Primary school Olympiad 1, to solve, it is best to use the primary school method to solve

It takes 50 seconds for a train to cross a 1,140-meter-long bridge with the front of the train on the bridge until the rear of the train leaves, and 80 seconds for the train to cross a 1,980-meter tunnel.

28 (ms).

As can be seen from the inscription, the second time I ran (1980-1140) meters more than the first time, and it took (80-50) seconds longer. So the velocity is (1980-1140) (80-50)=28. Car length: 28 * 50-1140 = 260

The speed of the vehicle is (1980-1140) (80-50) = 28 (meters and seconds).

The length of the car is 28 * 50-1140 = 260 (meters).

The length of the train is 260 meters, and the speed is 28 meters per second.

The train is 260 meters long and has a speed of 28 meters in 1 second.

In a time of 80-50 seconds, the train travels a distance of 1980-1140 meters, so the speed is 28 meters in seconds.

In this way, the train travels 1400 meters in 50 seconds, so the body length is 1400-1140 = 260 meters.

The length of the car is 260 meters, and the speed is 28 m s

1980-1140 = 840 (m) 80-50 = 30 (s) 840 30 = 28m s

50*28=1400m 1400-1140=260m

1980-1140 = 840 (m) 840 (80-50) = 28 (m-s).

28*50-1140=260 (m).

A: The speed of this train is 28 meters per second and the length of the train is 260 meters.

The zoom problem is a bit demanding for a third-grader in elementary school, but it's understandable as long as it's done properly.

The zoom problem is ultimately answered by the "difference factor problem".

The magnification problem is ultimately to calculate the size of each serving by comparing the differences between the two different states of the uncle.

The first state: according to the actual situation: that is, after 5 years, the increase of 5 years is 4 times, that is, 4 copies (not counting the red part).

The second state: according to the hypothetical situation: that is, the original 5 years of each copy is added, which is 7 times, that is, 7 copies (counting the red part).

Through the comparison of these two states, it can be seen that the uncle counts 7 1 6 5 years old, that is, 30 years old, and 7 4 3 more copies. So each copy (that is, Xiao Ming's age after 5 years) is: 5 (7 1) (7 4) 10 (years old).

Then Xiao Ming is 10 5 5 (years old) this year; Uncle this year: 5*7 35 (years old).

This is considered in the order of the title, i.e., this year as the initial state and five years from now as the final state. In this way, Xiao Ming's age after 5 years is calculated, and it needs to be restored to this year.

If we think backwards, i.e., 5 years from now as the initial state and this year as the final state, the calculation will be simpler: (the result is this year's, so there is no need to revert).

As can be seen from the schematic diagram: (Look at Xiao Ming's age this year as 1 copy).

In fact, if the uncle is reduced by 5 years, (from 5 years later to this year) it becomes a 7-fold relationship, that is, the uncle age is 7 parts.

Suppose you reduce each of your uncle's shares by 5 years, that is, reduce them by 4 5 years, which is still a 4-fold relationship.

As a result of the loss of 4 1 3 5 years, i.e. 5 (4 1) 15 years, the multiplier changes from 7 times to 4 times, and 7 4 3 copies less. So 1 copy (that is, Xiao Ming's age this year) is: 5 (4 1) (7 4) 15 3 5 (years old).

The age of the uncle is: 5 7 35 (years).

Through the schematic diagram, the calculation method was determined. However, drawing is very time-consuming, and after intuitively understanding the change law of the multiples relationship, it is easier to use the method of drawing a line segment diagram

For example, method 1 uses a line diagram:

Regard Xiao Ming's age after 5 years as 1 part. (You can not draw, but you should explain or know what 1 copy is.) Because in the end, it is not the age of the uncle and Xiao Ming that are compared, but the age of the two states of the uncle is compared)

Method 2, how to draw with line segments, watch and practice drawing a picture by yourself, primary school Olympiad learning, we must pay attention to the "combination of numbers and shapes", and turn the abstract quantitative relationship into a concrete and intuitive image, so that it is easy to understand and remember, and understanding is the best memory.

