College Algebra Problem...?

Question

Uniform Motion: Marlene rides her bicycle to Jon's house and returns home by the same route. Marlene rides her bike at constant speeds of 6 mph on level ground, 4 mph when going uphill, and 12 mph when going downhill. If her total time riding was 1 hour, how far is it to Jon's house?

From the book: College Algebra and Trigonometry Sixth Edition by Aufmann/Barker/Nation
ISBN 13#: 978-0-618-82517-2
ISBN 10#: 0-618-82517-7

(Problem #27, Page 97)

Answer 1

Suppose there are x miles of level ground and y miles of sloping ground.
The x miles of level ground are covered at 6mph each way, so going both ways occupy 2x/6 = x/3 hours.
The y miles of sloping ground are covered at 4mph one way and at 12mph the other way, so going both ways occupy y/4 + y/12 = y/3 hours.
So the total time is (x+y)/3 hours, and therefore the total distance is x+y = 3 miles.

Obviously you can't do this with arbitrary speeds, since in general the coefficients of x and y will be different and you couldn't find x + y.

If we let a and b be the speeds up and down hills, for a solution to be possible we need the speed on level ground to be 2ab / (a+b). In this case we have 2(4)(12) / (4+12) = 6 so we can solve the problem.

Answer 2

Let total distance be 2d, i.e. distance travelled while going and coming back.
Suppose she rode a distance of x miles on level ground and a distance (d - x) uphill while going. So, when coming back she rode (d - x) miles downhill and the same level ground of x miles.
Now time= distance/speed

Total time going=x/6 + (d-x)/4
Total time coming back=x/6 + (d-x)/12

=>x/6 + (d-x)/4 + x/6 + (d-x)/12=1

=>x/3 + (d-x)[1/4 + 1/12]=1

=>x/3 + (d-x)[4/12]=1

=>x/3 + d/3 - x/3=1

=>d/3=1

=>d=3 miles

Answer 3

d1=level ground
d2=uphill(while going)
d3=downhill(while going)
On her return journey d2 will be downhill and d3 will be uphill.
also distance/speed = time
time while going t1=d1/6 + d2/4 + d3/12
return time t2= d1/6 + d2/12 + d3/6
total time t=t1+t2 = 1
solving this equation we get
d1+d2+d3=3 =>total distance=3 miles :)