arora007 wrote:Bonny left Philadelphia at 10am Monday and headed for Los Angeles by car. At 3pm on the same day, Clyde left and followed the same route. If they each traveled at a constant speed, when did Clyde catch up with Bonny?
1)Clyde's car traveled 25% faster than Bonny's car
2)When he caught Bonny,Clyde had spent 20% less time traveling than Bonny had.

B left at 10 AM -- from philadelphia
C left at 3 PM
Assuming that he also started form Philadelphia:

1)Clyde's car traveled 25% faster than Bonny's car
if B's speed is S, then C's speed is 1.25S.
Assume they meet after T hours that C starts.
Meanwhile B covered a distance of S*5 = 5S

in T hours, B traveled a total of 5S + ST
in T hours, C travles 1.25ST
Therefore 5S+ST = 1.25ST
5+T = 1.25T
T=20
So they will meet after 20 hours.

2)When he caught Bonny,Clyde had spent 20% less time traveling than Bonny had
Assume they meet after T hours after C starts.
Total time B spent = 5 + T hours.
Total time C spent = T hours.

T = 4/5(T+5)
T=20 hours.

both are sufficient.
pick D.

OA is D.