Sender: Brian
Subject: word problems, algebra 1
Date: Thu, 4 Feb 1999 21:47:37 -0600


Help please
secondary 10 -12

Please answer these problems with a formula. Also explain in English step by step how these problems should be solved. I know the answer I just don't know how the answer was achieved Thank you very much for trying to help

  1. A ski lift carries a skier up a slope at the rate of 120 feet per minute and he returns from the top to the bottom on a path parallel to the lift at an average rate of 2640 feet per minute. How long is the lift if the round trip traveling time is 20 minutes?
  2. An airplane on a search mission flies due east from an airport, turns and flies due west back to the airport. The plane cruises at 200 miles per hour when flying east, and 250 miles per hour when flying west. What is the farthest point from the airport the plane can reach if it can remain in the air for 9 hours?

Hi Brian

For your first problem you need to realize that, for each trip, you are dealing with 3 things: distance, rate, time. The relationship among them is that distance = rate x time. Decide what you are trying to determine and what relevant information you have. In this case we know the rate up and the rate down, the combined time and the fact that the distance up = the distance down.

Let's do some preliminary thinking. For the up trip we know that distup = 120 x timeup For the down trip we know that distdown = 2640 x timedown We also know that timeup + timedown = 20

Since distup = distdown we know that 120 x tup = 2640 x tdown
We also know that tup + tdwn = 20.

To write this concisely let the time up be represented by c minutes, and the time down be represented by b minutes then

120 c = 2640 b.

c + b = 20

You can divide both sides of the 1st equation by 120 and you find that

c = 22 b

Replacing c by 22 b in the second equation gives

23 b = 20

and hence

b = 20/23 minutes

So the distance down can be obtained by multiplying this by the speed down.

Problem 2 can be solved in a similar fashion.

Jack and Penny

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