If Xiao Ming is now x years old, then his uncle is 7x years old this year.

Five years later, Xiao Ming will be (x+5) years old, and his uncle will be (7x+5) years old 7x+5=(x+5) 4

7x+5=4x+20

3x=15x=5A: Xiao Ming is 5 years old this year, and his uncle is 35 years old.

Since the age difference between Uncle and Xiao Ming remains the same, the following formula is obtained:

Solution: Original formula = (20-5) 3 This is the difference multiple problem in the Olympiad number.

5 (years) 5*7=35 (years).

It turns out that Xiao Ming is 5 years old this year and his uncle is 35 years old.

A: Xiao Ming is 5 years old this year, and my uncle is 35 years old.

Let's say this year, Xiao Ming's age is x, and his uncle's age is 7x; Five years later, Xiao Ming's age is (x+5), and his uncle's age is (7x+5).

From the inscription, I know that after five years, my uncle's age will be four times that of Xiao Ming, so there is (7x+5) (x+5)=4

Solve x on it.

Xie Shi is Xiao Ming a year old this year, so the uncle is 7A years old.

7a+5）=4（a+5）

a=5 then the uncle is 35 years old. Answer: Slightly.

This problem is easier for children to understand by using the equations learned in the sixth grade.

She Xiaoming is x years old this year, and his uncle is 7x years old this year. 7x+5=(x+5)x4 is fine.

Since you haven't learned, then you write Xiao Ming Sui directly, and the uncle is Xiao Ming Sui 7, (7x Xiao Ming + 5) = (4 x Xiao Ming + 5).

Broken down into: Xiao Ming = 5

Uncle years = 5x7 = 35 years old.

Let the total distance be s

Donald Duck's velocity is v Don.

The number of degrees of Mickey Mouse is v meters.

According to Mickey Mouse will get to the finish line 1 minute earlier than Donald Duck.

s(v Tang)-1=s (v m) is set to a formula.

According to the fact that before the race, Mickey Mouse and Doping increased his speed by 20%, Donald Duck put on a special magic shoe to increase his speed by 25%, and during the race, the magic shoe malfunctioned and repaired it on the spot for 2 minutes, and the final result of the game was: Donald Duck got it 1 minute earlier than Mickey Mouse.

s Tang +2 = s m -1 is set to two.

After that, the 2nd formula is turned into Tang + m).

Later we get s(v Tang)-1=Don+

Both sides are multiplied by V Don at the same time.

S-V Tang = Tang.

And then turn into Don.

Finally, it turns into 1 25s = Tang.

Then turn to s (v Tang) = 115

And because t Tang = s Tang +2

So t Tang = 115

Last = 94

Mickey Mouse original speed v0, Donald Duck original speed v1, according to the original speed: Mickey Mouse takes time t01, Donald Duck t11, after speeding up, meter takes time t02, don t12

Mickey Mouse arrives at the finish minute earlier than Donald Duck: T01+

Donald Duck arrives minutes ahead of Mickey Mouse to the finish minute: T12+

The relationship between the original speed of Mickey Mouse and the time of the race: t01 = the relationship between the time of Donald Duck's original speed and the time of the race: t11 = solution equation: t12 is equal to 70, t02 = 72

From the condition, it can be deduced that 270 2-312 = 228 and 211 2-270 = 152 are divisible by a, the reason is simple, the remainder obtained by dividing by 312 is twice the remainder obtained by dividing by 270, then, divided by twice the 270, that is, 540, the remainder obtained should be equal to the remainder of 312. Then 540-312

228, this number should be divisible by a).

In the same way, it can be deduced that 211 2-270=152 can be divisible by a) and the common divisor of the two numbers 228 and 152 is 76,38,19,4,2,1, because the remainder of a division by A is at least 4, so the exclusion of 2, 1, 76 and 38 is even, if the remainder of the three numbers divided must be even, even, and odd, that is, it can only be to meet the later conditions, and the verification is not right (so excluded), 19 after substitution of the remainder are , to meet the topic, then a = 19

If you buy a towel, you will lack 4 jiao, and if you buy a piece of soap, you will have 2 jiao, which means that the towel is 4 + 2 = 6 jiao more expensive than the unit price of soap.

