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16 Dec 2025, 11:01
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The answer is 145 mins :
Solution - Slow pace time taken : D = Speed x Time
10 = 4 x T
T = 2.5hrs ---> 150 mins, it's in the option but lets check the faster pace too,
10 = 5 x T
T = 2hrs ----> 120 mins, but 30% more oxyge used at any given distance, since slower pace only requires 20 can (1kms - 2 cans, 10km how much? (20 cans) 30% of 20 cans would be 20+6 = 26, therefore 25 mins stoppage time needed making total time = 145 mins
Solution - Slow pace time taken : D = Speed x Time
10 = 4 x T
T = 2.5hrs ---> 150 mins, it's in the option but lets check the faster pace too,
10 = 5 x T
T = 2hrs ----> 120 mins, but 30% more oxyge used at any given distance, since slower pace only requires 20 can (1kms - 2 cans, 10km how much? (20 cans) 30% of 20 cans would be 20+6 = 26, therefore 25 mins stoppage time needed making total time = 145 mins
16 Dec 2025, 11:11
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Bunuel
Calculating slow pace first:
\(10 km * 2 canisters = 20\)
He has 25 canisters, so no need to stop.
\(\frac{10km}{4kph}=2.5 hours=150 min\)
Now calculating fast pace:
Increased oxygen consumption by 30%
\(2 canisters * \frac{130}{100}=\frac{13}{5} canisters/km\)
\(\frac{13}{5} canisters/km * 10 km = 26 canisters\)
He will need to stop, add 25 min to the time.
Now, to calculate time:
\(\frac{10km}{5kph} = 2 hours + 25 min = 120+25 min = 145 min\)
\(145 min < 150 min\)
Thus fast pace is quicker than slow pace, and the minimum possible time is option C, 145 min.
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16 Dec 2025, 11:19
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For 10kms , at slow pace of 4 km/hr = 2.5 hrs needed (150 mins) with 20 canisters utilised without stopping
and at faster pace of 5km/hr = 2 hrs needed (120 mins) with 30% more utilization 26 cannisters are needed and he must stop at the stopping adding 25 mins to the time taken which adds upto 145 mins
Min time = 145 mins (option C)
and at faster pace of 5km/hr = 2 hrs needed (120 mins) with 30% more utilization 26 cannisters are needed and he must stop at the stopping adding 25 mins to the time taken which adds upto 145 mins
Min time = 145 mins (option C)
Bunuel
