First, multiple-choice questions (4 points for each small question, 40 points for * * *)
The absolute value of 1 -4 is ()
A.B.C.4D.﹣4
Test center: absolute value.
Analysis: It can be solved according to the fact that the absolute value of a negative number is its inverse.
Solution: The absolute value of 4 is 4.
So choose C.
Comments: This question examines the nature of absolute value, which requires mastering the nature and definition of absolute value and skillfully applying it to practical operation.
Summary of the law of absolute value: the absolute value of a positive number is itself; The absolute value of a negative number is its reciprocal; The absolute value of 0 is 0.
2. In the following figures, the equivalent value is ().
A.32 and 23b. -23 and (-2) 3c.3× 22 and (3× 2) 2d. -32 and (-3) 2
Test center: the power of rational numbers.
Analysis: According to the meaning of power, we can get the answer.
Solution: Solution: A32=9, 23=8, so the values of A are not equal;
B﹣23=﹣8, (﹣2)3=﹣8, so the values of b are equal;
C3×22= 12, (3×2)2=36, so the values of c are not equal;
D-32 =-9, (-3) 2 = 9, so the values of d are not equal;
Therefore, choose: B.
Comments: This question examines the power of rational numbers. Note that even powers of negative numbers are positive numbers and odd powers of negative numbers are negative numbers.
3.0.3998 is rounded to one percent, which is approximately equal to ().
A.0.39B.0.40C
Test sites: approximate number and effective number.
Analysis: Rounding 0.3998 to the percentile means rounding the percentile number.
Solution: 0.3998 is rounded to one percent, which is about 0.40.
So choose B.
Comments: This question examines the rounding method, which is something to remember.
4. If it is a cubic binomial, the value of a is ()
A.2B.﹣3C.2D。 three
Test center: polynomial.
Special topic: calculation problems.
Analysis: It is understood that the cubic binomial is a cubic term in a polynomial and has the sum of two monomials, so the result can be obtained.
Solution: Solution: Because there must be 3 single items,
So |a|=2
a= 2。
Because it is binomial, a-2 = 0.
a=2
So a =-2 (give up).
So choose a.
Comments: This question examines the understanding of the concept of cubic binomial. The key is to know that the degree of polynomial is 3, which contains two terms.
5. The result of simplification of p ﹣ [q ﹣ 2p ﹣ (p ﹣ q)] is ().
A.2pB.4p﹣2qC.﹣2pD.2p﹣2q
Test site: addition and subtraction of algebraic expressions.
Special topic: calculation problems.
Analysis: According to the mixed arithmetic of addition and subtraction of algebraic expressions, the parentheses are removed first, then the parentheses are removed, and finally the similar items are merged to get the answer.
Solution: original formula =p﹣[q﹣2p﹣p+q],
=p﹣q+2p+p﹣q，
=﹣2q+4p，
=4p﹣2q.
So choose B.
Comments: This topic mainly investigates the addition and subtraction of algebraic expressions. The key to solve this problem is to remove brackets correctly according to the rule of removing brackets (brackets are preceded by a ﹣, and all items will change signs after removing brackets).
6. If x=2 is the solution of the equation 2x+3m- 1 = 0 about x, then the value of m is ().
A.﹣ 1B.0C. 1D.
Test site: the solution of a linear equation.
Special topic: calculation problems.
Analysis: According to the definition of the solution of the equation, the value of m can be obtained by substituting x=2 into the equation 2x+3m- 1 = 0.
Solution: ∫x = 2 is the solution of equation 2x+3m1= 0 about x,
∴2×2+3m﹣ 1=0，
Solution: m =- 1.
So choose: a.
Comments: The key to this problem is to understand the definition of the solution of the equation, that is, the value of the unknown quantity that can make the left and right sides of the equation equal.
7. In the spring sports meeting of a school, the competitive strength of Class 8 (1) and Class 5 is equal. Regarding the results of the competition, classmate A said: The score ratio between class (1) and class (5) is 6: 5; Student B said that the score of class (1) is 40 points less than that of class (5). If class (1) gets an x score and class (5) gets a y score, the equation listed according to the meaning of the question should be ().
A.B
CD。
Test center: abstract binary linear equations from practical problems.
Analysis: The equivalent relationship of this question is: (1) Class score: (5) Class score = 6: 5; (1) Class score =(5) Class score × 2 ~ 40.
Answer: According to the score ratio of (1) and (5) of 6: 5, there are:
X: y = 6: 5，5x = 6y
According to (1) class, the score is 40 points less than that of (5) class, and x = 2y ~ 40.
