Question

A train travel 360km with uniform speed. the speed of the train is increased by 5km/hr, it taken 48minutes less to cover the same distamce. find the initial speed if the train.
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Answer 1

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

$\sf \: 360 km=\sf \dfrac{360}{x} hr$

•Now ,speed of the train =(x+5)km/h

Time taken by the train to travel the same

$\pink{ \sf \: distance =\sf \dfrac{360}{x}hr}$

By the given condition,we have

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

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

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

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

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

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

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

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

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

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$\sf \implies (x - 45)(x + 50) =5$

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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