Yesterday it took Robert 3 hours to drive from City A to City B. Today it took Robert 2.5 hours to drive back from City to City A along the same route. If he had saved 15 minutes in both trips, the speed for the round trip would be 60 miles per hour. What is the distance between city A and city B?
Total round trip time = 5.5 hours (330 minutes)
t=330
speed = distance/time
60 = 2d/t-2*(15)
60 = 2d/t-30
60t-1800 = 2d
30t - 900 = d
I messed up here two ways. First, I converted hours to minutes. After realizing this was a mistake, I plugged in the total number of hours (5) for t but still subtracted 30 minutes from t when I should have converted subtracted the fraction that 15 minutes takes off of each trip from the total time.
(subtracting 15 minutes from both trips)
AB - 2.75 hours
BA - 2.25 hours
Speed = distance/time
we are given the average speed if 15 minutes were shaved off of each leg of the trip and the time for each leg of the trip. With this information we can solve for d.
60 = 2d/(2.75 + 2.25)
60 = 2d/5
300 = 2d
150 = d
An alternative way of solving using d=r*t formula:
2d=s*t
2d=60*(2.75+2.25)
2d=60*(5)
2d=300
d=150
(A) 90
(B) 120
(C) 150
(D) 240
(E) 300
t=330
speed = distance/time
60 = 2d/t-2*(15)
60 = 2d/t-30
60t-1800 = 2d
30t - 900 = d
I messed up here two ways. First, I converted hours to minutes. After realizing this was a mistake, I plugged in the total number of hours (5) for t but still subtracted 30 minutes from t when I should have converted subtracted the fraction that 15 minutes takes off of each trip from the total time.
(subtracting 15 minutes from both trips)
AB - 2.75 hours
BA - 2.25 hours
Speed = distance/time
we are given the average speed if 15 minutes were shaved off of each leg of the trip and the time for each leg of the trip. With this information we can solve for d.
60 = 2d/(2.75 + 2.25)
60 = 2d/5
300 = 2d
150 = d
An alternative way of solving using d=r*t formula:
2d=s*t
2d=60*(2.75+2.25)
2d=60*(5)
2d=300
d=150
(A) 90
(B) 120
(C) 150
(D) 240
(E) 300