Elementary school math problems, the maker is a master, math problems, a master is advanced, element

I only thought of this method, so I'll talk to you.

A six-digit number, so the first digit can be taken from 1-9.

Assuming that the first digit is 1, and the number of the first and last digits is exchanged, the number obtained is three times the original number, so the last digit must be 7, but at this time, after the exchange, that is, a number of more than 700,000, it is impossible to be three times the number of more than 100,000, so it is not appropriate for the first digit to be 1.

Assuming that the first digit is 2, and the number of the first and last digits is exchanged, the number obtained is three times the original number, so the last digit must be 4, but at this time, after the exchange, that is, a number of more than 400,000, it is impossible to be three times the number of more than 200,000, so the first digit is not appropriate to be 2.

Assuming that the first digit is 3, and the number of the first and last digits is exchanged, the number obtained is three times the original number, so the last digit must be 1, but at this time, after the exchange, that is, a number of more than 100,000, it is impossible to be three times the number of more than 300,000, so it is not appropriate for the first digit to be 3.

Assuming that the first position is 4-9, and its three times is more than 1 million, it is 7 digits, which is impossible, so the first position of 4-9 is not appropriate.

To sum up, such a number does not exist.

You check it again to see if there are any errors in my analysis, and if there are any, tell me, and then I will think about how to solve them.

When the last 5 digits of this number are x, then this number is 100000+x, and after the exchange, it becomes 10x+1Hence the equation is obtained.

3(100000+x)=10x+1

10x-3x=300000-1

7x=299999

So x=299999 7

100000+x=142857

So this number is 142857

Let the middle 4 digits be b, and the first digit is a, and the single digit c

100000a+10b+c)*3=100000c+10b+a

299999a+20b=977777c

The guy up there is really annoying Do you want to do it so quickly I just made a list of plates

Solution: Let the middle 4 digits be b, the first digit is a, and the single digit c

3100000a+10b+c)*3=100000c+10b+a

I didn't even graduate from elementary school. So don't know!

1. The donation of class A is two-thirds of the donation of the other two classes, then A accounts for the total: 2 (2+3)=2 5

If the donation of class B is three-fifths of the donation of the other two classes, then B accounts for the total: 3 (3+5)=3 8

So C accounts for the total: 1-2 5-3 8=9 40

So the total: 72 (3 8-9 40) = 480 (yuan).

2. Wang accounts for the total sum: 1 (1+2)=1 3, in the same way, Li: 1 4;Zhao: 1 5

So the sum: 26 (1-1 3-1 4-1 5) = 120

So Wang: 120*1 3=40 (years).

3. The boss accounts for the total: 1 (1+2)=1 3;Second: 1 4;Three: 1 5;Fourth: 1-1 3-1 4-1 5

So the total: 91 (1-1 3-1 4-1 5) = 420 (yuan).

4. The third week of repair: 1200*(1-2 3)=400

So the second week: 700-400 = 300 (meters).

5. After 7 cups are filled, there are still seven-eighths left in the barrel, that is, 7 cups are equivalent to 1-7 8=1 8;Then 14 cups is (14 7) * 1 8 = 1 4

So there are a total of drinks: 3 (1-1 4) = 4 (catties).

6. One-third of the number of parts processed by the master is 10 more than one-quarter of the number of parts processed by the apprentice; Then the master's 1 3 * 3 = 1 is 10 * 3 = 3 4 more than the apprentice's 1 4 * 3 = 3 4, so the apprentice process: (170-30) (1 + 3 4) = 80 (pcs).

7. Age and + age difference * 2 = 72 zhang + li + (zhang - li) * 2 = 72 3 zhang - li = 72 6 zhang - 2 li = 72 * 2 = 144;Li - (Zhang - Li) = Zhang * 1 5 6 Zhang = 10 Li.

