Nephew s Olympiad problems, solving, Olympiad problems solving

If there are 8 people each divided into 6, and the rest of the people divided into 4 each, these peaches will have 60 left, so each person will divide 4 and the remaining 60 + 8 * 2 = 76.

If there are 10 people who are divided into 5 each, and the rest of the people are allocated 7 each, so that each person is divided into 7 less than 10 * 2 = 20.

Then 76+20 can be divided into 3 per person, the number of students = 96 3 = 32, the number of peaches = 32 * 7-20 = 204

Profit and loss issues. Let's sort out the conditions.

If each person divides 4, there is a total left: 8 (6-4) + 60 = 76.

If each person is divided into 7, the total is missing: 10 (7-5) = 20.

There are students: (76+20) (7-4)=32.

Peaches have: 32 4 + 76 = 204 pcs.

If there are 8 people with 6 each, and the rest of the people have 4 each, there will be 60 peaches left, and all the people will have 4 each, and these peaches will have 60 + (6-4) 8 = 76.

If there are 10 people who are given 5 each, and the rest of the people are given 7 each, it is exactly finished, and all the people are divided into 7, and these peaches are less (7-5) 10 = 20.

How many students are in a 3rd grade class? (76+20) (7-4) = 32 people.

How many peaches do the fruit stores deliver? 50+(32-10) 7=204.

There are x people.

It's too simple, the number of peaches is 8*6+(x-8)*4+60=10*5+(x-10)*7

Solve x and then get the number of peaches.

The equation is easy to solve. There are x people in this class.

According to the problem, (x-8) 4+48+60=(x-10) 7+50 is solved x=32

The fruit store sent peaches (32-10) 7 + 50 = 204 (pcs).

There are x people in this class.

x-8）×4+48+60=（x-10）×7+50

Solution x = 32 peaches from the fruit store (32-10) 7 + 50 = 204 (pcs).

One side of the opposite side of the square increases by 18 cm, the other side decreases by 12 cm, and the length of the two opposite sides changes by 18-12 6 cm, becoming 38 cm, so the sum of the original two sides is 38-6 32 cm, and the length of one side is 32 2 16 cm. The trapezoidal area is 38 16 2 38 8 304 square centimeters.

Suppose the side length of the square is a, (a+18)+(a-12)=38, and the side length of the square is 16; The area of the question type is 38 16 2 = 304

1.It turns out that the denomination of the check is yuan, and the exchanger mistakenly looks at it as yuan, and it should be returned to the yuan The method I used involves a binary equation, I don't know if it meets your requirements?

Let the digit of the yuan be x, the digit of the cent is y (the digit of the angle is represented by the ten digits of y), the original denomination of the check is x yuan y cent, the exchanger sees it as y yuan x cents, we may as well use cents as the unit of measurement.

Then the original denomination of the check was (100x+y) cents, and later the exchanger saw it as (100y+x) cents.

After Li Lin spent 350 points, that is, the remaining (100y+x-350) points, which was twice as much.

Therefore, the column equation is 100y+x-350=2*(100x+y) shift, y=(350+199x) 98=25 7+199x 98 because x, y are integers within 100, and 199 is a prime number, 98=7*14, so x must be a multiple of 14, and the value of x divided by 14, the sum of 25, is divisible by 7.

25 7 = 3 remaining 4

199 7 = 28 remainder 3

So the value of x divided by 14 must be 1 or 8

Substituting the verification, we get x=14, y=32

So it turned out that the cheque was denominated in yuan.

2.I don't understand the second question, please explain, why is it a and b in front and x in the back?

The problem is a matter of speed, and it is necessary to distinguish the difference between meeting head-on and chasing behind.

For this kind of switchback and running problem, think about when you will meet head-on, then think about when you will catch up from behind, and finally consider the answer to the original question.

Step 1: Think about the first question, what does the first 10 encounters between the two of you mean?

If the distance of the straight track is a, on the way back to the switchback, the first n times can meet head-on, and the meeting point is not at the end.

The distance difference between the two is always less than A.

Therefore, before the 10th encounter, the distance difference between Xiao Ming and Dazhu was less than A.

Step 2: Think about the second question, what can Dazhu say about catching up with Xiao Ming from behind on the 11th time?

If the distance of the straight track is A, and the 11th meeting point is B from the starting end of the big column, then by the 11th encounter, the running distance of the big column is 10A+B, and the running distance of Xiao Ming is 9A+B, and the big column happens to run more distance than Xiao Ming.

Step 3: Combine the above questions and consider the answers to the original questions.

The velocity ratio in the same time is equal to the distance ratio, and on the basis of the conclusion of step 2, the velocity of the large column can be found as 9*(10a+b) (9a+b).

The range of (10a+b) (9a+b) is considered below:

Note that b is greater than 0 and less than a, then 9a+b is greater than 9a and less than 10a, and (10a+b) (9a+b).

1+a (9a+b), must be greater than 11 10 and less than 10 9, so the range of 9*(10a+b) (9a+b) is greater than and less than 10.

So the speed of the big column is greater than kilometers per hour, but less than 10 kilometers per hour.

The process is a bit tedious, but the idea must be clear.

It was the 11th time to catch up from behind, which means that this time I caught up with Xiaopeng by 1 runway distance.

1.Meet at the beginning of the direction of pursuit, then:

9 10 (10+1)=km).

2.Meet at the end of the direction of pursuit, measure:

9 11 (11+1) = 108/11 (km) Therefore, the speed of the large column is between 108/11 and 108/108 (km).

If you can't see clearly, please draw correctly.

Let's take a picture of the topic directly, there should be a lot of conditions that should not be said.