It was a skiing resort in
Switzerland. I met a skiing enthusiast by the skiing slopes. He had a packed
lunch with him. He asked me to join him for lunch back in the spot where we
were standing at 1 p.m. after he had done a bit of skiing. I told him no, as I
had an appointment to keep at 1 p.m. But if we could meet at 12 noon, I told
him that perhaps I could manage.
Then he did some loud thinking, “I
had calculated that if I could ski at 10 kilometres an hour I could arrive back
at this pot by 1 p.m. That would be too late for you. But if I ski at the rate
of 15 kilometres an hour then I would reach back here at 11 a.m. And that would
be too early. Now at what rate must I ski to get beck here at 12 noon? ….. Let
me se.”
He got the right figure and he got
back exactly at 12 noon. We had an excellent lunch.
What do you think the figure was?
Answer: This problem would easily
lead one to think that the speed we seek is the mean result of 10 and 15
kilometres i.e. 12.5 an hour. But this is wrong.
In fact, if the distance the skier
covers is x kilometers, then going at 15 kilometres an hour he will require
x / 15 and at 12.5 kilometres x /12 ½ or 2x /25
So the equation: 2x /25 - 1/15 = 1/10
- 2x /25
Because each of these is equal to
one hour, when we simplify we obtain: 2/25 – 1/15 = 2/25
Or it can be expressed in
arithmetical proportion:
4/25 = 1/15 +1/10
But this equation is wrong hecause:
1/15 + 1/10 = 1/6 i.e. 4/24 and not 4/25
But it can be solved orally in the
following manner: If the speed is 15 kilometres an hour and was out for two
hours more he would cover an additional 30 kilometres. In one hour, as we
already know, he covers 5 kms more. Thus he would be out for 30:5 = 6hours.
This figure determines the duration of the run at 15 kilometres an hour: 6-2 =
4 hours. And now we can easily find the distance covered: 15x 4 4 = 60
kilometres. Now, again, without any difficulty we can see how fast he must ski
to arrive at the appointed place at 12 noon - - i.e. Five hours.
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