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: Don't set up algebra at all \(-\) plug the options into the clean numbers. The journey is $6$ hours, so for any initial speed $x$ the distance is $D=6x$; test which $x$ makes both stop-scenarios consistent.

Step 1: Try $x=15$ (option 2): $D=90$ km.

Step 2 (20-min stop, speed up by 3): New speed $=18$ km/h, so the whole $90$ km would take $90/18=5$ h. Add the $\tfrac13$-h stop $=5+\tfrac13=5\tfrac13$ h $-$ that overshoots $6$ h, which simply tells us he does not run the whole way at $18$; he runs only the part after the stop. He drives $4$ h at $15$ (=$60$ km), stops $20$ min, then $30$ km at $18$ takes $\tfrac{30}{18}=\tfrac53$ h. Total $=4+\tfrac13+\tfrac53=6$ h. On time.

Step 3 (30-min stop, speed up by 5): Same $60$ km in $4$ h, stop $30$ min, then $30$ km at $20$ takes $\tfrac{30}{20}=1.5$ h. Total $=4+0.5+1.5=6$ h. On time.

Step 4: Both conditions hold for $x=15$ with a consistent stop point ($60$ km in). Quick checks on $x=12,18,20$ fail to satisfy both at once.

Final answer: Initial speed $=15$ km/h.