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12 Days of Christmas GMAT Competition 2025 - Day 1: At 10:00 p.m.

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cann2406

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16 Dec 2025, 07:01

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From 10:00 PM to 10:48 PM, we have 48 mins and as cyclist is 10 miles away covering 1 mile in 3 mins. It will cover 10 miles in 30 mins.
So, 18 mins will be left to cover distance to and fro which will result in total of 6 miles.

So, away from home cyclist will cover 6/2= 3 miles.

whollyshiv

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16 Dec 2025, 07:24

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Total time from 10:00 PM to 10:48 PM = 48 minutes.
Speed is 3 minutes per mile, so the cyclist rides 1/3 mile per minute

Total distance traveled after 10:00 PM= 48x1/3 = 16miles
Let x be the additional miles the cyclist rides farther from home after 10:00 before turning around

So the total distance after 10:00 PM = x + (10+x) = 10+2x
And this distance is equal to 16miles from earlier.

Hence 10+2x=16 => 2x=6 => x=3

Bunuel

At 10:00 p.m., a cyclist is 10 miles from home on a straight road. From that time on, the cyclist continues riding farther from home at a constant speed of 3 minutes per mile, then turns around and rides back toward home at the same speed along the same road. If the cyclist must arrive back home at 10:48 p.m., how many additional miles can the cyclist travel away from home after 10:00 p.m. before turning around?

A. 2
B. 3
C. 6
D. 9
E. 12

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akrutij

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16 Dec 2025, 07:32

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Speed=3min/miles
total time = 48 minutes
total distance =possible = 48/3 = 16
let additional distance be x
So total distance back and forth = x + (10+x) = 16
x=3

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16 Dec 2025, 07:49

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Total time taken by cyclist from 10:00 PM to 10:48 PM is 48 mins.
Speed = 3 min/mile can be written as = 1/3 mile/min
Distance can be travelled in this would be 48*1/3 = 16 miles he will travel.
Since, he is 10 mile away from his home this means he have to travel 10 mile again back in order to reach his home. So, this leaves as 16-10 = 6 miles left.

Now, he has to go farthest and turnaround travel same towards his home. So, adding this should be equal to 6 mile & since both are equal the additional farthest distance he can go would be 6/3 = 3 miles.

Hence, Answer is 3 miles

Bunuel

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KhullarG

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16 Dec 2025, 08:04

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Bunuel

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B. 3 miles

SPEED = 1/3 miles pm.
Extra Mile = x
Total Distance = x + x + 10 = 2x + 10
D= S * T
2x + 10 = 1/3 * 48
2x + 10 = 16
x = 3

Leebash

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16 Dec 2025, 08:09

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Let the unknown distance covered be 'x'
Total distance covered 'D' is 10 + 2x
Speed is 3min per mile so in 48minutes he will travel 48/3 miles which is 16miles
Substitute above in D we get x=3
Ans:- 3miles

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16 Dec 2025, 08:42

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Give :

1 Start time, 10 AM
End time 10:48 AM

Subtracting both we get 48 minutes (Total time )

2. Cyclist Speed = 3 minutes / mile

3 . Is 10 miles away from home at 10 pm

Now let’s set up the distance equation

Distance travelled away from = x miles

Let D be initial distance from home
Time taken = T1
Time = Distance x speed

T1= x*3 = 3x

Distance Travelled towards home

D + x = 10 + x miles

Time taken
T2 = (10+x)* 3 min/ miles = 3(10+x)

Total time for the journey will be
T= T1 + T2

48 = 3x + 30 + 3x

Solve the equation

48 = 6x + 30

48 - 40 = 6x

18 = 6x

X = 18/6

X = 3

The cyclist can travel three additional 3 miles away from home after 10 PM before turning

corporisnobis

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16 Dec 2025, 09:15

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he must use 10 miles just to get back home from where he starts

Total Capacity: 16 miles
Mandatory Return: -10 miles
Leftover for Detour: 6 miles
so 6/2 = 3

Bunuel

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okHedwig

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16 Dec 2025, 09:43

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IMO Ans B

Forming the eq, assuming the distance travelled is 'd' from 10pm till the turning point
Then: d+d+10 = (1/3)* 48, where speed = 1mile/3min & time is 48mins
So, d= 3 !

Bunuel

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Tyler1

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16 Dec 2025, 10:52

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This is a problem where it can help to draw a quick diagram of distance.

Think about it like this C|-------------|H. When we start at 10pm, the cyclist is already 10 miles from home. This is the trick part of the question. The cyclist is traveling at a constant rate of 1 Mile per 3 Minutes. He has 48 minutes total until 10:48pm to ride back home, and that 10 mile return must be accounted for.

