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

	R	T	D
to station	45kmh	?	d
from station	r2	6h	d

1. d = 180km
R * T = D
45 * T = 180, can solve for T from this, sufficient

2. (45 + r2)/2 = 36kmh
can find d then find T, sufficient

Ans D

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18 Dec 2025, 00:43

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Bunuel

So we know :

Cart travel to monitoring station at average speed: 45 km/h
return trip : 6 hours

can we determine the time to reach monitoring station ?

statement 1 : so here we have distance to monitoring station : 180 and we already know the average speed : 45

Average speed = Distance / Time

= 180/45 = 4 hours

So statement 1 is sufficient

statement 2 : Average speed for entire trip is 36 km / hr

so total distance will be 2d ( d+d = round trip)
and time will be d/45 + 6 hours

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

Here we will get the value of d from which we can determine time taken

statement 2 is also sufficient ; so both statement is sufficient.

So Answer is D

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18 Dec 2025, 01:36

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The answer is both statements alone are sufficient to answer the question

If statement 1 is considered it says that distance is 180 km and in question it is given that 45km/hr is the onward speed therefore time to go monitoring station is 4hrs

If statement 2 is considered, it is given that avg speed is 78km/hr from which we get speed for the return journey is 33km/hr and from question we get that time taken for returnjourney is 6hr. Again from this we get distance is 33*6

From this distance we can get time to monitoring station by using the speed 45/hr.

Hence both statements individually answers the question

Bunuel

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18 Dec 2025, 01:41

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Given :
Speed on the way to station = 45 km /hr
Return Trip home = 6 hours
ST 1 :
Distance = 180 km
Time to reach station : 180/45 = 4 hours
ST 1 is sufficient

ST 2 :
Avg Speed for entire round trip = 36 km /hr
Balance of loss method
Avg Speed vs Going Speed
36 and 45 , losing 9 km/hr on avg , Must come from return leg as going leg is fixed at 45
Converting Speed loss into distance loss
If cart travelled 10 hours Total
at 45 km/h = 450 kms
at 36 km/hr = 360 kms
loss = 90 kms
time attributed = 6 hr
speed is lower than 45 by ;
90/6 = 15 km/hr
return speed = 45-15 = 30
Distance one way = 30x 6 = 180
Going Time = 180/45 = 4 hours
Therefore Sufficient and Ans is d ✅

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18 Dec 2025, 01:44

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Can't deny how deceptively simple DS questions look at times - like this one - but a bit of logic and alertness may help save us from the falling for any traps.

Here's what's given:

- Inspection cart travels from workshop to monitoring station and then returns from station to workshop.
- On the onward trip, the speed is 45 kilometers per hour.
- But the return trip takes 6 hours.
- We need to determine how many hours the cart'll take to reach the station (the onward trip).

Now, just knowing the onward speed won't give us the distance to the monitoring station. For instance, at 45kmph, if the station is only 45 kilometres away, it will take the cart an hour. But if it is, say, 135 kilometres away, that's a 3-hour journey.

Now, let us look at the statements:

Statement 1: This says the distance from the workshop to the monitoring station is 180 kilometers. Bingo. That's a 180 / 45 = 4-hour onward journey. Statement A is sufficient.

We can now see if Statement 2 is sufficient as well or not.

Statement 2: This states the average speed for the entire round trip is 36 kmph. The key thing here is to disregard the 180 kilometers given in Statement 1, as we may have a tendency to forget we can no longer apply that info. With us just knowing the onward speed is 45kmph and the overall speed is 36kmph, we could think that's not enough information to answer the question but think again.

Essentially, the average speed at which the cart covers the distance while headed to the monitoring station, reduces when the cart returns and the total journey is taken into account. So, 1/2 of the distance is covered at 45km / hr, and 1 /2 of the distance is covered at z km / hr, which calculates to 36 km / hr for the entire journey.

Now, at this point, one can apply algebra to find the answer and prove Statement 2 is sufficient too, but logic should be enough, and we don't need the exact answer anyway for DS questions.

So, if a vehicle heads to a destination at a known speed, and then returns from that destination at an unknown speed, but in a known duration, which then leads to the average speed reducing by 9 kilometers per hour, we will be able to know the distance to the location. How, though? Well, if it was 2 hours away at 45kmph, or 90kms away, then the return journey at 6 hours would mean that the total 90 + 90 or 180kms / hour journey has been covered in 8 hours, or at an average speed of 22.5km / hour. But we have an average speed of 36kmph.

