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
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
