SOLUTION: Bike Rides - Jo and Chris set out together. they rode their bikes from a to b, from b to c and from c back to a. A to B = 4 miles, B to C = 4 miles, C to A = 4 miles. jo rod

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: Bike Rides - Jo and Chris set out together. they rode their bikes from a to b, from b to c and from c back to a. A to B = 4 miles, B to C = 4 miles, C to A = 4 miles. jo rod      Log On

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Question 58898: Bike Rides - Jo and Chris set out together. they rode their bikes from a to b, from b to c and from c back to a. A to B = 4 miles, B to C = 4 miles, C to A = 4 miles.
jo rode at a constant speed all the way.
Chris rode from a to b at double jo's speed.
She rode from b to c at the same speed as jo
She rode from c back to a at half jo's speed.
Who got back to a first?
How far a head was she?

Answer by uma(370) About Me  (Show Source):
You can put this solution on YOUR website!
Let Jo's speed be xmph
Total distance travelled by Jo = 12 miles.
So time taken by Jo = 12/x hrs
Speed of Chris for travelling from A to B = 2x mph
A to B = 4 miles
So time taken = 4/2x
= 2/x hrs
Speed of Chris for travelling from B to C = x mph
B to C = 4 miles
So time taken = 4/x
= 4/x hrs
Speed of Chris for travelling from C to A = x/2 mph
C to A = 4 miles
So time taken = 4/(x/2)
= 8/x hrs
So total time taken = 2/x + 4/x + 8/x
= 14/x hrs
Comparing , we find that Jo reached A first as the time taken by her was only 12/x hrs
Distance covered by Chris in 12/x hrs = (12/x)/ (14/x) * 12
= 10.3 miles
So Jo is ahead of Chris by 12 - 10.3 = 1.7 miles