The countable equation is.
Therefore, choose: d.
Comments: The key of equations is to find the equivalence relation. At the same time, according to the basic properties of proportion, we can convert the proportion formula into the equal product formula.
8. In the plan below, the plan expansion of the cube is ().
Asian Development Bank.
Test site: geometric expansion diagram.
Analysis: Solving problems by folding plane figures and expanding cubes.
Solution: In options A, B and D, after folding, there is a line with two faces that cannot be folded, and it lacks a bottom face and cannot be folded into a cube.
So choose C.
Comments: Mastering the surface expansion diagram of a cube is the key to solving problems.
9. As shown in the figure, if ∠ AOB = ∠ COD = 90 and ∠ AOD = 170, the degree of ∠BOC is ().
A.40 B.30 C.20 D. 10
Test center: angle calculation.
Special topic: calculation problems.
Analysis: Let ∠BOC=x, because ∠ AOB = ∠ COD = 90, that is ∠ AOC+X = ∠ BOD+X = 90, so it is easy to find ∠ AOB+∠ COD.
Solution: let ∠BOC=x,
∠∠AOB =∠COD = 90，
∴∠AOC+x=∠BOD+x=90，
∴∠aob+∠cod﹣∠aod=∠aoc+x+∠bod+x﹣(∠aoc+∠bod+x)= 10，
X = 10.
So choose D.
Note: This topic examines the calculation and vertical definition of angles. The key is to express ∠AOD and ∠AOB+∠COD as the sum of several angles.
10. Xiao Ming uses the statistical chart as shown in the figure to express his weekly expenditure, so it can be seen from the figure ().
A. Total expenditure for one week
B. Percentage of various expenditures to total expenditures in one week
C. the cost of one week
D. Changes in the amount of various expenditures within one week
Test center: fan chart.
Analysis: Answer according to the characteristics of the fan-shaped statistical chart.
Solution: ∵ Fan chart uses a whole circle to represent the total, and the size of each sector in the circle indicates the percentage of each part in the total. Through the pie chart, we can clearly show the relationship between the number of each part and the total.
As can be seen from the figure, the amount of each expenditure accounts for the percentage of the total expenditure in a week.
So choose B.
Comments: This question examines the fan-shaped statistical chart. It is the key to answer this question to clearly know the relationship between the quantity of each part and the total amount from the fan-shaped chart.
Fill in the blanks (5 points for each small question, 20 points for * * *)
1 1. at (﹣ 1)20 10, (﹣1) 201,﹣ 20.
Test center: rational number comparison; Subtraction of rational numbers; The power of rational numbers.
Analysis: Calculate numbers according to the power law of rational numbers, find out the number and the smallest number, and then calculate.
Solution: ∵ (﹣1) 2010 =1,﹣1) 201= ﹣/kloc-0.
The number of ∴ is (-3) 2, and the smallest number is -23.
The difference between ∴ number and minimum number is equal to = 9 (8) = 17.
So the answer is: 17.
Comments: this question examines the comparison of rational numbers, calculates numbers according to the power law of rational numbers, and finds out the values and minimum values of this group of data is the key to this question.
12. Given m+n= 1, the algebraic expression -m+2-n = 1.
Test center: Algebra Assessment.
Special topic: calculation problems.
Analysis: To analyze the known problems, we can find the value of the algebraic expression by the whole substitution method, convert the algebraic expression-m+2-n into an algebraic expression containing m+n, and then substitute m+n= 1 for evaluation.
Solution: solution: -m+2-n =-(m+n)+2,
It is known that m+n= 1 is substituted into the above formula:
﹣ 1+2= 1.
So the answer is: 1.
Comments: This question examines students' mastery and application of the overall idea of mathematics and algebraic evaluation. The key is to change the algebraic expression -m+2-n into an algebraic expression containing m+n.
13. Since monomial and-﹣3x2n﹣3y8 are similar terms, the value of 3m-5n is-7.
Test center: similar projects.
Special topic: calculation problems.
Analysis: Because the monomial and ﹣3x2n﹣3y8 are similar terms, we can get m=2n﹣3, 2m+3n=8, and then we can get the values of m and n respectively, so we can get the values of 3m﹣5n.
Answer: solution: from the meaning of the question, m = 2n-3, 2m+3n=8,
Substitute m = 2n ~ 3 into 2m+3n=8,
2(2n﹣3)+3n=8，
The solution is n=2,
Substitute n=2 into m = 2n ~ 3,
m= 1，
So 3m ~ 5n = 3× 1 ~ 5× 2 = ~ 7.