So: 8 Li = 144 Li = 18;Zhang=10*18 6=30;Age and: 18 + 30 = 48

8. When Xiaogao completes a total of 1 3 + (1-1 3) * 4 5 = 13 15, Xiaoxin completes 1-5 9 = 4 9

So their speed ratio: 13 15:4 9=39 20 that is, when Xiaogao completes 1 3, Xiaoxin completes: 1 3 39 20=20 117

So in total: 97 (1-20 117) = 117 (Dao).

[3 8-(1-2 5-3 8)] = 480 yuan (1-1 3-1 4-1 5) = 120 years old, 120x1 3 = 40 years old.

(1-1 2-2 3-1 3 4 3-1 4 5 4) = 420 yuan.

1200x1 3 = 300 meters.

3 4 = 4 catties.

x4 7-40 = 80.

12 5x3 5+72 12 5=48 years old (1-1 3 39 20)=117 I have great mercy, I don't need your 10 points, you just need to adopt me.

Think of half an hour's journey as 1 portion.

Then B walks 2 parts an hour, starts 2 hours early, leads 2x2 = 4 parts A walks in one hour (the first half an hour runs, the second half an hour rests, and the time of each run of A is regarded as 1 time period), A catches up with each time period so in the last half an hour should catch up with the distance, then the first few time periods chase a mile, because the first few time periods need to have a rest time, so 1 hour can only chase the share It takes hours to catch up with the last share.

It will take a total of 5+ hours to catch up with B.

It should be at 9:00+ hours = 14:30 minutes.

A and B are two hours apart, although A runs at a speed of 2 times faster than B5 times, but every half hour of running requires half an hour of rest, in fact, A's speed is only twice the speed of B's walking.

So the time it takes for A to catch up with B is: 2 (hours.)

9+8=17 At 17 p.m., A catches up with B.

Let B travel 1 unit per hour.

Time 9:00 9:30 10:

B 2 3 4 5 6 7

A 0 5 5

It can be seen that at 14:30 A catches up with B.

From the topic, we might as well set the one-hour journey as 1

Then: when A starts chasing, the distance between A and B is 2, and the distance between A and B shortens in the first half hour of each hour

B's journey in half an hour) * A's speed is a multiple of B's, and in the same time, the distance is also an equal multiple of B's) = A's distance traveled in the first half an hour)

So in the first half hour of the hour, the distance between A and B is shortened.

In the second half hour, the distance between A and B increases.

So a total of 6 first half hours + 5 second half hours = hours.

So A catches up with B at 9+, i.e. at 14:30.

Let the speed of B be x, and time t A catches up with B.

From the question, it can be seen that B has already walked 2x more distance when A wants to start, and A starts to chase a uniform speed that can be calculated as *t=xt+2x to get t=8, and because A is running for half an hour and taking half an hour off, it is not really a constant speed.

Let's compare it an hour in advance, A's journey is B's journey is 2x+7x=9x7 hours, and after a few hours, it takes time for A to continue chasing B. t1 has t1="So the time for A to catch up with B is an hour later.

Set: B's walking speed = x1, A's running speed = x2The distance from A to B = ab, A takes y2, B takes y1, then, x2=, because: A needs to rest for half an hour every half an hour, then, x2=. Y1 Y2 2 (hours).

Solution: ab x1 y1 x2 y2 , x1 (y2 2) ,x1 y2 2x1 1 25x1 y2. 2x1 0 25x1 y2, y2 2 0 25 2 1 4 8 (hours).

A: It takes 8 hours for A to catch up with B. A can catch up with B at 17:00.

Let's assume that A is at a constant speed, then since A runs for half an hour and rests for half an hour, A's speed is equivalent to a double of B's. If A catches up with B, and B walks for hours t, then A's time is (t-2) hours; B's velocity is x, then A's velocity is. Equations can be listed:

tx=(t-2)*, t=10

But in fact, A is not at a uniform speed, so in the last hour of catching up with B, A actually only took half an hour, and the time for A to catch up with B is 16:30.

Maybe I'm thinking too much, but I'll say anyway.

