I won t be able to do two Olympiad numbers, and I have to pay it on Saturday! Everybody help.

Backwards... When the last three layers are equal, each has: 192 3 = 64 copies.

Before the third layer is placed in the first layer, the first layer has: 64 2 = 32 copies.

The second floor has: 64 copies.

The third layer has: 192-32-64 = 96 copies.

Before the second layer is placed in the third layer, the third layer has: 96 2 = 48 copies.

The first floor has: 32 books.

The second layer has: 192-48-32 = 112 copies.

Before the first layer is placed in the second layer, that is, originally, the second layer has: 112 2 = 56 copies.

The third floor has: 48 copies.

The first layer has: 192-56-48 = 88 copies.

Or backwards. If the quantity of the two warehouses is equal as the unit 1, then the total number of A and B is 2 Before B is transported to A, B has: 1 (1-1 4) = 4 3 A has: 2-4 3 = 2 3

Before A is transported to B, that is, originally, A has: 2 3 (1-1 4) = 8 9 B has: 2-8 9 = 10 9

It turns out that A is B's: 8 9 10 9 = 4 5

1.The three layers are a b c, 2(a-b)=2b-c=2c-(a-b)=192 3=64, and the solution is a=88 b=56 c=48

2.A and B x y, x*3 4+(y+x 4) 4=(y+x 4)*3 4, x y=4 5

Clause. 1. The rabbit and the cat did not meet and the difference of 4 kilometers was searched, which means that the distance traveled by the rabbit and the cat was 40-4 = 36 kilometers, and the speed and 36 were equal to 9 kilometers by 4.

If the speed of the cat is set to x, then the speed of the rabbit is 9-x, then: 2 (40 - 5x) = 40 - 5 (9 - x) The equation is solved as follows: 17 kilometers for the cat and 10 kilometers for the rabbit for 3 points.

Clause. 2. The difference between the rabbit and the cat after meeting is 4 kilometers, which means that the distance traveled by the rabbit and the cat is 40 + 4 = 44 kilometers, and the speed and 44 divided by 4 are equal to 11 kilometers.

If the speed of the cat is set to x, then the speed of the rabbit is (11-x) then: 2(40-5x)=40-5(11-x) The equation is solved: 19/3 km of the cat and 14/3 of the km of the rabbit.

The second question is done again.

The sum of any two numbers from 1 to 99 is less than 100.

When the first number is taken as 99, there are 0 ways to take it.

When the first number is taken as 98, there is 1 way to take it.

When the first number is taken as 97, there are 2 ways to take it.

When the first number is taken as 50, there are 49 ways to take it.

When the first number is taken as 49, there are 48 ways to take it.

When the first number is taken as 48, there are 47 ways to take it.

When the first number is taken as 47, there are 46 ways to take it.

When the first number is taken as 2, there is 1 way to take it.

So there is a total of: or the other way around:

When the first number is taken as 1, there are 97 ways to take it.

When the first number is taken as 2, there are 95 ways to take it.

When the first number is taken as 3, there are 93 ways to take it.

When the first number is taken as 49, there is 1 way to take it.

Total: 1+3+5+7+......95+97=2401

Classification calculation, the larger number is 99, and there are 0 ways to take it.

The larger number is 98, and there is 1 way to take it (98+1).

The larger number is 97, and there are 2 ways to take +2).

The larger number is 51, and there are 48 ways to take it (51+1......).51+48) The larger number is 50, and there are 49 ways to take it (50+1......50 + 49) The larger number is 49, and there are 48 ways to take it (49 + 1......49+48) The larger number is 2, and there is one way to take it (2+1).

The larger number is 1, and there are 0 ways to take it.

A total of 0+1+2+......48+49+48+……2 + 1 + 0 = 49 49 = 2401 ways to take it.

Method 2: Use and Classify:

The sum is 3:1 (1+2).

The sum is 4:1 (1+3).

The sum is 5:2 (1+4, 2+3).

Sum is 6:2 kinds (1+5, 2+4).

The sum is 97:48 (1+96......48 + 49) and 98:48 (1 + 97......48+50) and 99:49 (1+98......49+50) for 2 (1+2+......48) +49 = 2401 ways to take it.

Method: Starting from 1, 1 plus to 98 are all satisfied (1+2;1+3……There are 97 species in total.

Starting with 2, 2 plus to 97 is satisfied (2+3;2+4……There are 95 types in total.

Finally, 49+50, there is only one, which is equivalent to finding 1+3+5+......97=2401

Both are odd checks and balances and 3*3=9 kinds.

Both of them are 3*3=9 kinds of even jujubes.

There are 9+9=18 cases where the sum of the numbers on the upward side is even.

20 rabbits can be exchanged for 2 sheep, that is, 1 sheep can be exchanged for 10 rabbits;

9 sheep can be exchanged for 3 pigs, that is, 1 pig can be exchanged for 3 sheep;

8 pigs can be exchanged for 2 cows, that is, 1 cow can be exchanged for 4 pigs.

So, 5 cows can be exchanged for 5x4=20 pigs.

20 pigs for 20x3=60 sheep.

60 sheep can be exchanged for 60x10=600 rabbits.

Question 1, when my brother set off from home, my sister had already walked 3x2=6 minutes, which means that my sister had already walked 6x60=360 meters. Now it's a matter of catching up, the time it takes for the brother to catch up with the sister is 360 (90-60)=12 minutes. The distance that my brother walked in 12 minutes is 12x90=1080 meters.

At this time, it is still 100 meters from the zoo. So the home to the zoo is 1080 + 100 = 1180 meters.

In the second question, A took 15 minutes to repair the car, but arrived 5 minutes later than B, indicating that if A did not delay, it would take 10 minutes less than B to arrive. Let's assume that A will take x minutes to arrive without delay, then B should need x + 10 minutes. With the equation of distance equality, 200x=160(x+10), it is easy to solve x=40 minutes, so the distance is 200*40=8000 meters.

The formula for this: speed time = distance.

Distance, time, speed.

Distance, speed, time.

According to the title, A and B ride bicycles from school to the Children's Palace, with a difference of 15-5=10 minutes.

Then you might as well look at it as a catch-up problem, B sets off for 10 minutes first, and then both of them arrive at the same time.

160*10 (200-160)=40 points, 200*40=8000 meters;

or 160*(40+10)=8000 meters.

Here's the answer:

01.Width of the rectangle: 32, 4, 8 meters.

The length of the rectangle is: 75, 5, 25 meters.

The area of the original rectangle is 25 8 200 square meters.

02.The width of the rectangle: 42, 6, 7 meters.

The length of the rectangle is: 40 4 10 meters.

The area of the original rectangle is 10 7 70 square meters.

02 Solution: Let the original length of this rectangular iron plate be x meters and the width be y meters.

Then the original area is xy square meters, according to the title:

4y=325x=75

Solve this system of equations to get:

x=15y=8xy=15x8=120

A: The original volume of this rectangular iron plate was 120 square meters.

01 Solution: The original length of this rectangular vegetable plot is x meters and the width is y meters.

Then the original area is xy square meters, according to the title:

6y=424x=40

Solve this system of equations to get:

x=10y=7xy=10x7=70 square meters.

A: The original area of this rectangular vegetable plot was 70 square meters.

01 is 8 meters wide and 25 meters long, with an original area of 200 square meters.