Question:
Rajesh walks to and from a shopping mall. He spends 30 minutes shopping there. If he walks at a speed of 10 km/hour, he returns home at 19.00 hours. If he walks at 15 km/hour, he returns home at 18.30 hours. How fast must he walk in order to return home at 18.15 hours?
Rajesh walks to and from a shopping mall. He spends 30 minutes shopping there. If he walks at a speed of 10 km/hour, he returns home at 19.00 hours. If he walks at 15 km/hour, he returns home at 18.30 hours. How fast must he walk in order to return home at 18.15 hours?
Show Hint
The 30-minute shopping stop is constant, so compare only the walking times to find the one-way distance, then find Rajesh's fixed leaving time before solving for the new speed.
Updated On: Jul 10, 2026
- 17 km/hour
- 17.5 km/hour
- 18 km/hour
- None of the above.
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Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Let the one-way distance to the mall be \(d\) km.
At speed \(v\), the round-trip walking time is \(\frac{2d}{v}\) hours. His total time away from home is walking time plus 30 minutes (0.5 hour) of shopping, and he leaves home at the same fixed time every day.
Step 2: Use the difference between the two known trips.
The 30 minutes of shopping is the same in both cases, so it cancels when comparing arrival times. The difference in walking time equals the difference in arrival time:
\[ \frac{2d}{10} - \frac{2d}{15} = (19{:}00 - 18{:}30) = 0.5 \text{ hour} \]
\[ \frac{d}{5} - \frac{2d}{15} = 0.5 \implies \frac{3d - 2d}{15} = 0.5 \implies \frac{d}{15} = 0.5 \implies d = 7.5 \text{ km} \]
Step 3: Find the fixed time Rajesh leaves home.
At 10 km/hr, round trip walking time \(= \frac{2 \times 7.5}{10} = 1.5\) hours \(= 90\) minutes. Add 30 minutes shopping: total time away \(= 120\) minutes \(= 2\) hours. He returns at 19:00, so he left home at 17:00.
Check with 15 km/hr: walking time \(= \frac{15}{15} = 1\) hour \(= 60\) min, plus 30 min shopping \(= 90\) min \(= 1.5\) hours; leaving at 17:00 gives a return at 18:30, which matches.
Step 4: Find the speed needed to return at 18:15.
Total time away from 17:00 to 18:15 is 1 hour 15 minutes \(= 75\) minutes. Subtract 30 minutes shopping: walking time \(= 45\) minutes \(= 0.75\) hour. Round trip distance is \(2d = 15\) km, so the required speed is
\[ v = \frac{15}{0.75} = 20 \text{ km/hour} \]
Step 5: Compare with the choices.
20 km/hour is not equal to 17, 17.5 or 18 km/hour, so none of those three options is correct.
Final Answer:
Since the required speed of 20 km/hour is not among the first three choices, the correct choice is option E, none of the above. \[ \boxed{20 \text{ km/hour}} \]
At speed \(v\), the round-trip walking time is \(\frac{2d}{v}\) hours. His total time away from home is walking time plus 30 minutes (0.5 hour) of shopping, and he leaves home at the same fixed time every day.
Step 2: Use the difference between the two known trips.
The 30 minutes of shopping is the same in both cases, so it cancels when comparing arrival times. The difference in walking time equals the difference in arrival time:
\[ \frac{2d}{10} - \frac{2d}{15} = (19{:}00 - 18{:}30) = 0.5 \text{ hour} \]
\[ \frac{d}{5} - \frac{2d}{15} = 0.5 \implies \frac{3d - 2d}{15} = 0.5 \implies \frac{d}{15} = 0.5 \implies d = 7.5 \text{ km} \]
Step 3: Find the fixed time Rajesh leaves home.
At 10 km/hr, round trip walking time \(= \frac{2 \times 7.5}{10} = 1.5\) hours \(= 90\) minutes. Add 30 minutes shopping: total time away \(= 120\) minutes \(= 2\) hours. He returns at 19:00, so he left home at 17:00.
Check with 15 km/hr: walking time \(= \frac{15}{15} = 1\) hour \(= 60\) min, plus 30 min shopping \(= 90\) min \(= 1.5\) hours; leaving at 17:00 gives a return at 18:30, which matches.
Step 4: Find the speed needed to return at 18:15.
Total time away from 17:00 to 18:15 is 1 hour 15 minutes \(= 75\) minutes. Subtract 30 minutes shopping: walking time \(= 45\) minutes \(= 0.75\) hour. Round trip distance is \(2d = 15\) km, so the required speed is
\[ v = \frac{15}{0.75} = 20 \text{ km/hour} \]
Step 5: Compare with the choices.
20 km/hour is not equal to 17, 17.5 or 18 km/hour, so none of those three options is correct.
Final Answer:
Since the required speed of 20 km/hour is not among the first three choices, the correct choice is option E, none of the above. \[ \boxed{20 \text{ km/hour}} \]
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