Points A and B are 70 km apart on a highway. A car starts from A and another car starts from B simultaneously.

Let X and Y be the cars starting from points A and B, respectively and let their speeds be x km/h and y km/h, respectively. 

Then, we have the following cases: 

Case I: When the two cars move in the same direction 

In this case, let the two cars meet at point M. 

Distance covered by car X in 7 hours = 7x km 

Distance covered by car Y in 7 hours = 7y km 

∴ AM = (7x) km and BM = (7y) km 

⇒(AM – BM) = AB 

⇒(7x – 7y) = 70 

⇒7(x – y) = 70 

⇒(x – y) = 10 ……..(i) 

Case II: When the two cars move in opposite directions 

In this case, let the two cars meet at point N. 

Distance covered by car X in 1 hour = x km 

Distance covered by car Y in 1 hour = y km 

∴ AN = x km and BN = y km 

⇒ AN + BN = AB ⇒ x + y = 70 ………(ii) 

On adding (i) and (ii), we get: 

2x = 80 

⇒x = 40 

On substituting x = 40 in (i), we get: 

40 – y = 10 

⇒y = (40 – 10) = 30 

Hence, the speed of car X is 40km/h and the speed of car Y is 30km/h.

Let the speed of the cars be x km/h and y km/h.

Relative speed in same direction = (x - y) km/h

Relative speed in opposite direction = (x + y) km/h

Now, \(\frac{70}{x+y} = 1\)

=> x + y = 70 .....(i)

and \(\frac{70}{x-y} = 7\)

=> x - y = 10

Solving Eqs. (i) and (ii),

x = 40 km/h and y = 30 km/h