Aptitude Discussion

Q. |
Arun, Barun and Kiranmala start from the same place and travel in the same direction at speeds of 30, 40 and 60 km per hour respectively. Barun starts two hours after Arun. If Barun and Kiranmala overtake Arun at the same instant, how many hours after Arun did Kiranmala start? |

✖ A. |
3.0 |

✖ B. |
3.5 |

✔ C. |
4.0 |

✖ D. |
4.5 |

**Solution:**

Option(**C**) is correct

Let us assume that Arun started running at 10 AM and Barun started at 12 noon.

So, in these two hours distance travelled by Arun is 60 km and the relative speed of Barun w.r.t Arun is 10 km/hr.

So Barun will overtake Arun after $=\dfrac{60}{10} = 6$ hrs

So, Barun reaches there at 6 PM.

So, Kiranmala also overtakes Arun at 6 PM.

Let us assume Kiranmala takes $t$ time to overtake Arun and the relative speed of Kiranmala w.r.t Arun is 30 km/hr and Arun ran for 8 hrs.

So, distance travelled by Arun is:

$=\text{time}\times \text{speed}$

$=8\times 30$ -------- (1)

While Kiranmala's distance travelled is:

$=t \times 60$ -------- (2)

Since distance traveled by them is equal,

$\Rightarrow (1) = (2)$

$\Rightarrow 8\times 30 = t \times 60$

$t = 4$ hours

So, after $\textbf{4 hrs}$, Kiranmala will start running

**Edit:** For a shortcut alternative solution, check comment by **Pasanna.**

**Pasanna**

*()
*

Good one, thank you Pasanna :)

OR take LCM of Barun and Kiran's speed which equals a distance of 120km.

This implies Kiran passes Barun at a distance of 120km (irrespective of starting time) and also means BARUN and KIRAN can only pass ARUN at SAME INSTANT only after ARUN travelling 120 km.

=> Arun has to travel a distance of 120 km.

Hence, $\dfrac{120}{30}= \textbf{4 hrs}$.