Question

A train travelling at a uniform speed for 360 km would have taken 48 minutes less to travel the same distance if its speed was 5 km/ h more. Find the original speed of train.

Answer 1

Let the speed of train be x km/hr
Distance to be travelled  = 360km
We know that,
 begin mathsize 12px style s p e e d space equals space fraction numerator d i s tan c e space over denominator t i m e end fraction therefore space t i m e space equals space fraction numerator d i s tan c e over denominator t i m e end fraction end style
Time take by the train intially = begin mathsize 12px style 360 over x end style
If the speed was increased by 5km/hr
Time taken by train = begin mathsize 12px style fraction numerator 360 over denominator x plus 5 end fraction end style
Difference in the time taken is 48 minutes,
begin mathsize 12px style fraction numerator 360 over denominator x space end fraction minus space fraction numerator 360 over denominator x plus 5 end fraction space equals space 48 over 60 space space space space space left parenthesis c o n v e r t i n g space 48 m i n s space t o space h o u r s right parenthesis rightwards double arrow 360 open square brackets fraction numerator x plus 5 minus x over denominator x left parenthesis x plus 5 right parenthesis end fraction close square brackets equals 4 over 5 rightwards double arrow 90 open square brackets fraction numerator 5 over denominator x left parenthesis x plus 5 right parenthesis end fraction close square brackets equals 1 fifth rightwards double arrow 90 left square bracket 5 right square bracket 5 equals x squared plus 5 x end style
x2+5x -2250 = 0
(x+50)(x-45)
Speed = 45 km/hr as speed cannot be negative

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.