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

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arnab24

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16 Dec 2025, 19:33

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Good Question , here we have only two choices either adopt slower pace or adopt faster pace. We need to find time for each pace in order to solve the question. If the mountain climber has taken slower pace then , he will complete the climbing in 2.5 hrs (150 mins) with required 20 canisters ,well below 25. Now if he takes faster pace then , without break it will not be possible since 30% more oxygen is required in this case and mountaineer is having deficit of 1 canister. so He needs to stop at some point which will take more 25 mins. So , total time taken will be 120 Mins + 25 Mins which is equal to 145 Mins. So , minimum time taken is 145 Mins.

Bunuel

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?

A. 120 minutes
B. 140 minutes
C. 145 minutes
D. 150 minutes
E. 175 minutes

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topgmat25

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16 Dec 2025, 19:55

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At the slower pace the time is 10/4 = 2.5 hours = 150 minutes if he didn't need to stop. He use 10*2=20 oxygen canisters < 25 -> no stop -> total 150 minutes

At the faster pace the time is 10/5 = 2 hours = 120 minutes if he didn't need to stop. He use 1.3*20=26 oxygen canisters > 25 -> stop -> add 25 minutes -> total 145 minutes

145 minutes < 150 minutes

At the faster pace the time is shorter.

The answer is C

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16 Dec 2025, 20:18

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Oxygen canisters needed at the slower pace: 10x2=20
Oxygen canisters needed at the faster pace: 20 + 20*30/100 = 26

The climber has 25, so at the faster pace he needs to stop once.

Time needed at the slower pace: 10/4 = 2.5 hours = 150 minutes
Time needed at the faster pace: 10/5 * 60 + 25(STOP) = 145 minutes

Minimum time = 145 minutes

The correct answer is C

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16 Dec 2025, 20:19

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Bunuel

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Total journey : 10km
Available 25 cans

Slow Pace
4 km/h
2 can/km

For 10km, 150 mins, no stopping and use 20 cans

Faster pace
5km/h
2.6can/km

Without can restriction, For 10km, time required 120 mins

So for 10km he needs 26 cans, so he has to stop and get additional cans, which adds 25 mins

Total time for faster pace 145 mins

Minimum time is 145 mins

Correct Answer C

HarishChaitanya

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16 Dec 2025, 20:52

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From the question it is understood that only one type of peace is to be selected
Therefore if the climber selects fast pace he consumes 13 cylinders for 5 km and completes in 1 hour
If he has to complete 10km he will run out of the oxygen and he needs to stop for 25 mins at the rest area where he can pick oxygen as well, if this pace is followed the time for 10km is 145 min

If show pace is followed, 10km takes 150 min and cylinders consumed will be 20

The earliest time is 145 min and the option is c

Quote:

Bunuel]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?

A. 120 minutes
B. 140 minutes
C. 145 minutes
D. 150 minutes
E. 175 minutes

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16 Dec 2025, 21:19

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Slow Pace:
4km per hour. Has to travel 10 km -> 2.5 hours travel time
2 oxygen canisters per km. For 10 km -> 20 oxygen canisters which is less than (25 oxygen) the canisters climber started with.
So no need to stop.
Total time: 2.5 hours -> 150 minutes

Fast Pace:
5 km per hour. Has to travel 10 km -> 2 hours travel time
30% more -> 1.3 * 2 -> 2.6 canisters per km -> 26 canisters per 10 km.
Need one more oxygen cylinder. So the person has to stop at the stopping point to get the 1 extra canister.
Total time: 2 hours + 25 mins -> 120 + 25 -> 145 minutes

So min time is 145 mins

Ans: option C

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16 Dec 2025, 23:04

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SLOWER SPEED= 4KMPH + 8 CANITER
FAST SPEED = 5PKPH + 8+30%= 10.4 CANISTER

FOR MINIMUM TIME SPEED SHOULD BE MAX SO
10 KM @ 5KMPH TAKES 120 MINUTES+ A BERAK OF 25 MINUTES REGARDELESS OF SPEES

120+25= 145

Bunuel

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16 Dec 2025, 23:19

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In slower pace : climber covers 4Km in 1hr and 2 cannisters are needed per km from this 8 cannisters are needed per hour

In faster Pace: The climber covers 5Km in 1 hr and 1.3*2 cannisters are needed per km from this 13 cannisters per hour.

