Question

if a cyclist travel at speed 2km/h more than his usual speed ,he reaches the destination 2 hours earlier. If the destination is 35 km away, what is the usual speed of the cyclist? ​

Answer 1

Step-by-step explanation:

Given : distance = 35 km

1st case

let the usual speed be X km/h

Time = Distance/time = 35/x km/h

2nd case

Speed = (x+2)km/h

Time = 35/(x+2) km/h

according to the question,

35/x - 35/(x+2) = 2

=> 35(x+2) - 35x / x(x+2) = 2 (lcm)

=> 35x + 70 - 35x = 2x(x+2)

=> 70 = 2x^2 + 4x

or, 2x^2 + 4x = 70

=> 2x^2 + 4x - 70 = 0

=> 2 (x^2 + 2x - 35) = 0

=> x^2 + (7-5)x - 35 = 0

=> x^2 + 7x - 5x - 35 = 0

=> x(x+7) - 5(x +7) = 0

=> (x+7) (x-5) = 0

therefore speed of train = 5km/h in 1st case

n in 2nd case it's speed is = x+2 = 5 + 2 = 7km/hr

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Answer 2

Answer:

see the photo

I hope it helped you