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

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vikramadityaa

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Bunuel

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.

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Speed going to station = 45km/hr; Return trip time = 6 hours.
Statement 1: Time to reach the station = (180/45) = 4 hours. (Sufficient)
Statement 2: Car's average speed for the round trip = 36km/hr
Let the one-way distance = d
Time to station = (d/45)
Time back = 6
Total distance = 2d
Total time = (d/45)+6
Average spped: 36=(2d/(d/45+6))
36d/45+216=2d
4d/5+216=2d
d=180
Time=180/45=4 hours. (Sufficient)

Hence, OPTION D.

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17 Dec 2025, 10:59

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We are supposed to find the time to reach monitoring station or d/45

Statement 1 directly gives d and is therefore sufficient.

Statement 2 gives the avg speed for the entire trip

36= 2d/(d/45+6).

Solving for d, we get 36. Sufficient.

Therefore, Option D

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17 Dec 2025, 11:21

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Let oneway distance = D
Speed going to the station = 45 kmph
Time taken to reach station = D/45 hours =t
Time returning = 6 hours

(1) D=180 km
Time taken to reach=180/45=4 hours
Sufficient

(2) Average speed for entire round trip = 36 kmph
Total distance = D+D=2D
Total time for going and returning = t+6 hours
2D/(t+6)=36 (also, D=45t)
2(45t)=36(t+6)
t=216/54=4 hours
Sufficient

D

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time = d/45.
so distance required.

1. distance given. SUFFICIENT
2. 2d/((d/45)+6)) = 36...so, d=180..SUFFICIENT

Ans D

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Statement 1:
Total distance given = 180km
Constant speed available already which is 45km/hr.
T = 4 hrs
(Sufficient)

Statement 2:
Average speed = total distance / total time = 36
=> Distance to and from say 2d / (time to ) + (Time to return)
=> 2d /( (d/45) + 6) = 36
=> d= 180.
We can calculate the time to do which is 180/45 = 4 hours.
Sufficient

Hence D.

Bunuel

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Workshop - Station
D= 180km
S= 45km/hr
t= D/S= 180km/45km/hr
= 4hrs
Statement 1 alone is Sufficient.
A.S =36km/hr
A.D= 360km
A.T= 360/36= 10hrs
Time from Workshop to station= 10hrs - 6hrs= 4hrs
Statement 2 alone is Sufficient

so; Each statement aloe is Sufficient

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Given: From workshop to monitoring station, the speed is 45 km/hr and the return trip takes 6 hrs
We have to identify time taken to reach monitoring station.

(1) D = 180 km
T = D/S
Since we have both for the towards journey, the time can be computed.
Therefore, (1) is sufficient.

(2) Avg speed for round trip = 36 km/hr
Using this, we can identify the speed of return journey (let x).
Formula: 36 = (45+x)/2
Once we identify x, we can can identify distance using D = S*T = x*6
Now that we have D, the approach would be similar to (1)
Therefore, (2) is sufficient.

Since both options are individually sufficient, the answer is (D)

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Bunuel

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The inspection cart travels from point A (Workshop) to point B ( Remote Monitoring Station).

The cart moves from A to B, and then returns back along the same track. So, the distance is same.

Let’s consider the route (A to B): towards monitoring station.

Forward speed = 45 Kmph.

Forward time = Ta

Route (B to A) : from monitoring station.

Return speed = Sr

Return time = 6 hrs.

We need to find : Ta.

Statement 1:

The distance from workshop to monitoring station is 180 kilometers.

Ta = distance / speed forward

= 180/45

= 4 hrs.

Hence, Sufficient.

Statement 2:

The carts average speed for the trip = 36 kmph.

Average speed = Total distance / Total time taken

Total distance = 2d

Time taken = ta + 6

36 = 2d / (ta +6)

Ta + 6 = (d/18)

(d/45) + 6 = (d/18)

6 = d [(1/18) - (1/45)]

Solving, we get d =180

Substitute it in the initial equation, we get

Ta + 6 = 360/36

Ta = 4 hrs.

Hence, sufficient

Option D

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Bunuel

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From the stem, we need to find a definitive hour amount for the hours it takes to go for the initial trip from workshop to station.

Statement 1:

We're given the distance which makes life very easy! $\frac{180}{45}=4 hours$

We got a definitive value, therefore sufficient.

Statement 2:

Let t = time it takes to reach the monitoring station, and let r = the rate at which the cart travels on the return trip.

Using average rate formula:

$\frac{45t+6r}{t+6}=36$

Since this is a round-trip question, we know that $d_1=d_2$. Which we can use to reduce variables in the average rate formula.

$45t=6r$

$r=\frac{15t}{2}=7.5t$

Subbing into our average rate equation, we get:

$\frac{45t+6*7.5*t}{t+6}=36$

When we simplify this, we get:

$t=\frac{216}{54}=4$

Statement 2 is also sufficient and matches Statement 1 answer, therefore answer is D.

