a train takes one hour less for a journey of 150 km if its speed increased by 5 km/hr from its usal speed find speed of train
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Heyaa user!!!
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Here's your answer :-
Let the usual speed of the train be x kmph....
Distance =150km.....
Then,
Time taken to cover the distance with this sped =(Distance /speed)
=(150/x)hr...........
Now, Increased speed =(x+5)kmph...
Time taken to cover 150km with this speed =(150/x+5)hr...
Acc. To question,
150/x-150/(x+5)=1
=>150(x+5)-150x/x(x+5)=1
=>150x+750-150x/x²+5x=1
=>x²+5x=750
=>x²+5x-750=0
=>x²+30x-25x-750=0
=>x(x+30)-25(x+30)=0
=>(x+30)(x-25)=0
=>x=25,-30
Hence speed can't be in negative, we take the positive value of x....
Therefore usual speed of the train =25kmph.....
Hope this helps!!!
Good luck :-)
-------------------------------------------------------
Here's your answer :-
Let the usual speed of the train be x kmph....
Distance =150km.....
Then,
Time taken to cover the distance with this sped =(Distance /speed)
=(150/x)hr...........
Now, Increased speed =(x+5)kmph...
Time taken to cover 150km with this speed =(150/x+5)hr...
Acc. To question,
150/x-150/(x+5)=1
=>150(x+5)-150x/x(x+5)=1
=>150x+750-150x/x²+5x=1
=>x²+5x=750
=>x²+5x-750=0
=>x²+30x-25x-750=0
=>x(x+30)-25(x+30)=0
=>(x+30)(x-25)=0
=>x=25,-30
Hence speed can't be in negative, we take the positive value of x....
Therefore usual speed of the train =25kmph.....
Hope this helps!!!
Good luck :-)
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