12 Days of Christmas GMAT Competition 2025 - Day 2: A mountain climber

Reon

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17 Dec 2025, 04:12

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A mountain climber is 10 kilometers from the summit and carries 25 oxygen canisters. The climber can climb at one of two constant paces. At the slower pace, the climber gains 4 kilometers per hour and uses 2 oxygen canisters per kilometer. At the faster pace, the climber gains 5 kilometers per hour but uses 30 percent more oxygen to climb any given distance than at the slower pace. Along the route, there is a single stopping point where the climber may pick up extra canisters. Stopping there always adds exactly 25 minutes to the total climb time, regardless of how many canisters are taken. The climber must choose either the slower pace or the faster pace and maintain that pace for the entire climb. What is the minimum possible time, in minutes, needed for the climber to reach the summit?

Hello carries 25 oxygen canisters
Total distance from summit = 10 km

At slower pace,
Speed= 4 kmph
He uses 2 oxygen canisters per km.
Time taken to climb at slower pace = Distance/Speed =10/4= 2.5 hrs or 150 minutes
Oxygen canisters used = 2 per km, he climbed 10km, so he uses 10*2=20 oxygen canisters and hence didn't need to stop to pick extra canisters.

At faster pace,
Speed= 5 kmph
He uses 30% more oxygen at faster pace than at slower pace= 1.3*2= 2.6 oxygen canisters per km
For 10 km he will need 10*2.6=26 canister, so he will need to stop to pick more canister which will taken him 25 minutes more.
Time taken to climb = 10/5= 2 hours or 120 minutes
And total time taken including time to pick canister = 120+25= 145 minutes

Minimum possible time to reach the summit is 145 minutes at the faster pace.

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17 Dec 2025, 04:42

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Option 1
Speed = 4km/h
Time needed = 10km/4km/h = 2.5 hours

Cannisters usage rate= 2/km
Total Cannisters used = 10 * 2 = 20 cannisters

Time taken to reach the summit = 2.5 hours = 150 minutes

Option 2
Speed = 5Km/h
Time needed = 2 hours

Cannister usage = 2(1.3)/km
Total cannisters used = 2(1.3) * 10 = 26
Now requirement of 26 cannisters is more than he is carrying (25 cannisters) hence he will have to make a stop
Added time for stop = 25 minutes

Time taken to reach the summit = 2 hours + 25 minutes = 145 minutes

Comparing both options, Minimum possible time to reach summit = 145 minutes

Option C

Bunuel

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17 Dec 2025, 05:04

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Compute time and oxygen under each constant pace.

Slower pace

Speed = 4 km/hour
Distance = 10 km

Time = 10 ÷ 4 = 2.5 hours = 150 minutes

Oxygen use = 2 canisters/km
Total oxygen needed = 10 × 2 = 20 canisters
Available = 25 canisters → no stop needed

Total time (slow) = 150 minutes

Faster pace

Speed = 5 km/hour
Distance = 10 km

Time = 10 ÷ 5 = 2 hours = 120 minutes

Oxygen use per km = 2 × 1.3 = 2.6 canisters/km
Total oxygen needed = 10 × 2.6 = 26 canisters
Available = 25 canisters. Therefore he would need to stop at 1 stop required

Stop time = 25 minutes

Total time (fast) = 120 + 25 = 145 minutes

Since these are the only two possibilities, hence

Minimum possible time = 145 minutes & therefore the answer is Option C

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17 Dec 2025, 06:00

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IMO Option A 120 mins is minimum possible time
based on given info-
Slow pace 4km/hr 2 Oxy/km
=> 1km in 15mins (60/4) = 2Oxy tanks
Fast paced 5km/hr 2.6Oxy/km(30% inc)
=> 1km in 12 mins = 2.6 Oxy tanks

for minimum time we need to cover max distance in fast paced and no stopping time to be used
case1
for 10km in fast paced 10km 120 mins but 26 tanks so stopping time included it will take 120+25 =145mins
case2
9km fast 1 km slow 9km - 108mins 1km - 15mins total 123 mins but 25.4 tanks so with stoping time it will be 148 mins
case 3
8km fast 2km slow 8km-96mins 2km 30 mins total 126 mins and 24.8 tanks so this looks good
but we have an option that says 120 mins s lets try case 4
case 4
7km fast 3 km slow 7km=84mins 3 km =36mins total 120mins and 24.2 oxy tanks
hence answer 120 mins

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17 Dec 2025, 06:32

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Question Extraction:

Alright, two different paces & two different oxygen consumptions.

