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Answers to the Sixth Grade Final of the 12th Huajin Cup Mathematics Invitational Tournament
Examination questions and answers of the final exam of the 12 China Cup

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The four horns of 1. Hua, Bei and Sai are "2440", "4 199" and "3088" respectively. The code of "Huabei" is 24404 1993088. If this code starts from the left,

2. Calculation: = _ _ _ _ _.

3. As shown in the figure, the side lengths of the two squares ABCD and DEFG are all integer centimeters, the point E is on the line segment CD, and CE < DE and the line segment CF = 5cm, then the area of the pentagonal ABCFG is equal to _ _ _ _ _ square centimeters.

4. Ranking, from small to large, the third number is _ _ _ _ _ _.

5. Figure A below is a cross-sectional view of a sealed water bottle. The upper part is conical and the lower part is cylindrical. Bottom diameter 10 cm, water bottle height 26 cm, and liquid level in the bottle height 12 cm. After the water bottle is inverted, as shown in Figure B below, if the liquid level in the bottle is 16 cm, the volume of the water bottle is equal to _ _.

6. Determine the number of a column according to the following conditions: the first column is 3, the second column is 6, and the third column is 18. After that, each number is twice the sum of all the previous numbers, so the sixth number is equal to _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

7. A natural number whose sum of the largest divisor and the second largest divisor is11,and this natural number is _ _ _ _ _ _.

8. Put some small cube codes with a side length of 1 into a solid, and look at the solid from top to bottom, as shown in Figure A below, and look at the solid from the front, as shown in Figure B below, so the surface area of the solid is at most _ _ _ _ _ _ _ _ _.

Secondly, answer the following questions briefly (write a short process)

9. As shown in the figure, in the triangle ABC, point D is on BC, and ∠ABC = ∠ ACB, ∠ ADC = ∠ DAC, ∠ DAB = 2 1, so the number of times to find ∠ABC; And answer: which triangles in the picture are acute triangles.

10. Liu Yun was sitting by the window in a 60-kilometer-per-hour train when he saw a truck with 30 carriages approaching. When the front of the truck passed the window, he began to time until the last car passed the window, and the recorded time was 18 seconds. It is known that the car length of a truck is 15.8m, the car spacing is 1.2m, and the front of the truck is 10m. What's the speed of the truck?

1 1. The following figure is a 9×9 grid diagram, which is divided into nine "small nine palaces" with thick lines, with three grids in each direction, and some small squares are filled with numbers from 1 to 9. Xiaoqing fills in the natural number of 1 to 9 in the blank of the fourth column, so that the numbers of each row, column and "Xiao Jiu Gong" are not repeated. Then Xiaoqing writes the number in the fourth column as a 9-digit number from top to bottom. Please write down this 9-digit number and briefly explain the reason.

12. In a class math exam, the average score of all students with excellent grades is 95, and the average score of students with poor grades is 80. It is known that the average grade of the whole class is not less than 90. What is the percentage of students who get excellent questions in the class?

Three. Answer the following questions in detail (write down the detailed process)

13. As shown in the figure, connect the vertices of a regular hexagon and ask how many isosceles triangles (including equilateral triangles) are there in the graph?

14. There are seven empty boxes on the circumference, which are numbered 1, 2, 3, 4, 5, 6 and 7 in turn clockwise. Xiao Ming first puts the 1 th white chess piece in the 1 th box, then puts the second white chess piece in the third box, and then puts the third white chess piece in the sixth box ... After Xiao Ming puts the K- 1 th white chess piece, count the K- 1 th box clockwise, and put the k-th. Subsequently, Xiaoqing started from the box of 1 and put 300 red chess pieces in these boxes according to the same rules in the counterclockwise direction. Please answer: How many white pieces are there in each box? How many pieces are there in each box?

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1. Solution: Even numbers are 4, 0, 1, 9, 0, 8 from left to right, and their complements about 9 are 5, 9, 8, 0, 9, 1 from left to right, so the new code of "China Cup" is: 254948900.

2. Solution: The original formula = [20.75+1.24×] ÷ 41.75 = [20.75+0.125] 2475438+0.75 = 20.875 sad.

3. solution: CF = 5, CD and DF are integers. According to Pythagorean Theorem, CE = 3, DF = 4 and CD = 7.

So the area of pentagonal ABCFG = 16+49+6 = 7 1 (square centimeter).

4. Solution: = 0.524, = 0.525, so:, the third smallest number is

5. Solution: If part of the liquid in the bottle is taken out, the upright height is 1 1 cm and the inverted height is 15 cm. At this time, the liquid in the bottle is just half the volume of the bottle, so the volume of the bottle is equivalent to the volume of a cylinder with a height of 22 cm (the bottom area is unchanged), that is, the volume of the bottle is:

3. 14×× 22 = 1727 (cubic centimeter)

6. Solution: The first number in this column is 3, the second number is 6, the third number is 18, the fourth number is (3+6+ 18) × 2 = 54, and the fifth number is (3+6+18+54 )× 2.

