Doing more problems and reading more examples is a good way to learn mathematics. I believe all the students are doing well. Next, I'll bring you the final exam paper and answers about junior one mathematics, hoping to help you!

The final papers and answers of the first grade mathematics are all multiple-choice questions.

1.(4 points) Determine the position of this point in the plane rectangular coordinate system as ()

A. A real number B. An integer C. A pair of real numbers D. Ordered real number pairs

Test center: coordinates determine the location.

Analysis: For example, real numbers 2 and 3 can't indicate a definite position, but ordered real number pairs (2, 3) can clearly indicate that the abscissa of this point is 2 and the ordinate is 3.

Solution: Solution: Make sure that the position of the point in the plane rectangular coordinate system is an ordered real number pair, so choose D. 。

Note: This topic examines the concept of using ordered real number pairs to represent a point in a plane rectangular coordinate system.

2.(4 points) If the following equation is a binary linear equation, it is ().

A.x2+x = 1 b . 2x+3y﹣ 1=0 c . x+y﹣z=0 d . x ++ 1 = 0

Test site: the definition of binary linear equation.

Analysis: According to the definition of binary linear equation, that is, an integral equation with only two unknowns, its degree is 1.

Solution: Solution: A, x2+x= 1 are not binary linear equations, because their highest degree is 2, and they only contain one unknown number;

B, 2x+3y- 1 = 0 is a binary linear equation;

C, x+y-z = 0 are not binary linear equations because they contain three unknowns;

D, x++1=0 are not binary linear equations because they are not integral equations.

So choose B.

Remarks: Note that the binary linear equation must meet the following three conditions:

Equation (1) contains only two unknowns;

(2) Unknown items at most once;

(3) The equation is an integral equation.

3.(4 points) If it is known that point P is located on the right side of Y axis, 3 unit lengths away from Y axis and 4 unit lengths away from X axis, the coordinate of point P is ().

A.(﹣3,4) B. (3，4) C. (﹣4,3)

Test center: coordinates of points.

Analysis: According to the meaning of the question, the point P should be in the first quadrant with positive abscissa and ordinate, and then the coordinates of the point should be determined according to the distance from the point P to the coordinate axis.

Solution: the point ÷P is located on the right side of the Y axis and above the X axis.

? Point p is in the first quadrant,

Point \p is 3 unit lengths away from the y axis and 4 unit lengths away from the x axis.

? The abscissa of point P is 3 and the ordinate is 4, that is, the coordinate of point P is (3,4). So I chose B.

Comments: This question examines the method of judging the position of a point and the geometric meaning of the point.

4.(4 points) Connect three line segments with the following length end to end to form a triangle ().

A.4cm, 3cm, 5cm b. 1 cm, 2cm, 3cm C. 25cm,12cm,1/cm D.2cm, 2cm, 4cm.

Test center: triangular trilateral relationship.

Analysis: See which option the sum of the two smaller edges is greater than the largest edge.

Answer: solution: a, 3+4 > 5, can form a triangle;

B, 1+2=3, and cannot form a triangle;

c、 1 1+ 12 & lt； 25, can not form a triangle;

D, 2+2=4 cannot form a triangle.

So choose a.

Comments: This topic mainly examines the understanding and application of the triangular relationship. To judge whether a triangle can be formed, it is only necessary to judge that the sum of two smaller numbers is less than the maximum number.

5.(4 points) If the solution of equation 2a-3x = 6 about x is non-negative, then the condition that A meets is ().

A.a> three bachelors? 3 c.a. & lt D.A. three

Test site: the solution of a linear equation; Solve a linear inequality of one variable.

Analysis: this problem can be expressed as the value of x by a, and then according to x? 0, you can get the range of a.

Solution: Solution: 2A-3x = 6

x=(2a﹣6)？ three

∫x again? 0

? 2a﹣6？ 0

? Answer? three

So choose d

Comments: This question examines the value range of the root of a linear equation. X is expressed by the expression of A, and then it can be judged according to the value of X, so this problem can be solved.

6.(4 points) The school plans to buy a batch of exactly the same regular polygon floor tiles to pave the ground, but what cannot be inlaid is ().

A. regular triangle B. regular quadrangle C. regular pentagon D. regular hexagon

Test center: plane mosaic (dense shop).

Special topic: geometric problems.

Analysis: Which regular polygon's sum of several internal angles at the same vertex cannot be 360? Do it.