Slower pace: 4km/hr & 2 cans/km

Faster pace: 5km/hr & 2.6 cans/km

Stop for new canisters = 25mins

Question: Time taken for the fastest route.

Soln:

Time taken with slow pace: 10/4 = 2.5hours (no can stop required because number of cans consumed = 10*2 = 20)

Time taken for fast pace: 10/5 = 2hours (cans = 10 * 2.6 = 26, so we need a PIT STOP for 25mins)

Now post 9kms let me take a stop => 9/5 = 1.8hours + 25mins + 0.2hours for the last kilometer.

Total = 108mins + 25mins + 12mins
= 145mins

Hence the answer is 145mins (Option C)

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17 Dec 2025, 06:49

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Option C, bcz
as per slow option
10 km to be covered at 4km/per hour means 150 minutes
as per fast option
10 km to be covered at 5km/per hour that means 120 minutes + extra 25 minutes for cannister= 145 minutes

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17 Dec 2025, 06:50

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Considering the slower speed
-> Total canisters required = 2*10 = 20 -> Enough canisters available hence no break
-> Total time required = 10 km/ 4 km/h = 2.5 hours = 150 mins

Considering the faster speed
-> Canisters per kilometers = 130/100 * 2 (30% more than the slow speed) = 2.6 canisters
-> Total Canisters required = 2.6*10 = 26 -> Not enough for the whole climb -> break required
-> Total time required = 10km/5km/h + 25 mins = 120 min + 25 min = 145 minutes.

Minimum time required = min (time slow speed, time fast speed) = 145 mins
Correct answer => C

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17 Dec 2025, 07:26

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What we know is that,
Distance to summit = 10km
Starting oxygen canisters = 25

Slow Pace,
Speed = 4Km/hr
Oxygen use = 2 canisters /km

Fast Pace,
Speed = 5km/hr
Oxygen use = 1.3 * Slower use = 1.3* 2 = 2.6 canisters/km

Also, during the climb there is atmost 1 stop and it adds 25 minutes to the climb regardless of how many canisters are refiled. It is also mentioned that a constant pace is maintained the entire climb.

We check for slower pace initially,
Oxygen usage = 10km * 2 canisters/km = 20 canisters used which is doable in 1 climb since 25 is the initial stock
Time = 10 / 4 = 2.5 hrs = 150 mins

We check for faster pace now,
Oxygen usage = 10km * 2.6 canisters/km = 26 canisters which is more than 25, hence we need to make a stop in the middle which will add 25 minutes to the climb
Time = 10 / 5 = 2hours = 120 minutes + 25 mins (due to the stoppage)
Time = 145 minutes

Hence,
Since we need the minimum possible time in minutes for the climber to reach the summit, going the faster pace at 145 minutes is correct as it is feasible

C. 145 minutes

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17 Dec 2025, 07:33

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2 canisters per km at 4 km/h. 30% more if climbing at 5 km/h = 2*(13/10)= 2.6 canisters per km, i.e., 13 canisters per hour. Since the climber has 25 canisters, they'll have to pick up another one, adding 25 minutes. So for 10 km at 5 km/h, the time taken would be 120+25 =145 min.

