Math rational number examination paper of junior one.

Junior high school Olympic mathematics test 1

First, multiple-choice questions (each question 1 point, * * *1point)

1. If both A and B represent rational numbers and A+B = 0, then ()

A.a and b are both 0.

One of b.a and b is 0.

C.a and b are opposites.

D.a and b are reciprocal.

Answer: c

Analysis: Let a=2, B =-2, and satisfy 2+(-2) = 0, so A and B are reciprocal.

2. The following statement is true ()

A. The sum of a monomial and a monomial is a monomial.

B. the sum of a single term and a single term is a polynomial.

C. the sum of polynomials is polynomials.

D. the sum of algebraic expression and algebraic expression is algebraic expression.

Answer: d

Analysis: x? , x3 are all monomials. Two monomials x3, x? The sum is x3+x? Is a polynomial, excluding a. Two monomials x? The sum of 2x2 is 3x2 is a monomial, and B is excluded. The sum of the two polynomials x3+x2 and x3-x2 is 2x3, which is a monomial, excluding C, so D is chosen.

3. The following statement is incorrect ()

A. there is the smallest natural number

B. There is no minimum positive rational number

C. there is no largest negative integer

D. there is no largest non-negative number

Answer: c

Analysis: The largest negative integer is-1, so C is wrong.

4. If A and B represent rational numbers, and the value of a+b is greater than that of a-b, then ().

A.a and b have the same number.

B.a and b are different numbers.

C.a>0

D.b>0

Answer: d

5. An integer greater than-π and not a natural number is ()

A.2

B.3

C. IV

D. countless

Answer: c

Analysis: On the number axis, it is easy to see that the integers to the right of-π and to the left of 0 (including 0) are only -3, -2,

-1,0 * * * 4. Choose C.

6. There are four kinds of statements:

A. the square of a positive number is not necessarily greater than itself;

The cube of b positive number is not necessarily greater than itself;

C. the square of a negative number is not necessarily greater than itself;

The cube of a negative number is not necessarily greater than itself.

Among these four statements, the incorrect statement is ()

A.0

B. 1

C.2

D.3

Answer: b

Analysis: the square of a negative number is a positive number, so it must be greater than itself, so C is wrong.

7.A stands for rational number, so the relationship between A and -A is ().

A.a is greater than-a.

B.a is less than-a.

C.a is greater than -a or a is less than-a.

D.a is not necessarily greater than-a.

Answer: d

Analysis: let a=0, you can immediately exclude a, b, c, and choose D.

8. In the process of solving the equation, in order to make the obtained equation and the original equation have the same solution, you can add () on both sides of the original equation.

A. multiply by the same number

B. Multiply by the same algebraic expression

C. Add the same algebraic expression

D. add both 1.

Answer: d

Analysis: For the deformation of the same solution of the equation, both sides of the equation are required to be multiplied by a number not equal to 0, so A is excluded. When we consider the equation X-2 = 0, it is easy to know that its root is x=2. If both sides of the equation are multiplied by an algebraic expression X- 1, we get (X- 1) (X-2) = 0, its root is x= 1, and X = 2. If it is not the same solution of the original equation, B is excluded. In fact, a constant is added to both sides of the equation, and the new equation has the same solution as the original equation. For d, the constant added here is 1, so choose d.

9. There was more than half a glass of water in the cup, which decreased by 10% on the second day and increased by 10% on the third day. So, the comparison result between the water in the cup on the third day and the first day is ().

A. also

Too much

C.it's gone

D.it may be more or less.

Answer: c

Analysis: Let the original amount of water in the cup be a, which can be obtained according to the meaning of the question.

The next day, the amount of water in the cup is a× (1-10%) = 0.9a;

On the third day, the amount of water in the cup is (0.9a) × (1+10%) = 0.9 ×1.1× a;

The ratio of that wat in the third cup to the water in the first cup is 0.99:1,

So on the third day, there was less water in the cup than on the first day, so C was chosen.

10. This ship goes back and forth between two docks of a river. If the speed of the ship itself in still water is fixed, then when the flow speed of the river increases, the time it takes for the ship to make a round trip will be ().

A. Improve

B. Reduce

C. constant D. possible increase or decrease

A: A.

II. Fill in the blanks (each question 1 point, * * *1point)

1. 1989 1990? - 1989 1989? =______。

Answer: 1989 1990? - 1989 1989?

=( 1989 1990+ 1989 1989)×( 1989 1990- 1989 1989)

=( 1989 1990+ 1989 1989)× 1=39783979。

Analysis: use formula a? -B? =(a+b)(a-b) calculation.