If climber chooses slower pace he can cover 10 km in three parts i.e 4km,4km,2km i.e he needs total 1+1+0.5 hrs = 2.5hrs =150min, the total no of cannisters = 8+8+4=20 so there is no need for break.

If climber chooses faster Pace he can cover 10km as 5km,5km.=2hrs of time=120min, total no of cannisters needed = 26. But he have only 25 cannisters. so he need to stop for a break for loading an extra oxygen cannister . this adds extra 25 mins to the time. therefore the total time is 145 min.

Minimum time needed=145min.

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16 Dec 2025, 23:20

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Dist to cover 10Km

Condition 1 slower pace
speed 4km/hr; time to cover 10km is 2.5 Hr=150min
can used at a rate of 2can/km=20 can, No need to stop to pick up can

Condition 2 faster pace
speed 5km/hr; time to cover 10km is 2hr=120min
can used is 30% more than slower pace=2.6/km=26 can. Hence need 1 can to pick up.
Now, it adds 25min to pick up can at stopping point, Now the total time to reach becomes 145min

since 145<150, hence as per question option C is correct answer

Bunuel

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16 Dec 2025, 23:20

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The fastest possible travel time, not factoring for oxygen canisters is to go at the faster pace without stopping which would take 2hrs or 120 mins. However, you would need 2*1.3*10 = 26 oxygen canisters to do that. So eliminate answer A.

If we did the fastest possible travel time and did stop at the stopping point, the oxygen canisters would not be something we have to factor for. Therefore the total travel time would be 120mins + 25mins = 145mins.

See if traveling at the slower pace without stopping is less than 145 mins.

Traveling at the slower pace without stopping, not factoring for oxygen canisters, would take 10/4hrs = 2.5hrs = 150 mins. Without even factoring for whether this is possible or not based on oxygen canisters, this option is longer than 145 mins and thus, would be incorrect even if it were possible.

Answer is C.

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16 Dec 2025, 23:32

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My favorite kind of questions - more words, less variables, less math, more logic

Here's the simplest approach I can think of - the mountain climber is 10 kilometers away from the summit. At the slower pace, he averages 4 km / hr, which means he needs about 2.5hrs or 150 minutes to summit. He'll use 2 oxygen canisters per hour, or a total of 20 out of the 25 canisters he is allotted. That's pretty much all for the slower method - we can keep the 150 minutes aside for now.

Now, the faster method is at an average speed of 5 km / hr, which means if oxygen is taken out of the equation, the climber only needs 2 hours to summit. However, this method implies 30% more oxygen consumption for "any given distance", which can be taken to assume 1.3x the oxygen requirement every hour in comparison to the slower method. Hence, where 20 oxygen canisters were enough in the slower method, the faster one will need 20*1.3 or 26 canisters.

Anything more than 25 will prompt the climber to stop and add 25 minutes to the duration (irrespective of him collecting 1 or 100 canisters). Still, this is faster, the climber needing 120 + 25 = 145 minutes in total, which is the answer.

Bunuel

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remdelectus

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16 Dec 2025, 23:35

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p=4km/h
d=10km
t=10km/4km/h=2.5 150minutes
O2 rate=2 can/km
total=10kmx2can/km=20 can
can carried=25 can

faster pace
time=10/5hrs=2hrs 120minutes
Add stop time:120+25=145 minutes

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16 Dec 2025, 23:37

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Since the question asked 'the minimum time possible to reach the summit' = the fastest way

since the fastest way to reach the summit is to use the 'faster pace'

if the climber use the 'faster pace', the time needed

10 km : 5km/h = 2 hour = 120 mins (A)

Bunuel

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16 Dec 2025, 23:41

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If faster pace, the cans required = 2*1.3 can/km = 2.6can/km => 26can required for whole trip. That means a 25min.s stoppage is required.
Total time (faster pace) = 2hrs + 25mins =2 hrs 25 mins = 145 minutes

