right-angled triangle and DA  AC.
∴ A is the orthocentre of △ADC.
∴ II and III must be true.
∴ I, II and III must be true.
given
base s, isos. △
base s, isos. △
 sum of △
 sum of △
ratio of 2 sides, inc. 