Question

A van left Town A for Town B at 07 42 and arrived at 08 54.In the afternoon, the same van traveled back to Town A at 15 33. However, its average speed decreased by 11 km/h.If the van's initial average speed was 35 km/h, what time did the van arrive back at Town A?

Answer 1

The van arrived back at Town A at 17:18 (5:18 PM)

Explanation:

Let x be the distance between Towns A and B.

Lets calculate the distance travelled from the first trip to Town B. We have the time travelled and the van's intital speed of 35 km/h. We know that:

Distance(or x)= (Speed)*(Time)

Speed is in units of km/hour, so time must also be in hours. The time between 7:42 and 8:54 is 72 minutes. Convert that to hours using the conversion factor (1 h/60 min)

(72 min)*(1 h/60 min) = 1.2 hour

x = (Speed)*(Time)

x = (35 km/h)*(1.2 h)

x = 42 km [The distance between Towns A and B]

Returning to Town A, the van's speed decreased by 11 km/h. making the speed 24 km/h for the return trip. Use the same calculation as above to calcuate the time for travel on the return from B to A:

Distance = (Speed)*(Time)

We now know the disance, 42 km, and the speed, 24 km/h, so we can solve for time:

Time = (Distance)/(Speed)

Time = (42 km)/(24 km/h)

Time = 1.75 hours or 1 hr and (0.75)*(60 min)

= 1 hr and 45 min

The van left at 15:33 or 15 + 33 minutes

Add the return travel time to the leaving time:

15 hr + 33 minutes

+ 1 hr + 45 minutes

16 hr + 78 minutes

or 17 hr + 18 minutes

The van will arrive at 17:18 (5:18 PM)