The second exam of the second day of junior high school
(8: 30 on February 26th, 2004, 1 1: 00)
1. Multiple choice questions (7 points for each question, ***56 points) Only one of the following four conclusions is correct. Please fill in the English letters of the correct answer in the brackets after the question.
1. Chen Shengshen, an outstanding mathematician, died in Tianjin on February 3, 2004. Professor Chen Shengshen has made outstanding contributions in the field of differential geometry, and is the only China person who won the Wolff Prize. He once pointed out that there are two important theorems in plane geometry, one is Pythagorean theorem, and the other is triangle interior angle sum theorem. The latter shows that a plane triangle can be ever-changing, but the sum of three internal angles is.
(1) parallelogram ABCD with definite side length, when ∠A changes, the sum of any set of diagonals remains unchanged;
(2) When the number of sides of a polygon increases, the sum of its outer angles remains unchanged;
(3) When △ABC rotates around vertex A, the internal angle of △ABC remains unchanged;
(4) Observing under a magnifying glass, when the figure containing the angle A is enlarged, the size of the angle A remains unchanged;
(5) When the radius of the circle changes, the ratio of the circumference to the radius of the circle remains unchanged;
(6) When the radius of a circle changes, the ratio of the perimeter to the area of the circle remains unchanged.
Among them, the narrative error is ()
2 (B)3 (C)4 (D)5。
2. In the process of cell division, each cell divides into two cells at a time. 1 Cell divides into 2 cells at 1, 4 cells at the second time, and 8 cells at the third time ... The number of cells after the 50th division is the closest ().
(A) 10 15(B) 10 12(C)lO8(D)lO5
3. As shown in the figure, in the pentagonal ABCDE, BC‖AD, BD‖AE, AB ‖ EC. The triangle with the area equal to △ABC in the figure is ().
1 (B)2 (C)3。 (d) 4。
4. As shown in the figure, the quadrilateral ABCD is square, and straight lines l 1, l2 and l3 pass through points A, B and C respectively, and L1/L2//L3. If the distance from l 1 to l2 is 5 and the distance from l2 to l3 is 7, then the area of square ABCD is equal to ().
70(B)74(C) 144(D) 148
5. On the ABCD of the rectangular billiard table, click a ball from somewhere near AB and hit BC, CD,
1 DA and then return to the starting point p. Every time the ball hits the edge of the table, the included angle between the route before and after the impact and the edge of the table is equal (for example, ∠ α = ∠ β in the figure). If AB=3 and BC=4, the total length of the route taken by the ball (regardless of the size of the ball) is ().
(a) Uncertainty (b)12 (c)11(d)10.
6. Algebraic expression 2x2-6xy+5y2, where x and y can be arbitrary integers, then the value of this algebraic expression is not greater than lO ().
(A)6 (B)7 (C)8 (D) 10。
7. In 2004, 2005, 2006 and 2007, the number that cannot be expressed by the square difference of two integers is ().
2004 (B)2005 (C)2006 (D)2007
8. It is known that the inequality group about X:-1, 0, l, 2 has only four integer solutions, so the number of all possible integer pairs (a, b) suitable for this inequality group is ().
1 (B)2 (C)4 (D)6
Fill in the blanks (7 points for each small question, ***56 points)
9. There are many yellow sand supply stations along the highway, and there is a construction site between every two yellow sand supply stations. A truck loaded with yellow sand set out from the company and arrived at 1 yellow sand supply station to load sand, increased the yellow sand on the truck by 1 times, and unloaded 2 tons of yellow sand at 1 site. After that, every time the yellow sand on the truck was loaded at the yellow sand supply station, it increased by 65438+.
10. There are 20 teams, and each team and other teams only play L games. Every game is won by the team with 1, that is, there is no draw. If 1 match wins 1 point and the loser gets zero point, then the sum of the scores of any eight teams is at most _ _ _ _ _.
1 1. In the trapezoidal equation table as shown in the figure, the equation in row is _ _ _ _ _ _ _ _ _ _ _ _.
12. Ordinary dice are cubes with a point on each side of 1, 2, 3, 4, 5 and 6. There are two common dice, a and b, and the sum of the points on each side of a and b is shown in table 1, from which we can see that the sum is 2, 3, 4, ...
Now, two special cube dice, C and D, are designed, and the points on each side of the C dice are required to be compared with those on each side of the D dice.
The sum obtained by adding the numbers is still 2, 3, 4, …, 12, and the number of occurrences of the same sum is exactly the same as that of two ordinary dice A and B, that is, 2 appears 1 time, 3 appears twice, …, 12 appears 1 time. It is known that two dice c and d are on one side.
13. As shown in the figure, connect four pieces of wood with screws to form a quadrilateral ABCD (movable at A, B, C and D). Now, fix AB and change the shape of the quadrilateral. When point C is on the extension line of AB, ∠ C = 900; When point D is on the extension line of BA and point C is on the line segment of AD, it is known that AB=6cm and DC= 15cm, then AD = _ _ _ _ _ cm and BC = _ _ _ _ _ cm.
14. The length, width and height of a cuboid are prime numbers. The length-width product ratio is 8 higher and the length-width difference ratio is 9 lower. The volume of this cuboid is _ _ _ _ _.
15. As shown in the figure, two rectangles ABCD and EFGH intersect, EH and DC intersect at point M, EF and DA intersect at point P, FG and AB intersect at point N, and GH and BC intersect at points Q, MN‖DA, PQ ‖ Eh. It is known that MN=lO, PQ=9, rectangular EFGH.
