A bicyclist started on his trip from city A to city B. In 1 hour and 36 minutes, a biker also left A and headed towards B, and he arrived there at the same time as the cyclist. Find the speed of the bicyclist if it is less than the speed of the biker by 32 km/hour, and the distance between the two cities is 52 km.

20km/hr

Step-by-step explanation:

Distance between the two Cities=52km

If the biker's Speed=S

And the biker's travel time = t

Distance = Speed X Time

52 = St ... [1]

For the bicyclist

Speed = S-32

Time = t + 1 hour and 36 minutes = (t+1.6) hours

Distance = Speed X Time

52 = (S-32) (t+1.6) ... [2]

Therefore: Combining equations 1 and 2.

St = (S-32) (t+1.6) = 52

St=St+1.6S-32t-51.2

1.6S-32t=51.2

From equation 1, t=52/S

1.6S-32 (52/S) = 51.2

1.6S - (1664/S) = 51.2

(1.6S²-1664) / S=51.2

Cross multiplying

1.6S²-1664=51.2S

1.6S²-51.2S-1664=0

Solving for S gives S=-20 or 52

S=52km/hr

Therefore the bicyclist speed, S-32=52-32=20km/hr