Let the first one in this series be a, then the second one is 2a, the third one is 6a = 2× 3× a, the fourth one is 18a = 2×A, the fifth one is 54a = = 162 A = 2×A, and the nth one is 2×A because A = 3. 2007 ÷ 486 > 3, and 2007 ÷ 3 < 9, we can see that from the eighth number, each number is greater than 2007.

7. Solution: Because11is an odd number, and odd number = odd number+even number, the maximum approximate number and secondary divisor of the number must be even and odd. The approximate number of a number is itself. If a number has an even divisor, it must be even, and the divisor of even number should be even. Let this secondary approximation number be a, then the maximum approximation number is 2a, a+2a =11,a = 37, 2A = 74, that is, the required number is.

8. Solution: According to the given view, we can draw a direct view of this solid as follows:

It can be seen that the upper and lower areas are 8× 2 = 16 (square centimeter), the front and rear areas are 8× 2 = 16 (square centimeter), and the left and right areas are 8× 2 = 16 (square centimeter). The surface area of this solid is * * 48 square centimeters.

Secondly, briefly answer the following questions

9. Solution: ∫ DAC+∠ ADC+∠ C =, and ∠ DAC = ∠ ADC = ∠ B+2 1, ∠ B = ∠ C,

∴∠b=46 ∴3×∠b+2 1 = 180

∠DAC=46 +2 1 =67,∠BAC=67 +2 1 =88

∴△ABC and△△△ ADC are both acute triangles.

10. solution: the bus speed is 60 km/h, and the distance of 18 second = 300 (m).

The length of the truck is (15.8+1.2) × 30+10 = 520 (m).

18 second, the distance that the truck passes is 520-300 = 220 (meters).

Truck speed = 44 km/h

1 1. Solution:

Use (a, b) to represent the squares in row A and column B. The fourth column has the numbers 1, 2, 3, 4, 5, and the sixth row has the numbers 6, 7, 9, so the square (6, 4) = 8; The numbers in lines 3 and 5 are 9, so (7,4) = 9; The middle "Xiao Jiugong" already has a number 7, so it can only be (3,4) = 7; At this time, only (5,4) is left in the fourth column, and only the number 6 in this column is empty, so (5,4) = 6. So the number in the fourth column is written as 9 digits from top to bottom: 327495 1.

12. Solution: In order to make the average grade of the whole class reach 90 points, it is necessary to match two students with excellent grades and/kloc-0 students with poor grades into a group, that is, the students with excellent grades should be at least twice as many as those with poor grades, so as to ensure that the average grade of the whole class is not lower than 90 points, so the proportion of students with excellent grades in the whole class should be at least.

Three. Answer the following questions in detail.

13. Solution:

Firstly, according to whether it is an equilateral triangle, there are three types of equilateral triangles in Figure A, Figure B and Figure C, including six red triangles, six blue triangles, two yellow triangles and *** 14 equilateral triangles. There are three isosceles triangles in Figure D, six in green, six in purple, 24 in brown 12 and * *. So there are 38 isosceles triangles (including equilateral triangles).

14. Solution: If the seven boxes are numbered consecutively clockwise, the available numbers of the boxes are 1, 8, 15, … 7k+1; The available numbers of Box 2 are 2, 9, 16, … 7k+2; …; The seventh box can have the numbers 7, 14, 2 1, …7k+7(k is an integer).

According to the rules, Xiao Ming put 1 in the box of 1, the second in the box of 3, the third in the box of 6, the fourth in the box of 10, the fifth in the box of 15 and the sixth in the box of/. According to this rule, starting from the eighth chess piece, the boxes put by the above chess pieces will be repeated, that is, the eighth chess piece will be put in the box 1 and the ninth chess piece will be put in the box 3, …, that is, every seven chess pieces will be a cycle. Moreover, of these seven pieces, two were put in the box of 1, two in the box of No.3, two in the box of No.7, and 1 was put in the box of No.6, and there were no pieces in the boxes of No.2, No.4 and No.5.. The number of Bai Zi in each box is as follows.

After 200 = 7× 28+4.28 cycles, the 1 97th chess piece is put into the1box, the198th chess piece is put into the No.3 box, the199th chess piece is put into the No.6 box and the 200th chess piece is put into the No.3 box.

Xiaoqing rolled the dice counterclockwise, so we numbered the boxes in a row counterclockwise. Similarly, every 7 dice are a cycle, 300 = 7× 42+6, and the number of red dice in each box is shown in the table below.

Case number 1 2 3 4 5 6 7

Bai Zi 57 058 029 56

Hongzi 86 85 43 0 86 0

The total number of chess pieces is143851010011556.