SOLUTION: A single-engine plane and a commercial jet leave an airport at 10 A.M. and head for an airport 930 mi away. The rate of the jet is four times the rate of the single engine plane. T

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Question 1114108: A single-engine plane and a commercial jet leave an airport at 10 A.M. and head for an airport 930 mi away. The rate of the jet is four times the rate of the single engine plane. The single-engine plane arrives 5 h after the jet. Find the rate of each plane
Found 2 solutions by ankor@dixie-net.com, greenestamps:
Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
A single-engine plane and a commercial jet leave an airport at 10 A.M. and head for an airport 930 mi away.
The rate of the jet is four times the rate of the single engine plane.
The single-engine plane arrives 5 h after the jet.
Find the rate of each plane
:
let s = the speed of the prop plane
then
4s = the speed of the jet
:
Write a time equation; time = dist/speed
Prop time - jet time = 5 hrs
- = 5
multiply equation by 4s, cancel the denominators and we have:
4(930) - 930 = 5(4s)
3720 - 930 = 20s
2790 = 20s
s = 2790/20
s = 139.5 mph is the prop plane
then
4(139.5) = 558 mph is jets speed
:
:
Check this by finding the actual time of each
930/139.5 = 6 hrs
930/558 = 1 hrs
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travel time dif: 5 hrs

Answer by greenestamps(13203) (Show Source):
You can put this solution on YOUR website!

Here is a very different way you could solve this problem....

The rate of the jet is 4 times the rate of the single engine plane; since the distances are the same, the time required for the small plane is 4 times the time required for the jet.

let t = time for jet
then 4t = time for small plane

The difference in flight time is 5 hours:

The jet takes 5/3 hours for the flight of 930 miles; its speed is mph.

The jet's speed is 4 times the small plane's speed; the speed of the small plane is mph.