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Thu 30 Dec, 2004 11:26 am
An Acara traveled from Meridell to Brightvale, 50 miles away, with his pet Pawkeet. When he arrived in Brightvale, he let the pawkeet fly free while he traveled back to Meridell, at a speed of 10 miles per hour. The pawkeet, who flies at 25 miles per hour, arrived at Meridell and promptly turned around until he got to the Acara. Once he got to the Acara, he turned around and returned to Meridell. He continued to fly back and forth between the Acara and Meridell until the Acara reached Meridell.
How many miles total did the Pawkeet travel on this trip?
The acara would have taken 5 hours to return home, if his pet Pawkeet flew all the time his owner travelled back home, he would have flown 125 miles.
If you include the original journey to Brightvale it would be 175 miles.
Carditel,
I've been struggling with this for an hour or more, and a math expert I aint, but before I try to explain where I'm up to, can I ask you if you are good at math because if you are I'll stop right here and take two Anadin
5 hours x25 mph = 125
by car = 50
125+50 = 175
but it's mid way home, it must finish.
175+25= 200 miles
Francis,
If this sounds too simple for words then by all means tell me to shut up, but you are taking into account that when the bird is flying back to where Acara is he is STILL traveling at 10 miles per hour WHILST the bird is flying back, aren't you?
I'm checking the whole business...
Francis wrote:5 hours x25 mph = 125
by car = 50
125+50 = 175
but it's mid way home, it must finish.
175+25= 200 miles
It's not mid way home. It flies until the the guy gets home which puts it home as well.
It seems I misunderstood something...
Don1 wrote:Francis,
If this sounds too simple for words then by all means tell me to shut up, but you are taking into account that when the bird is flying back to where Acara is he is STILL traveling at 10 miles per hour WHILST the bird is flying back, aren't you?
Don1:
Although the problem seems like it would require calculus or infinite sums to solve, the aha! solution is recognizing that the speed of the man and the distance he travels determines the time the bird is flying. Combine that with the speed of the bird to get the distance the bird flies.
markr wrote:Don1 wrote:Francis,
If this sounds too simple for words then by all means tell me to shut up, but you are taking into account that when the bird is flying back to where Acara is he is STILL traveling at 10 miles per hour WHILST the bird is flying back, aren't you?
Don1:
Although the problem seems like it would require calculus or infinite sums to solve, the aha! solution is recognizing that the speed of the man and the distance he travels determines the time the bird is flying. Combine that with the speed of the bird to get the distance the bird flies.
Markr, I dont know why I torture myself with these things, excuse me whilst I have a soothing cup of tea, with a small tonic of course.
I made it but I've too many decimals so I'm going to put it in more understable way..
Just returned home and re-looked at the problemn,
5 hours to return home, parkeet would be flying for 5 hours,
5 x 25 = 125 + original journey = 125+50 = 175
appologies for error in first answer.
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