joyseychow wrote:
It takes the high-speed train x hours to travel the z miles from Town A to Town B at a constant rate, while it takes the regular train y hours to travel the same distance at a constant rate. If the high-speed train leaves Town A for Town B at the same time that the regular train leaves Town B for Town A, how many more miles will the high-speed train have traveled than the regular train when the two trains pass each other?
(A) z(y – x)/x + y
(B) z(x – y)/x + y
(C) z(x + y)/y – x
(D) xy(x – y)/x + y
(E) xy(y – x)/x + y
Variables in the answer. Lets use plug in
z=100 miles
x=1 hr
y=2 hrs
high speed = 100 miles/hr
regular speed = 50 miles /hr
combined speed 150 miles/hr
time taken to meet= d/s =100/150 = 2/3 hrs
in 2/3 hrs, high speed would have traveled 66 2/3 miles
regular speed would have traveled 33 1/3 miles
difference is 33 1/3 miles
Choice A gives this value when plugged in with the above values
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Srinivasan Vaidyaraman
Magical LogiciansHolistic and Holy Approach