A train traveling at a uniform speed for 360km would have taken 48 min less to travel same distance if speed was 5km/hr more. Find original speed
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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
Hope This helps :)
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
Hope This helps :)
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