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.

Given:

Menu travels from point A to B, a distance of 105 km, at speeds of 10 km/h.

Daya travels from point A to B, a distance of 105 km, at speeds of  25 km/h.

Formula used:

\(Time={Distance\over Speed}\)

Calculation:

Let Daya takes time t to reach at point B, then

⇒ t = \({105\over25} = {21\over5}\)

Menu reach at point O in same time t and travel distance AO. then,

⇒ AO = \(10\times{21\over5}=42\) km

Remaining distance OB = 105 - 42 = 63 km 

Now, 

Daya and Menu meet at point C and they take the same time. then,

⇒ Time_Menu = Time_Daya 

⇒ \({d\over10}={63 - d\over25}\)

⇒ \({d\over2}={63 - d\over5}\)

⇒ 5d = 126 - 2d

⇒ 7d = 126 

⇒ d = 18 km

So, the distance from point A to point C,

⇒ AC = AO + OC

⇒ AC = 42 + 18 = 60 km.

∴ The distance from point A to point C is 60 km.

 Alternate Method

Let's assume that they meet at point C, and CB = x then AC = (105 - x) [ the distance of AB = 105 km ]

Daya covers a distance of (105 + x) and Menu covers (105 - x) at the same time, then

⇒ \(\frac{105 + x}{25} = \frac{105 - x}{10}\) 

⇒ 210 + 2x = 525 - 5x 

⇒ 7x = 315 

⇒ x = 45

So, distance from AC = (105 - 45) = 60 km