SOLUTION: An elevator went from the bottom to the top of a tower at an average speed of 4 m/s, remained at the top for 90 s, and then returned to the bottom at 5 m/s. If the total elapsed

Question 25194: An elevator went from the bottom to the top of a tower at an average speed of 4 m/s, remained at the top for 90 s, and then returned to the bottom at 5 m/s. If the total elapsed time was 4 1/2 min., how high is the tower?
Can you please also give the the equation you used to solve this problem?
Thank you!
Answer by Paul(988) (Show Source): You can put this solution on YOUR website!
The elvator went at 4m/s at a distacne of x
The elvator reutrned at 5m/s at a distance of x
Total time in SECONDS. because its meters per second. (4*2+1)/2=4.5*60=270s
270s-90s=180s

4*5 =20
5*4=20
Hence the common multiple is 20

Simplfy the expression.
5x+4x=3600
9x=3600
x=400

Hence the tower is 400m tall.
Paul.
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