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Bill spends two days driving from point a to point

b. on the first day, he drove 2 hours longer and at an average speed 5 miles per hour faster than he drove on the second day. if during the two days he drove a total of 680 miles over the course of 18 hours, what was his average speed on the second day, in miles per hour?

1 Answer

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Answer: his average speed on the second day is 35 miles per hour

Explanation:

Bill spends two days driving from point a to point b

Let x = his speed on the first day.

Let y = his speed on the second day

Speed = distance / time

Time = distance / speed

Total time spent on the trip is 18 hours

Let time that he spent on the first day be t hours

Time that he spent on the second day will be 18 - t

Bill drove 2 hours longer on the second day than on the first day. This means

t = 18 - t + 2

t+t = 18+2

2t = 20

t = 10

He spent 10 hours on the first day.

He spent 18-10 = 8hours on the second day.

Bill drove at an average speed of 5 miles per hour faster than he drove on the second day. This means that

x = y + 5 - - - - - - - -- 1

Distance travelled on the first day = speed on the first day × time spent. This becomes

10 × x = 10x miles

Distance travelled on the second day = speed on the second day × time spent. This becomes

8 × y = 8y miles

He drove a total of 680 miles over the course of the 18 hours. This means that

10x + 8y = 680 - - - - - - - - 2

Substituting equation 1 into equation 2, it becomes

10(y + 5) + 8y = 680

10y + 50 + 8y = 680

10y + 8y = 680 - 50

18y = 630

y = 630/18 = 35 miles per hour

x = y + 5 = 35 + 5

x = 40 miles per hour

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