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A. The distance between two places B is 20km. A advances from a to B, while B advances from B to A. two hours later, they meet on the way. After meeting, a returns to a, and B still advances to A. when a returns to a, B is 2km away from A. find out the speed of a and B
Let a's speed be x km / h, B's speed be y km / h, from the meaning of the question, 2 (x + y) = 202x − 2Y = 2, the solution is: x = 5.5y = 4.5, a: A's speed is 5.5 km / h, B's speed is 4.5 km / h
A. The distance between B and B is 20km. Party A and B travel from a and B respectively. It is known that the speed of Party A is 3km per hour, and that of Party B is 2km per hour,
When a starts, he takes a dog with him. The dog goes in the same direction with a at the speed of 5 kilometers per hour. After meeting B, he immediately returns. In this way, he goes back and forth with a and B until they meet. He asks the dog how much distance he has run
Lie equation
Similar time: 20 / (3 + 2) = 4 hours
The dog ran at a speed of 5 kilometers per hour for 4 hours, so the dog ran:
5 (km / h) × 4 (H) = 20 km
The distance between a and B is 30km. A and B walk from a and B respectively. A travels 20km per hour and has already traveled 15km per hour
(1) Two people set out at the same time, a few hours later plus catch up with B?
(2) If B starts 20 minutes first, how many hours after a starts, will the two be 20 km apart?
Do it with linear equation of one variable
1. Suppose x hours catch up with (20-15) x = 305x = 30x = 62, suppose B starts 20 minutes first, then the distance between a and B is 20km after y hours. First, a has not caught up with B (20-15) y + 20 = 30 + 15 × (20 △ 60) 5Y + 20 = 355y = 15y = 3. Second, a catches up with B and exceeds B 20km (20-15)
The distance between AB and B is 500km. Party A and Party B respectively walk from a and B. the speed of Party A is 20km, and that of Party B is 30km. Party B starts for 2 hours first. How long does it take for Party A to start
When did they meet?
（500-30×2）÷（20+30）
=440÷50
=8.8 hours
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