Question

A train travelling at a uniform speed for 360 km would have taken 48mins less to travel the same distance if it's speed were 5km more find original speed

Answer 1

let original speed of train = x km/h

we know,

time = distance/speed

first case

———————

time taken by train = 360/x hour

second case

——————————

time taken by train its speed increase 5 km/h = 360/( x + 5)

question says that

time taken by train in first - time taken by train in 2nd case = 48 min = 48/60 hour

360/x - 360/(x +5) = 48/60 = 4/5

360{ 1/x - 1/(x +5) } = 4/5

360 ×5/4 { 5/(x² +5x )} =1

450 x 5 = x² + 5x

x² +5x -2250 = 0

x = { -5±√(25+9000) }/2

=(-5 ±√(9025) )/2

=(-5 ± 95)/2

= -50 , 45

but x ≠ -50 because speed doesn't negative

so, x = 45 km/h

hence, original speed of train = 45 km/h

pls mark it as brainliest

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.