2. 1-2+3-4+5-6+7-8+…+4999-5000=______。

Answer:1-2+3-4+5-6+7-8+…+4999-5000.

=( 1-2)+(3-4)+(5-6)+(7-8)+…+(4999-5000)

=-2500。

Analysis: This question uses the associative law in the operation.

3. When A =-0.2 and b=0.04, the algebraic expression a? The value of -b is _ _ _ _.

Answer: 0

Analysis: Original formula = = (-0.2)? -0.04=0。 Just substitute the known conditions into algebraic calculation.

4. There is 60 kilograms of salt water with 30% salt, which is evaporated on the scale. When salt water becomes 40% salt, the weight of salt water is _ _ _ _ _ grams.

Answer: 45,000 (grams)

Analysis: 60 kg of salt water with 30% salt content is 60×30% (kg).

Assuming that the weight of water evaporated to 40% salt is x grams,

That is, 0.00 1x kg, at this time 60×30%=(0.00 1x)×40%.

Solution: x=45000 (grams).

When we encounter this kind of problem, we need to find an invariant. The salt content in this problem is an invariant, and we can calculate it by listing the equations.

Third, answer questions.

1. Both parties have the same annual income. If Party A saves the annual income every year, Party B will spend 100 yuan more than Party A every month. Debt to 600 yuan after three years. What is the annual income of each person?

Answer:

：

Solution, x=5000

A: Each person earns 5000 yuan a year.

So the sum of the last four digits of S is 1+9+9+5 = 24.

4. A person goes uphill at a speed of 3 km/h and downhill at a speed of 6 km/h, and the journey 12 km * * * It takes 3 hours and 20 minutes to try to find the distance between uphill and downhill.

Answer: Let's assume that the uphill journey is X kilometers and the downhill journey is Y kilometers.

From ② 2x+y=20, ③

If ① has y= 12-x, substitute ③ to get 2x+ 12-x=20.

So x=8 (km), so y=4 (km).

Answer: 8 kilometers uphill and 4 kilometers downhill.

5. Total:

A: item n is

therefore

6. Prove that the remainder obtained by dividing the prime number p by 30 must not be a composite number.

Prove: let p = 30q+r, 0 ≤ r < 30,

Because p is a prime number, r≠0, that is, 0 < r < 30.

Assuming that R is a composite number, since R < 30, the minimum prime factor of R can only be 2,3,5.

From P = 30q+R, we know that when the minimum prime factor of R is 2, 3 and 5, P is not a prime number, which is contradictory.

Therefore, r must not be a composite number.

Solution: Suppose.

(2p- 1)(2q- 1)=mpq of formula ①, i.e.

(4-m)pq+ 1=2(p+q).

It is known that m < 4. It can be seen from ① that m > 0 is an integer, so m= 1, 2,3. Let's study p and q separately.

(1) If m= 1, there is

The solutions are p= 1 and q= 1, which are inconsistent with the known ones and are discarded.

(2) If m=2, there is

Because 2p- 1=2q or 2q- 1=2p is impossible, there is no solution when m=2.

(3) If m=3, there is

Get a solution

So p+q = 8.

Junior high school olympiad math test 2

First, multiple choice questions

1. 1 This number is ()

A. Minimum integer

B. Minimum positive number

C. Minimum natural number

D. minimum rational number

Answer: c

Analysis: there is no minimum number of integers, excluding a; Positive numbers have no minimum number, excluding b; There is no minimum number for rational numbers, except that d. 1 is the minimum natural number, which is correct, so C is chosen.

2. If A is a rational number, the relationship that must be established is ().

A.7a>a

B.7+a>a

C.7+a>7

D.|a|≥7

Answer: b

Analysis: If a=0, 7×0=0 excludes A; 7+0=7 does not include c; | 0 | < 7 Exclude D. Actually, because 7 > 0, there must be 7+A > 0+A = A. Choose B.

The value of1416× 7.5944+3.1416× (-5.5944) is ().

A.6. 1632

B.6.2832

C.6.5 132

D.5.3692

Answer: b

Analysis: 3.1416× 7.5944+3.1416× (-5.5944)

=3. 14 16(7.5944-5.5944)=2×3. 14 16

=6.2832, choose B.

4. Among the five numbers -4,-1, -2.5, -0.0 1 and-15, the product of the largest number and the number with the largest absolute value is ().

A.225

B.0. 15

C.0.000 1

D. 1

Answer: b

Analysis: -4,-1, -2.5, -0.0 1 and-15, the maximum number is -0.0 1, and the maximum absolute value is-15, (-0.0/kloc.