The cyclist will be able to travel a total of 16 miles in 48 minutes (48 minutes/3 minutes = 16 miles). Now we take 16 miles and subtract the 10 needed to get home. We have 6 miles left the cyclist can ride. He can ride 3 miles away and must also ride 3 miles in return. The question asks how many miles can he ride away from point C and that is 3 miles. Answer choice B

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16 Dec 2025, 19:48

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I visualized the question. The total trip from the 10pm starting point is 2x+10. Which is x additional on the first way and x +10 coming back. The trip is 48 minutes. I filled in the table and solved for the total minutes. x=6

Bunuel

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Shivam23shukla

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17 Dec 2025, 04:19

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in 48 mins the rider can ride 16 miles (48/3)
10 + x + x = 16
x =3

007Phoenix007

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17 Dec 2025, 06:53

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The cyclist covers 1 mile in 3 mins.
And he has to cover the entire distance in 48 mins.
So let's take the extra miles m.
So, 10+m+m= (48/3) miles
10+2m= 16
2m=6
m= 3
Therefore cyclist travels 3 more miles.
So option B is correct

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17 Dec 2025, 22:35

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The total trip took 48 minutes, which is equal to total 16 miles ride.
The return trip has additional 10 miles compared to the initial trip, so the additional miles the cyclist travel after 10pm before returning is 6/2=3 miles.

Answer: B.

Bunuel

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Duttonide

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18 Dec 2025, 07:11

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Calculate the remaining time: 10 to 10:48 -> 48 mins
Speed: 3 miles/min

In 1 min -> 3 miles
In 48 mins -> 16 miles

Given he was already 10 miles away, so need to go back those 10 miles, essentially left with 6 miles round trip, hence one side 3 miles(additional miles)

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18 Dec 2025, 20:04

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Given the cyclist is at 10 miles from home. Taking x as the extra distance he travels
Hence total distance traveled is 2x + 10 before reaching back home
total time is 48 min
speed = 1/3 mpmin
Hence time = distance / speed => 2x + 10 / (1/3) = 48
Solving we get extra distance x = 3 (Option B)

dolortempore

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18 Dec 2025, 23:13

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It can be solved without pen and paper!

So the solution is simple:

Cyclist is 10 miles away from home, now he will go some additional distance and back and for doing this event he has 48 min.

so now total distance he will travel is (X + X + 10)....X is the additional distance and X + 10 is total distance away from home.

Now we know that. 3 min per mile, so 48 min means 16 mile.

10 + 2X = 16

Hence X = 3 miles.

Bunuel

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Anish27

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20 Dec 2025, 02:43

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I used the hit and try method
A) 3mins/mile *2 mile =6 mins , so total distance covered is 12 miles and time will be 48 mins -6mins =42 mins, 3mins/mile*12mile = 36 mins. So incorrect
B) 3mins/mile*3mile =9mins, so total distance covered is 13 miles and time will be 48 mins -9mins =39 mins, 3mins/mile*13mile=39mins. So option B

Bunuel

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SSBhargavaReddy

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22 Dec 2025, 04:26

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|home|---------10miles-----------|--(X - 10)--|
*****|-------------------------X----------------|

let us consider the cyclist traveled for X miles away from home. Cyclist is already 10miles away by 10:00 pm, he/she traveled, { (X - 10) + X }, after 10:00 pm and got home by 10:48 pm, which means it took him/her 48 mins to travel {(X - 10) + X} miles. if his/her speed is 3mins/mile i.e for each mile He/She took 3 mins, how many miles covered in 48 mins?

Minutes Miles
3 --------- 1
48 ---------

=> 48/3 = 16 miles

=> X - 10 + X = 16
=> 2X = 26
=> x = 13

but we need additional miles apart from 10 miles given, so X - 10 = 13 - 10 = 3, So 3 Miles

Bunuel

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Prakruti_Patil

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23 Dec 2025, 07:37

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I find diagraming the best way to tackle TSD questions and not get caught up in the double calculations and considerations required. And I didn't diagram and tried mental math and got this one wrong, because I missed considering that when the cyclist moves ahead again, she needs to come back the same distance!

so total time spent cycling - 48 mins. At 3 mins per mile, 30 mins will be spent travelling back the 10 miles. This leaves 18 mins, which at 3 min per mile means 6 miles. This is TWICE the distance the cyclist can go, because she needs to head back. So possible further distance = 6/2 = 3 miles.

Option B