De thi Toan Tieng anh 8

PEOPLE’S COMMITTEE OF LE CHAN DISTRICT

THE DISTRICT-LEVEL TEST OF
EXCELLENT STUDENT
SCHOOL YEAR: 2017 - 2018
SUBJECT: MATHEMATICSGRADE 8

EDUCATION AND TRAINING DEVISION
THE OFFICIAL TEST

Time: 120minutes (not count the time of the examination)
Note: Examination consists of 09 pages. Students write directly in the examination
Mark

Full name, signature

Code

...................................................................................
.

................... ...
................ .......
............

Examiner 1:

Examiner 2:
...................................................................................
.

................... ...
................ .......
.............

CODE 1
PART 1: MULTIPLE-CHOICE (150 marks)
(Choose the bold word part is the answer to each question)
f  x  5 x 3  x  3
Question 1. The remainder when the polynomial
is divided by
g  x  x  3
is
A. 135
B. –135
C. x  3
D. x  3
Question 2.Suppose that 108n is asquare number. Find the least positive integer n.
A. 1
B. 3
C. 12
D. 27
5
2
Question 3.Which of the following is a factor of the polynomial x  x  28
A. x  2
B. 4 x  2
C. x  28
D. 4  2x
 x3 y  6 x2 y 2  2 x    13 x 2 y  
Question 4.Multiply:

A.
C.



1 5 2
2
x y  2 x 4 y 3  x3 y
3
3



1 5 2
2
x y  2 x 4 y 3  x3 y
3
3

B.
D.



1 5 2
2
x y  2 x 4 y 3  x3 y
3
3



1 5 2
2
x y  2 x4 y 2  x2 y
3
3

5
3
3
Question 5.The product of polynomial 3x  x and monomial  2x is
15
9
8
6
8
6
8
6
A.  6 x  2 x
B. 3x  x
C.  6 x  2 x
D. 6 x  2 x
2
Question 6. The smallest real number x satisfying the inequality x  18 0 is
A.  3 2
B. 3 2
C.  2 3
D. 2 3

Question 7. 60% of the pupils in the school are girls. 40% of the girls and 80% of the
boys in this school can swim. What is the percentage of the pupils in the school who can
swim?

1


A. 80%
B. 76%
C. 56%
D. 64%
Question 8. Nguyen is 25 years younger than his mother. In 21 years time, Nguyen will
be half of his mother’s age. What is the product of their ages now?
A. 112
B. 116
C. 136
D. 166
P  x   1  2 x   1  4 x   1  6 x  ...  1  2016 x   1  2018 x  .
Question 9.Given the expression
The coefficient of x in the expansion of P(x) is:
A. 1007
B. 1008
C. 1009
D. 1010
P  x   m  3 x3  x 2   2m  n  x  n  1
Question 10.The polynomial
isdivisible by x and
x + 1 if and only if
1


m 
m  1
 m 0
m 3
3




n 1
n 0
n 1


n 1
A.
B.
C.
D.
Question 11. The remainder when 2x3 + kx2 + 7 is divided by x – 2 is half of the
remainder when the same expression is divided by 2x – 1. Find the value of k.
A. –5
B. –10
C. –15
D. –20
Question 12. Consider the sequence: 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6..............
What is the 2018th term?
A. 62
B. 63
C. 64

D. 65
Question 13. Out of 80 children, 60 can swim, 54 can play chess and only 10 can do
neither. How many children can swim and play chess?
A. 34
B. 44
C. 54
D. 24
Question 14. Nancy and Ellen each start reading a copy of Gone with the Wind on the
first day of their summer vacation. Nancy decides to read 7 pages each day, but Ellen
only wants to read 5 pages each day. When Nancy is on page 84, what page is Ellen
reading?
A. 35
B. 40
C. 50
D. 60
Question 15. A girl group baked a batch of cookies to sell at the annual of bake sale.
They made between 100 and 150 cookies. One fourth of the cookies were lemon crunch
and one fifth of the cookies were chocolate macadamia nut. What is the largest number of
cookies the group could have baked?
A. 110

B. 120

C. 130

D. 140

Question 16. Two positive integers a and b differ by 5. Suppose that the sum of their
square roots is also 5. What is the value of 100(a + b)?
A. 1300

B. 3600

C. 7200

D.10800

ab  2b 2
1
4a 2

2
2
2
Question 17.Given that a  3ab  15b 3 , find the value of b .

A. 25

B. 36

C. 49

2

D. 64


Question 18. Given that
CODE
 2

1  5 x 2 

x
9  2x
and 1 2 y 6
2
. Find range of the value of y .

 1 
  ;5 
B.  15 


  5 ;5
A.

 2 5
 ; 
C.  5 3 

 2 5
 ; 
D.  5 6 

2
Question 19. The smallest value of A 2 x  6 x is:

A.



9
4

B.



