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

chasing725

Joined: 22 Jun 2025

Last visit: 13 Jan 2026

Posts: 176

Own Kudos:

173

[1]

Given Kudos: 5

Location: United States (OR)

Schools: Stanford

Posts: 176

Kudos: 173

[1]

16 Dec 2025, 10:05

Kudos

Bookmarks

Bunuel

Slower Pace

Speed = 4km / hr
Oxygen consumed = 2 can per km

So in this pace the climber doesn't have to refill

Time taken = time of climb only
= 120 min + 30 min = 150 mins

Faster Pace

Oxygen Consumed = 1.3 * 2 = 2.6 cans / km

So in the very last leg, the climber will need to pick up extra can.

Time taken = Time of climb + Time to pick up extra can

= 120 + 25 = 145 mins

Option C

Kinshook

Major Poster

Joined: 03 Jun 2019

Last visit: 20 Apr 2026

Posts: 5,985

Own Kudos:

5,855

[1]

Given Kudos: 163

Location: India

Schools: HBS '26 CBS '26 Wharton '26

GMAT 1: 690 Q50 V34 Verified score

WE:Engineering (Transportation)

Products:

Schools: HBS '26 CBS '26 Wharton '26

GMAT 1: 690 Q50 V34 Verified score

Posts: 5,985

Kudos: 5,855

[1]

16 Dec 2025, 10:06

Kudos

Bookmarks

Option 1: Slower pace

Time required to climb the mountain = 10/4 = 2.5 hours = 150 minutes
Oxygen canisters required = 10*2 = 20 oxygen canisters
Since he carries 25 oxygen canisters and need to use only 20, he need not stop at the Stopping point to pick them

Time in minutes = 150 minutes

Option 2: Faster pace

Time required to climb the mountain = 10/5 = 2 hours = 120 minutes
Oxygen canisters required = 10*2*1.3 = 26 oxygen canisters
Since he carries only 25 oxygen canisters and need 26 of them, he need to stop for 25 minutes at the Stopping point to pick 1 extra.

Time in minutes = 120 + 25 = 145 minutes

Minimum possible time, in minutes, needed for the climber to reach the summit = min (150,145) = 145 minutes

IMO C

obedear

Joined: 05 Sep 2024

Last visit: 20 Apr 2026

Posts: 61

Own Kudos:

[1]

Given Kudos: 11

Products:

Posts: 61

Kudos: 39

[1]

16 Dec 2025, 10:11

Kudos

Bookmarks

I selected C - 145 minutes.

We want to minimize the time that it takes for the hiker to travel 10km.

Let's have him start at the faster pace, but lets make sure we understand how many oxygen cannisters he is using per kilometer and per hour.
Slow Pace: 4 km/hr; 2 cannisters/km, effectively: 8 cannisters/hour
Fast Pace: 5 km/hr; 1.3*2 cannisters/km = 2.6 cannisters / km, or 13 cannisters / hour.

Our constraint is the number of cannisters which is 25, which prevents us from traveling the entire 10 kms in 2 hours at the faster pace because we do not have enough cannisters.

Lets travel the first 5kms in 1 hr at the faster pace, leaving us with 12 cannisters.

Let's say we continue to travel at the faster pace and use as many cannisters as we can to go 4 more kms at the faster pace, spending 48 more minutes and 10.6 cannisters, leaving us with 1.4 cannisters and 1 km to go. At this point our slowest option requires 2 cannisters / km so we need to rest in order to refill on cannisters. After the 25 min rest, we are at 60 + 48 + 25 = 133 minutes, with 1 km to go. We can travel again at the faster pace, using 12 minutes to get there, for a total of 145 minutes.

However, what if, after 5kms, we used the slower pace? With 12 cannisters and 5 km to go, we could travel for 1.5 hrs at the 4 km/hr pace and it would only take us 75 minutes to travel 5kms while using 10 cannisters. 60 + 75 = 135 minutes. This is not an answer choice. Can someone explain why?