Answer: Solution: A. Every inner angle of a regular triangle is 60? It can be inlaid with six planes, which does not meet the meaning of the question;

B, regular quadrilateral every internal angle is 90? It can be inlaid with four planes, which does not meet the meaning of the question;

C. Each internal angle of a regular Pentagon is 108? , can't mosaic plane, in line with the meaning of the question;

D, each internal angle of a regular hexagon is 120? It can be inlaid with three planes, which does not meet the meaning of the question;

So choose C.

Comments: Investigate the plane splicing problem of a graphic; The knowledge points used are: regular polygon mosaic plane, can the degree of an internal angle of regular polygon be divisible by 360? .

7.(4 points) The sum of the following angles that can become the inner angles of a polygon is ()

A.270? B. 1080？ C. 520？ In 780 ad?

Test center: polygon inner corner and outer corner.

Analysis: According to the formula of polygon interior angle sum, the polygon interior angle sum is an integer multiple of 180 degrees, from which the answer can be found.

Solution: Solution: Because the sum of the internal angles of a polygon can be expressed as (n-2)? 180? (n？ 3 and n is an integer), the sum of the internal angles of the polygon is an integer multiple of 180 degrees,

Of these four options, only 1080 is an integer multiple of 1080 degrees.

So choose B.

Comments: This question mainly examines the interior angles and theorems of polygons, which is something that needs to be memorized.

8.(4 points) (2002? Nanchang) set? ●▲■? Represents three different objects, and the current balance is weighed twice, as shown in the figure, so? ■▲●? The order of these three objects in descending order of mass is ()

A.■●▲ B. ■▲● C. ▲●■ D. ▲■●

Test site: the application of one-dimensional linear inequality

Topic: the finale.

Analysis: this question is mainly obtained by observing the graph? ■▲●? These three objects are arranged in descending order of mass.

Solution: Solution: Because as you can see from the picture on the left? ■? Than? ▲? Heavy,

Can you see one from the picture on the right? ▲? Weight = 2? ●? Weight,

So the order of these three objects in descending order of mass is ■▲●,

So choose B.

Comments: This question mainly investigates the application of one-dimensional linear inequality, and the key to solving the problem is to use inequality principle and lever principle to solve the problem.

The answer to the final examination paper and the second fill-in-the-blank question of junior one mathematics.

9.(3 points) If point A is known (1,-2), then point A is in the fourth quadrant.

Test center: coordinates of points.

Analysis: Answer according to the coordinate characteristics of each quadrant.

Solution: Solution: Point A (1, ﹣2) is in the fourth quadrant.

So the answer is: four.

Comments: This question examines the symbolic characteristics of the coordinates of each quadrant. It is the key to remember the symbol of the coordinates of the points in each quadrant. The symbolic features of the four quadrants are: the first quadrant (+,+); The second quadrant (-,+); The third quadrant (-,-); The fourth quadrant (+,|).

10.(3 points) As shown in the figure, in the right triangle ACB, CD is the center line on the hypotenuse AB. If AC=8cm and BC=6cm, the circumference difference between △ACD and △BCD is 2 cm, and S△ADC= 12 cm2.

Test center: the center line on the hypotenuse of a right triangle.

Analysis: pass c as CE? For AB in E, find CD= AB, find AB according to Pythagorean theorem, and find CE according to triangle area formula, and you can get the answer.

Answer: Solution: Pass C as CE? AB in e,

∫D is the midpoint of the hypotenuse AB,

? AD=DB= AB，

∫ AC = 8cm, BC = 6cm.

? The circumference difference between △ACD and △BCD is (AC+CD+AD)-(BC+BD+CD) = AC-BC = 8cm-6cm = 2cm;

In Rt△ACB, we can know from Pythagorean theorem that AB= = 10(cm).

∫S triangle ABC= AC? BC= AB？ CE，

8? 6= ? 10? CE，

CE=4.8 (cm),

? S-delta ADC= AD CE= 10cm？ 4.8cm= 12cm2，

So the answer is: 2 12.

Comments: This paper examines the Pythagorean theorem, the nature of the midline on the hypotenuse of the right triangle, the area of the triangle and other knowledge points. The key is to find the length of AD and CE.

1 1.(3 points) As shown, like a chessboard? Will. At point (1,-2),? Elephant? At point (3, -2), then what? Cannon? The coordinates of are (-2, 1).

Test center: coordinates determine the location.

