Menu and Daya travel from point A to B, a distance of 105 km, at speeds of 10 km/h and 25 km/h, respectively. Daya reaches point B first and returns immediately and meets Menu at point C. Find the distance from point A to point C.

60 km

Speed of Menu = 10 km / hr.

Speed of Daya =  25 km/ hr.

Total distance b/w A & B = 105 km

Daya covered 105 km and returns immediately.

Daya reaches the other end of B ie

 the time taken to travel 105 km is = \(\frac{105}{25}\) = 4.2 hours

So , in 4.2 hours distance traveled by Menu=  10 x 4.2 = 42 km.

Remaining distance to be covered by Menu =  105 – 42 = 63 kms.

As , both now travel in opposite direction . So relative speed  25 + 10 = 35 km/h

So, the time taken to meet each other = \(\frac{63}{35}\) =1.8 hrs

Distance covered by Menu  in 1.6 hours is 10 x 1.8 = 18 km

Now the distance AC is 42 + 18 = 60 kms.