A passenger train takes one hour less for a journey of 150 Km. if its speed is increased by 5 km /h from its original speed. find the original speed of train.
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Answered by
22
Heyy mate ❤✌✌❤
Here's your Answer....
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Let the usual speed of train be x km/hr
Then, increased speed of the train = (x + 5)km/hr
Time taken by the train under usual speed speed to cover 150 km = 150/x hr
Time taken by the train under increased speed to cover 150km = 150(x +5)hr
So,
150/x - 150/(x + 5) = 1.
=> [150(x+5) - 150x]/x(x+5) = 1
=> 150x + 750 - 150x / x^2 + 5x = 1.
=> 750 = x^2 + 5x
=> x^2 + 5x - 750 =0
=> x^2 +( -25 - 30)x - 750 = 0
=> x^2 - 25x + 30x - 750 =0
=> x(x - 25) +30 ( x - 25)=0
=> (x-25) (x+ 30)
=> x= 25 and x= -30( Neglect)
✔✔✔
Here's your Answer....
⤵️⤵️⤵️⤵️⤵️⤵️⤵️
Let the usual speed of train be x km/hr
Then, increased speed of the train = (x + 5)km/hr
Time taken by the train under usual speed speed to cover 150 km = 150/x hr
Time taken by the train under increased speed to cover 150km = 150(x +5)hr
So,
150/x - 150/(x + 5) = 1.
=> [150(x+5) - 150x]/x(x+5) = 1
=> 150x + 750 - 150x / x^2 + 5x = 1.
=> 750 = x^2 + 5x
=> x^2 + 5x - 750 =0
=> x^2 +( -25 - 30)x - 750 = 0
=> x^2 - 25x + 30x - 750 =0
=> x(x - 25) +30 ( x - 25)=0
=> (x-25) (x+ 30)
=> x= 25 and x= -30( Neglect)
✔✔✔
Answered by
0
Answer:
-30
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