So the answer is: -7.
Comments: This question mainly examines students' understanding and mastery of similar projects. The key to solve this problem is that the monomial and ﹣3x2n﹣3y8 are similar terms, and it is concluded that m=2n﹣3, 2m+3n=8.
14. Given a straight line AB=8cm, a point C on the straight line AB, BC=4cm, and m is the midpoint of the straight line AC, the length of the straight line AM is 2cm or 6cm.
Test center: the distance between two points.
Special topic: calculation problems.
Analysis: It is necessary to consider the various possibilities of the positional relationship between A, B and C, that is, point C is on the extension line of AB line or point C is on AB line.
Solution: Solution: ① When point C is on the extension line of line segment AB, AC=AB+BC= 12cm, and ∫M is the midpoint of line segment AC, then AM = AC = 6cm.
② When point C is on the AB line, AC = AB-BC = 4cm, and ∵M is the midpoint of the AC line, then AM=AC=2cm.
So the answer is 6cm or 2cm.
Comments: This question mainly examines the knowledge points of the distance between two points. It is the key to solve the problem to transform the overlapping relationship between line segments by using the nature of midpoint. Flexible use of its different representations in different situations is conducive to the simplification of problem solving. At the same time, it is also very important to flexibly use the sum, difference, multiplication and division of line segments to transform the quantitative relationship between line segments.
Third, the calculation problem (this problem is ***2 small questions, 8 points for each small question, *** 16 points)
15.
Test center: mixed operation of rational numbers.
Special topic: calculation problems.
Analysis: in the mixed operation of rational numbers, we should first pay attention to the operation order, first calculate the operation of the superior, then calculate the operation of the subordinate, that is, multiply first, then divide, then add and subtract. Operations at the same level are carried out from left to right, and the operations in brackets are counted first. Second, we should pay attention to observation, flexibly use the algorithm for simple calculation, and improve the operation speed and ability.
Answer: Solution:,
=﹣9﹣ 125×﹣ 18÷9,
=﹣9﹣20﹣2,
=﹣3 1.
Comments: This question examines the comprehensive operation ability of rational numbers, and we should also pay attention to how to remove the absolute value when solving problems.
16. Solve the equation:
Test site: solving binary linear equations.
Special topic: calculation problems.
Analysis: according to the properties of the equation, simplify the equation in the equation and then solve it.
Solution: Solution: Simplify the original equations.
①+②: 20a=60，
∴a=3，
Substituting into ① gives: 8×3+ 15b=54,
∴b=2，
Namely.
Comments: This topic is the method of addition, subtraction and elimination when investigating the properties of equations and solving binary linear equations.
Four, (this question ***2 small questions, each small question 8 points, *** 16 points)
17. It is known that ∠ α and ∠βare complementary angles, and the ratio of ∠βis larger than ∠α 15. Find the complementary angle of ∠ α.
Test center: complementary angle and complementary angle.
Special topic: application problem.
Analysis: According to the definition of complementary angle, the sum of two complementary angles is 180. We can get ∠ α by listing the equation according to the meaning of the question, and then we can get the result according to the definition of complementary angle.
Solution: According to the meaning of the question and the definition of complementary angle,
∴,
Solve,
The complementary angle of ∴ ∠ α is 90 ∠α = 90 ∠ 63 = 27.
So the answer is: 27.
Comments: This question mainly examines the definition of complementary angle and complementary angle and solves binary linear equations, with moderate difficulty.
18. As shown in the figure, C is the midpoint of line AB, D is the midpoint of line CB, and CD= 1cm. Find the sum of the lengths of AC+AD+AB in the graph.
Test center: the distance between two points.
Analysis: First, the length of BC is calculated according to D as the midpoint of CB and CD= 1cm, and then the lengths of AC and AB are obtained from C as the midpoint of AB, so that the length of AD can be obtained and the conclusion can be drawn.
Solution: solution: ∫CD = 1cm, d is the midpoint of CB,
∴BC=2cm，
∵C is the midpoint of AB,
∴AC=2cm,AB=4cm，
∴AD=AC+CD=3cm，
∴AC+AD+AB=9cm.
Comments: This question examines the distance between two points, and it is the key to solve this question to know the sum, difference and multiple relationship between each line segment.
