Let’s consider the car starting from point A as X and its speed as x km/hr. 

And, the car starting from point B as Y and its speed as y km/hr. 

It’s seen that there are two cases in the question: 

# Case 1: Car X and Y are moving in the same direction 

# Case 2: Car X and Y are moving in the opposite direction 

Let’s assume that the meeting point in case 1 as P and in case 2 as Q. 

Now, solving for case 1: 

The distance travelled by car X = AP 

And, the distance travelled by car Y = BP 

As the time taken for both the cars to meet is 8 hours, 

The distance travelled by car X in 7 hours = 8x km [∵ distance = speed x time] 

⇒ AP = 8x 

Similarly, 

The distance travelled by car Y in 8 hours = 8y km 

⇒ BP = 8Y 

As the cars are moving in the same direction (i.e. away from each other), we can write 

AP – BP = AB 

So, 8x – 8y = 80 

⇒ x – y = 10 ……… (i) [After taking 8 common out] 

Now, solving for case 2: 

In this case as it’s clearly seen that, 

The distance travelled by car X = AQ 

And, The distance travelled by car Y = BQ 

As the time taken for both the cars to meet is 1 hour and 20 min, 

⇒ 1 + (20/60) = 4/3 hr

The distance travelled by car x in 4/3 hour = 4x/3 km 

⇒ AQ = 4x/3 

Similarly, 

The distance travelled by car y in 4/3 hour = 4y/3 km 

⇒ BQ = 4y/3 

Now, since the cars are moving in the opposite direction (i.e. towards each other), we can write 

AQ + BQ = AB 

⇒ 4x/3 + 4y/3 = 80 

⇒ 4x + 4y = 240 

⇒ x + y = 60 …………… (ii) [After taking LCM] 

Hence, by solving (i) and (ii) we get the required solution 

From (i), we have x = 10 + y……. (iii) 

Substituting this value of x in (ii). 

⇒ (10 + y) + y = 60 

⇒ 2y = 50 

⇒ y = 25 

Now, using y = 30 in (iii), we get 

⇒ x = 35 

Therefore, 

– Speed of car X = 35 km/hr. 

– Speed of car Y = 25 km/hr.