12 Days of Christmas GMAT Competition 2025 - Day 3: An inspection cart

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Bunuel

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17 Dec 2025, 10:00

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GMAT Club Official Explanation:

An inspection cart travels from a main workshop to a remote monitoring station and then returns to the workshop along the same track. On the trip to the monitoring station, the cart travels at a constant speed of 45 kilometers per hour. If the return trip takes 6 hours, how many hours does the cart take to reach the monitoring station?

(1) The distance from the workshop to the monitoring station is 180 kilometers.

If the one way distance is 180 kilometers and the speed on the first leg is 45 kilometers per hour, then the first leg time is 180/45 = 4 hours. Sufficient.

(2) The cart’s average (arithmetic mean) speed for the entire round trip is 36 kilometers per hour.

Let the time to reach the monitoring station be t hours. The distance one way is 45t kilometers. The total round trip distance is 90t kilometers. The total time for the round trip is t + 6 hours. Average speed is total distance divided by total time, so 90t/(t+6) = 36. Solving gives t = 4. Sufficient.

Answer: D.

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paragw

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17 Dec 2025, 09:14

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We need the time taken to go from the workshop to monitoring station.
Let the one way distance by D km.
Speed on the onward trip = 45 km/hr
Time to reach: D/45

I. Distance is 180 km
Time = 180/45 = 4hrs
Sufficient

II. Return trip takes 6 hrs
=> return speed = D/6
Average speed for round trip = total distance/total time

36 = (2D)/([D/45] + 6)
Solve:
D = 180
Time to reach station = 180/45 = 4 hrs
Sufficient

Answer: D

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17 Dec 2025, 09:27

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Quote:

(1): The amount of time taken for the cart to travel is 180/45=4 hours. Sufficient
(2): Suppose that the amount of time it takes for the cart to travel to the station is t
From the question stem: d = 45*t (distance = speed * time). With the average speed of both trips being 36, we also have 2d = 36*(t+6)
=> 2*45*t = 36*t + 216 => 54t = 216 => t = 4 => It takes 4 hours for the cart to travel. Sufficient.
The answer is D

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17 Dec 2025, 09:32

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	Speed	Time	Distance
MW - RMS	45	T	D
RMS - MW	D/6	6	D
Total	36	T+6	2D

T=?
D = 45T

S1
D = 180
S=D/T
45 = 180/T
T =4
Sufficient

S2
S = D/T
36 = 2D/(6+T)
D = 18T + 108
D = 45T
45T = 18T + 108
T = 108/27 = 4
Sufficient

Answer D

Bunuel

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17 Dec 2025, 09:32

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Bunuel

Deconstructing the Question
We need to find the time taken for the outbound trip ($t_{out}$).
Given: 1. Outbound trip (Workshop $\to$ Station):
Speed = 45 km/h.
Distance $D = 45 \times t_{out}$.
From this, $t_{out} = D / 45$.
So, finding $D$ is sufficient. 2.
Return trip (Station $\to$ Workshop):
Time = 6 hours.
Distance is the same $D$.

Analyze Statement (1)
"The distance from the workshop to the monitoring station is 180 kilometers."
$D = 180$.
Substitute into our formula:
$t_{out} = 180 / 45 = 4$ hours.
We have a unique value. SUFFICIENT

Analyze Statement (2)
"The cart’s average speed for the entire round trip is 36 km/h."

Theory: Average Speed = Total Distance / Total Time

Total Distance = $2D$.
Total Time = $t_{out} + 6$.
Equation: $36 = \frac{2D}{t_{out} + 6}$
We know from the prompt that $D = 45 \times t_{out}$.
Let's substitute $D$ into the equation:
$36 = \frac{2(45 \cdot t_{out})}{t_{out} + 6}$ $36 = \frac{90 \cdot t_{out}}{t_{out} + 6}$
Divide both sides by 18 to simplify:
$2 = \frac{5 \cdot t_{out}}{t_{out} + 6}$ $2(t_{out} + 6) = 5 \cdot t_{out}$ $2t_{out} + 12 = 5t_{out}$ $3t_{out} = 12$ $t_{out} = 4$.

We found a unique value for time. SUFFICIENT

Answer: D

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17 Dec 2025, 09:36

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1)
given
ws to rms 45 km/hr
rms to ws=6 hour
time to reach rms?
1)
D=180km
t=d/s
=180/45
we get a definite answer for time sufficent
2)
avg speed=36
avg=2s1s2/s1+s2
s1 =45km/hr
36/2=45s1/s1+45
we get s2 and we have coming back time at 6 we get distance
diastance is same
so
time=dist/speed(given in question) sufficent
therefore D

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17 Dec 2025, 09:48

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(1) you have distance and speed -> distance / speed = time -> 180/45=4h

(2) avg speed = total distance / total time
avg speed = 36
total distance = 2x
total time = x/45 + 6
solve -> x=180

180/45 = 4h

each statement alone is sufficient

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17 Dec 2025, 09:53

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From workshop to monitor : 45*T1 = S
From monitor to workshop : 6*V2 = S

(1) S = 180 => sufficient to find T1
(2) v = (45+V2)/2 = 36 => V2 = 27 => S = 162 => sufficient to find T1

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(1) Obviously, t=S/v= 180/45 = 4 hours. This statement is SUFFICIENT.
(2) Lets call the distance from workshop to the station (and vice versa) is A
Round trip: 2A=36*6 => A = 108km => t = 108/45 = 2,4 hours. This statement is SUFFICIENT.

