Question

A train covers a distance of 360 km at a uniform speed. Had the speed been 5km/hour
more, it would have taken 48 minutes less for the journey. Find the original speed of the
train.​

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 = 45km/hr

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Answer 2

Answer:

let the orginal speed be x km/h

there for increased speed= (5+x) km / h

thn the time taken in increased speed be 48min = 48/60

then the original time = 360/x

there for the equation formed is

360/X - 360/5+X = 48/60

after taking the LCM

we get

1800+360x-360x/5x+5x^2=4/5

4(5x^2+5x)=1800x5

divide both the sides by 4

5x^2+5x=2250

5x^2+5x-2250=0

divide by 5

x^2+x-450

then solve using quadratic equton

x=5m/s (approx)

Step-by-step explanation: