Question

Two planes make a 900 mile flight, one flying 50 miles per hour faster than the other. The quicker plane makes the trip 3 hours faster. How long did it take the slower plane to complete the flight?
I am totally confused with this question i got it took 6 hours.

Answer 1

\[distance = speed\space x \space time\]Both planes travel the same distance so you can create 2 equations with 2 unknowns:\[900 = (s+50)(t-3)\]\[900 = s * t\]
Rearrange the 2nd equation to get:\[s=\frac{900}{t}\]Substituting that into the first equation and you get:
\[900 = (\frac{900}{t}+50)(t-3)\]
Solving for t you will get two values...one will be negative so you can ignore it. The other answer will be 9. Therefore, it takes the slower plane 9 hours to get there.

Answer 2

Another approach.
Let s be the speed of the slower plane.
r*t = d
\[s\left(\frac{900}{s+50}+3\right)= 900\]s = 100
t = d/r
t = 900/100
t = 9 hours

Answer 3

@robtobey: Nice thanks for posting that method...a bit simpler

Answer 4

Thank you for your comment.

Answer 5

Thank you it was the right answer and it helped me answer the rest of the answers.

Answer 6

Glad the solution was helpful.
Thank you for the medal.

Answer 7

No problem.