Question:
Ankita walks from A to C through B, and runs back through the same route at a speed that is 40% more than her walking speed. She takes exactly 3 hours 30 minutes to walk from B to C as well as to run from B to A. The total time, in minutes, she would take to walk from A to B and run from B to C, is
Ankita walks from A to C through B, and runs back through the same route at a speed that is 40% more than her walking speed. She takes exactly 3 hours 30 minutes to walk from B to C as well as to run from B to A. The total time, in minutes, she would take to walk from A to B and run from B to C, is
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When speed changes by a fixed percentage, keep everything in terms of one speed (like \(w\)), express all distances using that speed and given times, then recompute the required times with the new speed.
Updated On: Jul 7, 2026
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Correct Answer: 444
Approach Solution - 1
Approach: Two given times each isolate one leg, so use them to express both leg-distances in terms of the walking speed \(w.\) The actual value of \(w\) cancels later, so don't worry about it.
Step 1: Let walking speed \(=w,\) running speed \(=1.4w\) (\(40\%\) faster).
Step 2: "Walk B to C in \(3\) h \(30\) min \(=3.5\) h": \(\dfrac{d_{BC}}{w}=3.5\Rightarrow d_{BC}=3.5w.\)
Step 3: "Run B to A in \(3.5\) h": \(\dfrac{d_{AB}}{1.4w}=3.5\Rightarrow d_{AB}=3.5\times1.4w=4.9w.\)
Step 4: Required trip is walk A to B, then run B to C:
\[t=\frac{d_{AB}}{w}+\frac{d_{BC}}{1.4w}=\frac{4.9w}{w}+\frac{3.5w}{1.4w}=4.9+2.5=7.4\text{ h}.\]
Step 5: \(7.4\times60=444\) minutes. (Notice \(w\) vanished entirely.)
Final answer: \(444\) minutes.
Step 1: Let walking speed \(=w,\) running speed \(=1.4w\) (\(40\%\) faster).
Step 2: "Walk B to C in \(3\) h \(30\) min \(=3.5\) h": \(\dfrac{d_{BC}}{w}=3.5\Rightarrow d_{BC}=3.5w.\)
Step 3: "Run B to A in \(3.5\) h": \(\dfrac{d_{AB}}{1.4w}=3.5\Rightarrow d_{AB}=3.5\times1.4w=4.9w.\)
Step 4: Required trip is walk A to B, then run B to C:
\[t=\frac{d_{AB}}{w}+\frac{d_{BC}}{1.4w}=\frac{4.9w}{w}+\frac{3.5w}{1.4w}=4.9+2.5=7.4\text{ h}.\]
Step 5: \(7.4\times60=444\) minutes. (Notice \(w\) vanished entirely.)
Final answer: \(444\) minutes.
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Approach Solution -2
Alternate approach — using the inverse time-speed ratio directly:
Running speed is \( 1.4 \) times the walking speed, so for the same distance, running time is \( \frac{1}{1.4} = \frac{5}{7} \) of the walking time.
Walking B to C takes 210 minutes, and running B to A also takes 210 minutes. Since running takes \( \frac{5}{7} \) of the walking time for the same stretch, the walking time for A to B is \( 210 \times \frac{7}{5} = 294 \) minutes.
Similarly, since walking B to C takes 210 minutes, the running time for B to C is \( 210 \times \frac{5}{7} = 150 \) minutes.
So, walking A to B and running B to C together take \( 294 + 150 = \) 444 minutes.
Running speed is \( 1.4 \) times the walking speed, so for the same distance, running time is \( \frac{1}{1.4} = \frac{5}{7} \) of the walking time.
Walking B to C takes 210 minutes, and running B to A also takes 210 minutes. Since running takes \( \frac{5}{7} \) of the walking time for the same stretch, the walking time for A to B is \( 210 \times \frac{7}{5} = 294 \) minutes.
Similarly, since walking B to C takes 210 minutes, the running time for B to C is \( 210 \times \frac{5}{7} = 150 \) minutes.
So, walking A to B and running B to C together take \( 294 + 150 = \) 444 minutes.
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