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
x, y, and z? No thank you!!! Let's Plug In.
High-speed travels 4 mph. Regular travels 2mph. Total distance is 6 miles.
If they both start at the same time heading toward each other, in one hour, high-speed will have gone 4 miles and regular will have gone 2 miles, so they meet.
High-speed has traveled 2 miles farther than regular.
High-speed can cover the 6 miles in 6/4 hours, so x = 1.5.
Regular can cover the 6 miles in 6/2 hours, so y = 3.
z = 6
(A) z(y – x)/x + y = \(\frac{6(3-1.5)}{1.5+3} = \frac{6(1.5)}{4.5} = 1 \)Keep it.
(B) z(x – y)/x + y = Different numerator from (A) but same denominator, so it can't also be right. Eliminate.
(C) z(x + y)/y – x = \(\frac{6(3+1.5)}{3-1.5} = \frac{6(4.5)}{1.5}\) Eliminate.
(D) xy(x – y)/x + y = \(\frac{(1.5)(3)(1.5-3)}{1.5+3}\) Numerator is negative and denominator is positive. Eliminate.
(E) xy(y – x)/x + y = \(\frac{(1.5)(3)(3-1.5)}{1.5+3} = \frac{4.5*1.5}{4.5} = 1.5\) Eliminate.
Answer choice A.
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