A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Form the quadratic equation to find the speed of the train.
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SOLUTION :
Given : Distance travel by train = 360 km
Let the usual speed of the train = x km/h.
Time taken to cover a distance of 360 km = 360/x hours.
[ Time = Distance /speed]
If the speed is increased by 5 km/h then the New speed of the train = (x + 5) km/h.
Time taken to cover a distance of 360 km at new speed = [360/(x+5)] h
A.T Q
360/(x + 5 ) = (360/x) - 1
360/x - 360/ (x+5) = 1
[360 (x + 5 ) - 360 (x)] / x(x + 5) = 1
360x + 1800 - 360 x = 1 × (x² + 5x)
1800 = x² + 5x
x² + 5x = 1800
x² + 5x - 1800 = 0
Hence, the required quadratic equation is x² + 5x - 1800 = 0 .
By factorisation :
⇒x² + 5x - 1800 = 0
⇒x² + 45x - 40x - 1800 = 0
⇒x (x+ 45) - 40( x + 45) = 0
⇒(x+ 45) (x- 40) = 0
⇒x = - 45 or x = 40
But speed cannot be in negative. x ≠ -45
∴ x = 40 km/hr.
Hence, the usual speed of the train is 40 km/h.
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Answered by
5
Answer is given below!
Given : Distance travel by train = 360 km
Let the usual speed of the train = x km/h.
Time taken to cover a distance of 360 km = 360/x hours.
[ Time = Distance /speed]
If the speed is increased by 5 km/h then the New speed of the train = (x + 5) km/h.
Time taken to cover a distance of 360 km at new speed = [360/(x+5)] h
A.T Q
360/(x + 5 ) = (360/x) - 1
360/x - 360/ (x+5) = 1
[360 (x + 5 ) - 360 (x)] / x(x + 5) = 1
360x + 1800 - 360 x = 1 × (x² + 5x)
1800 = x² + 5x
x² + 5x = 1800
x² + 5x - 1800 = 0
Hence, the required quadratic equation is x² + 5x - 1800 = 0 .
By factorisation :
⇒x² + 5x - 1800 = 0
⇒x² + 45x - 40x - 1800 = 0
⇒x (x+ 45) - 40( x + 45) = 0
⇒(x+ 45) (x- 40) = 0
⇒x = - 45 or x = 40
But speed cannot be in negative. x ≠ -45
∴ x = 40 km/hr.
Hence, the usual speed of the train is 40 km/h.
Given : Distance travel by train = 360 km
Let the usual speed of the train = x km/h.
Time taken to cover a distance of 360 km = 360/x hours.
[ Time = Distance /speed]
If the speed is increased by 5 km/h then the New speed of the train = (x + 5) km/h.
Time taken to cover a distance of 360 km at new speed = [360/(x+5)] h
A.T Q
360/(x + 5 ) = (360/x) - 1
360/x - 360/ (x+5) = 1
[360 (x + 5 ) - 360 (x)] / x(x + 5) = 1
360x + 1800 - 360 x = 1 × (x² + 5x)
1800 = x² + 5x
x² + 5x = 1800
x² + 5x - 1800 = 0
Hence, the required quadratic equation is x² + 5x - 1800 = 0 .
By factorisation :
⇒x² + 5x - 1800 = 0
⇒x² + 45x - 40x - 1800 = 0
⇒x (x+ 45) - 40( x + 45) = 0
⇒(x+ 45) (x- 40) = 0
⇒x = - 45 or x = 40
But speed cannot be in negative. x ≠ -45
∴ x = 40 km/hr.
Hence, the usual speed of the train is 40 km/h.
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