Iift 2010 Question 26

Let us assume that Mukesh and Suresh started on bike from Delhi to Mathura and Dinesh started walking at the same time. Mukesh drops Suresh in a mid-way such that Suresh will take same time to reach Mathura with Mukesh and Dinesh. In the same time Mukesh will turn back and will pick up Dinesh.
Let us assume that '4D' is the distance covered by by Mukesh and Suresh. Then Suresh must have covered remaining 300 - 4D kms by walking at 15 km/hr. When Mukesh dropped Suresh by that time Dinesh have already covered D kms and now Mukesh and Dinesh and would have 3D kms apart. Since Mukesh is 4 times faster than Dinesh, Mukesh would have covered (4/5) of 3D kms.
In addition to this, Mukesh and Dinesh have to cover 2.4D+300-4D kms to reach Mathura at 60 km/hr.
By equating the time taken by Mukesh and Suresh to reach Mathura after the point when Mukesh dropped Suresh,
$$\dfrac{300-4D}{15}=\dfrac{2.4D+2.4D+300-4D}{60}$$
$$\Rightarrow$$ 16.8*D = 900
$$\Rightarrow$$ D = $$\dfrac{375}{7}$$ kms
Therefore, minimum time taken by them to finish the journey = $$\dfrac{4D}{60}+\dfrac{300-4D}{15}$$
$$\Rightarrow$$ $$20 - \dfrac{D}{5}$$
$$\Rightarrow$$ $$20 - \dfrac{375}{7*5}$$
$$\Rightarrow$$ $$20 - \dfrac{75}{7}$$
$$\Rightarrow$$ $$\dfrac{65}{7}$$ or $$9\dfrac{2}{7}$$ hours. Hence, option B is the correct answer.