If slower pace, the cans require = 2 can/km => 20can required for whole trip. That means no storage required.
Total time (slower pace) = 10/4 =2.5hrs = 150 minutes

B

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16 Dec 2025, 23:47

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Oxygen consumption at each place ;
Slower pace = 10 km x 2 canisters per km = 20
Faster pace = 10 x 2.6 = 26 Canisters .(30 percent more canisters are needed in faster pace)
Stops 25 minutes regardless of oxygen taken
Slower pace
Speed = 4 km/h , Time = 10/4 = 2.5 hrs = 150 min.
Faster Pace
Speed = 5 km/h , Time = 10/5 = 2 hrs +25 min break = 120+25 = 145 mins. Correct ans C ✅

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17 Dec 2025, 00:08

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Bunuel

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Given,
Distance = D = 10km
Number of oxygen canisters => O = 25
Speeds and consumption of Oxygen => s1 = 4 kmph, o1 = 2 / km
s2 = 5 kmph, o2 = o1 ( 1 + 30/100) = 1.3 (o1) = 2.6 / km

Considering s1 and o1,

Oxygen consumption for D (10km) => o1 * D = 2 * 10 = 20 Canisters (Stop NOT required)

Speed = Distance / Time => Time = Distance / Speed

t1 = D/s1 => 10/4 = 5/2 hrs = 150 minutes

Considering s2 and o2,

Oxygen consumption for D (10km) => o2 * D = 2.6 * 10 = 26 (Stop required)

t2 = D/s2 => 10/5 = 2 hrs = 120 minutes

t2 = t2 + 25 minutes (since stop is needed to collect canisters)
t2 = 120 + 25 = 145 minutes

Therefore, answer C) 145 minutes

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17 Dec 2025, 00:45

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At 4 kph, the climber would take 10km / 4kph = 2.5 hours or 150 mins to complete the climb. He won't need to stop for canisters because he has 25 and he would have used only 20 @ 2 cans/km

At 5kph, he requires 30% more cans which is (1.3 * 2 cans/km) 2.6 cans/km. Therefore, he will need 26 cans for 10 km and hence, he will need to stop for more.
Time taken for travelling = 10 / 5 = 2 hours or 120 mins. With an additional 25 mins for collecting more cans, the total time comes to 145 mins, which is lower than the first case.

Answer = 145 mins

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17 Dec 2025, 01:57

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Distance = 10km
Vslow = 4km/h ---- 2 can/km
Vfast = 5km/h ---- 1.3*2 can/km

Slower pace:
Time = 10/4 = 2.5 hours = 150 minutes
Canisters used = D * can/km = 10 *2 = 20
20<25 - No stopping required
Total time taken = 150 minutes

Faster pace:
Time = 10/5 = 2 hours = 120 minutes
Canisters used = 10 * 1.3*2 = 26
26>25 - Stop is required
Total time taken = 120 +25 = 145 minutes.

145<150,
Therefore, the minimum time required = 145 minutes
Answer. C

Bunuel

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17 Dec 2025, 02:44

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Easily to see that if climber choose slower pace, it takes 2.5 hour to complete 10 km. if climber choose faster pace, it takes 2 hour. But we don't have enough oxygen, (the amount of oxygen needed is 2x1.3.10=26 oxygen canisters), so we need to add 25 minutes at stopping point. minimum possible time will be 145 minutes

Bunuel

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17 Dec 2025, 02:52

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At the slower pace: the climber's pace is 4km/h and use 2*4=8 oxygen can/km => The total oxygen can the climber needs is 8*10/4= 20 < 25 cans the climber had initially, therefore the climber doesn't need to stop to take extra cans.
Total time = 10/4 = 2.5h = 150 min

At the faster pace: the climber's pace is 5km/h and use 5*2*1.3=13 oxygen can/km => The total oxygen can: 13*10/5= 26 > 25, hence the climber needs to take extra cans.
Total time = 10/5 + 25 min = 2h + 25 min = 145 min

Answer: C

Bunuel

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