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A and B two cars from a, B two cities at the same time, 10 hours meet, the speed ratio of the two cars is 3:2, when a car arrives at a city, B car how many hours to arrive at B city?
(30-10 × 23 × 2) / - 2, = (30-403) / - 2, = 503 × 12. = 253 (hours). Answer: it will take 253 hours for bus B to reach city B
Party A and Party B are 300 meters apart. If they are going in opposite directions, they will meet in 3 minutes. If they are going in the same direction, Party A will catch up with Party B in half an hour to find their speed
Suppose the speed of a and B is x and Y respectively, then:
3x+3y=300
30x-30y=300
The solution is: x = 55 (M / min), y = 45 (M / min)
Party A and Party B set out from a and B with a distance of 1500 meters on the two days, facing each other and meeting three minutes later. We already know that the speed of Party B is 5 meters per second, so we can find the speed of Party A
B OK
180*5=900
A line 1500-900 = 600
The speed of a
600 / 3 = 200m / min
It can also be expressed in seconds
So the speed of a is 200 m / min
Party A and Party B are walking towards each other from a and B, which are 200 meters apart. They meet 4 minutes later. It is known that the speed of Party B is 5 meters per second. What is the speed of Party A?
Let a's speed be x per second
（5+X)240=200
5 + x = five sixths
X = twenty five sixths of the negative
Because x = 25 / 6 of negative, it doesn't fit the meaning of the question
So x = minus twenty-five sixths
(there is an obvious problem in this problem. If we can change the meeting time from four minutes to four seconds, we can solve it.)
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