This only checks out if the distance is 180kms, as then that'll take 4 hours onwards (at 45kmph), 6 hours back (at 180 / 6 or 30kmph), to get a round trip journey time of 10 hours, which when divided by 360kms (total journey distance), gives us the required 36kmph.

Hence, the answer is D.

Bunuel

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18 Dec 2025, 01:47

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S1 T=D/S
180/45 = 4hr Sufficient
S2 Average speed =Total distance divide by total time
assume t is the time taken to get to the station. That means distance is 45Xt. Back and forth we get 2x45xt= 90t
Total time = t+6
Total average speed = 90t/t+6=36. From this we can easily find t hence sufficient
Ans D

Bunuel

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18 Dec 2025, 02:32

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From all the data we need, we have total distance 2S (way there and back), T2 = 6 hours (and T1 is unknown, which is the way to the station), and V1=45 km/h (and V2 is unknown, which is the speed on the way back). The question asks specifically for T1.

(1) If S=180 and since V1=45, then $T1=S/V1=180/45=4$ hours. Sufficient.

(2) If the average speed is 36, it equals the ratio between total distance and total time. Therefore, $2S/(T1+T2)=36$, and we can insert T2=6 there.
On the other hand, $ S=V1*T1$, or $2S=2V1*T1$, and V1 is 45. Therefore:
$2S = 36(T1+6) = 90*T1$
=> $36T1 + 216 = 90T1$, and $54T1=216$, so T1=4 (hours). Sufficient.

Therefore, the answer is D.

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18 Dec 2025, 03:29

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S1 : Distance given. Already speed is available in question. Hence, time can be deduced. (S)
S2 : Avg of round trip given. We can find distance. Hence, time can be deduced. (S)

D

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18 Dec 2025, 03:35

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Let's explore the options:

1) The distance Workshop - Monitoring Station is 180 km => Immediate Time = Distance/Speed = 180/45 = 4 hours => SUFFICIENT

2) The aritmetic mean of the speed is 36 km/h for the whole trip. => Let's keep in mind that in general the aritmetic speed does not correspond to the Average speed (computed like Vmean = total track/ total time).
BUT : Assuming the cart has not constant speed in the trip back (Let it be stepwise form), it means that the aritmetic mean is computed over an unknown number of steps, hence bringing to the non sufficiency. V_aritmeticMean = Average of the speeds in different segments of the total trip, hence not knowing the length of the segments we have no chance to know the average speed. ==> NOT SUFFICIENT

IMO A!

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18 Dec 2025, 04:18

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Speed=45km/h
time=t hr
dist=d km
Speed returning=D/6
R trip distance= 2D
r trip time=t+6

d=180km
time = D/45=180/45=4hrs
Statement one alone is sufficient.

Average speed=total distance/total time
Total distance=2d Total time=t+6
36=2d/t+6
but t=d/45
36=2d/d/45+6
36=2d/(d+270/45)=2dx45/d+270=90d/d+270
36(d+270)=90d
36d+9720=90d
9720=54d
d=180

t=180/45=4hours
Statement 2 alone is also sufficient

Each statement alone is sufficient D

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18 Dec 2025, 04:47

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Onward Speed S1 = 45 km/h
Return Time T2 = 6 h

Let distance be D and we have to find out T1 = D/45. Essentially, we need value of D

1) D = 180
Sufficient

2) Average speed for entire round trip = 36 km/h
=> 36 = 2D/[(D/45) + 6]
=> 18 = 45D/[D + 270]
=> 2 = 5D/(D+270)
=> 2D + 540 = 5D
=> 3D = 540
=> D = 180
Sufficient

Option D

Bunuel

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18 Dec 2025, 05:01

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Distance travelled in one stint = D. Distance travelled during both stints are equal.

Stint 1: V1 = 45 km/h, T1 = ?

Stint 2: T2 = 6 hours

1. D = 180 km
T1 = 180/45 = 4 hours
Sufficient

2. Vavg = 36 = Total Distance/ Total Time = 2D/(T1+T2)
36 = 2D/(6+D/45)
On solving => D = 180km
T1 = 180/45 = 4 hours
Sufficient

Answer D

Bunuel

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18 Dec 2025, 05:05

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Let first trip be from Workshop to Station and second trip be from Station back to workshop.

S1=D/T1
45=D/T1
T1=D/45

T2=D/S2
6=D/S2

equating D

6 x S2=T1 x 45

any single info on Distance, Time or Speed of unknown qty will give us answer.

Statement 1: Dist is given. Hence it is sufficient.
Statement 2: Arithmetic mean is given. We can find out S2 with this and overall time. Hence sufficient.

Answer is D

Bunuel