Question

Atrain travels 360km with uniform speed. The speed of the train is increased by 5km/hr, it takes 48 minutes less to cover the same distance, Find the initial speed of the train.

Answer 1

Step-by-step explanation:

360/5

= 72

72 - 48

= 24 answer

Answer 2

SOLUTION:-

•Total distance travelled by the train =360km/h

•Let the speed of the train be x km/h

Time taken by the train to travel 360 km=$\sf \dfrac{360}{x} hr$

•Now ,speed of the train =(x+5)km/h

Time taken by the train to travel the same distance =$\sf \dfrac{360}{x}hr$

By the given condition,we have

$\longrightarrow \sf \dfrac{360}{x} - \dfrac{360}{x + 5} = \dfrac{48}{60}$

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$\sf \implies \dfrac{360(x + 5) - 360x}{x(x + 5)} = \dfrac{4}{5}$

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$\sf \implies \dfrac{360x + 1800 - 360x}{x {}^{2} + 5x} = \dfrac{4}{5}$

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$\sf \implies \dfrac{1800}{ x{}^{2} + 5x } = \dfrac{4}{5}$

⠀⠀⠀⠀

$\sf \implies 9000 = 4 x{}^{2} + 20$

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$\sf \implies 4x {}^{2} + 20x - 9000 = 0$

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$\sf \implies x {}^{2} + 5x - 2250 = 0$

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$\sf \implies x {}^{2} + 50x - 45x - 2250 = 0$

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$\sf \implies x(x + 50) - 45(x + 50) = 0$

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$\sf \implies (x - 45)(x + 50) = 0$

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$\sf \implies x = 45 \: or \: x = - 60$

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But speed of train cannot be negative

Therefore,x=45

Hence ,original speed of the train is 45km/hr