Question

A train travellings at uniform speed for 360km would have taken 48 min less to travel the same distance if its speed were 5km/hr more. Find the original speed of the train

Answer 1

let original speed be x km/hr
distance = 360 km
time = 360/x

if speed would be 5 km/hr more than the original speed (x) than time taken would be 48 min less than original.....

time (t1) = (360/x)-48
speed (s1) = x+5


A/Q
x(360/x). =. (x+5){(360/x)-48}
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=>48x² + 288x - 1800 = 0

solve this equation..............
you will get your answer.......

Answer 2

Answer:

Step-by-step explanation:

Let initial speed of train be x km/hr.

Distance travelled = 360 km.

//we know that Speed = Distance/time => Time = Distance /Speed.

Time taken by train Initially = 360/x.

If speed is increased by 5 km/hr,

Time taken by train = 360/x+5.

Difference in time taken = 48/60 hr.

=> 360/x - 360/x+5 = 48/60

=> 360(1/x - 1/x+5) = 48/60

=> 360[x + 5 - x / x(x+5)] = 48/60

=> 360 * 5/x(x+5) = 48/60

=> x(x+5) = 360 * 5 * 60 / 48

=> x² + 5x =  2250

=> x² + 5x - 2250 = 0

=> x² + 50x - 45x - 2250 = 0

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

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

=> x = 45 or -50.

Since x cannot be negative, x= 45 km/hr.

Thus original speed of train = 45 km/hr.