Analysis: First of all, according to? Will. And then what? Elephant? The coordinates of the plane rectangular coordinate system, and then write further? Cannon? The coordinates of.

Solution: solution: as shown in the figure, then what? Cannon? The coordinates of are (-2, 1).

So the answer is: (-2, 1).

Comments: This topic examines the establishment of plane rectangular coordinate system and the representation of point coordinates.

12.(3 points) (2006? Heze) black and white hexagonal floor tiles are spliced into several patterns according to the rules shown in the figure: then there are 4n+2 white floor tiles in the nth pattern (represented by an algebraic expression containing n).

Test center: general type: diversity of graphics.

Topic: Grand finale; Ordinary type.

Analysis: Through observation, the number of white floor tiles in the first three patterns is 6 10 and 14 respectively, so it will be found that there are 4 more white floor tiles in the back pattern than in the front pattern, and the nth pattern has 4n+2 white floor tiles.

Answer: Solution: Analysis shows that there are white floor tiles in the 1 pattern? 1+2=6 pieces. The second pattern has white floor tiles. 2+2= 10. The nth pattern has 4n+2 white floor tiles.

Comments: This question examines students' ability of observation and induction. This question belongs to the conventional topic. Pay attention to the analysis method from special to general. The rule of this problem is: the nth pattern has 4n+2 white floor tiles.

The final examination paper and answer 3 of junior one mathematics.

13.(5 points) Solve the equation with method of substitution.

Test site: solving binary linear equations.

Analysis: sort out the second equation to get Y = 3x ~ 5, then substitute it into the first equation to get the value of x, and then substitute it back to get the value of y to get the solution.

Answer: Solution:,

From ②, y = 3x-5 ③,

③ Substitute into ①, 2x+3 (3x-5) = 7,

The solution is x=2,

Substitute x=2 into ③, y = 6-5 = 1,

So the solution of the equation is.

Comments: This topic examines the substitution of elimination method for solving binary linear equations, and it is the key to get the equation in the form of y=kx+b from one of the two equations.

14.(5 points) Solve the equation by addition, subtraction and elimination.

Test site: solving binary linear equations.

Special topic: calculation problems.

Analysis: According to the same coefficient of X, it can be solved by addition, subtraction and elimination.

Answer: Solution:,

①-②， 12y =-36，

The solution is y =-3,

Substitute y =-3 into ① to get 4x+7? (﹣3)=﹣ 19,

The solution is x=,

So the solution of the equation is.

Comments: This question examines the solution of binary linear equations by adding, subtracting and eliminating. The key to solving the problem lies in finding or constructing unknowns with the same or opposite coefficients.

15.(5 points) Solve inequality:? .

Test center: Solve linear inequality of one variable.

Analysis: Using the basic properties of inequality, the denominator is removed first, then the term is moved, and similar terms are merged, and the coefficient is converted into 1, so that the solution set of the original inequality can be obtained.

Solution: Solution: Divide by the denominator to get: 3(2+x)? 2(2x﹣ 1)

Without brackets, you get: 6+3x? 4x﹣2，

Move items and get: 3x ~ 4x? ﹣2﹣6,

Then-x? ﹣8,

x？ 8.

Comments: This question examines the ability to understand simple inequalities. Students who solve such problems often make mistakes because they don't pay attention to changing numbers when solving problems.

Solving inequalities should be based on the basic properties of inequalities:

(1) The direction of addition and subtraction of the inequality of the same number or algebraic expression on both sides remains unchanged;

(2) The direction of multiplication or division by the same positive number on both sides of inequality remains unchanged;

(3) When both sides of the inequality are multiplied or divided by the same negative number at the same time, the direction of the inequality changes.

16.(5 points) Solve the inequality group, find its integer solution number and indicate that the solution set is on the number axis.

Test site: solving a set of linear inequalities; Represent the solution set of inequality on the number axis; Integer solution of linear inequality in one variable.

Analysis: Find the solution set of each inequality, then find its common solution set, and then find the integer solution of X that satisfies the conditions in its common solution set.

Solution: solution: obtained from ①, X.

Therefore, the solution set of the inequality group is: ﹣2? X< 1, expressed as:

So the integer solution of the inequality group is: -2,-1, 0.

Comments: This question examines the solution of a group of linear inequalities. Knowing the difference between solid points and hollow points is the key to solve this problem.

17.(5 points) If the solutions of equations X and Y are equal, find the value of k. 。

Test site: the solution of binary linear equations.

