Journey 1:
Speed = 20km/hr = 1/3km/min
Time = 20 mins late
Journey 2:
Speed = 30km/hr = 1/2km/min
Time = 15 mins early
Journey 3:
Speed = 25km/hr = 5/12km/min
y = the distance to work
speed (s) = distance (d) / time (t)
If the distance to work was (1/3 x 20)km shorter for journey 1 and (1/2 x 15)km longer for journey 2 both journeys would take the same time. Therefore, using s=d/t:
t = (y - 1/3 x 20)/(1/3) = (y + 1/2 x 15)/(1/2)
=> The distance to work (y) = 35km.
=> Journey 1 took 35/(1/3) = 105 mins.
105 mins is 20 mins late therefore the journey should take 85 mins.
Journey 3 takes 35/(5/12) = 84 mins. Therefore he's 1 minute early for work.
Add a Comment or Suggest an Answer
Rate this puzzle
No votes yet β be the first!Score: 0
β€οΈ 0
π 0
π§ 0
π 0
π 0
Comments (1)
Journey 1:
Speed = 20km/hr = 1/3km/min
Time = 20 mins late
Journey 2:
Speed = 30km/hr = 1/2km/min
Time = 15 mins early
Journey 3:
Speed = 25km/hr = 5/12km/min
y = the distance to work
speed (s) = distance (d) / time (t)
If the distance to work was (1/3 x 20)km shorter for journey 1 and (1/2 x 15)km longer for journey 2 both journeys would take the same time. Therefore, using s=d/t:
t = (y - 1/3 x 20)/(1/3) = (y + 1/2 x 15)/(1/2)
=> The distance to work (y) = 35km.
=> Journey 1 took 35/(1/3) = 105 mins.
105 mins is 20 mins late therefore the journey should take 85 mins.
Journey 3 takes 35/(5/12) = 84 mins. Therefore he's 1 minute early for work.
Add a Comment or Suggest an Answer