Question

Next door to me lives a man with his son. They both work in the same factory. I watch them going to work through my window. The father leaves for work ten minutes earlier than his son.one day I asked him about it and he told me that it takes 30 minutes for him to walk to his factory, whereas his son is able to cover the distance in only 20 minutes. If the father were to leave the house 5 minutes earlier then his son, then the son would catch up with his father in 2k minutes. where k is-----​

Answer 1

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Answer:

  k = 5

Step-by-step explanation:

Suppose the distance to the factory is 1 unit. Then the speed the father walks is ...

  speed = distance/time = 1 / 30 . . . . units per minute

When the father leaves 10 minutes earlier than the son, his position t minutes after the son starts walking is ...

  d = 1/30(t +10)

Similarly, when the father starts 5 minutes earlier than the son, his position t minutes after the son starts walking is ...

  d = (1/30)(t +5)

We want that to be the same as the distance the son covers t minutes after he starts walking. The son's progress is ...

  d = 1/20t

To find the value of k such that the distance covered is the same 2k minutes after the son starts, we will use t=2k and d=d, so ...

  (1/30)(2k +5) = (1/20)(2k)

  4k +10 = 6k . . . . . multiply by 60

  10 = 2k . . . . . . . . . subtract 4k

  5 = k . . . . . . . . . . . divide by 2

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The attached graph shows the different scenarios, where the blue line is the son's progress. When the father starts 5 minutes earlier, the son meets him half-way to the factory after 10 minutes, so k = 10/2 = 5.