If Gideon had driven for 20 minutes less than the time he would have driven if he had driven 20 kilometers less than he did drive but at two-thirds the speed at which he drove, he would have driven 10 kilometers less than he did.

However, if he had driven 20 minutes longer than the time he would have driven if he had driven 10 kilometers less than he did drive but at three-quarters the speed at which he drove, he would have driven 20 kilometers further.

How far did Gideon drive?

The basic equation here is d = vt  (distance = time * velocity).

d = distance traveled (km)
v = velocity traveled (km/min)
t = duration of travel (min)
t1 = duration of travel in first hypothetical (min)
t2 = duration of travel in second hypothetical (min)

----------------------------------------------------------

For the first hypothetical, two equations can be derived:

t1 = [3(d-60)/2v]-20
t1 = (d-10)/v

Setting them equal to each other and reducing yields:

v = (d-40)/40

----------------------------------------------------------

For the second hypothetical, two equations can be derived:

t2 = [4(d-10)/3v]+20
t1 = (d+20)/v

Setting these equal to each other and reducing yields:

v = (100-d)/60

------------------------------------------------------------

Initial inspection shows that 60<d<100.  Setting these two equations for v equal to each other leaves:

(d-40)/40 = (100-d)/60

----------------------------

Solution:  d = 64 km

Posted by hoodat on 2007-05-15 13:47:32