A man travels 300 km partly by train and partly by car. He takes 4 hrs. If he travels 60kms by train and rest by car. If he travels 100 km by train and tge remaining by car he takes 10 mins longer . Find the speed of train and car
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hi there !!!
here is ur answer
Total distance travelled by man=300
Let speed of train be x
Let speed of can be y
Travelling 60km by train and rest by car he took 4 hours to cover 300km
Travelling 100km by train and rest by car he took 4 hours and 10 minutes.
Putting this relations into eqn, we get,
60/x+240/y=300
100/x+200/y=300
Rearranging the above eqns
60y+240x=300xy→(1)
100y+200x=300xy→(2)
Divide first eqn by 60
y+4x=5xy→(3)
Divide second by 50
2y+4x=6xy→(4)
Subtracting (3) from (4)
2y+4x=6xy
-y-4x=-5xy
Gives x=1km/min
=60km/hr
here is ur answer
Total distance travelled by man=300
Let speed of train be x
Let speed of can be y
Travelling 60km by train and rest by car he took 4 hours to cover 300km
Travelling 100km by train and rest by car he took 4 hours and 10 minutes.
Putting this relations into eqn, we get,
60/x+240/y=300
100/x+200/y=300
Rearranging the above eqns
60y+240x=300xy→(1)
100y+200x=300xy→(2)
Divide first eqn by 60
y+4x=5xy→(3)
Divide second by 50
2y+4x=6xy→(4)
Subtracting (3) from (4)
2y+4x=6xy
-y-4x=-5xy
Gives x=1km/min
=60km/hr
Answered by
0
Let the speed of train and bus be u km/h and v km/h respectively.
According to the question,
....(i)
....(ii)
Let
The given equations reduce to:
60p + 240q = 4 ....(iii)
100p + 200q =
600p + 1200q = 25....(iv)
Multiplying equation (iii) by 10, we obtain:
600p + 2400q = 40....(v)
Subtracting equation (iv) from equation (v), we obtain:
1200q = 15
q =
Substituting the value of q in equation (iii), we obtain:
60p + 3 = 4
60p = 1
p =
:. p = , q =
u = 60 km/h , v = 80 km/h
Thus, the speed of train and the speed of bus are 60 km/h and 80 km/h respectively.
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