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Question 32114: (Sir i am in confussion with this question & i need an answer fastly as possible) Q. A person travels 600 km to his home , partly by train and partly by car . He takes 8 hours if he travels 120km by train abd the rest by car . He takes 20 minutes longer if he travels 200 km by train and the rest by car . Find the speed of the train and the car ?
Found 2 solutions by mukhopadhyay, stanbon:
Answer by mukhopadhyay(490)   (Show Source):
You can put this solution on YOUR website! Let the spped of the train be x km/hr and that of the car is y km/hr
Time taken to travel 120 km by train and 480 km by car is 120/x + 480/y
Per question, 120/x + 480/y = 8
=>15/x + 60/y=1
=>15y+60x=xy....... (Eq 1)
Time taken to travel 200 km by train and 400 km by car is 200/x + 400/y
Per question, 200/x + 400/y = 8 1/3=25/3
=> 8/x + 16/y = 1/3
=> 24y+48x=xy....... (Eq 2)
From both equations:
15y+60x = 24y+48x
=>9y=12x
=>y=4x/3
Substituting this in (Eq 1)
15y+60x=xy
=>20x+60x=4x^2/3
=> 80x=4x^2/3
=>x^2-60x=0
=>x=60 (because speed Cannot be negative)
y=80
The speed of the train is 60 km/hr and the speed of the car is 80 km/hr
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! On the 1st 600km trip:
Train Data: distance=120km, time=x hrs. rate=120/x
Car Data: distance=480km, time=8-x hrs. rate=480/(8-x)
On the 2nd 600km trip:
Train Data: distance=200km, time=y hrs, rate=200/y
Car Data: distance=400km, time=25/3-y, rate= 400/[(25/3)-y]
Equations:
Train Rate on 1st trip = Train rate on 2nd trip
120/x=200/y
Then x=(3/5)y
Car Rate on 1st trip = Car Rate on 2nd trip
480/(8-x)=400/(25/3)-y]
Substitute x=(3/5)y into this second equation
and reduce the numerators, to get:
6/(8-x)=5/[(25/3)-y]
6[25/3 - y] = 5[8-3y/5)]
50-6y=40-3y
10=3y
y=(10/3) hrs (time for car on 2nd 200km trip)
Then x=(3/5)(10/3)= 2 hrs (time for train on 1st 600km trip)
Speed of train = 120/x=120/2= 60km/hr
Speed of car = 480/(8-x)= 480/(8-2)=80km/hr
Cheers,
Stan H.
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