BTGmoderatorRO wrote:Car X is 40 miles west of Car Y. Both cars are traveling east, and Car X is going 50% faster than Car Y. If both cars travel at a constant rate and it takes Car X 2 hours and 40 minutes to catch up to Car Y, how fast is Car Y going?
a) 18
b) 25
c) 30
d) 12
e) 40
OA is c
How can I set up the formula to use here? can any expert help me?
Thanks
Hello.
This is how I would do it:
Since Car X is going 50% faster than Car Y, then if the speed of the Car Y is "y" then the speed of the Car X is "1.5y". The relative speed is $$1.5y\ -\ y\ =\ 0.5y.$$
Now, the distance between Car X and Car Y is 40 miles. The time that took Car X to reach Car Y was 2 hours and 40 minutes; we know must convert it to hours: $$160\min\ =160\ \min\cdot\frac{1\ hour}{60\ \min}=\frac{8}{3}hours.$$
Now, using that d=v*t we get: $$40=0.5y\cdot\frac{8}{3}\ \Rightarrow\ \ 40=\frac{4}{3}y\ \Rightarrow\ \ y=\frac{120}{4}=30\ mph.$$ Therefore, the answer is the option
C.
I hope it helps.