Special topic: calculation problems.

Analysis: from y=x, substitute into the equation and find the values of x and k.

Solution: solution: from the meaning of the question: y=x,

Substituting into the equation system leads to,

Solution: x=, k= 10,

Then the value of k is 10.

Comments: This question examines the solution of a binary linear equation. The solution of the equation is an unknown number that can make two equations in the equation hold.

18.(2 points) As shown in the figure, in △ABC, D is on the extension line of BC, and D is DE? AB in E, AC in F, known? A=30？ ,? FCD=80？ , beg? D.

Test site: triangle interior angle sum theorem.

Analysis: From the theorem of triangle interior angle sum, we can find that? D into the request? Which is CFD? AFE, and then solve it in △AEF.

Answer: solution: ∵DE? AB (known),

FEA=90？ (Vertical definition).

∵ In △AEF, FEA=90? ,? A=30？ (known),

AFE= 180？ ﹣? FEA﹣？ A (the sum of the internal angles of a triangle is 180)

= 180? ﹣90? ﹣30?

=60? .

Again? CFD=？ AFE (equal vertex angle),

CFD=60？ .

? In △CDF,? CFD=60FCD=80？ (known)

? D= 180？ ﹣? CFD﹣？ Function circuit diagram

= 180? ﹣60? ﹣80?

=40? .

Comments: Mastering the interior angle and the theorem of interior angle of triangle is the key to solving problems.

19.(2 points) Known: As shown in the figure, E is a point on the extension line of CA beside △ABC, F is a point on AB, and D is a point on the extension line of BC. Try to prove it? 1 & lt； ? 2.

Test site: the nature of the outer corner of the triangle.

Special topic: proving the problem.

Analysis: What are the properties of the external angles of triangles? 2=? ABC+？ BAC，？ BAC=？ 1+? AEF to obtain the certificate.

Answer: Proof: ∵? 2=? ABC+？ BAC，

2 & gt？ BAC，

∵? BAC=？ 1+? AEF，

BAC & gt？ 1,

1 & lt； ? 2.

Comments: This question mainly examines students' understanding and mastery of the nature of the outer corner of the triangle, which is not difficult and belongs to the basic question.

The final examination paper and the answer to the fourth drawing question of junior one mathematics.

20.(6 points) As shown in the figure, in △ABC,? BAC is obtuse, please draw according to the following requirements.

( 1)? The bisector ad of BAC;

(2) The center line is on the AC side;

(3) The high CF is on the side of 3)AB.

Test center: drawing? Complex drawings.

Special topic: drawing problems.

Analysis: (1) Take point A as the center, draw an arc with any length as the radius to intersect with both sides of AB side and AC side at a point, then draw an arc with a radius greater than the distance between these two points as the center to intersect at a point, and draw an angle bisector AD with point A after this point;

(2) The middle vertical line of line segment AC with vertical foot E can be connected;

(3) Take C as the center and any length as the radius, draw an extension line of an arc that intersects with BA at two points, then draw an arc with these two points as the center and the length greater than these two points as the radius, intersect at one point, and then make it high.

Solution: Solution: (1) As shown in the figure, what is AD for? The bisector of BAC; (2) As shown in the figure, BE is the center line of the AC edge to be made; (3) As shown in the figure, CF is the height on the side of AB.

Comments: This topic examines complex drawing methods, including bisector method of angles, perpendicular bisector method of line segments, and drawing a vertical line after a known straight line passes a little, all of which are basic drawing methods and need to be mastered skillfully.

The final examination paper and answer five of junior one mathematics (2 1, 5 points)

2 1.(5 points) refers to the following points A (0 0,3), B( 1, ﹣3), C (3 3,5), d (﹣ 3,5) and e in the plane rectangular coordinates.

The distance from the point (1)A to the origin o is 3.

(2) Translate point C to the negative direction of X axis by 6 units, which coincides with point D. 。

(3) When connecting CE, the positional relationship between straight line CE and Y axis is parallel.

(4) The distances from point F to X axis and Y axis are 7 and 5 respectively.

Test center: coordinate and graphic changes-translation.

Analysis: First, draw points in plane rectangular coordinates.

(1) According to the distance formula of two points, we can find the distance from point A to origin O;

(2) Finding the point where point C translates 6 units in the negative direction of X axis is the solution;

(3) The straight lines of two points with the same abscissa are parallel to the Y axis;

(4) The distances from point F to the X axis and Y axis are equal to the absolute values of the ordinate and abscissa, respectively.

