Scheduled time:1/2 ÷1/7+1/2 = 4 hours.

When returning, it takes 20/60= 1/3 hours less than the original time.

120km distance, according to the original speed:1/3 ÷1/6+1/3 = 7/3 hours.

Drive at the original speed 120km, time: 4-7/3=5/3 hours.

Original speed:120 ÷ 5/3 = 72 km/h.

Distance: 72× 4 = 288km

2.857+9-64)×30÷ 12=2005

3.(4,6)= 12

From left to right, every four people are a group, and then from right to left, every six people are a group. It is concluded that only when the last person reports 1 at the time of reporting 6 can someone report 3 at the same time. After understanding, they were divided into the same two groups. From left to right, every 12 person is a group. But there was only one person in the first group because the last person reported 1 at 6 o'clock.

Moreover, because the last number of 1 in the group of 6 can only be 1 (starting from 1), you only need to add 6 after completing five groups. However, don't forget to add 1 person to the 1 group.

12*5+6+ 1=67 (person)-the largest number of soldiers in this column.

A: This train has a maximum of 67 soldiers.

4. The two cubes drawn must be on the edge, not at the vertex, but in the middle;

At present, only four cubes are drawn on two faces, and a cuboid has *** 12 sides. The key point here is that you have to think of "four on the side, four on one floor", only 1 floor. This cuboid has ***3 layers, and the first layer and the third layer each have 4 corner blocks, so this cuboid is 2× 2.

5.

What was sold in the morning was 15 and 18, a * * * 33.

I sold 16, 19, 3 1, a * * * 66 in the afternoon;

The remaining barrels weigh 20 kilograms.

The multiple of 6 ..2 is 2006/2 = 1003, 3 is 2006/3 = 668, 5 is 2006/5 = 40 1, 2 and 3 are 2006/(2 * 3) = 334, and 2 and 5 are multiples.

66 people pulled it three times, (334-66)+(200-66)+(133-66) = 268+134+67 = 469 people pulled it once.

There are 936+66 = 1002 lights off and 2006- 1002 = 1004 lights on.

7.

Analysis: Let the water speed be V km/h, A sailed X km from the departure to the discovery of the falling object, and the total time was T.

The sailing speed of Party A and Party B is 6v in still water, 5v in countercurrent and 7v in downstream.

Vt+5vt=60, that is, the goods meet the ship B,

Available vt= 10,

That is, the goods drifted along the river 10km, so the B ship sailed 50km.

x/5v+(x+ 10)/7v=50/5v，

That is, the sailing time of a is equal to the sailing time of b,

You can get x=25km by removing the V on both sides.

8 .. Solution: 480/(55+65)=4 minutes.

4 * 55 = 220m

Then Mr. Wang met another teacher and walked 220 meters at a time. He met 10 and walked 10 * 220 = 2200 meters.

2200/480 = 4.280m.

So 480-280=200 (meters)

A: Miss Wang will walk 200 meters and then return to the starting point.