Rahul starts on his journey at 5 pm at a constant speed so that he reaches his destination at 11 pm the same day. However, on his way, he stops for 20 minutes, and after that, increases his speed by 3 km per hour to reach on time. If he had stopped for 10 minutes more, he would have had to increase his speed by 5 km per hour to reach on time. His initial speed, in km per hour, was

✅ Correct Option: 2

Total journey time: 5 PM to 11 PM = 6 hours

Let initial speed = $v$ km/hr, so total distance = $6v$ km

Let $t$ = time (in hours) traveled at initial speed before stopping

When Rahul stops for 20 minutes (= $\frac{1}{3}$ hour), his new speed becomes $(v + 3)$ km/hr.

Remaining time after the stop $= 6 - t - \frac{1}{3} = \frac{17}{3} - t$

Distance before stop $= vt$

Distance after stop $= 6v - vt$

$6v - vt = (v + 3)\left(\frac{17}{3} - t\right)$

$18v - 3vt = (v + 3)(17 - 3t)$

$18v - 3vt = 17v - 3vt + 51 - 9t$

$v = 51 - 9t \quad \cdots (1)$

When Rahul stops for 30 minutes (= $\frac{1}{2}$ hour), his new speed becomes $(v + 5)$ km/hr.

Note: "10 minutes more" means $20 + 10 = 30$ minutes total, not 10 minutes alone.

Remaining time after the stop $= 6 - t - \frac{1}{2} = \frac{11}{2} - t$

$6v - vt = (v + 5)\left(\frac{11}{2} - t\right)$

$12v - 2vt = (v + 5)(11 - 2t)$

$12v - 2vt = 11v - 2vt + 55 - 10t$

$v = 55 - 10t \quad \cdots (2)$

From (1) and (2):

$51 - 9t = 55 - 10t$

$10t - 9t = 55 - 51$

$t = 4$ hours

$v = 51 - 9(4)$

$v = 51 - 36$

$v = 15$ km/hr

Rahul's initial speed was $15$ km/hr.

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