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test series

by thephoenix » Sat Feb 20, 2010 12:36 am
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?

z(y - x)/x + y

z(x - y)/x + y

z(x + y)/y - x

xy(x - y)/x + y

xy(y - x)/x + y

my way of solution was lengthy , suggest some shorter approach
oa will follow soon
source MGMAT test series
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by shashank.ism » Sat Feb 20, 2010 12:41 am
thephoenix 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?

z(y - x)/x + y

z(x - y)/x + y

z(x + y)/y - x

xy(x - y)/x + y

xy(y - x)/x + y

my way of solution was lengthy , suggest some shorter approach
oa will follow soon
source MGMAT test series
phoenix please post ur solution too...so that we can think of shorter approach otherwise we might lead in the same direction as you..
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by bitsho » Sat Feb 20, 2010 12:50 am
OA is A ?? here is the approach ,which i followed ... assuming distance between town A and town B is 300 miles , the speed of high speed train 100 miles/hr and speed of regular train 50 miles/hrs .
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by bitsho » Sat Feb 20, 2010 12:51 am
OA is A ?? here is the approach ,which i followed ... assuming distance between town A and town B is 300 miles , the speed of high speed train 100 miles/hr and speed of regular train 50 miles/hrs .
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by vijay_venky » Sat Feb 20, 2010 1:54 am
The velocity of HST=z/x and of NT=z/y

Now because the trains are head-on the relative velocity=z(x+y)/(xy)

Now the time taken for them to meet=z/(relative velocity)=xy/(x+y)

So, in this time the extra distance traveled by HST = time(diff in velocity)=xy/(x+y)[z/x-z/y]

which is z[y-x]/[x+y]
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by shashank.ism » Sat Feb 20, 2010 8:13 am
thephoenix 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?
z(y - x)/x + y
z(x - y)/x + y
z(x + y)/y - x
xy(x - y)/x + y
xy(y - x)/x + y
my way of solution was lengthy , suggest some shorter approach
oa will follow soon
source MGMAT test series
Just go like this let p be distance covered by HST and z-p by RT when they cross each other so their time taken is equal
--> p/(z/x)= (z-p)/((z/y) --> px = yz-py --> p=yz/(x+y)
so extra distance covered = p-(z-p) = 2p-z = 2yz/(x+y) -z = [spoiler]z(y-x)/(x+y) Ans A[/spoiler]
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