V. (There are ***2 small questions in this question, each small question 10, ***20 points)
19. Given a = a3-a2-a, b = a-a2-a3 and c = 2a2-a, find the value of a-2b+3c.
Test site: addition and subtraction of algebraic expressions.
Special topic: calculation problems.
Analysis: Substitute the values of A, B and C into A-2B+3C, remove the brackets, and then combine the similar items to get the answer.
Answer: solution: a ﹣ 2b+3c = (a3 ﹣ a2 ﹣ a) ﹣ 2 (a ﹣ a2 ﹣ a3)+3 (2a2 ﹣ a).
=a3﹣a2﹣a﹣2a+2a2+2a3+6a2﹣3a，
=3a3+7a2﹣6a.
Comments: This topic examines the addition and subtraction of algebraic expressions. The key to solve this kind of problem is to memorize the rules of removing brackets and skillfully use the rules of merging similar items, which is a common test center in senior high school entrance examinations everywhere.
20. The sum of ten digits and one digit of two digits is 7. If you add 45 to this two-digit number, it will just become a two-digit number after the single-digit number and the ten-digit number are reversed. Find this two-digit number.
Test center: the application of one-dimensional linear equation.
Special topic: digital problems; Equation thinking.
Analysis: Let the digits of 10 and 1 0 of this two-digit number be x and 7-x respectively, and list the equations according to the meaning of the question to solve this two-digit number.
Solution: let the ten digits of this two-digit number be x, then the one digit is 7-x,
According to the meaning equation,10x+7-x+45 =10 (7-x)+x,
The solution is x= 1,
∴7﹣x=7﹣ 1=6，
This two-digit number is 16.
Comments: This question examines the problem of numbers, and the idea of equations is a very important mathematical idea.
Six. (The full mark of this question is 12)
2 1. Take a rectangular piece of paper, as shown in Figure ①, and fold one corner. Note that the falling position of vertex A is A' and the crease is CD, and then fold another corner as shown in Figure ②, so that DB falls in the direction of DA' and the crease is DE. Try to judge the size of ∠CDE and explain your reasons.
Test center: calculation of angle; Folding transformation (folding problem)
Special topic: geometric problems.
Analysis: According to the folding principle, we can know that ∠ BDE = ∠ A ′ de, ∠ A ′ DC = ∠ ADC. The reuse angle is 180, and it is easy to get ∠ CDE = 90.
Answer: Solution: CDE = 90.
Reason: ∫∠BDE =∠A' de, ∠A'DC=∠ADC,
∴∠cda′=∠ada′,∠a′de=∠bda，
∴∠cde=∠cda′+∠a′de，
= ∠ Ada '+∠BDA,
= (∠ Ada '+∠BDA'),
=× 180 ,
=90 .
Comments: This topic examines the calculation and folding transformation of angles. To solve this problem, we must understand that the two angles in the fold are equal, and then use the implicit condition that the right angle degree is 180.
Seven. (The full mark of this question is 12)
In order to "make all children can afford to go to school and study hard", the state has issued a series of policies of "subsidizing poor students" since 2007, including the policy of providing free textbooks to students with financial difficulties. In order to ensure the smooth implementation of this work, the school needs to investigate the students' family situation. The following are the survey results of Class A and Class B of a middle school in a suburb, which are arranged into tables (1) and graphs (1).
Non-minimum living security in type towns
Number of registered permanent residence rural registered permanent residence urban registered permanent residence
Total subsistence allowance
Jiaban 20550
Class b 28224
(1) Fill in the blanks in table (1) and figure (1).
(2) Book textbooks for the next semester in 2009, with a total amount of 100 yuan. Rural registered students can be exempted, urban low-income students can be exempted, and urban registered students (non-low-income students) can pay in full. How much do I have to pay for class B books? What is the proportion of students in Class A who get state-funded textbooks?
(3) During the May 4th Youth Day, the Youth League Committee presented a batch of books on popular science and literature to Class A and Class B free of charge, including literature books 15, and the ratio of the three books is shown in Figure (2). How many art books are there?
Test center: bar chart.
Analysis: (1) According to statistics, the number of rural households in Class A is 50-20-5 = 25; The total number of Class B is 28+22+4 = 54;
(2) According to the meaning of the question, Class B has 22 rural hukou, 28 urban hukou and 4 urban subsistence allowances, which can be solved according to the charging standard;
Class A rural registered permanent residence students and urban low-income students can receive state-funded textbooks. The total number of textbooks funded by the state is 25+5=30, and the total number of students in the class is 50, which can be obtained.
