Solution: suppose the dog ran x kilometers.

Equivalence: the time for a dog to run = the time for two people to meet.

x/ 10=30/(6+4)

x/ 10=3

x=30

The dog ran 30 kilometers.

2.

Solution: If the average speed from Beijing to Tianjin is X kilometers per hour, then the average speed from Tianjin to Beijing is x+40 kilometers per hour.

Equidistant relation: equidistant

(0.5+6/60)x=0.5(x+40)

0.6x=0.5x+20

0. 1x=20

x=200

A: The average speed from Beijing to Tianjin is 200 kilometers per hour.

3.

Downstream speed: 8+2 = 10km.

The countercurrent speed is per hour: 8-2 = 6 kilometers.

1) If C is between AB, for example: A |-C |-B.

Ac downstream, time: 2÷ 10=0.2 hours.

CB round-trip time: 3-0.2=2.8 hours.

The round-trip speed ratio is 10:6=5:3.

CB Downlink Time: 3/(5+3)*2.8= 1.05 hours.

CB distance:10×1.05 =10.5km.

AB distance: 2+10.5 =12.5km.

2) if a is between cbs, for example, c |-b.

Ac countercurrent, time: 2÷6= 1/3 hours.

AB round trip, time: 3- 1/3=8/3 hours.

The round-trip speed ratio is 10:6=5:3.

AB downlink time: 3/(5+3)*8/3= 1 hour.

AB distance: 10× 1= 10 km.

4.

Solution: Suppose you need to add X people.

Equal relationship: everyone's work efficiency is the same.

3/5÷(9× 14)=( 1-3/5)÷[4 *(9+x)]

( 1-3/5)*(9× 14)= 3/5 * 4 *(9+x)

2*9* 14=3*4*(9+x)

9+x=2 1

x= 12

A: You need to add 12 people.

5.

Solution: let this material be X.

Equivalence relationship: the remaining part, the original time -20 minutes = the time spent after improving efficiency.

x *( 1-2/5)/30-20 = x *( 1-2/5)/[30 *( 1+50%)]

Multiply by 90 at the same time to get:

3x*0.6- 1800=2x*0.6

0.6x= 1800

x=3000

A: The material is 3,000 words.