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?

• A. 17 km/hour
• B. 17.5 km/hour
• C. 18 km/hour
• D. None of the above.

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.

Answer 1

The Correct Option is D

Step 1: Notice that shopping time is constant and drops out.
Since he always shops for 30 minutes, only the walking time changes between scenarios. Compare walking times directly instead of full time-away-from-home totals.

Step 2: Express round-trip walking time as $\frac{2d}{v}$.
At 10 km/hr, walking time is $\frac{2d}{10}$ hours. At 15 km/hr, it is $\frac{2d}{15}$ hours. The 30 minute gap between the two return times (19:00 versus 18:30) is entirely due to the change in walking time.
\[ \frac{2d}{10} - \frac{2d}{15} = \frac{1}{2} \]
Multiply through by 30:
\[ 6d - 4d = 15 \implies 2d = 15 \implies d = 7.5 \text{ km} \]

Step 3: Anchor the timeline using one full scenario.
At 15 km/hr: walking time $= \frac{2(7.5)}{15} = 1$ hour, plus 0.5 hour shopping $= 1.5$ hours total, arriving back at 18:30. So he left home at 18:30 minus 1.5 hours $= 17{:}00$.

Step 4: Work out the time budget for the new target.
To be back by 18:15, the total time away from 17:00 must be 1 hour 15 minutes. Removing the fixed 30 minutes of shopping leaves 45 minutes, or 0.75 hour, for walking both ways.

Step 5: Convert the walking time into a speed.
Round trip distance is $2 \times 7.5 = 15$ km, walked in 0.75 hour, so
\[ v = \frac{15 \text{ km}}{0.75 \text{ hr}} = 20 \text{ km/hour} \]

Final Answer:
20 km/hour is not one of 17, 17.5 or 18 km/hour, so the answer must be option E, none of the above. \[ \boxed{20 \text{ km/hour}} \]