SBN

Joined: 12 Jun 2023

Last visit: 15 Mar 2026

Posts: 22

Own Kudos:

[1]

Given Kudos: 68

Posts: 22

Kudos: 16

[1]

16 Dec 2025, 10:14

Kudos

Bookmarks

Ans is ( C ) imo

First ignoring the 25 min refill pause, the slower approach would take 150 mins (10 div by 4 to get hrs hiked and then times 60 to convert to mins)
Similarly the faster approach would take 120 mins

now to check which approach requires a refill. the slower approach needs 2 cans per km, which means the hiker would have consumed 20 cans over the 10km hike and the faster approach which used 30% more oxygen(or 2.6 cans per KM) would need 26 cans to reach the summit, since the hiker has only 25 cans, the faster approach time now gets another 25 mins added to it. So Final times :
Slower : 150 mins
Faster : 120+25 = 145mins (options C)

gchandana

Joined: 16 May 2024

Last visit: 20 Apr 2026

Posts: 191

Own Kudos:

139

[1]

Given Kudos: 169

Location: India

Products:

Posts: 191

Kudos: 139

[1]

16 Dec 2025, 10:15

Kudos

Bookmarks

Two options are available here,
Option one:
4 kmph and 2 canisters per km.
This means the climber will use 20 canisters. Time taken is 10/4, 2.5, which is 150 minutes.

Option two:
5 kmph and 2.6 canisters per km.
This means the climber will need 26 canisters, but only 25 are available, so he would have to stop for 25 minutes to load up the extra canisters. And the time taken to climb is 10/5, 2 hours plus 25 minutes for the stop, which gives a total of 145 minutes.

So the minimum possible time is 145 minutes, option C

Bunuel

Rahilgaur

Joined: 24 Jun 2024

Last visit: 26 Jan 2026

Posts: 162

Own Kudos:

125

[1]

Given Kudos: 47

GMAT Focus 1: 575 Q81 V82 DI72 Verified score

Products:

GMAT Focus 1: 575 Q81 V82 DI72 Verified score

Posts: 162

Kudos: 125

[1]

16 Dec 2025, 10:18

Kudos

Bookmarks

Bunuel

D= 10 Km, Oxygen Can = 25

S1 = 4 km/hr, 2 oxygen per Km ---> Time taken to cover 10km = 10/4 = 2.5 Hrs = 150 minutes ---> Oxygen Canister used = 20 ---No extra canisters are required

S2= 5km/h, 2.6 Oxygen Can. per Km ---> Time taken to cover 10 Km= 10/5 = 2 hrs = 120 minutes ---> Oxygen Canister used = 2.6 *10 = 26 ---> To pick up additional one Canister 25 Minutes are required ---> Still total time taken = 120+25 =145 minutes Answer C.

Archit3110

Major Poster

Joined: 18 Aug 2017

Last visit: 20 Apr 2026

Posts: 8,625

Own Kudos:

5,190

[1]

Given Kudos: 243

Status:You learn more from failure than from success.

Location: India

Concentration: Sustainability, Marketing

GMAT Focus 1: 545 Q79 V79 DI73 Verified score

GMAT Focus 2: 645 Q83 V82 DI81 Verified score

GPA: 4

WE:Marketing (Energy)

Products:

GMAT Focus 2: 645 Q83 V82 DI81 Verified score

Posts: 8,625

Kudos: 5,190

[1]

16 Dec 2025, 10:20

Kudos

Bookmarks

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

total distance is 10 km
cannisters available are 25

slow pace time taken will be
4 kmph ; 10/4 ; 2.5 hours
2 o2 cannister per minute ; 2* 10 ; 20 canisters
total time taken 120+ 30 ; 150 minutes + 25 min ( stoppage ) = 175 minutes

fast pace time taken will be
10 / 5 = 2 hours
1.3* 2 = 2.6 kms per canister
10*2.6= 26 canisters
net time is 120 + 25 ( stoppage ) = 145 minutes

OPTION C ; 145 minutes is correct

JiriNovacek

Joined: 14 Dec 2025

Last visit: 07 Feb 2026

Posts: 17

Own Kudos:

Posts: 17

Kudos: 9

16 Dec 2025, 10:28

Kudos

Bookmarks

C) 145 min is the minimum time. He should go faster (4 km/hours) and to one stop. So he goes 2hours and 25 minutes

Abhishake04

Joined: 08 Jan 2025

Last visit: 20 Apr 2026

Posts: 12

Own Kudos:

[1]

Posts: 12

Kudos: 9

[1]

16 Dec 2025, 10:29

Kudos

Bookmarks

Taking 1st condition when 10km with a speed of 4kmph and 2 O2 canister/km. So for 10km total O2 canister used is 20 which is within limit as we have 25 Nos. So time taken= 10/4 hrs= 150mins.
Now when 10km with a speed of 5kmph and 2.6 [2*1.3 bcz 30% increase] O2 canister/km. So, for 10km O2 canister is 26 hence, we need to take 1 more canister and spend 25mins.
Time taken to travel= [10/5]*60 + 25= 145 mins
Hence, shortest duration is 145 mins.

harishg

Joined: 18 Dec 2018

Last visit: 09 Apr 2026

Posts: 176

Own Kudos:

174

[1]

Given Kudos: 31

GMAT Focus 1: 695 Q88 V84 DI81 Verified score

Products:

GMAT Focus 1: 695 Q88 V84 DI81 Verified score

Posts: 176

Kudos: 174

[1]

16 Dec 2025, 10:29

Kudos

Bookmarks

Distance to be covered is 10km with 25 oxygen canisters.