9
2

9
C. 4

9
D. 2

x2 y 2
 2 
2
Question 20. If xy = 3 and x + y = 5 then y x
253
343
243
A. 9
B. 9
C. 9

553

D. 9

Question 21. The area of the isosceles trapezoid ABCD is 144cm2. If AB = 12cm and
CD = 24cm then AD =
A. 7cm

B. 8cm

C. 9cm

D. 10cm

Question 22.ABC is a triangle with AB = 5cm, BC = 3cm, and AC = 4cm. Let G be the
centroid of triangle ABC. What is the length of AG?
5
cm
2,5cm
5cm
3
A.
B.
C.
D. 2cm
Question 23. Given the equilateral triangle ABC, AB = 16cm. The area of the triangle
ABC is:
A. 64cm

2

2

B. 36 3cm

2
C. 64 3 cm

2
D.128 3cm

0

Question 24. Given a right trapezoid MNPQ (MN // PQ, NMQ 90 ), PQ = MN + MQ.
The measurement of the angle N is:

A. 900

B. 600

C. 450

D. 1350

Question 25.The sum of the exterior angles of a polygon is:
A. 1800

B. 3600

C. 5400

D. 7200

Question 26. Given a parallelogram ABCD. Let H be orthogonal projection of A on the
side DC. If AH = 4cm, AB = 100cm, BC = 5cm then HC =
A. 97cm

B. 96cm

C. 95cm

D. 103cm

Question 27. The quadrilateral ABCD has two diagonals that are perpendicular lines. If
AB = 16cm, BC = 14cm and AD = 8cm then CD =
A. 4cm

B. 3cm

C. 2cm

D. 1cm

Question 28. Two squares, each with side 8m, are placed such that a vertex of one lies at
the centre of the other. Find the area of the overlapping region.
A. 48m2

B. 16m2

C. 32m2

D. 8m2

Question 29. ABCD is a rectangle. Find the perimeter of ABCD if its area is 66cm 2 and
AB – BC = 5cm.
3


A. 34cm

B. 17cm

C. 48cm

D. 66cm


BAC
90 , AB 6cm, AC 9cm 

Question 30. Given a right angle ABC
. Let E, D be the
0

midpoint of AB and AC respectively. Let G be the intersection of BD and EC. Find the
area of the quadrilateral AEGD.
A. 4,5cm2
B. 9cm2
PART 2: ESSAY (150 marks)
Question 31.

C. 18cm2

D. 6cm2

5x2
P  x  6
x  x 5  x 3  5 x 2  4 x  1 . Prove that there is a polynomial
Given a fraction

Q  x

with integer coefficients such that

polynomial

R  x  x 8  4 x 4  1

Q  x0  P  x0 

for every x0 which is a root of the

.

…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..

…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..

4


…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
CODE

…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..

…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
………………………………………………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………………………………………………….
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..

5

…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
………………………………………………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………………………………………………….

Question 32.
Given an isosceles at A triangle ABC. Let H be the midpoint of BC. Let E be the
the opposite ray of the ray CB such that CE = CA. Let I be the point on the ray
point on
AB such that AI = HE. Prove that the line IH passes through the midpoint of AE.
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..

…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..

6


…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..

…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
CODE
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..

…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
………………………………………………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………………………………………………….
…………………………………………………………………………………………………………………………………………………………..

7


…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
………………………………………………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………………………………………………….

Question 33.
Let a, b, c be positive real numberssuch that 
a
b  c 10
b)


a) b  c 3a
bc
a
3

3a  2b   3a  2c  16bc

. Prove that

…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..

…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..

8


…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
………………………………………………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………………………………………………….
…………………………………………………………………………………………………………………………………………………………..
CODE
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..

…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
………………………………………………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………………………………………………….
…………………………………………………………………………………………………………………………………………………………..

9


…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..

…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………………..

----- TheEnd ----LE CHAN DIVISION OF
EDUCATION AND TRAINING

LE CHAN DISTRICT MATHEMATICS COMPETITION
School Year 2017 – 2018

ANSWERS AND MARKS
Questio
n

Mark
s

Answers

1
A
1 to 30 11
CODE1 A
21
D
1

2
B
12

C
22
C

3
D
13
B
23
C

4
B
14
D
24
D

5
C
15
D
25
B

6
A
16
A
26

A

7
C
17
B
27
C

8
B
18
A
28
B

9
D
19
B
29
A

10
A
20
B
30
B

2

3

4

5

6

7

8

9

10

50
50
50

50
1 to 30 11
CODE2
21

12

13

14

15

16

17

18

19

20

50
22

23

24

25

26

27

28

29

30

50
We have : x  4 x  1  x  x  x  5 x  4 x  1  x  x  1  5 x
8

4

6

 x 6  x 5  x3  5 x 2  4 x  1 

31

5

3

2

2

20

x8  4 x 4  1  5 x
x 2  x 1

5 x 2  x 2  x  1  5 x   x 3  x 2  x 

5x2
P  x  6

 8
x  x 5  x 3  5 x 2  4 x  1 x8  4 x 4  1  5 x
x  4 x4 1  5 x
Therefore,
8
4
R x x 8  4 x 4  1
P x  x03  x02  x0
If x0 is a root of  
then x0  4 x0  1 0 . Thus,  0 
.

So there is a polynomial

Q  x   x 3  x 2  x

10

that

Q  x0  P  x0 

20
10


Questio

n

Answers

Mark
s

Suppose that IH intersects AE at M
We have AI = AB + BI = HE = CE +
HC = AC + HB so BI = BH
Therefore,
ACB ABC 2BHI 2MHC  1

25

32

Then, we also have ACB 2MEC (2)
From (1) and (2), we have MHC MEC , so MHC is an isosceles triangle at M,
Because the triangle AHE is right at H, we have MA = ME.
We have

 3a  2b   3a  2c  16bc 

2
 3a  2a  b  c   b  c 

33

3a 2  2ab  2ac 4bc

2

  a  b  c   3a  b  c  0  3a  b  c 0  3a b  c
bc
a
bc
1 t 2 1
t

t  
a ( t 3 ). We have b  c
a
t
t
Let

 t  3  t 



t

1

3  10 10
 
3
3

25

10
10
10
15

3
b c  a
2 .
The equality holds when

5

11