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

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crystxl

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15 Dec 2025, 21:36

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The cyclist has 48 mins to cover some additional distance twice PLUS 10 miles (she is already 10 miles away from home, she will go an additional distance one way and then come back on the same way again).

Consider the additional distance to be x miles.

If you set up the Speed, Time, Distance table, it would look like this.

Speed	Time	Distance
1 mile / 3mins	48 mins	10 + 2x

[1 mile][/3 mins] * 48 mins = 10 + 2x

--> 16 = 10 + 2x
--> x = [16-10][/2]
--> x = 3

Therefore, she can travel an additional 3 miles before turning around.

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

prepapr

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15 Dec 2025, 22:04

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Let d be the distance that cyclist travels before turning back.
Then, the total distance to be covered by cyclist onward and back home = d + d +10 = 2d+10
The time taken for cyclist to reach back = 48 minutes.
Velocity of cyclist = 1 mile per 3 min

Appying the distance time velocity formula:
Distance/Time=Velocity
(2d + 10)/48 = 1/3
2d + 10 = 48/3 = 16
2d = 16-10=6
d=3miles

Bunuel

bhnirush

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15 Dec 2025, 22:22

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Bunuel

Given,
Distance from home - 10 miles
Constant speed of cyclist - 3 minutes per mile -> 20 mph
So,
Time needed to cover the 10 miles to home - 10/20 Hrs -> 1/2 hrs -> 30 mins ( remaining time -> (48mins - 18mins) = 18 mins )
Now, he has 18 minutes remaining as he needs to be back home by 10:48pm -> he can cover 6 miles in 18 minutes (3 miles away from home and 3 miles back towards home)

Therefore, answer B) 3 miles

jai169

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15 Dec 2025, 22:24

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The first few keyword to notice in this question is Cyclist continues to ride Farther(opposite direction) also speed in th equestion is given in 3 minutes per mile
so we will solve in same way

first lets find out how much he can travel in time 10 pm to 10:48 pm in 48 minutes
so we can cross-multiplication in this scenario

3 minute ------ 1mile
48 minut-------- X Mile

x=(48*1)/3= 16 miles

so now we know he can travel 16 miles total in which he is already farther from home 10 miles so now we can get additional miles he can travel is 16-10= 6
but in 6 miles he has to travel half a miles in each direction farther an again when he has to come back to home. so additional miles he can travel is 6/2= 3 miles
Answer is 3 miles (B)

Bunuel

ayaazansari97

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15 Dec 2025, 22:28

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Imagine a straight line:
|--x--|--------10 Miles--------|Home|

Given: 3 mins/miles => 1/3 miles/mins
Total Distance travelled between 10PM to 10:48PM -> x (away from home) + x (towards home after turning around) + 10 Miles
Total Time: 48 Mins

Total Distance (Miles) = Speed (miles/mins) * Time (Mins)
2x + 10 = (1/3)*48
2x + 10 = 16
2x = 6
x = 3

Option: B

harsha12344556

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15 Dec 2025, 22:31

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Firstly the distance between house and cyclist is 10 miles and Now in the question it is mentioned he takes 3 minutes per mile, and he returns back at 10:48pm which means 48 minutes to be back so , which means 16 miles in totally from 10:00pm, he can travel 3 more miles away from where he is right now which means 3 away from 10 miles.

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15 Dec 2025, 22:39

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Outward Trip = x miles post 10 pm
Return Trip = x+10 miles (since the initial distance is 10 meters away)

Total Distance travelled = x+x+10 = 2x+10
Now, total time taken = Distance/Speed (in miles per minute) or Distance* Speed (minutes/mile)
Speed = 3 minutes per mile
Total time = (2x+10)*3 = 6x+30 = 48
6x=18
x=3 miles

So the cyclist can go an additional 3 miles to fulfill the conditions. Hence the answer is Option (B) 3

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15 Dec 2025, 22:56

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Let the Distance travelled from home after 10 be x miles
Time taken to travel x miles= 3x min
For travelling back home the total distance covered = 10+x miles
Time taken for travelling back = 3(10+x)
Total distance travelled after 10:00 clock = 2x+10
Total time = 3x+3(10+x)= 6x+30
6x+30=48
x=3 miles.

gmat147

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15 Dec 2025, 22:56

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The cyclist has to be back by 10:48 giving him a travel time of 48mins.
rate is given by 3min per mile and let "x" be the extra distance travelled
=> 3x + 3(x+10) = 48
=> x = 3

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15 Dec 2025, 23:11

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Let x be the additional distance the cyclist can travel before turning back home,
Pt. A - point when the cyclist is 10M away from home
Pt. B - point of U-turn
Distance b/w pt. AB - X

H-----------------A---------------->B
10m @10PM X |
H<----------------------------------B
@10.48pm (10+X)

total distance the cyclist will cover from 10pm to 10:48pm = 2x+10

total distance covered in 48 mins speed of 1mile/3mins = 48x1/3=16miles
2x+10=16
solving for x, x=3miles.

additional distance covered before making the u-turn is 3miles.

