Problem Solving

Greetings,

This is a Manhattan CAT question.

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)

Following is my method of solving:
Let the distance covered by fast train in t hrs be a.
So the distance covered by regular train in t hrs will be z-a.
No since both the distances are covered in t hrs
a/x = (z-a)/y
=> a = xz/(x+y)
z-a = yz/(x+y)

Now we assume that in this case a>z-a
So the difference in the two distances will be z(x-y)/(x+y). So I get the answer as B.

However following is the solution:
Since the trains traveled the z miles in x and y hours, their speeds can be represented as z/x and z/y respectively.

We can again use an RTD chart to evaluate how far each train travels when they move toward each other starting at opposite ends. Instead of using another variable d here, let's express the two distances in terms of their respective rates and times.

High-speed | Regular | Total
R z/x | z/y |
T t | t |
D zt/x | zt/y | z


Since the two distances sum to the total when the two trains meet, we can set up the following equation:

zt/x + zt/y = z divide both sides of the equation by z
t/x + t/y = 1 multiply both sides of the equation by xy
ty + tx = xy factor out a t on the left side
t(x + y) = xy divide both sides by x + y
t = xy /(x + y)

To find how much further the high-speed train went in this time:
(ratehigh × time) - (ratereg × time)
(ratehigh - ratereg) × time
(z/x - z/y) * xy/(x + y)
(zy - zx)/xy * xy/(x + y)

z(y - x)/(x + y)
The correct answer is A.
May someone (experts) throw light on the confusion?

Thanks and Regards
Prashant

Prashant Ranjan 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 both of the two individual rates and of 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 we 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.

As for your approach:

Let the distance covered by fast train in t hrs be a.
So the distance covered by regular train in t hrs will be z-a.

With your reasoning:
a = the distance traveled by the fast train WHEN THE TWO TRAINS MEET.
z-a = the distance traveled by the regular train WHEN THE TWO TRAINS MEET.

Now since both the distances are covered in t hrs
a/x = (z-a)/y

According to the problem:
x = the time for the fast train to travel the ENTIRE Z MILES.
y = the time for the regular train to travel the ENTIRE Z MILES.

Thus, the expression above implies the following:
(distance traveled by the fast train WHEN THE TWO TRAINS MEET)/(time for the fast train to travel the ENTIRE Z MILES) = (distance traveled by the regular train WHEN THE TWO TRAINS MEET)/(time for the regular train to travel the ENTIRE Z MILES).

This relationship is not valid.

Thanks Mitch. The solution was very helpful. I realized the mistake i was doing.

Thanks Anurag for the tip.

perfect way of explaining !!!

GMATGuruNY wrote:

Prashant Ranjan 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 both of the two individual rates and of 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 we 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.

As for your approach:

Let the distance covered by fast train in t hrs be a.
So the distance covered by regular train in t hrs will be z-a.

With your reasoning:
a = the distance traveled by the fast train WHEN THE TWO TRAINS MEET.
z-a = the distance traveled by the regular train WHEN THE TWO TRAINS MEET.

Now since both the distances are covered in t hrs
a/x = (z-a)/y

According to the problem:
x = the time for the fast train to travel the ENTIRE Z MILES.
y = the time for the regular train to travel the ENTIRE Z MILES.

Thus, the expression above implies the following:
(distance traveled by the fast train WHEN THE TWO TRAINS MEET)/(time for the fast train to travel the ENTIRE Z MILES) = (distance traveled by the regular train WHEN THE TWO TRAINS MEET)/(time for the regular train to travel the ENTIRE Z MILES).

This relationship is not valid.