How do you calculate the math problems of a certain grade?

First of all, there are 50 people playing Bumper Chen, so there is 650-50 3 = 400 yuan left.

In order to make the most people who can play, first consider rowing, because rowing is cheaper, then there can be up to 200 people to play, now there are 50 people playing bumper cars, so there are 50 2 = 100 yuan left, and because the roller coaster is more expensive than rowing 2 yuan people, so you can take out 50 people to play the roller coaster of 150 rowing, then there are 100 people rowing, 50 people playing roller coaster, the total cost of 100 2 + 50 4 = 400 yuan, just run out.

According to the title, there are 50 people playing bumper cars, x people playing roller coasters, and 200-50-x people rowing.

4x+50*3+(200-50-x)*2=650

x=100 So: 100 people on the roller coaster, 50 people on the bumper car, and 50 people on the boat.

Solution: If there are x people playing roller coasters, then there are 200-50-x rowers, according to the title.

4x+2*(200-50-x)+3*50=650 to solve the equation to get x=100

So there are 200-50-x = 200-50-100 = 50 A: 50 people play bumper cars, 100 people play roller coasters, and 50 people play rowing.

Set up a roller coaster x people, rowing 200-50-x people.

There are 4x+50*3+(200-50-x)*2=650

x = 100 roller coasters for 100 people and 50 people for rowing.

50 people play bumper cars, remove 50 * 3 = 150 yuan, there is 500 yuan left, 150 people and 150 people if they all go to play roller coaster, it will cost 150 * 4 = 600 yuan, which is 100 yuan more than the remaining 500 yuan.

If someone out of 150 people goes rowing instead of a roller coaster, they can save 4-2=2 yuan. Because it exceeds 100 yuan, it takes 50 people to play rowing to save 50*2=100 yuan.

So there were 50 people rowing and 100 people going on a roller coaster.

Don't use algebra, elementary school students can't understand it.

Solution: The people who play roller coasters and rowing boats are x and y respectively, according to the meaning of the title.

x+y+50＝200

4x+2y+50*3 650 solves the equation to get x=100 y 50, so there are 50 people playing bumper cars, roller coasters, and 50 rowing people, respectively.

Solve the roller coaster x people, rowing y people.

x+y=200-50

4x+2y=650-50*3

x=100y=50 100 roller coasters, 50 rowers. Complete.

If there are x people on the roller coaster, then there will be (200-50-x) people rowing.

50*3+x*4+(200-50-x)*2=650x=100, so there are 100 people playing roller coasters.

50 people rowing.

The number of people is x, y

x+y=200-50=150

50*3+3x+2y=650

x=100y=50 roller coaster 100, bumper car 50, rowing 50

The calculation is 30-8, the first step of the counting grinder is 10-8 = 2, and the second blind touch step is 3-1 = 2

The third step is 20 + 2 = 22

Vertical: 3 0

<> this question should be the question of the 4th-5th grade of Dabu Primary School.

This is a math problem for grades 4-6 in primary school, the knowledge involved can be judged in junior high school, a little difficult, but the answer can also be tried out by slow adjunct ruler and slow test, and the logic must be analyzed slowly, and it is better for children's logical thinking training or sock height! It is suitable for development capacity training.

Elementary school grades.

The money for 2 clothes + 2 hats + 2 pairs of pants: 80 + 60 + 36 = 176 yuan.

The money for 2 pieces of clothing: 176-2 36 = 104 yuan, the unit price of clothes: 104 2 = 52 yuan.

Unit price of pants per mu: 80-52 = 28 yuan.

Hat unit price: 60-52 = 8 yuan.

Clothes wheel stove clothes + pants = 80

Hat + Clothes = 60 = > Clothes = 60 - Hat.

Hat + pants = 36 = > pants = 36 - hat.

Clothes + pants = 80

60 - hat + 36 - hat = 80

96-2 hat = 80

2 hats = 80-96

2 hats = -16

Hat = 8 yuan.

If you know the hat, you will know everything else.

Clothes + pants + hat = (80 + 60 + 36) 2 = 88 (yuan), hat Ling infiltration = 88-80 = 8 (yuan), spine yard.

Pants = 88-60 = 28 (yuan), clothes = 88-36 = 52 (yuan) cherry blossoms.

Use the first one to subtract the second pants to subtract the wisdom of this hat is equal to 20 plus the third hail, which is two source blue size pants is equal to 56, so a pair of pants is equal to 28, a hat is equal to 8, and a top is equal to 52

It feels like an improvement problem, before learning a one-dimensional equation, there will usually be this kind of problem, and it is possible in the sixth grade.

I don't see the attachment, this one really can't be judged.

I'm sorry, but I can't see it here, so I'd better ask my teacher.

The skill required is the "split term", that is, the difference between one item and two items, and the subtraction of the latter term is equal to the subtraction of the remaining missing term of the previous guess, so that it can be eliminated one by one.

<>So: spike rot a=b, 2a-b=2. Solution: a=b=2. Namely:

Therefore: <>

If it is a junior high school or elementary school, then this is an Olympiad problem.

If Qihe is a high school student, then this is a regular question.

The solution is as follows: <>

For reference, please smile. Bright branches.

This is a math problem for summing the silver sequence. It should be a question above high school and blind.