Under slower pace, 20 oxygen canisters are required. Therefore, no stoppage time

Time taken = 10/4 hrs =150 minutes

Under faster pace, 26 oxygen canisters are required. Therefore, stoppage time of 25 mins will apply.

Time taken = 10/5 hrs +25 mins = 145 mins (min possible time)

Therefore, Option C

martina3750

Joined: 24 Apr 2025

Last visit: 05 Jan 2026

Posts: 6

Own Kudos:

[1]

Given Kudos: 212

Posts: 6

Kudos: 3

[1]

16 Dec 2025, 10:29

Kudos

Bookmarks

Distance = 10 km

Slow Option
4 km/h for 10 km is 2.5h = 150 minutes
Oxigen = 2 * 10 km = 20 (sufficient)

Fast Option
5 km/h for 10 km is 2h = 120 minutes
Oxigen = + 30% of slow option = (30/100 * 2) + 2 = 2.6 Oxigen / km -> 2.6 * 10 = 26 (not sufficient)
Must add the stop = +25 minutes
Total time = 120 + 25 = 145 minutes

Solution = 145 minutes in the slower pace

Bunuel

ikan444

Joined: 31 May 2025

Last visit: 19 Apr 2026

Posts: 18

Own Kudos:

[1]

Given Kudos: 23

Posts: 18

Kudos: 12

[1]

16 Dec 2025, 10:30

Kudos

Bookmarks

At slower pace

Covers 4 km/h and uses 2 oxygen canisters/km, total canisters needed = 2*10 = 20 (he already has 25, so he won't need to stop)
In 1 hour, he travels 4 km.
In 2 hours, we traveled a total of 8 km.
Finally, to complete the remaining 2 km, he will take 30 minutes more.

Total time to reach the summit = 60+60+30 = 150 mins

At faster pace

Covers 5 km/h and uses 1.3*2 = 2.6 cannisters/km, total canisters needed = 2.6*10 = 26 (we fall short of 1 cannister, so the climber will have to make a stop to get an extra cannister)

In 1 hour, he travels 5 km.
After this, he needs to make a stop to collect an extra canister, which adds 25 minutes.
And then starts again, in the next 1 hour he completes another 5 km (completes 10 km!)

Total time to reach the summit = 60+25+60 = 145 mins (C)

surisoni

Joined: 28 May 2021

Last visit: 22 Jan 2026

Posts: 3

Own Kudos:

[1]

Given Kudos: 8

Posts: 3

Kudos: 2

[1]

16 Dec 2025, 10:37

Kudos

Bookmarks

Bunuel

Ans: Option C.
If travelling at slower pace:
Time taken = 10/4 hours = 2.5 hours = 150 minutes
Canisters used = 2*10 = 20.

If travelling at faster pace:
Time taken = 10/5 hours = 2 hours = 120 minutes
Canisters used = 20*1.3(30% more oxygen than slower pace) = 26.
But climber has only 25 canisters. So, he would have to stop to take more canisters.
Waiting time for more canisters = 25 minutes.
Total time = 120 + 25 = 145 minutes.
Thus, Option C.

MANASH94

Joined: 25 Jun 2025

Last visit: 11 Apr 2026

Posts: 88

Own Kudos:

[1]

Given Kudos: 16

Location: India

Schools: IIM IIM ISB

GPA: 2.9

Schools: IIM IIM ISB

Posts: 88

Kudos: 63

[1]

16 Dec 2025, 10:38

Kudos

Bookmarks

Lets calculate separately:
Using the 4 Kms /hr speed and 2 canisters per km

The bifurcation will be : 4 Km in 1st hour (8 Cans), Another 4 Kms in 2nd (Another 8 cans) and remaining 2 kms in 0.5 hours which is 30 mins ( 4 cans)
Total time in this case : 2.5 hours and 20 cans. So no stops needed here.