8 pieces of soap, if you change to buy a towel, you have to make up 8 * 6 = 4 yuan, 8 jiao, then the price of each towel is (20 + yuan.

Set the average score of the first prize x and the average score of the second prize y.

Then the average score of the first prize is x+3, and the average score of the second prize is y+1

Since the total score of 30 people remains the same, 10x+20y = (10-4) (x+3) + (20+4) (y+1).

So there is 4x-4y=42

That is to say, x-y=, that is, the second method of the original average score of the first prize is higher than the average score of the second prize, and the Olympiad idea is not used in the equation

4 people were transferred from the first prize to the second prize, and the result increased by a total of 42 points, of which the average score of the first prize increased by 3 points, a total increase of (10-4) * 3 = 6 * 3 = 18 points, the average score of the second prize increased by 1 point, a total increase of (20 + 4) * 1 = 24 * 1 = 24 points, so the average score of the first prize was 42 = points higher than the average score of the second prize.

2001/2002/2001 + 1/2003/12003 and 1/2003

1*4 and 21/97 + 9/19

3. Set up X hectares of land in A.

x/3+2(

4. There are 2 classes of x people, 1 class of 4x people who do not participate, and the number of people in each class is m.

m-4x=1/3(m-x)

m=11x/2

The realization that class 1 did not participate is a fraction of the number of people who did not participate in class 2:

4x/(m-x)=4x/(11x/2-x)=8/9

Question 1: 1-2=-1;

There are a total of 98 2 = 49 terms above, so the sum of the first 98 numbers is -49 plus 99+100, so the sum is 150

The second question extracts the subtraction in the equation and fills in the number of the corresponding position, which is simplified as follows:

Calculate the sum of the previous additive equation first.

where 1+99=100;

Add 50 and there are 49 100s in front, so the sum of the addition formula is 4950;

The sum of the subtracted terms of the latter term is then calculated.

Add 17 for a total of 16 34s, and multiply 17 to get the sum of the subtracted terms to be 561

So the equation becomes: 4950-6x561=1584 The second question is complicated, you can do it according to the second question on the first floor, but the second question on the first floor is wrong, it should be.

-2+3-4+5-6+..97-98+99+100。

Analysis, two numbers in a group, the sum of each group is -1, from 1-98 for a total of 49 groups, it is -49.

Original -49+99+100 150

Three groups were analyzed, a total of 33 groups, which were 0, 3, 6, 9....96, and is the answer.

a. The sum of the difference series (0+96)*33 2 1584.

b. If you have not learned the equal difference series, you can also sum it by adding the beginning and the end. 0+96 96,3+93 96,So there are (33-1) 2 16 groups, the middle number is 48, the original formula 16*96+48 1584

Hope it helps.

-2+3-4+5-6+..97-98+99+100=100+99－98+97－96+..7－6+5－4+3－2+1=100+（99－98）+（97－96）+.

2-3+4+5-6+7+8-9...97+98-99 first calculate the sum of 1 99 (1 + 99) * 99 2 4950 and then calculate 3 + 6 + 9 ...The sum of 99 (3 + 99) * (99 3) 2 = 1683

Last 4950 1683*2 1584

1-2+3-4+5-6+..97-98+99+100=100+(99-98)+(97-96)+(95-94)+.3-2) +1 (inverted, parentheses).

1+2-3+4+5-6+7+8-9...97+98-99=(1+2-3)+(4+5-6)+(7+8-9)..97+98-99) (in parentheses).

3*32*(1+32) 2 (inverted add up to 2) = 1584

In fact, it is to find the law, in the previous question, except for the first and last two numbers, the middle 98 numbers are two in a group. The following is still a group of three.