1. "Every half hour of running needs to rest for half an hour" If you rest or reverse half an hour after running in the first half hour, this problem is tangled enough, and it can still be solved by hypothetical methods and equations. Assuming that A's speed is 2, then B's running speed is 5If B runs half an hour before, it will take hours, and B will run at 16:

30 points to catch up with A or 17 points to catch up with A.

2. It's much easier to rest and run before and after. Need to catch up with A at 10:20.

Assuming B velocity 1

A speed defaults to first.

Let the time be x1*x= (x-2)*

x=x=10

i.e. 10 hours.

But A actually needs to rest for half an hour every half hour he runs.

So subtract half an hour hour.

A's actual speed is only twice the speed of B's walking.

So the time it takes for A to catch up with B is: 2 (hours.)

9+8=17 At 17 p.m., A catches up with B.

Hours i.e. 9:00 + hours = 1430 minutes to catch up.

This is to catch up with the problem, and grasping the key A and B distance must be the same to solve the problem.

You're all masters, it's a mess.

1.A cylindrical container, measured from the inside, the radius of the bottom surface is 1 decimeter, the high decimeter, what is its volume? If you put a liter of water in it, what is the distance between the water surface and the mouth of the container?

Volume cubic decimeter.

Dm. 2.The water depth in a rectangular glass tank is 628cm, pour this water into a cylindrical empty glass tank with a bottom diameter and height of 10cm, does the water overflow? If it does not overflow, what is the depth of water in the cylinder at this time?

The box does not have a length, width, or base area and cannot be calculated.

3.The volume of the cylinder is cubic centimeters, the diameter is expanded by 4 times, and the height remains the same, so how many cubic centimeters is the volume of the cylinder?

Cubic centimetre.

1. The volume is equal to the volume of the cylinder, equal to the bottom area multiplied by the height, and then divided by the bottom area to obtain the height of the water in the cylindrical container, and the distance between the water surface and the container mouth is obtained by subtracting the height.

2. You check the question. Can't do it.

3. Because the diameter is expanded by four times, the radius is expanded by four times, and the area at the bottom of the cylinder is expanded by 16 times, and the height remains the same, so the volume is expanded by 16 times and then multiplied by 16 cubic centimeters to draw conclusions.

Volume is volume, base area * height, pi* r 2 * h = cubic decimeter.

Ideas, this kind of topic mainly clarifies the relationship between quantities. Topic 1 is the volume of water compared to the volume of a cylindrical container. The key is to understand that water is poured into a container as volume, and the bottom area is the same as that of the container.

When you think about this, do you have a way of thinking? The same goes for the second question. The idea of the third question is that the bottom area of the cylinder is expanded by 4x4 times, and the height remains the same, and the volume is also expanded by 16 times.

Let me start by listing a few key figures:

The total milk volume is 1400 cubic centimeters.

The base area of the circle is square centimeters, and the number of people is 5.

1400 5 = 280, if you just pour 5 full cups is 280 ml (cubic centimeters) per cup, that is, the cup height is 280 cm.

The answer to this question is very open. Because there is no rule on whether there can be any leftover after pouring 5 full cups.

If there can be a surplus, Lele said that it can't be completed, that is, the milk is definitely not enough to pour 5 full cups, then the cup height can be infinitely high, and the cup volume is infinite.

If there is no leftover, Lele said that it cannot be completed, that is, one is not enough to pour 5 full cups, and the other is that 5 full cups can be poured but there will be leftovers.

According to the second question, the author of this question wants to say that there can be no surplus, only in this way can we know the maximum volume of the cup, and the maximum volume must be retained in the whole milliliter, which is only 280-1=279 milliliters.

So:1A total of 1400 ml of kefir.

2.The maximum volume of the glass is 279 ml (the whole milliliter number must be retained).

3.The sour milk in the box can be filled with 5 full cups (according to the author's idea that there is no rest), because if it can't be filled, the glass volume can be infinite, and the second question cannot be answered.

Either the landlord typed less, or the questioner himself didn't think it through.

I really don't understand that today's children would rather spend so much time typing out questions than spend so much time thinking about them. Tragic...