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17 Dec 2025, 07:34

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Bunuel

slow pace = 4kmph = 2.5h to summit using 20 oxygen canisters.
fast pace = 5kmph = 2h to summit using 26 oxygen canisters, so have to make that stop of 25mins

=> min time required is 2h + 25mins = 145 mins

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17 Dec 2025, 07:39

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150 mins 20 oygen
120 + 25 = 145 mins 26 oxygen

so 145 going fasteer route and breaking

Bunuel

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17 Dec 2025, 08:00

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At slower pace the climber uses 2 oxygen canisters per km
For 10 Kms , oxygen needed = 10*2 = 20 canisters
Since she has 25 canisters, no stop will be needed and total time taken will be 10/4 km/hr = 2.5 hrs = 150 mins

At Fast pace, 30% more oxygen is used as compared to slow paced (2*1.3=2.6 canister per km)
For 10 kms oxygen needed = 10*2.6 = 26 canisters
Sicne she has only 25 canisters , one stop will be needed for additional canister adding 25 mins to the total time
So total time will be 10/5km/hr = 2 hrs = 120 mins + 25 mins = 145 mins

Therefore minimum possible time will be 145 mins.

Bunuel

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17 Dec 2025, 08:30

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Conceptual refresh: Time needed for climber to reach the summit and oxygen needs follow directly from distance and rate. Distance = Rate x Time

Interpretation: We are asked to compare two options: a slower pace with sufficient oxygen and a faster pace that may require a stop due to higher oxygen usage.

Slower pace climb: 10km / 4 = 2.5 hours = 150 mins since oxygen is 2 canisters per km therefore 20 canisters required. Total oxygen available is 25 canisters hence sufficient oxygen.

Faster pace climb: 10 km / 5 = 2 hours = 120 mins since oxygen is 30% more than the slower pace means 3/10 x 20 = 6 more canisters needed for a total of 26 canisters leaving climber out of oxygen by 1 canister, a stop is required, adding 25 mins. Total time = 120 + 25 = 145 mins therefore the minimum possible time.

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17 Dec 2025, 08:40

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IMO C

We have to see the use from both paces

at slow pace, 4kmph and 2 cannisters/km. So total time= 10/4= 2.5 hrs= 150 minutes and around >=20 cannisters
at faster pace, 5kmph and 1.3 times faster consumption/km, so 10kms would require= 20 *1.3= 26 cannisters (1 MORE than available)
but at faster pace, it requires a time of= 10/5= 120 minutes, but since we need an extra cannister and stopping requires time of 25 minutes, we will take 120+25 min= 145 mins.

Hence min time required is 145 minutes after one stop.

Bunuel

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17 Dec 2025, 08:52

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According to the problem,
Climber can climb either at a slower pace or faster pace.
Total Distance needs to travel = 10 km
Available Canisters = 25
Extra time for taking rest = 25 mins.

Now, checking for a slower pace.
Speed: 4 km/hr
Required canisters: 2 per km.
Total Canisters needed = 10*2 = 20 --> Available, so that no need stop in between
Total time(in mins.) = total distance / speed = 10/4 = 2.5 hrs. = 2.5 * 60 = 150 mins.

Then, checking for a faster pace.
Speed: 5 km/hr
Required canisters: 30% more than 2 = 2.6 per km.
Total Canisters needed = 10*2.6 = 26 --> Not available.
Need 1 extra canister, for that climber needs to stop for once. which will add extra 25 mins.
Total time(in mins.) [not including rest time] = 10/5 = 2 hrs. = 2 * 60 = 120 mins.
Total time taken inluding extra time for rest = 120 + 25 = 145 mins.

Therefore, minimum possible time taken will be 145 mins.

OA: C

Bunuel

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18 Dec 2025, 18:35

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Slower pace: 4 kms/hr
total time for 10 km: 2 1/2 hrs = 150 mins
O2 canisters: 2 X 10= 20

Faster pace: 5 kms/hr
O2 canisters: 30% extra over slower pace: 20 * 1.3 = 26
So he has to stop for 25 mins for picking up extra canister
Total time for 10 kms: 2hrs + 25 mins : 145 mins

So, Minimum possible time: 145 mins

Bunuel