help in math thank you

Question

A train traveling from station A to station B travels exactly 1/3 of the distance when it is held up and stops for 16 minutes. When the train continues its journey, the engineer increases its average speed by 15 mph so that the train will arrive at station B on schedule. Find the average speed of the train before it stopped for 16 minutes

Answer 1

what is the average speed before the train stopped for 16 minutes. It went one third the distance from A to B. after stopping it sped up 15 mph faster to arrive on time as scheduled.

the train went x miles per hour for 1/3 the distance

then x+15 mph for 2/3 the distance

16 minutes = 16/60 hours = 4/15 hr.

distance = rate of speed times time

d=xt where x = the speed

d also = xt/3 + 2xt/3 = d/3 + 2d/3

16 minute stop = 16/60 = 4/15 hour

time spent on the first third = t/3

time spent on the last 2 thirds = t-t/3-4/15 = (6/15)t = 2t/5

xt/3 + (2t/5)(x+15) = xt divide by t

x/3 +(2/5)(x+15) = x

(11/15)x + 6 = x

4x/15 = 6

x = 6(15/4) = 45/2= 22.5 mph for the 1st 3rd of the trip

x+15 = 37.5 mph for the last 2/3 of the trip

(t/3)22.5 + (2t/5)37.5 = 22.5t

22.5t/3 + 75t/3 = 22.5t multiply by 15

112.5t +225t = 337.5t

337.5t = 337.5t