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A Car passes as many Lamptposts in 2 minutes as its speed is in miles per hour. What is the distance between the Lamposts in feet?

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Posted · Report post

176 Feet

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Posted · Report post

176 Feet

Can you please say, how you calculated this?

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The car's speed is 30 lamps/hour = speed in miles/hour.
30 lamps = 1 mile = 5280 feet
1 lamp = 176 feet.

That is the lamppost spacing.

Edit - check this.

speed = x mph = 5280x fph = 5280x/60 fpm = 88x fpm.

Thus, in 2 minutes the car travels 176x feet.

We are told that in 2 minutes he passes x lamp posts.

Thus there is 176 feet per lamp post.
That is the spacing.

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Car's speed is say x miles per hour.


So it passes x lampposts in 2 minutes.
Say the distance between two lamposts is d miles.
So the distance between 30 lampposts is d(x-1).
So the car passes d(x-1)miles in 2 minutes.
Therefore, the car's speed is 30d(x-1) miles per hour
Therefore x = 30d(x-1) which can be written as x = 30d/(30d-1)
Value of x in the above eq. should be a whole number,
So the value acceptable for d is 2/30 miles.
So d = 2/30 miles = 2/30 * 5280 Feet = 352 feet

I hope this is the reasonable answer.

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bhramarraj, we differ by an interpretation of OP, which is ambiguous.

If "2 minutes" means "every 2 minutes" or "a particular 2-minute interval" the answer could be what jordge and I posted or your answer or anything in between. That is, in the second case the first pole is passed (sometime) in the first minute and the second pole is passed (sometime) during the second minute. That would permit your answer if pole 1 happened at the beginning of minute 1, and the second pole happened at the end of minute 2.

Or am I missing something?

So, is John Wilson still around to clarify?

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Posted · Report post

bhramarraj, we differ by an interpretation of OP, which is ambiguous.

If "2 minutes" means "every 2 minutes" or "a particular 2-minute interval" the answer could be what jordge and I posted or your answer or anything in between. That is, in the second case the first pole is passed (sometime) in the first minute and the second pole is passed (sometime) during the second minute. That would permit your answer if pole 1 happened at the beginning of minute 1, and the second pole happened at the end of minute 2.

Or am I missing something?

So, is John Wilson still around to clarify?

You are right bonanova.

Interestingly there would be muliple answers with the conditions given in OP, as can be seen by the formula:

30d/x = 30d - 1 where x = no. of poles or speed of the car in mph, and d is the distance between the poles in miles.

We can select value of x as from 1, 2, 3, 4,.........etc. and can have different values for d in feet.

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