(c) 35 km/hr, 25 km/hr
Let X and Y be two cars starting from points A and B respectively.
Let the speed of the car X be x km/hr and that of car Y be y km/hr.
Case I : When the two cars move in the same direction.
Suppose the two cars meet at point P after 10 hours.
Then, Dist. travelled by car X = AP = 10x km
and Dist. travelled by car Y = BP = 10y km
Given, AP – BP = AB
\(\Rightarrow\) 10x – 10y = 100 \(\Rightarrow\) x – y = 10 ...........(i)
Case II: When the two cars move in the opposite direction :
Suppose the two cars meet at point Q of 1 hrs
40 min = 1\(\frac{40}{60}\)hrs = 1\(\frac{2}{3}\)hrs= \(\frac{5}{3}\)hrs.
Then, dist. travelled by car X = AQ =\(\frac{5X}{3}\)km
and dist. travelled by car Y = BQ = \(\frac{5y}{3}\)km
Given, AQ + BQ = 100
\(\Rightarrow\) \(\frac{5X}{3}\) + \(\frac{5y}{3}\) = 100
\(\Rightarrow\) (x+y) = \(\frac{300}{5}\) = 60 ...........(ii)
Adding (i) and (ii)
2x = 70
\(\Rightarrow\) x = 35
\(\therefore\) From (i), 35 – y = 10 \(\Rightarrow\) y = 25 km/hr
