Multiple choice Each questions worth 1 mark A1.. Sir Lance has a lot of tables and chairs in his houseA. Each rectangular table seats eight people and each round table seats five people.
Trang 1SỞ GD&ĐT VĨNH PHÚC
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ĐỀ CHÍNH THỨC
KỲ THI CHỌN ĐT DỰ THI HOMO NĂM 2011
ĐỀ THI MÔN TOÁN THCS (JUNIOR) Thời gian làm bài: 150 phút, không kể thời gian giao đề.
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Đề thi có 02 trang Thí sinh không được sử dụng máy tính, từ điển.
Làm bài vào tờ giấy thi.
Part I Multiple choice (Each questions worth 1 mark)
A1 If the net shown is folded to make a cube, which letter is opposite X ?
A2 Mr Owens wants to keep the students quiet during a Mathematics
lesson He asks them to
multiply all the numbers from 1 to 99 together and then tell him the last-but-one digit of the
result What is the correct answer?
A3 At the Marldon Apple-Pie-Fayre bake-off, prize money is awarded for 1st, 2nd and 3rd places in
the ratio 3 : 2 : 1 Last year Mrs Keat and Mr Jewell shared third prize equally What
fraction of the total prize money did Mrs Keat receive?
A 1
1
1
1
1 12
A4 In a triangle with angles , ,x y zo o o the mean of y and z is x What is the value of x ?
A 90o B 80o C 70o D 60o E 50o
A5 Which of the following is the longest period of time?
A 3002 hours B 125 days C 171
2 weeks D 4 months E
1
3 of a year.
A6 Sir Lance has a lot of tables and chairs in his house Each rectangular table seats eight people and
each round table seats five people What is the smallest number of tables he will need to use to seat 35 guests and himself, without any of the seating around these tables remaining unoccupied?
A7 Kiran writes down six different prime numbers, p q r s t u, , , , , , all less than 20, such that
p q r s t u+ = + = + What is the value of p q+ ?
A8 Nicky has to choose 7 different positive whole numbers whose mean is 7 What is the largest
possible such number she could choose?
A9 Sam's 101st birthday is tomorrow So Sam's age in years changes from a square number (100) to a
prime number (101) How many times has this happened before in Sam's lifetime?
A10 One of the examination papers for Amy’s Advanced Arithmetic Award was worth 18% of the
final total The maximum possible mark on this paper was 108 marks How many marks were
available overall?
C D
A X B
E
Trang 2Part II Short questions (Each question worth 3 marks).
B1 ABC is a triangle with sides 3,4 and 5 units A′ is the mirror image of the point A across line
BC Similarly, B′ and C′ are mirror image of B and C across lines CA and AB Find the area of triangle A B C′ ′ ′
B2 ABC∆ is right-anged at A∠ and AB AC< Let BCDE be square contruct outward of the triangle.
G is on AC such that ∠AGE=90 ,o F is on EG such that ∠BFE=90 o Prove that ABFG is a
square
B3 Find all the solutions of the simultaneous equations
19 11 14
x y xy
y z yz
z x zx
+ + =
+ + =
+ + =
B4 Find all the integer solutions (x y to the inequalities; )
x − <y x + <y
B5 McVees sell chocolate snacks in packets of 6 whereas Jays sell the same type of snack in packets
of 5 Mr Scrooge is running a conference and wants to provide exactly one snack per person at the coffee break Can he do this if he has to provide for 58 people?
If he has to provide for N people, what is the largest value of N where he will not be able to avoid buying more snacks than are needed?
B6 Let ABC be a triangle Let , D E be the points out side of the triangle so that AD AB AC= , =AE
and ∠DAB= ∠EAC=90 o Let F same side of the line BC with A so that FB FC= and
90
BFC
∠ = o Prove that triangle DEF is right-isoceles triangle.