In the second case: Oxygen required per km will be 2*1.3 = 2.6 canisters / km.
1st Hour travels 5 km and takes 13 canisters.
2nd Hour there has to be a stop in between because the canisters will needed for refill/restock.
But get this that the total distance to be travelled will be always 10 kms and there will be an intermediate stop. So, the climber will reach by 2 hours (@5km/hr speed) and 25 mins stop. so total is 145 minutes. This is still lower than the trip 1 time. Hence Ans is C 145 mins.

Bunuel

Tyler1

Joined: 17 May 2025

Last visit: 19 Dec 2025

Posts: 4

Given Kudos: 4

Posts: 4

Kudos: 0

16 Dec 2025, 10:39

Kudos

Bookmarks

We have two rates here, the slower and faster.

Slower:
4KM per hour . 2 Oxygen per KM. That comes out to 8 Oxygen used per hour.

It's 10km to the peak, meaning at 4km/1Hr rate it will take 2.5 hours to gt to the peak, or 150 minutes. Since only 20 Oxygen will be used, there is no need to take the stop to refuel the oxygen supply.

Now let's look at the faster rate

Faster:
5Km per hour. Oxygen is consumed at 130% of the slower pace, meaning 2 oxygen/1km is multiplied by 1.3 to get 2.6 Oxygen/1Km. Since the trek is 10KM long to the summit, 26 oxygen canisters will be used at this Faster rate, so the refuel stop will be necessary and will add 25 minutes to the trip.

So at a speed of 5km per hour it will take two hours to traverse 10KM to the summit. 2 hours is 120 minutes plus the 25 minute stop to refuel on oxygen = 145 Minutes. Answer Choice C

Sumimasen

Joined: 21 Jan 2024

Last visit: 20 Apr 2026

Posts: 36

Own Kudos:

[1]

Given Kudos: 11

Products:

Posts: 36

Kudos: 33

[1]

16 Dec 2025, 10:39

Kudos

Bookmarks

Total distance to be covered is 10 km & available O2 canisters are 25 nos.
Slower pace = 4 km/h
Total minimum time at slower pace = 10/4 = 2.5 hour = 2.5*60 = 150 Minutes
Total O2 canisters required at slower pace = 10 * 2 = 20 nos. (<25 nos of available canister, so no stop require to pick more canisters)

Faster pace = 5 km/h
Total time at faster pace = 10/5 = 2 hour = 2*60 = 120 Minutes
Total O2 canisters required at faster pace = 30% extra from slower pace = 20 * (1+30/100) = 26 nos (>25, so stop require to pick more O2 canisters which will add 25 minutes as per question statement. Hence overall minimum time at faster pace will be 120+25 = 145 Minutes. (C)

gemministorm

Joined: 26 May 2025

Last visit: 19 Apr 2026

Posts: 143

Own Kudos:

110

[1]

Given Kudos: 57

GMAT Focus 1: 565 Q82 V79 DI73 Verified score

GMAT Focus 2: 605 Q84 V83 DI73 Verified score

Posts: 143

Kudos: 110

[1]

16 Dec 2025, 10:40

Kudos

Bookmarks

Given: 25 O2 can present, 10km to travel
slow pace: 4km/hr + 2 O2/km -> time = 10km/4km/h = 2.5 h = 150 min and 10*2 = 20 O2 consumed < 25 hence possible.

with fast pace: 5km/hr + 1.3*2 = 2.6 O2/km -> time = 10km/5km/h = 2 h = 120 min and 2.6*10 = 26 O2 consumed > 25 hence need to stop for refill + 25 min, total time = 120+25 = 145 min

min. time is with fast pace -> 145 min

aka007

Joined: 15 Nov 2019

Last visit: 19 Mar 2026

Posts: 42

Own Kudos:

[1]

Given Kudos: 84

Posts: 42

Kudos: 37

[1]

16 Dec 2025, 10:44

Kudos

Bookmarks

Let C be the number of O2 Canisters carried.

Slow Pace is 4kmph @ 2C/ km or 8C/hr
Fast pace is 5kmph @ 2.6C/km or 13C/hr

total time in slow pace= 10km/4kmph= 2.5hrs or 150 mins @ 20C consumption
total time in fast pace= 10km/5kmph= 2 hrs or 120 mins @ 26C consumption

However, we need to add pit stop break to cater to refilling of o2 canisters i.e. 120 mins + 25 mins= 145mins

Bunuel