Solution: The distance from point (1)A to origin O is 3-0 = 3.

(2) Translate point C to the negative direction of X axis by 6 units, which coincides with point D. 。

(3) When connecting CE, the positional relationship between straight line CE and Y axis is parallel.

(4) The distances from point F to X axis and Y axis are 7 and 5 respectively.

So the answer is: 3; d； Parallel; 7,5.

Comments: Examine the concept of coordinates of points in the plane, the change law of coordinates during translation, and the distance formula of two points on the coordinate axis. This problem is a comprehensive one, but it is not difficult.

Solution (7 points)

22.(7 points) When a batch of goods is to be transported to a certain place, the owner intends to rent two kinds of trucks, A and B, from the automobile transportation company. It is understood that the two types of trucks were rented twice in the past as follows:

The first time and the second time.

Number of vehicles loaded with Class A goods (vehicles) 2 5

Number of vehicles loaded with Class B goods (vehicles) 3 6

Cumulative freight tonnage (ton) 15.5 35

At present, three A-class trucks and five B-class trucks of this company are leased, and the goods have just been delivered at one time. If the freight per ton is calculated in 30 yuan, what is the main freight for the goods?

Test site: the application of binary linear equations.

Topic: Chart types.

Analysis: This question needs to know the tonnage of 1 A class freight cars and 1 B class freight cars at one time. The equivalent relationship is: tonnage of two A-class trucks+tonnage of three B-class trucks =15.5; Tonnage of 5 Class A trucks+Tonnage of 6 Class B trucks =35.

Solution: Assume that each truck of Class A has x(t) and each truck of Class B has y(t).

Yes,

Solve.

30? (3x+5y)=30？ (3? 4+5? 2.5)=735 yuan.

A: The freight payable by the owner is 735 yuan.

Comments: It is necessary to know whether the unknown quantity to be set is a direct unknown quantity or an indirect unknown quantity according to conditions and problems. The key to solving the problem is to understand the meaning of the topic and find out the appropriate equivalence relationship according to the conditions given in the topic: the tonnage of two A-type trucks+the tonnage of three B-type trucks =15.5; Tonnage of 5 Type A trucks+Tonnage of 6 Type B trucks =35. List the equations and then solve them.

23.(7 points) Explore:

(1) as shown in figure ①. 1+? 2 and? B+？ What does c matter? Why?

(2) Fold the graph ①△ABC along DE to get the graph ②. Fill in the blanks: 1+? 2 = ? B+？ C (fill in? & gt& lt=? ), when A=40? What time? B+？ C+？ 1+? 2= 280? ;

(3) As shown in Figure ③, it is obtained by folding △ABC in Figure ① along DE. What if? A=30？ , then x+y=360? ﹣(? B+？ C+？ 1+? 2)=360? ﹣ 300? = 60? , guess? BDA+？ CEA and? What is the relationship of A? BDA+？ CEA=2？ Answer.

Test center: folding transformation (folding problem)

Topic: inquiry style.

Analysis: According to the fact that the internal angle of the triangle is 180 degrees, it can be concluded that? 1+? 2=? B+？ C, in order to find out when? A=40？ What time? B+？ C+？ 1+? 2= 140? 2=280? Through the above calculations, we can sum up the general rules:? BDA+？ CEA=2？ A.

Solution: Solution: (1) According to the triangle, the internal angle is 180? Know: 1+? 2= 180? ﹣? First,? B+？ C= 180？ ﹣? One,

1+? 2=? B+？ c； (2)∵? 1+? 2+? BDE+？ CED=？ B+？ C+？ BDE+？ CED=360？ ,

1+? 2=? B+？ c；

What time? A=40？ What time? B+？ C+？ 1+? 2= 140? 2=280? ; (3) What if? A=30？ , then x+y=360? ﹣(? B+？ C+？ 1+? 2)=360? ﹣300? =60? ,

So what? BDA+？ CEA and? The relationship of a is:? BDA+？ CEA=2？ A.

Comments: This topic examines the folding transformation of graphics and the interior angle theorems of triangles and quadrilaterals. In the process of solving problems, we should pay attention to the fact that folding is a symmetrical transformation and belongs to axial symmetry. According to the nature of axial symmetry, the shape and size of the figure are unchanged before and after folding, for example, the angles before and after folding in this question are equal.