16. A paper square "cactus" should be constantly growing, and the newly grown leaves are "squares with missing corners". The center of these "squares" is at the corner of the previous square, and the side length is half that of the previous square (as shown in the figure). If the side length of the 1 th square is 1,
Third, answer questions.
17. A rectangle with the ratio of long side to short side of 2: 1 is called a "standard rectangle". The agreed short sides are a 1, a2, a3, a4 and a5 (where a 1
(1,2,2.5,4.5,7). Answer the following questions:
(1) Write out the values of a4 and a5 in the rectangle (1, 2,5, a4, a5) and the corresponding rectangles with different areas (represented by the annotations of the above rectangles), and draw two schematic diagrams of rectangles that meet the requirements.
(2) What are the maximum values a4 and a5 of all these rectangles (1, 2,5, A4, A5)?
Five people, 18. A, b, c, d, e, went to the store to buy things, and each spent an integer yuan. The difference between A and B in 56 yuan (that is, the absolute value of the difference between two people's money, the same below) is 19 yuan, and the difference between B and C is the difference between C and D in 7 yuan. Why?
19. When x=20, the value of the quadratic trinomial about x is equal to 694. If all the coefficients and constant terms of the quadratic trinomial are integers with absolute values less than 10, find all the quadratic trinomials that meet the conditions.
In order to better improve the quality of the magazine, the editorial department of Time Mathematics Learning invited 20 students to sit around for a meeting.
There was a symposium on the table. At the symposium, there are enough magazines for everyone to choose at will, and everyone can take as many as they want. At the end of the symposium, the number of magazines held by each person was counted, and it was found that there were always some people sitting together (1 or all), and the sum of the number of magazines they held was an integer multiple of 20. Why?
The 19th Junior High School Mathematics Competition in Jiangsu Province (Test 2 of Grade 2)
Reference answers and grading standards
First, multiple choice questions
Title 1 2 3 4 5 6 7 8
Answer a A A C B. D B C D
Second, fill in the blanks
9. 1.75 The original yellow sand on the truck was x tons. According to equation 2[(2x-2)-2]-2=O, the solution is x= 1.75 (ton).
10. 124. 1 1 . N2+(N2+ 1)+(N2+2)+…+(N2+n)=(N2+n+ 1)+(N2+n+2)+…+(N2+n+n)。
12.C 1 2 2 3 3 4 D 1 3 4 5 6 8
Note: the number of surface points has nothing to do with the arrangement order. As long as the number of points filled in is correct, you can score.
13.39, 30. 14.273, ( 105) 15.34.4. 16.
Third, solve the problem.
17. According to the conditions, there are five kinds of rectangles.
( 1) ( 1,2,5,6, 12), ( 1.2,5,5,6), ( 1,2,5, 12,65433)
Note: The question (1) is 10. If you answer 1 correctly in 5 rectangles, you will get 1 * * 5 points. Draw 1 picture, get 3 points, and draw 2 pictures, get 5 points.
(2) Sub-item (2) is 2 points. According to (1), the rectangle (1, 2,5,12,29) has the largest area, and its area is 29x (2+ 10+58) = 29× 70.
18. Orders A, B, C, D and E respectively indicate the amount of each personalized fee of A, B, C, D and E, which is obtained from the meaning of the question.
..... 4 points
After adding the above equations, the left side is zero, so the sum on the right side must be zero. Because 4+5+7+11+19 = 46, we divide the five numbers A, B, C, D and E into two parts, one is 23 and the other is 23.
Here, we get the score of the system of equations (1) (2)...6.
According to the equations (1):
So (e+11)+(e-8)+(e-1)+(e+4)+e = 56, 5e+6 = 56, 5e=50, e = 10. ...
According to the equation set (2):
Get (e-11)+(e+8)+(e+1)+(e-4)+e = 56.5e-6 = 56, e = (non-integer, rounded),
Therefore, the project cost is RMB 65,438+00 ... 12.
19. substitute x=20 into ax2+bx+c to get 400a+20b+C = 694. ①.
So 400a = 694-(206+C)...5 points.
To-10
A is an integer, so a = 2...7 points.
Substituting a=2 into ① gives 20b+c= 106, ②.
So 20b=- 106-c and-10.
And b is an integer, so 6=-5. If you replace ②, you get C =-6... 10.
If x=20 is substituted into 2x2-5x-6, the value is 694. Therefore, the quadratic trinomial that meets the conditions is only 2x2-5x-6... 12.
20.20 people form a circle. Starting from any one person, the numbers taken by 20 people clockwise (or counterclockwise) are a 1, a2, a3, a4, …, a20 respectively.
Let S 1=a 1, S2 = a1+a2 ...
Sk = Al+A2+A3+A4+…+AK (k = 1, 2,3, …, 20) ……… 3 points.
If the number of 1 in Sk is a multiple of 20, this conclusion holds.
If there is no number in Sk where 1 is a multiple of 20, there must be a remainder when Sk is divided by 20, so the remainder is the divisor of the nonzero remainder where Rk (k = 1, 2, …, 20)20 is 1, 2, 3, …, 19, etc. * * 655.
Therefore, at least two of r 1, r2, r3, …, r20 are the same. Let it be ri=rj,1≤ i.
At this time, Si-SJ = Ai+ 1+Ai+2+…+AJ is a multiple of 20, that is, the sum of the numbers taken by neighboring people from i+ 1, i+2, … to J is a multiple of 20 … 12.