Jim can travel the 9.5 km from downtown to his home either by riding for 30 min and walking for 20 min, or by riding 10 min and walking for 1hr 10min. Find his average rates of riding and walking.

Answer 1

speed x time = distance

Let "r" be the speed of riding and let "w" be the speed of walking.

I would like to use speed in units of km/hr so I'm going to convert the times into hours. 30 minutes is 1/2 hr, 20 minutes is 1/3 hr, 10 minutes is 1/6 hr and 1hr 10 minutes is 7/6 hr.

So, for "riding for 30 min and walking for 20 min" we can say:

(A) 1/2r + 1/3w = 9.5

For "riding 10 min and walking for 1hr 10 min" we can say:

(B) 1/6r + 7/6w = 9.5

Solving equation (A) for r, we get:

r = 2(9.5 - 1/3w) or

r = 19 - 2/3w

Substituting that into equation (B) we get:

(1/6)(19 - 2/3w) + 7/6w = 9.5

19/6 - 2/18w + 7/6w = 9.5

-2/18w + 7/6w = 19/2 - 19/6

-2/18w + 21/18w = 57/6 - 19/6

19/18w = 38/6

w = (38/6)(18/19) [cancelling 6 and 18 we get:

w = (38/1)(3/19) [cancelling 38 and 19 we get:

w = (2/1)(3/1)

w = 6

Since r = 19 - 2/3w, then r = 19 - 2/3(6) = 19 - 4 = 15

So Jim's average walking rate is 6 km/hr and his average riding rate is 15 km/hr