Primary school math problems challenge intelligence, these 4 primary school math puzzles are difficu

If the donkey gives the mule the same bag of his goods, it means that they have two bags of goods on them. And if a bag of goods from a mule is given to a donkey, the donkey's cargo is twice that of the mule, and the mule can be set to carry the goods x bag, and the donkey is to carry the goods x+2, which can be listed:

x-1)*2=x+2+1

2x-2=x+3

2x-x=2+3

x=5x+2=5+2=7 (bag).

A: A mule carries 5 bags and a donkey carries 7 bags.

Isn't it allowed to be solved by equations? According to the second article, it is known that the donkey has one less bag, the mule has one more bag, and the goods of the two people are equal, so the donkey has two more bags than the mule, according to the first article, the mule is one bag less, and the donkey has one more bag, and the donkey = 2 The number of mule bags At this time, the difference between the two people is 4 bags (drawing the line segment, it can be easily seen that the donkey has 8 bags and the mule has 4 bags), and returns 1 bag, 7 bags of donkeys, and 5 bags of mules.

Set donkey x bag mule y bag.

x+1=2*（y-1） （1）

x-1=y+1 （2）

From (1) (2) we get 2*x=3*y-1 (3) from (2) we get x=y+2 (4).

4) Substantiation (3) solution y=5

Substituting y=5 into (4) results in x=7

Let the mule pack x bag and the donkey pack y bag, from the problem 2 (x-1)=(y+1), x+1)=(y-1) solve the equation 2x-y=3,y-x=2, solve: x=5, y=7 So the mule pack 5 bags, the donkey pack 7 bags.

I will do so. But I'm an elementary school student. Question 2 has a total of 1360 legs.

Yes. It's just a primary school math puzzle, which can be solved according to the requirements and logic, or by solving equations.

Of course, I will solve them, but it will take a long time because these 4 puzzles are situational mode questions, and they need to be converted into equations to be solved.

In fact, these 4 primary school math puzzles are not mastered for the knowledge I have learned now, and I will not be able to solve them, and I have no idea at all.

I don't either, although I did well in math when I was studying, but I didn't seem to learn these 4 math problems when I was studying, so I can't solve them.

Set up test questions x questions. (1 6) *x less than or equal to 5

X-5 is greater than or equal to. (1/2)*x

x is between 10 and 30.

When x=12, Wang Li answered 3 questions incorrectly, Zhao En answered 5 questions incorrectly, and answered 2 questions incorrectly, Wang Li answered 1 question incorrectly, Zhao En answered 3 questions incorrectly, and 12-2-1-3=6 did not exceed half of the test questions.

When x=24, Wang Li answered 6 questions incorrectly, Zhao En answered 5 questions incorrectly, all answered 4 questions incorrectly, Wang Li answered 2 questions incorrectly, Zhao En answered 1 question incorrectly, 24-2-1-4=17 is more than half of the test questions. After testing, it is in line with the topic. They all answered 17 questions correctly.

Both of them made 1/6 of the total number of questions, and they are known to have incorrectly typed more than half of the total number of questions.

Don't you think that's.

From the meaning of the question, the total number of questions must be a common multiple of 4 and 6, all wrong standing 1 6, and 36 one-sixth is greater than 5, rounded, are correct more than half of the total, if it is 12, it is correct 6, not greater than half, rounded. Therefore, the total number is 24, the number of errors is , and the number of errors is 4, so the number of two wrong collections is 7, and the correct number is 24-7=17.

Set: Wang Li mistyped the question as question x, and the total question is question y.

Then: {5+4 1x=6 1y.}

5+4/1x》2/1y

Isn't there a contradiction in the title? It's all wrong whether it's 1 6 or more than half.

（2007-1）（2007-1）…(2007-1) There are a total of 2008 (this number is not very meaningful, just look at whether it is odd or even).

If you look at the equation, you will find that all (-1) multiplication terms will contain 2007, so they are all divisible by 2007. All (-1) multiplication terms have 2008 (-1) multiplication, even number, so it is 1, then the whole formula, the remainder is 1

If it's 20062006....The remainder of 2006 (2007 of 2006) divided by 2007 is several (obviously not 0).

The analysis process is the same as above, except that all the terms (-1) are multiplied and the result is -1 (2007, odd number).

The remainder won't be negative, so what is his previous one? 2007*（-1）*(1)…(1) There are 2006-1, so this item is 2007Subtracting the -1 without the 2007 terms is equal to 2006, that is, the remainder of the lion is 2006

In short, the number of 2006 is an odd number, and the remainder is 2006;

The number of 2006 is even, and the remainder is 1

Apparently the previous 2007 2006 divided by 2007 is divisible and equals 2006

And the last 2006 divided by 2007 remainder is 2006

I don't know if you understand o( oha!

The first 2007 2006 divided by 2007 is divisible and equals 2006, while the last 2006 divided by 2007 remainder is 2006

You just put 2008-2007 = 1 1 2006, because it is multiplied by 2008 and divided by 2007, so directly subtracting 2007 is the difference between them, the difference is multiplied by the number to be multiplied, listen to the brother, absolutely yes.

The question is: how many seconds did it take for the second half of the journey? It's the second half of the journey.

The total runway is 720, and the second half of the distance is of course 360, and he only runs 320 in the second half of the time, which is still 40, and the poor 40 is of course running at a speed of 5.

Because the question asks about the second half of the "distance", not the time, the second half of the distance is half of 720, which is 360, but the last 320 is run with 4, which is the second half of the "time", and there are 40 meters that are in the first half of the "time" to run, with 5

Don't bother with it.

5 meters per second for the first half of the time.

The second half of the time is 4 meters per second.

The average speed of the whole journey is (5+4) 2=meters per second.

Running time: 720 seconds.

The first half of the journey took 720 2 5 = 72 seconds.

The second half of the journey took 160-72=88 seconds.

A total of 160 seconds has been calculated; Then the first half of the time is 80 seconds, running 5 meters per second, running 5 8 = 400 meters; In the second half of the time, 80 seconds, 4 meters per second, 4 80 = 320 meters.