Question

If a cyclist travels at a speed 2km/hr more than his usual speed, he reaches the destination 2 hours erlier. If the destion is 35km away, what is the usual speed of the cyclist ?

Answer 1

Let the original speed be 'v' km/hr and original time be 't' hrs

Given destination is 35 Km away

So vt=35
Also given

(v+2)(t-2)=35
vt-2v+2t-4=35
vt-2v+2t=39
35-2v+2t=39
-2v+2t=4
t-v=2
t=v+2

We know that vt=35

(35/v)=v+2
35=v^2+2v

On solving the quadratic equation ,we get v = 5 and -7,but we ignore -7 since speed can't be negative

So v=5km/hr

t=7 hours