Question

two places aurangabad and ellora are 120 km apart.two cyclist leave Ellora for aurangabad such that second leaves 4 hours after the first.both cyclist arrive at aurangabad at the same time.find the time taken by the slower cyclist on the trip if the speed of one cyclist is 5km/h more than the other.

Answer 1

you can solve this my relative velocity

Answer 2

Answer:

12 Hrs.

Step-by-step explanation:

Let the speed of faster cyclist be X km/ Hr.

So,

The speed of slower cyclist = (X - 5) km/ Hr.

then,

=> $\frac{120}{X-5} -\frac{120}{x}= 4$

=> $\frac{120x-120(x-5)}{(x-5)(x)}=4$

=> $\frac{120x-120x+600}{x^{2}-5x}=4$

=> x² - 5x - 150 = 0

=> (x - 15)(x + 10) = 0

=> X = -10 or 15     [#Speed not in negative]

=> X = 15 km/ hr

Speed of slower cyclist = 15 -5 = 10 km/hr

The time required for cover distance is 120/10 = 12 hrs.