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20 1 1 junior high school mathematics test paper 1-6.rar with the answer.
Key middle school entrance simulation test questions and analysis 10

1 Fill in the blanks:

1. 168.54+368.54+568.54+768.54+968.54=_______.

Solution: 2842.7

2

There are pentagons and hexagons on the football surface (see the right). Each pentagon connects five hexagons, and each hexagon connects three pentagons. Then the simplest integer ratio of pentagon and hexagon is _ _ _ _ _.

Answer 3: 5.

This solution has x pentagons. Each pentagon is connected with five hexagons, so there should be 5X hexagons, but each hexagon is connected with three pentagons, that is, each hexagon counts three times, so there are six hexagons.

6 Cut out a figure with an area of 4 with square paper, which can only have the following seven shapes:

If four of them form a square with an area of 16, then the sum of the four numbers is the largest _ _ _ _.

Answer 19.

Solution In order to get the maximum sum of numbers, we must first use numbers with large numbers, so that we can spell out: (7), (6), (5), (1); (7),(6),(4),( 1); (7), (6), (3) and (1) form a square with an area of 16:

Obviously, the largest sum of numbers is the number 1, and the sum of numbers is 7+6+5+ 1 = 19. Check again, there are no other spellings.

Pay attention to the thinking method based on results. Let's draw a square with an area of 16, and color it first (6), (7) and then (5). After proper transformation, we can see that only (1) can be used.

In other cases, if (6), (7) and (4) are used, only (3) and (5) are considered.

On 10, assuming that the answer number is A, the unit number of A is B, and seven consecutive natural numbers are filled in seven circles, so that the sum of the numbers in every two adjacent circles is equal to the known number on the line, then the circle that writes A must be filled in _ _ _ _ _ _.

Answer a = 6

The solution is as shown in the figure:

B=A-4,

C = B+3, so c = a-1;

D = c+3, so d = a+2;

And a+d =14;

So a = (14-2) ÷ 2 = 6.

It is suggested that the main point of this problem is to deduce the difference between two circles separated by a circle.

So as to get the final sum-difference relationship to solve the problem.

13 If a natural number is divided by 187 and then by 188, it is also divided by 52, then the remainder of this natural number divided by 22 is _ _ _ _ _.

Answer 8

After subtracting 52 from this natural number, it can be divisible by 187 and 188. For convenience of explanation, the number obtained by subtracting 52 from this natural number is expressed as m, because 187 =17×11,and m can be18. Because m can be divisible by 188 and m can also be divisible by 2, m can also be divisible by 1 1× 2 = 22. The original natural number is M+52, because m can be divisible by 22. When considering the remainder of M+52 divided by 22, we only need to consider 52 divided by 22.

10. There are two numbers A and B. If the decimal point of A is shifted one place to the right, it is the number B. Then, the number A is _ _ _ _ times the number B. 。

answer

Let the number a be a and the number b be b,

Get 10a= b

So a:b=: 10=

Tip: Set without seeking the law.

1 1.

14. _ _ minutes later, the hour hand and the minute hand are at right angles for the first time.

At 4 o'clock, the difference between the hour hand and the minute hand is 20 squares, so the minute hand needs to catch up with the hour hand 20- 15 = 5 squares, the speed of the scoring hand is "1", and the speed of the hour hand is "",so it takes (20- 15) ÷ = minutes to get the minute hand.

Expand the time from 4 o'clock to 5 o'clock, and the hour hand and the minute hand are at right angles. What time?

This is a combination of clock and travel, which can be solved by the first method of the original problem. Not difficult. But it should be noted that there are two answers to the question, that is, when the pointer is at right angles to the minute hand, the minute hand is located on both sides of the hour hand.

8. There is a three-digit number. After dividing by 7, 8 and 9 respectively, the sum of the remainder is 2 1. This three-digit number is

Answer 503.

Solution Because the sum of the remainder is 2 1, the remainder can only be 6, 7, 8. It is inferred that this number plus 1 should be a common multiple of 7, 8 and 9.

=789=504.

Considering that this number is three digits, it is 504- 1=503.

Second, answer questions:

1. Xiaohong went to the store and bought a box of flowers and a box of white balls. The number of two boxes of balls was equal. The original price of flower balls is 3 in 2 yuan and 5 in 2 yuan. When the new year's goods were on sale, the price of both balls was 4 yuan and 8, and as a result, Xiaohong spent less 5 yuan. So, how many balls did she buy?

150 answers

solve

Use a rectangular diagram to analyze, as shown in the figure.

Easy to get,

Solution:

So 2x= 150.

2.22 Parents (father or mother, they are not teachers) and teachers accompany some pupils to take part in a math contest. It is known that there are more parents than teachers, more mothers than fathers, more female teachers than mothers, and at least one male teacher. So how many of these 22 people have fathers?

Answer 5 people

Know parents and teachers ***22 people, parents are more than teachers, parents are not less than 12 people, teachers are not less than 12 people, mothers and fathers are not less than 12 people, mothers are more than fathers, and mothers are not less than 7 people. There are two more female teachers than mothers, and the number of female teachers is not less than 7+2 = 9 (people). But the title points out that male teachers have at least 1, so male teachers have 1 and female teachers do not exceed 9. It has been concluded that there are no fewer than 9 female teachers, so there are 9 female teachers and 7 mothers, so the number of fathers is: 22-9- 1-7 = 5 (.

Wonderful tip, this question uses the thinking method of maximum problem many times, and skillfully borrows the half-difference relationship to get the range of inequality.

The method of combining positive and negative discussions is also reflected.

3. The sum of the ages of Party A, Party B and Party C is 1 13 years old. When Party A is half the age of Party B, Party C is 38 years old. When Party B is half the age of Party C, Party A is 17 years old. How old is Party B now?

The answer is 32 years old

The solution is shown in the figure.

X years later, A is 17 years old, so:

The solution is x= 10,

At a certain moment, A is 17- 10=7 years old, B is 7×2= 14 years old, C is 38 years old and 59 years old.

So now everyone has to add (113-59) ÷ 3 =18 (year).

So b is now 14+ 18=32 years old.

4. It is known that S =1+1+11+…+,so the last four digits of S are.

Answer 7890.

The solution S is the sum of 65,438+000 terms. Among these 65,438+000 terms, there are 65,438+000 L's in the unit, 99 L's in ten, 98 L's in hundreds, and 97 L's in thousands. The last four digits of s are only related to numbers below 1000.

100 1+99× 10+98 100+97 1000

= 100+990+9800+97000= 107890。

The last four digits of s are 7890.