Answer: D

Bunuel

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17 Dec 2025, 09:57

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An inspection cart travels from a main workshop to a remote monitoring station and then returns to the workshop along the same track. On the trip to the monitoring station, the cart travels at a constant speed of 45 kilometers per hour. If the return trip takes 6 hours, how many hours does the cart take to reach the monitoring station?
(1) The distance from the workshop to the monitoring station is 180 kilometers.

(2) The cart’s average (arithmetic mean) speed for the entire round trip is 36 kilometers per hour.

speed to monitoring is 45 kmph
return time is 6 hours
#1
distance to monitoring station is 180 km
180/45 = 4hours
sufficient
#2
2 * 45*b = ( (45+b)) * 36
solve for b we get 30
distance is 30*6 = 180 km
time it takes to reach monitoring station = 180/45 ; 4 hours
sufficient
OPTION D is correct

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17 Dec 2025, 09:59

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Bunuel

1) Distance is know, and speed is known. Hence we can find the time.

Time = 180 / 45

Sufficient

2) Lets assume that the distance between Main workshop and monitoring station is d

average speed = 2d / (d/45 + 6) = 36

We can find the value of d and thereby find the average speed.

This statement is also sufficient.

Option D

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17 Dec 2025, 10:01

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IMO D

St 1: Since D is given & in the Q stem we have speed, we can find T. Sufficient
St 2: Avg speed = Tot dist/Tot time = 2(45*t)/(t+6), t can be solved. Sufficient

Hence D

Bunuel

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17 Dec 2025, 10:04

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From main workshop to remote monitoring station: -
Distance = x km
Speed = 45 mph
Time taken = x/45 hours

From remote monitoring station to main workshop (return trip): -
Distance = x km
Time taken = 6 hours
Speed = x/6 mph

Average speed = 2x/(x/45+6)

What is time taken to reach the monitoring station = x/45 = ?

(1) The distance from the workshop to the monitoring station is 180 km.
x = 180
x/45 = 180/45 = 4 hours
The time taken to reach the monitoring station = 4 hours
Sufficient

(2)The cart's average (arithmetic mean) speed for the entire round trip is 36 mph
2x/(x/45+6) = 36
2x = 36x/45 + 36*6 = 4x/5 + 216
2x - 4x/5 = 6x/5 = 216
x = 216*5/6 = 180
x/45 = 180/45 = 4 hours
The time taken to reach the monitoring station = 4 hours
Sufficient

IMO D

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answer D. both are sufficient.

1) 180km distance, speed 45km/hour. t=distance/velocity 180km/45km/h = 4 hours

2) average speed 36km/hour. speed from workshop to monitoring station 45km/hour. average speed = total distance/time 2x(there and back)/time 1 plus time 2.

time 1= distance x / velocity 45km/hour time 2= distance x / velocity y

therefore average speed 36km/hour = 2x / x/45 + x/velocity y

x/45 + x/velocity y = 2x/36
divide by x equals 1/45 + 1/velocity y = 2/36
1/velocity y = 1/18 - 1/45
1/velocity y = 5/90 - 2/90
1/velocity y = 3/90 = 1/30
velocity y = 30

return distance x = return velocity X return time
x = 30km/h x 6 hours (given in question)
x = 180 km

time to monitoring station = 180 km / 45km/h (given in question)
time to monitoring station = 4hours

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17 Dec 2025, 10:12

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We know the distance is the same both ways.

(1) Gives distance, we have speed, we can calculate the time taken. Sufficient.

(2) Average speed formula, when the distance covered is the same, is Avg speed = 2*s1*s2 / (s1 + s2). We have s1, we can s2, which then gives us the distance, since we have the time taken in the return trip. We can then find the time taken for the onward journey. Sufficient.

Option D

Bunuel

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17 Dec 2025, 10:13

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Bunuel

AD ---> Total distance = 180 Km , Speed = 45 KMPH , TIME = distance/speed= 180/45=4 hours. ---> Sufficient

D ---> Average speed = 36kmph = Total distance/ total time = 2x/(6+t), also t=x/45 (given )

36(6+x/45) = 2x ---> 18*6 + 18*x/45 = x ----> 18*6 + 2x/5 = x ---> x(1-2/5) = 18*6 --> x = 18*6*5/3 = 180 Km

tIME REQUIRED = 180/45 = 4 Hours -- Sufficient D answer.

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17 Dec 2025, 10:14

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Given: Distance (D) is same from workshop to monitoring station and station to workshop and t2 = 6hours
Speed from workshop to monitoring station is V1 (45km/hour) and Time t1.

Statement (1)
D is 180 km. So, t1 = D/V1 = 180/45 = 4 hours. (Sufficient)

Statement (2)
Average speed for round trip is 36km/hour.
Average speed formula = 2D/(t1+t2) = 2D/(D/V1+t2) = 36
=> 2D/((D/45)+6) =36
After solving, D=180. t1=D/V1 = 180/45 = 4hours (Sufficient)

Ans. D

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17 Dec 2025, 10:15

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Speed from workshop to station = 45kph
Time taken to workshop from station = 6hrs

(1) => Distance = 180kms => Time =180/45 = 4hrs. Sufficient
(2) => Average = 36 => Total Distance/Total Time =36 => 2d/(d/45+6) = 36. Sufficient.

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17 Dec 2025, 10:37

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Bunuel