Two buses leave at the same time towards each other from points A and B and meet at 12 noon.

| September 12, 2021 | education

Two buses leave at the same time towards each other from points A and B and meet at 12 noon. If the speed of the first bus is doubled, and the speed of the second is left the same, then the meeting will take place 56 minutes earlier. If we double the speed of the second bus, leaving the same speed of the first, then the meeting will take place 65 minutes earlier. Determine the meeting time if the speed of both buses is doubled.

To solve the problem, we introduce the variables x and y.
Let’s take that the speed of the first is x, and the second is y, then, by condition, they meet at 12 o’clock in the afternoon: x + y = 12.
Since if the speed of the first bus is doubled, and the speed of the second is left the same, then the meeting will take place 56 minutes earlier, namely: 2x + y = 12 – 0.56 = 11.04.
We also know that if we double the speed of the second bus, leaving the same speed of the first, then the meeting will occur 65 minutes earlier, namely: x + 2y = 12 – 0.65 = 10.55.
Then, to find out the meeting time, if the speeds of both buses are doubled, then we transform the first equation: (2x + y) + (x + 2y) – (x + y) = 2x + 2y = 11.04 + 10.55-12 , 00 = 9.59.
Answer: 9 hours 59 minutes.



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