Question

The distance between 2 places a and b is 45km. Two cyclists ride from aplace a to place b. The first cyclist arrives 30 min earlier than the second one. While returning from b to a, the first cyclist gives the second a start of 3 km and yet reaches the destination 10 min earlier. Find the speed of the cyclists in km/hr

Answer 1

Let ,

v1 be the velocity of 1st cyclist

v2 be velocity of 2nd cyclist

So,

I am considering the time required in km/hr

Time = Displacement/Velocity

45/v2 - 45/v1 = 30/60

Let 1/v1 be y and 1/v2 be x

Now,

45x - 45y = 1/2...................1)

By the second condition,

42/v2 - 45/v1 = 1/6

42x - 45y = 1/6...................2)

Substracting equation 1) by 2)

3x = 1/2 - 1/6

3x = 4/12

x=1/9

So,

v2 = 9 km/hr

Substituting value of x in equation 2)

42( 1/9) - 45 y = 1/6

45y = - 1/6 + 42 /9

y = 1/10

So,

value of v1 = 10 km/hr

Hence, the speed of 1st cyclist is 10km/hr and the second cyclist is 9km/hr