It takes the high-speed train x hours to travel the z miles

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.

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

We have a converging rate problem in which:

Distance(1) + Distance(2) = Total Distance

We are given that 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. Thus, we know the following:

rate of the high-speed train = z/x

rate of the regular speed = z/y

We are also given that they leave at the same time, so we can let the time when both trains pass each other be t. We can substitute our values into the total distance formula.

Distance(1) + Distance(2) = Total Distance

(z/x)t + (z/y)t = z

zt/x + zt/y = z

We can divide the entire equation by z and we have:

t/x + t/y = 1

Multiplying the entire equation by xy gives us:

ty + tx = xy

t(y + x) = xy

t = xy/(y + x)

Now we can calculate the distance traveled by both trains for time t, using the following formula:

distance = rate x time

distance of high-speed train = (z/x)[xy/(y + x)] = zy/(y + x)

distance of regular-speed train = (z/y)[xy/(y + x)] = zx/(y + x)

Now we need to calculate the difference between the distance of the high-speed train and the distance of the regular-speed train:

difference of distance traveled = distance of high-speed train - distance of regular-speed train

difference of distance traveled = zy/(y + x) - zx/(y + x)

difference of distance traveled = (zy - zx)/(y + x) = z(y - x)/(x + y)

Answer: A