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A train covered A certain distance at a uniform speed if the train would have been 10 km/h it would have taken 2hrs less than the scheduled time and if the train were slower by 10 km/h it would have taken 3 hours more than the scheduled time find the distance covered by the train ?
claßẞ => 10th
chäp => Pair of linear equations into two variables .
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speed = distance /time
distance = speed•time
distance = 50kmph • 12 hours
distance = 600Km
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Let speed of train be x km/hr
Time taken = y hrs
We know that,
Distance = Speed × Time
So, Distance = xy....... (1)
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If the train had been 10 km/hr faster :-
Speed = (x + 10) km/hr
Time = (y - 2) [As it would take 2 hours less time]
So, Distance = (x +10)(y - 2)
Putting this from equation 1 :-
=> xy = x (y - 2) + 10 (y - 2)
=> xy = xy - 2x + 10y - 20
=> xy - xy = 2x - 10y + 20
=> 2x - 10y + 20 = 0...... (2)
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If the speed had been 10 km/hr slower,
Speed = (x - 10) km/hr
Time = (y + 3) hrs [As it is given that the time will increase by 3 hrs]
So, Distance = (x - 10)(y +3)
Putting this from equation 1:-
xy = (x - 10)(y +3)
=> xy = x (y +3)- 10(y +3)
=> xy = xy + 3x- 10y - 30
=> xy - xy = 3x - 10y - 30
=> 3x - 10y - 30 = 0....... (3)
Now,
From (2),
Taking 2 common,
2 ( x - 5y +10)= 0
=> x = 5y - 10........ (4)
Putting (4) in (3) :-
3 ( 5y - 10)10y - 30 = 0
=>15 y - 30 - 10 y - 30 = 0
=> 5y - 60 =0
=> y = 12
Putting y = 12 in (4)
X = 5 × 12 - 10 = 50
Hence, the answer will be :-
SPEED OF TRAIN = 50 km/hr
TIME TAKEN = 12 hrs
So, DISTANCE = 50 × 12
=
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