Saks116

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15 Dec 2025, 23:17

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Total didtance = 10 + X +X +10 = 2X+10
I'm only intetrested in, after 10 miles journey. Therefore before turning and after turning same distance will be cover and than, again 10 miles.
Those last 10 miles will take 3o mins to cover as speed is given.
reminaing 18 mins will be consumed in before turning and after turning.
3 miles/min, therefore he will cover 3 miles and takes U turn again 3 miles --> Total time 18 mins .
Thus ans is 3 Miles

B

sureshkumark

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15 Dec 2025, 23:17

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The answer is choice B as the person needs to travel another 3 miles as for a total of 48 minutes, he needs to travel 16 miles. So, a total of 26 miles will be travelled. Half the distance is 13 miles.. So, 13-10 =3 miles.

gcshekar08

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15 Dec 2025, 23:21

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Time taken to travel = 10:48 - 10 = 48 minutes= 4/5 hr
Distance to find (D) = d+d+10 = 2d + 10
Speed = 3 miles/ minutes= 20 mph
D = s*T
D = 20 * 4/5 = 16
16 = 2d + 10
2d = 6
d = 3 miles

SwethaReddyL

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15 Dec 2025, 23:26

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since the total time taken is 48 mins and speed are also given as minutes per mile, divide the time by speed to get the distance
48/3 = 16
since it is given that the cyclist is 10 miles away from home, excluding that the cyclist has travelled 6 miles and that is included of both from and to
6/2 = 3
So, the additional miles that cyclist travelled away from home after 10.00 pm is 3 miles

Option B

Bunuel

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16 Dec 2025, 00:14

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Answer: B

If the cyclist takes 3 minutes per mile and has to be home by 10:48 pm, ie, in 48 minutes, he can travel a total distance of 48/3= 16 miles. Since he is 10 miles from home, he can travel an additional distance of 3 miles [(16-10)/2] before returning home.

Bunuel

malvikasachdev124

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16 Dec 2025, 00:17

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------------|------------ <home> (forward journey)
{x} {10}

------------|-----------<home> (returning back home journey)
{x} {10}

3 minutes for 1 mile
48 minutes for 48/3=16 miles

2x+10=16
x=3

sanjitscorps18

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16 Dec 2025, 00:28

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Bunuel

Let the cyclist travel 'a' miles from the point he is at 10 miles away from home

Then distance to cover back home = 10 + a + a = 10 + 2a

Time to travel this distance is 48 minutes

The cyclist speed = 1 miles / 3 minutes

Hence, we get

(10 + 2a) = (1/3) * 48
=> 10 + 2a = 16
=> a = 6/2 = 3

Additional miles post 10 miles would be 3 miles.

Option B

UditVarma777

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16 Dec 2025, 00:30

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in 48 minutes he has travel the remaining distance with farther away+ turnoaround+ 10 miles
he is covering 1 mile in 3 minutes so, in 48 mins he will cover 16 miles;

3+3+10 miles

3 miles is the ans

Sujithz001

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16 Dec 2025, 00:56

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Qn. Asked: How many more miles in excess to 10miles the cyclist can travel, if he can reach back home in 48mins.

Solution:

Speed: 3 minutes/mile. For easiness, let's take it as 1/3 miles/minute.

W.K.T, Rate * Time = Distance.

Cyclist has to complete the 10miles in-order to reach home. So, time taken for 10 miles: 3*10 = 30 minutes.

Left-out time: 18 mins.

R * T = D
1/3 * 18 = 6 miles.

Answer: 6 miles

Tejasethb

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16 Dec 2025, 00:58

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Speed= 1/3 miles in 1 min=> 48 min he traveled=> 16 miles and before turning around he need to travel 3 miles So, that he can reach at 10:48 back at his home. PFA file for diagrammatic representation of the question.

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