25.General term an = 1 (1+2+3+..)n+1) =1/[(1/2)(n+1)(n+2)]

2 Calling the Sail [(n+1)(n+2)] 2[1 (n+1) -1 (n+2)].

Original = 2 [1 2-1 3 + 1 3-1 4 + 1 4-1 5 + 1 100-1 101].

This is a question in the first handicap, the second year of the first year is enough, since the first n terms of the Gao Ran number series and the number of deformations, each term = 2 (n+1)(n+2).

Then the split item can.

This is the sum problem of the number series of prudent paragraphs, and the content of the first year of high school Yanshan grade.

Let's look at 1+2+3+...... firstn=n(n+1) 2, which is 1 (1+2+3+......n）

2/n(n+1)=2[1/n-1/(n+1）]。

The sum of the sought forgiveness and auspiciousness is.

2/2x3+2/3x4+……2/100x101

This should be a primary school math Olympiad question, and it is suitable for elementary school students from the fifth grade onwards.

This is a math Olympiad problem, for the sixth grade.

This is the fifth grade Olympiad - fraction addition and subtraction practice questions.

Primary School Olympiad is the fractional split-term elimination method, 1 collapse [1 + 2 + 3 + ......n+1)]=1/[(n+1)(n+2)/2]

2 (n+1)(n+2)=2{1 (n+1)-1 (n+2)], original = 2(1 2-1 3)+2((1 3-1 book oak 4)+2(1 state 4-1 5)+....2{1/(n+1)-1/(n+2)],2[1/2-1/3+1/3-1/41/4-1/5+……1/(n+1)-1/(n+2)]

2[1/2-1/(n+2)]=n/(n+2)

This is the sum of fractions, but it is not possible to solve the line block by the method of general fractions.

Let's first turn the original formula into 1 3+1 6+1 10+...1 (1+100) 100 Zheng takes 2 and turns it into 2 (shouting 1 6+1 12+1 20+...).1 100x101), using the method of difference, you can make it into 2 (1 2-1 3+1 3-1 4+1 4-1 5+...).

1 100-1 101), at which point the middle items in parentheses are canceled out, leaving only 2 (1 2-1 101), and the answer is 99 101.

High school questions, almost the second and third years of high school.

<> Yingqin is arguing, and the old Dansheng can understand it.

Elementary Mathematics: Deformation Crack Term: Differential Simplified State.

Original = 1 3 + 1 6 + 1 10 + ...1 (1+100) x100 2 (Gaussian sum deformation).

2x（1/6+1/12+1/20+..1 100x101)2x(1 2-1 3+1 3-1 Tachibana Zen 4+1 4-1 5+...1 100-1 101) (split term).

2x（1/2-1/101）

1+2 (2 3)+2 guess (3 4)+2/n(n+1)1+2×[1/(2×3)+1/(3×4)+.1 n(n+1)]1+2 [1 2-1 3+1 nano 3-1 4+.

1/n-1/(n+1)]

1+2 [1 Dong Xun 2-1 (n+1)].

1+1-2/(n+1)

2-2/(n+1)

2n (n+1), so, the original formula = 2*100 (100+1) = 200 101

Now there is also such a problem in the Olympiad of Primary Mathematics.

This is a math problem from the fourth grade onwards.

Equation puzzles are not only about being able to calculate, but also about analyzing.

For example, in the first problem, the divisor is a single number, and the last digit of its product (two digits) with a single digit can be 1 or 9, so it is an odd number, but it cannot be 1, 3, 5, 9, so it is 7, the quotient is 137, and the dividend is 959

In the second question, it is easy to know that the last digit of the divisor is 9, and the remainder is 97, so the first place of the divisor is 9...So the quotient is 31 and the dividend is 99 31 + 97 = 3069 + 97 = 3166...

This is an Olympiad and can be completed in grade 4 and above.

Draw a line diagram and take a look.

When they walk for 100 minutes, A catches up with B, i.e. for 100 minutes, A travels 1,200 meters more than B.

Per minute, the speed difference between A and B is: 1200 100 = 12 meters After 10 minutes, the distance between the two people and B is equal.

Explain that 10 minutes, the sum of the distance of two people, is equal to 1200 meters per minute, and the sum of the speed of A and B is: 1200 10 = 120 meters and then according to the sum difference formula:

A per minute: (120 + 12) 2 = 66 meters.

[Analysis].

After walking for 100 minutes, A catches up with B" A is 1200 100 12 meters faster than B per minute;

After 10 minutes of walking, the distance from place B is equal".

The conclusion of direct application, "after 10 minutes": A walked 12 10 120 meters more than B;

The distance from place B is equal" The distance A takes from place B ( distance A 120 m) The distance of B ( 1200 120 ) 2 540 m B speed 540 10 54 m min.

Column: (1200 1200 100 10) 2 10 = 54

Solution: Let the velocity of A be x and the velocity of B y. (Unit: m min) {1200 10x=10y.}

100x=100y+1200

Solution: {x=66.}

y=54, so A travels 66 meters per minute.

1.What is the distance between Xiao Ming's house and the school?

7 35-7 30 = 5 (minutes).

85x(1050÷75-5)=85x9

A: The distance from Xiao Ming's house to the school is 765 meters.

2.If two people walk from each other's house at the same time through the school, estimate how many minutes after departure?

7 35-7 30 = 5 (minutes).

85x(1050÷75-5)=85x9

The distance from Xiao Ming's house to the school is 765 meters.