Second, fill in the blanks

1. Calculation: (-1)+(-1)-(-1) × (-1) ÷ (-1) = _ _.

Answer: (-1)+(-1)-(-1) × (-1) ÷ (-1) = (-2)-(-/kloc-0)

2. Evaluation: (-1991)-| |-31|| = _ _ _.

Answer: (-1991)-| 31| =-1991-28 =-2019.

3.n is a positive integer, and the last four digits of1990n-1991are arranged in order, and the four digits are 8009. Then the minimum value of n is equal to _ _ _ _.

Answer: 4

Analysis: the last four digits of 1990n should be the last four digits of 199 1+8009, that is, 0000, that is, the last digit of 1990n should have at least four zeros, so the minimum value of n is 4.

4. Do not exceed (-1.7)? The largest integer of is _ _ _ _.

Answer: 2

Analysis: (-1.7)? =2.89, and the largest integer not exceeding 2.89 is 2.

5. A prime number is a two-digit number, and the difference between one digit and ten digits is 7, so this prime number is _ _ _ _ _.

Answer: 29

Analysis: There is18,29. The number of units with two digits is 7 larger than that with ten digits, of which only 29 is prime.

Third, answer questions.

1. Given 3x2-x= 1, find the value of 6x3+7x2-5x+2000.

A: Primitive form

= 2x(3 x2-x)+3(3 x2-x)-2x+2000 = 2x× 1+3× 1-2x+2000 = 2003 .

2. A store sells 100 items a day, and each item can make a profit in 4 yuan. Now they increase profits by raising prices and reducing purchases. According to experience, for every price increase of 1 yuan, the commodity is sold less 10 pieces every day. How much can I increase the price of each commodity to get the maximum profit? What is the maximum profit?

Answer: It turns out that you can make a profit of 4× 100 yuan every day. The price of each piece is raised by X yuan, and each piece makes a profit of (4+X) yuan, but (100- 10x) pieces are sold every day.

If the daily profit is y yuan,

Then y = (4+x) (100- 10x)

=400+ 100x-40x- 10x2

=- 10(x2-6x+9)+90+400

=- 10(x-3)2+490 .

Therefore, when x=3, the maximum value of y =490 yuan, that is, if the price of each piece is increased by 3 yuan, the maximum profit per day is 490 yuan.

3. As shown in figure 1-96, it is known that CB⊥AB, CE ∠BCD, DE ∠CDA, ∠ 1+∠ 2 = 90. Verification: DA⊥AB.

It is proved that ∫CE bisects ∠BCD, DE bisects ∠ADC and ∠ 1+∠ 2 = 90.

∴∠ADC+∠BCD= 180，

BC 100 ∴.

And AB⊥BC,

∴AB⊥AD。

4. Find the integer solution of the equation | xy |-| 2x |+y | = 4.

Answer: | x || y |-2 | x |+y | = 4, that is | x | (| y |-2)+(| y |-2) = 2.

So (| x |+ 1) (| y |-2) = 2.

Because | x |+ 1 > 0, and both x and y are integers, so

5. Wang Ping bought a three-year treasury bill with an annual interest rate of 7. 1 1% and a five-year treasury bill with an annual interest rate of 7.86% * * 35,000 yuan. If the three-year Treasury bill matures, he will deposit the principal and interest into two consecutive one-year time deposits. Five years later, the total principal and interest with the five-year treasury bonds is 477,665,438+0 yuan. Let Wang Ping buy three-year treasury bonds. (The annual interest rate of one-year time deposit is 5.22%)

Answer: Suppose Wang Ping buys three-year and five-year government bonds with X yuan and Y yuan respectively, then

Because y=35000-x,

So x (1+0.0711× 3) (1+0.0522) 2+(35000-x) (1+0.0786× 5) = 4777.

So1.3433x+48755-1.393x = 47761,

So 0.0497x=994,

So x=20000 (yuan), y=35000-20000= 15000 (yuan).

6. For what values of k and m, the equations have at least one set of solutions?

Answer: because (k- 1) x = m-4, ①

When m is a real number, the equations have a unique solution. When k= 1 and m=4, the solutions of ① are all real numbers, so the equations have infinite solutions.

When k= 1, m≠4, ① there is no solution.

Therefore, when k≠ 1, m is any real number, or k= 1, m=4, the equations have at least one set of solutions.

Junior high school Olympic mathematics test III

First, multiple choice questions

1. Among the four pairs of monomials given below, the similar pair is ().

A.x？ Y and -3x? z

B.3.22m？ N3 and n3m?

C.0.2a？ B and 0.2ab?

D. 1 1abc and ab

Answer: b

Analysis: A formula with two identical letters and the same index of the same letter is called a similar term.

2. (x-1)-(1-x)+(x+1) equals ().