Question

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

• A. \(20\)
• B. \(15\)
• C. \(12\)
• D. \(18\)

Hint:

In time–speed–distance problems with multiple scenarios:
Express the total distance using the original (planned) speed and time.
Write separate time equations for each scenario, keeping the same distance.
Look for a common expression (like the remaining distance) to reduce the number of variables and solve systematically.

Answer 1

The Correct Option is B

Approach: The trap is to make him speed up over the whole journey \(-\) that gives contradictory equations. Reality: he drives normally, stops at one fixed point, then speeds up over the remaining distance to land on time. Keep two unknowns: speed \(x\) and the driving time before the stop.

Step 1: Scheduled time \(5\) pm to \(11\) pm \(=6\) hours, so total distance \(D=6x\). Let him drive \(T\) hours at speed \(x\) before stopping; remaining distance \(=6x-xT=x(6-T).\)

Step 2 (20-minute stop): Wall-clock left to cover the remaining distance \(=6-T-\tfrac13\) hours at speed \(x+3\):
\[\frac{x(6-T)}{x+3}=6-T-\frac13.\]

Step 3 (30-minute stop): Same remaining distance, but now speed \(x+5\) and \(10\) more minutes lost:
\[\frac{x(6-T)}{x+5}=6-T-\frac12.\]

Step 4: Solving the two equations together gives \(T=4\) and \(x=15.\)

Step 5 (sanity check): \(D=90\) km. In \(4\) h he covers \(60\) km; \(30\) km remain. With a \(20\)-min stop, time left \(=6-4-\tfrac13=\tfrac53\) h, needed speed \(=30\div\tfrac53=18=15+3.\) With a \(30\)-min stop, time left \(=\tfrac32\) h, needed speed \(=30\div\tfrac32=20=15+5.\) Both fit.

Final answer: Initial speed \(=15\) km/h.

Answer 2

Step 1: Define variables and total travel time. Rahul travels from 5 pm to 11 pm, so the total scheduled travel time is: \[ 6 \text{ hours}. \] Let: \[ v = \text{initial speed (km/h)}, \quad S = \text{total distance (km)}. \] If he goes at constant speed \(v\) for 6 hours: \[ S = 6v. \] Suppose he travels for \(t\) hours at speed \(v\) before stopping.
Step 2: First scenario (20-minute stop, speed increased by 3 km/h). He:
travels for \(t\) hours at speed \(v\),
stops for 20 minutes \(= \dfrac{1}{3}\) hour,
then travels the remaining distance at speed \(v + 3\) km/h. Distance covered before stop: \(vt\). Remaining distance: \(S - vt = 6v - vt\). Time for remaining distance: \[ \frac{6v - vt}{v + 3}. \] Total time must still be 6 hours: \[ t + \frac{1}{3} + \frac{6v - vt}{v + 3} = 6. \] Rewriting: \[ \frac{6v - vt}{v + 3} = 6 - t - \frac{1}{3} = \frac{17}{3} - t. \quad \cdots (1) \]
Step 3: Second scenario (30-minute stop, speed increased by 5 km/h). Now he stops 10 minutes more, so the stop is 30 minutes \(= \dfrac{1}{2}\) hour, and then increases speed by 5 km/h (to \(v + 5\)). Similarly, \[ t + \frac{1}{2} + \frac{6v - vt}{v + 5} = 6 \Rightarrow \frac{6v - vt}{v + 5} = 6 - t - \frac{1}{2} = \frac{11}{2} - t. \quad \cdots (2) \]
Step 4: Use a common expression. Let: \[ K = v(6 - t) = 6v - vt. \] Then from (1) and (2): \[ \frac{K}{v + 3} = \frac{17}{3} - t, \qquad \frac{K}{v + 5} = \frac{11}{2} - t. \] Equating the two expressions for \(K\): \[ \left(\frac{17}{3} - t\right)(v + 3) = \left(\frac{11}{2} - t\right)(v + 5). \] Solving this equation for \(t\) in terms of \(v\) gives: \[ t = \frac{21}{4} - \frac{v}{12}. \]
Step 5: Substitute back to find \(v\). Using (1) with \(K = v(6 - t)\) and the expression for \(t\), we get: \[ v = 15. \] So the initial speed of Rahul was: \[ \boxed{15 \text{ km/h}}. \]