B7 Alice, Betty and Chris are wearing hats which have 1 or 0 written on them (chosen at random) in
such a way that each person can only see the numbers which the other two are displaying (There are no mirros for them to use and they cannot talk to each other) Each person casts a vote on whether or not there are more 1s than 0s based on what they observe the other’s hats are showing If in doubt, they flip a coin Work out whether the majority vote will be right more often than not
— Hết —
Giám thị coi thi không giải thích gì thêm
Họ tên thí sinh ……….… SBD ………
Trang 3Vinhphuc’s Training and Education Service
Pre – HOMO 2011 Junior Section Solutions.
Part I Multiple choice.
Part II Short question.
B1 Let line AA′ intersect BC at H, B C′ ′ at H′ Then A H′ ⊥BC AH, ′⊥B C′ ′ Also in lengths,
, 3
B C′ ′ = BC A H′ ′ = AH
So the area of ∆A B C′ ′ ′ is 3 times the area of ∆ABC, which is 3 × 6 =18.
B2 By hypothesis, AGFB is a rectangular and ACB∠ = ∠FEB Since BE BC= , ACB∆ = ∆FEB Therefore BF =BA Thus ABFG is a square
B3 The system of equations is equivalent to
( ) ( ) ( ) ( ) ( ) ( )
1 1 20 (1)
1 1 12 (2)
1 1 15 (3)
We see that , ,x y z≠ −1
Dividing (3) by (2), we get 1 5
1 4
x
y+ = + then multiplying by (1) we get ( )2
1 25
x+ = So x=4 or 6
x= −
If x=4, then y=3,z=2 If x= −6, then y= −5,z= −4
Remark Multiply (1),(2) and (3) together and then divide by the square of (2), we get ( )2
1 25
x+ =
B4 Adding the inequalities, 2x2 <5. So x can only be −1, 0 or 1.
The originals can be written x2− <1 y and y< −4 x2
For x = −1 or x = 1, 0 < y and y < 3 ; for x = 0, −1 < y and y < 4
So (−1;1 , 1;2 , 0;0 , 0;1 , 0; 2 , 0;3 , 1;1 , 1; 2) (− ) ( ) ( ) ( ) ( ) ( ) ( ) are solutions
Remark Use graphical method drawing 2 2
1, 4
y x= − y= −x would work
B5 Ans : 3 packets of McVees and 8 packets of Jays
We can see that there are values of N such as 13 where he can not buy the exact number
Suppose he buys x packets of McVees and y packets of Jays Then 6x + 5y = N
If N is a multiple of 5, then x = 0
Starting from solutions for N = 5k, we keep x to a
minimum To increase N by one, we increase x by
1 and decrease y by 1 as long as y has not been
reduced to 0 or less
For example, for N = 10, (x, y) = (0, 2) Thus we
can get N = 11 with (1, 1) and N = 12 with (2, 0)
but can not solve it for N = 13 or N = 14
This situation last occurs at N = 19
For all higher numbers of people, Scrooge can
avoid wasting any snacks!
B6 Since AD AB= ,∠DAC = ∠BAE AC, = AE,
∆ = ∆ Consequently, BE CD= and
BE⊥CD
Let G be intersection point of BE CD We have,
F
E
D
A
Trang 490 90
o o Thus ∠GBF = ∠GCF i.e ∠EBF = ∠DCF
On the other hand FC FB CD BE= , = , hence ∆DFC= ∆EFB It’s follows FE FD= and
B7 Answer we would expect the majority vote to be correct 13 times out of 16 (reduced from 26 out
of 32)
Arrangements for A, B, C are : (0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1), (0, 1, 1), (1, 0, 1), (1, 1, 0), (1, 1, 1)
For (0, 0, 0) all will vote against more 1s than 0s which is correct
For (1, 0, 0), A will vote against more 1s than 0s, B’s and C’s vote will be random The majority will only be incorrect if both B and C vote incorrectly This will only happen 1 time out of 4
Similarly for (0, 1, 0) and (0, 0, 1)
The same argument applies when there are two 1s and one 0 but voting for more 1s than 0s
For (1, 1, 1) all will vote for more 1s than 0s which is correct