Kỳ thi chọn ĐT dự thi HOMO năm 2014 Đề thi môn toán THCS (JUNIOR) Trường Thcs Lập Thạch

TRƯỜNG THCS LẬP THẠCH
KỲ THI CHỌN ĐT DỰ THI HOMO NĂM 2014
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.



C
D
A
X
B



E


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 ?
A. A
B. B
C. C
D. D
E. E

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?
A. 0
B. 1
C. 2
D. 8
E. 9

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. 
B. 
C. 
D. 
E. 

A4. In a triangle with angles the mean of and is What is the value of ?
A. 
B. 
C. 
D. 
E. 

A5. Which of the following is the longest period of time?
A. 3002 hours
B. 125 days
C. weeks
D. 4 months
E. 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?
A. 4
B. 5
C. 6
D. 7
A. 8

A7. Kiran writes down six different prime numbers, all less than 20, such that . What is the value of ?
A. 16
B. 18
C. 20
D. 22
E. 24

A8. Nicky has to choose 7 different positive whole numbers whose mean is 7. What is the largest
possible such number she could choose?
A. 7
B. 28
C. 34
D. 43
E. 49

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?
A. 1
B. 2
C. 3
D. 4
E. 5

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?
A. 420
B. 480
C. 540
D. 560
E. 600

Part II. Short questions (Each question worth 3 marks).
B1. is a triangle with sides 3,4 and 5 units. is the mirror image of the point across line Similarly, and are mirror image of and across lines and Find the area of triangle 
B2. is right-anged at and Let be square contruct outward of the triangle. is on such that is on such that Prove that is a square.
B3. Find all the solutions of the simultaneous equations

B4. Find all the integer solutions to the inequalities

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 be a triangle. Let be the points out side of the triangle so that and Let same side of the line with so that and Prove that triangle 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 ……………Lap thach’s Training and Education Service
Pre – HOMO 2014
Junior Section
Solutions.
Part I. Multiple choice.
A1. D
A2. A
A3. E
A4. D
A5. A
A6. C
A7. E
A8. B
A9. D
A10. E

Part II. Short question.
B1. Let line intersect BC at H, at . Then Also in lengths, 

So the area of is 3 times the area of ∆ABC, which is 3 × 6 =18.
B2. By hypothesis, is a rectangular and . Since , . Therefore Thus is a square.
B3. The system of equations is equivalent to

We see that 
Dividing (3) by (2), we get then multiplying by (1) we get So or 
If then If then 
Remark. Multiply (1),(2) and (3) together and then divide by the square of (2), we get 


B4. Adding the inequalities, So x can only be −1, 0 or 1.
The originals can be written and .
For x = −1 or x = 1, 0 < y and y < 3 ; for x = 0, −1 < y and y < 4 .
So are solutions.
Remark. Use graphical method drawing 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 , . Consequently, and 
Let be intersection point of We have

Thus i.e. .
On the other hand , hence . It’s follows and So 
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.