(3) According to the fan-shaped statistical chart, there are 15 literary books, accounting for 30%, so we can get the total number of books, and then get the percentage of art books.
Solution: Solution:
(1) Add the following numbers:
(2) Expenses payable for Class B: 28×100+4×100× (1) = 2,900 yuan;
The percentage of students in Class A who get state-funded textbooks is ×100% = 60%;
(3) Total number of photo albums: 15÷30%=50 (this),
Art book * * * is: 50× (1-30%-44%) =13 (volume).
Comments: This topic examines the comprehensive application of bar chart and fan chart. Reading charts and getting the necessary information from different charts is the key to solving the problem. Bar charts can clearly show the data of each project. The fan-shaped statistical chart directly reflects the percentage of local parts in the whole.
Eight, (full mark for this question 14)
23. As shown in figure ∠ AOB = 90, ∠BOC = 30, and OM bisects ∠AOC. On bisecting ∠BOC, find the degree of ∠MON.
(2) If ∠AOB=α in (1) and other conditions remain unchanged, the number of times to find ∠MON.
(3) If ∠BOC=β(β is an acute angle) in (1), and other conditions remain unchanged, find the degree of ∠MON.
(4) What laws can be seen from the results of (1)(2)(3)?
(5) The calculation of line segment and angle are closely related, and we can learn from each other's solutions. Please imitate (1) ~ (4), design a calculation problem with the line segment as the background, and write down the rules?
Test center: angle calculation.
Special topic: ordinary type.
Analysis: (1) First, calculate the degree of angle AOC according to the number of two angles known in the question. Then, according to the definition of the bisector, we can know that the two angles divided by the bisector are equal to half of its big angle, and the angle MOC and the angle NOC are calculated respectively, and the difference between them is the degree of the angle MON.
(2) The calculation method of (3) is the same as that of (1).
(4) Through the first three questions, we can find that the degree of angle MON is equal to half the degree of angle AOB.
(5) The calculation of line segments is closely related to the calculation of angles. We also designed the lengths of two line segments and two midpoints under known conditions to find out the length of the line segment between the midpoints.
Solution: (1) ∵∠ AOB = 90, ∠ BOC = 30,
∴∠AOC=90 +30 = 120，
OM divides equally ∠AOC,
∴∠MOC=∠AOC=60，
And ∫ON∠BOC,
∴∠NOC=∠BOC= 15
∴∠mon=∠moc﹣∠noc=45；
(2)∫∠AOB =α，∠BOC=30，
∴∠AOC=α+30，
OM divides equally ∠AOC,
∴∠MOC=∠AOC=+ 15，
And ∫ON∠BOC,
∴∠NOC=∠BOC= 15
∴∠mon=∠moc﹣∠noc=；
(3)∫∠AOB = 90，∠BOC=β，
∴∠AOC=90 +β，
OM divides equally ∠AOC,
∴∠MOC=∠AOC=+45，
And ∫ON∠BOC,
∴∠NOC=∠BOC=
∴∠mon=∠moc﹣∠noc=45；
(4) From the result of (1)(2)(3), it is known that ∠ mon = ∠ AOB;
(5)
① Given that the length of line AB is 20, the length of line BC is 10, point M is the midpoint of line AC, and point N is the midpoint of line BC, find the length of line MN;
② If the length of straight line AB becomes a, and other conditions remain unchanged, find the length of straight line MN;
③ If the length of line BC is changed to B, and other conditions remain unchanged, find the length of line MN;
What laws can you find from ① ② ③?
The law is: MN=AB.
Comments: This topic examines the understanding of the concept of angle bisector, the degree of finding angles, and the ability to learn to summarize laws and design experiments according to the close relationship between angles and line segments.
The second volume of the seventh grade mathematics examination paper and related articles are accompanied by reference answers;
★ The answers to the review questions in the second volume of seventh grade mathematics.
★ People's Education Edition seventh grade mathematics textbook exercise answers
★ Reference answer for the exercise book of the second volume of seventh grade mathematics.
★ 2020 Seventh Grade Mathematics Volume II Exercise Book 3 Answers
★ 2020 seventh grade second volume math review questions
★ The answer to the seventh grade math workbook
★ Answers to the Math Classroom Exercise Book for the next semester of Grade 7
★ The people's education press seventh grade mathematics final examination paper.
★ Examination questions for the second final exam of seventh grade mathematics volume.
★ Key examination questions for seventh grade mathematics review in 2020.