Question

i need ans for 2nd q plzz

Answer 1

Hey!!!

Good Afternoon

As promised I am here to help you

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let the original speed of the train be x km/h

Given distance = 360km

Then we know

=> time = distance/speed

ATQ,

$\frac{360}{x} - \frac{360}{x + 5} = \frac{48}{60}$

$360 \frac{(x + 5) - x}{x(x + 5)} = \frac{4}{5}$

$\frac{5}{ {x}^{2} + 5x } = \frac{1}{450}$

Cross Multiplying

=> x² + 5x = 2250

=> x² + 50x - 45x - 2250 = 0

=> x(x + 50) - 45(x + 50) = 0

=> (x - 45)(x + 50) = 0

Thus x = 45km/h (speed is positive)

Thus the original speed of the train is 45 km/h

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Hope this helps ✌️