SOLUTION: Two airplanes left the same airport and arrived at the same destination at the same time. The first airplane left at 8:00 a.m. and traveled at an average rate of 496 miles per hour

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Question 1125317: Two airplanes left the same airport and arrived at the same destination at the same time. The first airplane left at 8:00 a.m. and traveled at an average rate of 496 miles per hour. The second airplane left at 8:30 a.m. and traveled at an average rate of 558 miles per hour. Let x represent the number of hours that the first plane traveled.
How many hours did it take the first plane to travel to the destination?
Found 3 solutions by ikleyn, MathTherapy, greenestamps:
Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let  t = "How many hours did it take the first plane to travel to the destination".


Then (t-0.5) is the time in hours for the second plane (it started 30 minutes = 0.5 of an hour later).


The equation is


    496*t = 558*(t-0.5),


saying that the distance is the sane.


Simplify and solve


    496t = 558t - 558*0.5

    558*0.5 = 558t - 496t = (558 - 496)*t = 62t  ====>  t = %28558%2A0.5%29%2F62 = 4.5.


Answer.  4.5 hours.


Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Two airplanes left the same airport and arrived at the same destination at the same time. The first airplane left at 8:00 a.m. and traveled at an average rate of 496 miles per hour. The second airplane left at 8:30 a.m. and traveled at an average rate of 558 miles per hour. Let x represent the number of hours that the first plane traveled.
How many hours did it take the first plane to travel to the destination?
First plane took highlight_green%28matrix%281%2C2%2C+4.5%2C+hours%29%29

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Here is an alternative to the usual algebraic solution method used by the other tutor.

For many (probably most) students the algebraic method is easier to use. But try this one and see if it works for you.

The ratio of the speeds of the two planes is 496:558 = 8:9. Since the distances are the same, the ratio of times is 9:8.

So we can let 9x be the amount of time the first plane took and 8x be the amount of time the second plane took.

Then since the difference in flight times was half an hour, 9x-8x=x = .5 hours.

Then the flight time for the first plane was 9x=4.5 hours and the time for the second plane was 8x=4 hours.