Problem Solving

BTGmoderatorDC 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

OA A

Source: Manhattan Prep

Laborious algebraic approach.

T= time each travels until they meet. Faster train rate of speed is (z/x) and slower speed's (z/y).

So distance high speed travels is (z/x)*T and distance slower train travels is (z/y)*T. The sum of these two distances is Z:

so TZ(1/x + 1/Y) = Z

so T = (1/(1/x)+(1/y)) = xy/(x+y) . This is the amount of time they each travel.

Distance the high speed train travels in T is T(z/x) and distance slower train travels is T(z/y)

The difference in their distances is TZ(1/x - 1/y) = TZ((y-x)/xy)

Substituting for T: z( xy/(x+y)) ((y-x)/xy)
= [spoiler] z(y-x)/(x+y), A[/spoiler]

BTGmoderatorDC 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

Plug in easy numbers for the two rates.
Then plug in a distance that is a multiple of both the two individual rates and the COMBINED rate.

Let the high speed rate = 3 miles per hour.
Let the regular rate = 2 miles per hour.
Combined rate for the trains = 2+3 = 5 miles per hour.
Let z = 30 miles.
Time for the high speed train to travel 30 miles = x = 30/3 = 10 hours.
Time for the regular train to travel 30 miles = y = 30/2 = 15 hours.
Time for the trains to meet = 30/5 = 6 hours.
Distance traveled by the high speed train in 6 hours = r*t = 3*6 = 18 miles.
Distance traveled by the regular train in 6 hours = r*t = 2*6 = 12 miles.
Distance for the high speed train - distance for the regular train = 18-12 = 6. This is our target.

Now plug x=10, y=15, and z=30 into the answers to see which yields our target of 6.

Only answer choice A works:
z(y - x)/(x+y) = 30*(15-10)/(10+15) = 6.

The correct answer is A.

Algebra:

Rule:
If A takes x hours to do a job, and B takes y hours to a job, the combined rate for A and B working together = (x+y)/(xy), and the total time for A and B working together = (xy)/(x+y).

In the problem above:
When the two trains travel toward each other, they WORK TOGETHER to cover the z miles between them.
Since the time for the high-speed train = x, and the time for the regular train = y, the total time for the two trains working together = (xy)/(x+y).
Thus:
Since the rate for the high-speed train = d/t = z/x, the distance traveled by the high-speed train = rt = (z/x) * (xy)/(x+y) = (zy)/(x+y).
Since the rate for the regular train = d/t = z/y, the distance traveled by the regular train = rt = (z/y) * (xy)/(x+y) = (zx)/(x+y).
Difference between the distances = (zy)/(x+y) - (zx)/(x+y) = [(z)(y-x)]/(x+y).

The correct answer is A.