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Answer = D

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.
=> Since we know the distance and constant speed while traveling to monitoring station, time can be calculated = 180/45 = 4 hours. Hence, sufficient

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

=> Since we know average speed and distance is constant,
avg. speed = 2ab/(a+b)
36 = 2*45*x/(45+x)

we can find x = constant speed from monitoring station back to workshop. (not required to calculate x since we just need to prove sufficiency and not find an exact value). Once we get that, we know the hours = 6, we multiply 6*x to get total distance. Now we can simply divide distance by 45 km per hour and get the time from workshop to monitoring station. Hence, sufficient.

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Bunuel

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using s1) time = 180/45 ( this is sufficient)
using s2) total distance = avg speed * total time = 216 KM
now time to reach monitor station would be (216/2) distance -> 108/45 so this is also sufficient

amswer is D each statement suffice

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17 Dec 2025, 18:37

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Speed of the cart is constant between workshop to station (v1) = 45 km/h
return trip time (t2) =6 hours

Statement 1:
Distance main workshop to station = 180 km.
Time can be found

Sufficient

Statement 2:
Avg speed for round trip = 36 km/h

total distance / total time = 36

$\frac{(45 * t1 + v2 * 6) }{ (t1+6)}$ = 36

we can't find t1 without knowing v2 or total distance here.

Insufficient

Ans: option A

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Solution:
Given

Workshop-Monitoring	45		D
Monitoring - Workshop		6	D

Statement 1 Sufficient

	Speed	Time	Distance
Workshop-Monitoring	45	4	180
Monitoring-Workshop		6	180

Statement 2 Insufficient

	Speed	Time	Distance
Workshop-Monitoring	45	D/45	D
Monitoring-Workshop		6	D

Average speed - Total Distance/Total time = D+D/D/45+6 = 90D/D+270

Hence, Option A

Bunuel

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Bunuel

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Let d be the distance between cart and monitoring station.
Given on trip to monitoring station cart travels at constant speed of 45kmph, return trip takes 6 hours.
To find time taken to reach monitoring station = d/45
So if we know d, we have the time.

Statement 1:
(1) The distance from the workshop to the monitoring station is 180 kilometers.
we know the d.
Sufficient. AD/~~BCE~~

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

avg speed = total distance / total time
2d/(d/45 + 6) = 36.

No need to solve, one equation in one vaiable, we will know d.

Sufficient AD/~~BCE~~

Correct Answer: D

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I selected D.

I decided to set this up using an RT = D chart.

Since the distance is the same for each leg of the journey, set each distance to D/2. In general, if we are given one value of an RT=D chart and another as a variable, I rewrite the third variable in terms of the other two.
So the journey to the workshop becomes:
45 * D/90 = D/2
Journey from the workshop:
D/12 * 6 = D/2.

So the rephrase is, D/90 = ?

(1) Gives us D. Sufficient.
(2) Gives us the equation 36 = D/(D/90+6). This is enough to solve for D. Sufficient.

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Let dist be d km, s1=45 km/hr, t2= 6hr, t1=?

1. d=180km
since S*T=D
s1*t1=d
45*t1=180 sufficient

2. Sa=36

Sa=Total dist/Total time=2d/t1+t2
36=2*45t1/t1+6

sufficient Option D

Bunuel

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Take a look at statement 1 , distance between workshop to monitoring station is provided as 180 Kms. We also know that on the trip to monitoring station , the cart travels at a constant speed of 45 Km/Hr. Applying speed distance time formula , we will get time to reach monitoring station. SO , 1st Statement is sufficient. Take a look at second statement , Average speed for the whole trip is provided as 36 Km/hr. On applying average speed formula , we get 36 = (2d)/(t+6) , now replace d with 45t , we will get t value which is time to reach monitoring station. So , 2nd Statement is sufficient. So Answer is D.

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from given we have: t = d/45 and 6 = d/v where d is the distance and v is the return velocity, we need to find t?

A -> d= 180 =>t = 180/45 -> enough
B -> avg velocity = 36 = 2d/(d/45 + d/v) -> we can get v and then get d using eq 6 = d/v and then can get t = d/45 hence enough
therefore D -> each is enough.

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Let the velocity of the cart from workshop to monitoring station be Va.= 45km/hr
The velocity of the cart from monitoring station to workshop be Vb.
The time taken by cart to travel from workshop and back be t1 and t2 respectively.
Let the distance between these two points be D
Given,
Va = 45 km/hr
t2 = 6 hrs
We have to find t1

Statement 1:
D = 180 km
t1 = D/v1 = 180/45 = 4 hrs
Statement A is sufficient

Statement 2:
Average speed, v1t1+v2t2 / ( t1 + t2) = 36km/hr
45*t1 + v2*6 / ( t1 + 6) = 36km/hr
We have two unknowns t1 and v2. Hence this is not sufficient

Hence answer is A

Bunuel

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to find time From W to M (given speed i.e. require distance) and for M to W we are given time taken as 6hrs
1. Directly given D
2.Given avg speed of 36 for round trip i.e. we know entire distance is 2D and times of (D/45 hrs & 6hrs), can setup eqn and solve for D