SOLUTION: I am not sure how to set this up. An express and a local train leave Seattle at 3pm and head for Arlington 50 miles away. The express travels twice as fast as the local and arrives
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Question 29463: I am not sure how to set this up. An express and a local train leave Seattle at 3pm and head for Arlington 50 miles away. The express travels twice as fast as the local and arrives 1 hour ahead of the local. Find the speed of the train.
I have tried this several ways.... trying to use elimination, or substitution. I just can't get it. Help!
50 = x + y
50 = 2y + x Is this even close? Thanks for your help!
Found 2 solutions by longjonsilver, venugopalramana:
Answer by longjonsilver(2297) (Show Source): You can put this solution on YOUR website!
DEFINITIONS
Let x = speed of the local train
Let 2x = speed of the Express train.
Let t = time taken for the local train to travel.
REQUIRED INFO
Formula: speed = dist/time
distance = 50
CREATING THE MATHS
For Express train: 2x = 50/(t-1)
--> t-1 = 50/2x
--> t = (50/2x) + 1
For Local train: x = 50/t
--> t = 50/x
SOLUTION
Equate both together for t, giving
(50/2x) + 1 = 50/x
25/x + 1 = 50/x
multiply everything by x... this doesn't change anything, just "scales everything". Reason... to cancel the fractions out.
--> 25 + x = 50
x = 25 mph
and so we have:
Speed of Local train = 25mph
Speed of Express train = 50mph
CHECK
Express train would take 1 hour to travel 50 miles at 50mph.
Local train would take 2 hours to travel 50 miles at 25mph.
jon.
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
SEE BELOW MY COMMENTS......
I am not sure how to set this up. An express and a local train leave Seattle at 3pm and head for Arlington 50 miles away. The express travels twice as fast as the local and arrives 1 hour ahead of the local. Find the speed of the train.
I have tried this several ways.... trying to use elimination, or substitution. I just can't get it. Help!
50 = x + y......FIRST ASSUME LOCAL TRIN SPEED =X.MPH..HENCE EXPRESS SPEED IS =2X..MPH
DISTANCE TO BE COVERED =50 MILES....
TIME TAKEN BY LOCAL =DISTANCE/SPEED = 50/X................................I
TIME TAKEN BY EXPRESS = 50/2X =25/X....................................II
DIFFERENCE = (50/X) - (25/X) = (50-25)/X = 25/X....BUT THIS IS 1 HOUR..SO
25/X=1...OR ......X=25 MPH
SO SPEED OF LOCAL = 25 MPH
SPEED OF EXPRESS = 2*25 = 50 MPH...
50 = 2y + x Is this even close? Thanks for your help!
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