Last visit was: 24 Apr 2026, 05:38

It is currently 24 Apr 2026, 05:38

Toolkit

Practice Tests

Data Insights Questions

Data Sufficiency (DS)

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Go to My Error Log Learn more

Request Expert Reply

Confirm Cancel

Apr 25
From ‘GMAT Is Not for Me’ to a 695— Beating Her Target by 50 Points
12:30 AM EDT
-
01:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Apr 26
Score High on the GMAT with Target Test Prep
10:00 AM EDT
-
11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 27
Try OnDemand GMAT Masterclass
11:00 AM EDT
-
12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 28
Join - GMAT Day!
08:00 AM PDT
-
11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.

Back to Forum

Create Topic

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

1 2 3 4

vikramadityaa

Joined: 28 Jul 2025

Last visit: 23 Dec 2025

Posts: 55

Own Kudos:

[1]

Given Kudos: 1

Posts: 55

Kudos: 41

[1]

17 Dec 2025, 10:58

Kudos

Bookmarks

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.

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

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.

harishg

Joined: 18 Dec 2018

Last visit: 09 Apr 2026

Posts: 176

Own Kudos:

174

[1]

Given Kudos: 31

GMAT Focus 1: 695 Q88 V84 DI81 Verified score

Products:

GMAT Focus 1: 695 Q88 V84 DI81 Verified score

Posts: 176

Kudos: 174

[1]

17 Dec 2025, 10:59

Kudos

Bookmarks

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

Ayeka

Joined: 26 May 2024

Last visit: 22 Apr 2026

Posts: 528

Own Kudos:

402

[1]

Given Kudos: 158

Location: India

Schools: ISB

GMAT Focus 1: 645 Q82 V83 DI80 Verified score

GPA: 4.2

Schools: ISB

GMAT Focus 1: 645 Q82 V83 DI80 Verified score

Posts: 528

Kudos: 402

[1]

17 Dec 2025, 11:21

Kudos

Bookmarks

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

AviNFC

Joined: 31 May 2023

Last visit: 10 Apr 2026

Posts: 306

Own Kudos:

366

[1]

Given Kudos: 5

Posts: 306

Kudos: 366

[1]

17 Dec 2025, 11:34

Kudos

Bookmarks

time = d/45.
so distance required.

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

Ans D

MANASH94

Joined: 25 Jun 2025

Last visit: 23 Apr 2026

Posts: 89

Own Kudos:

[1]

Given Kudos: 16

Location: India

Schools: IIM IIM ISB

GPA: 2.9

Schools: IIM IIM ISB

Posts: 89

Kudos: 63

[1]

17 Dec 2025, 12:12

Kudos

Bookmarks

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

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

linnet

Joined: 11 Dec 2025

Last visit: 22 Jan 2026

Posts: 81

Own Kudos:

[1]

Given Kudos: 1

Posts: 81

Kudos: 42

[1]

17 Dec 2025, 12:17

Kudos

Bookmarks

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

canopyinthecity

Joined: 12 Jul 2025

Last visit: 23 Apr 2026

Posts: 92

Own Kudos:

[1]

Given Kudos: 19

Posts: 92

Kudos: 61

[1]

17 Dec 2025, 12:17

Kudos

Bookmarks

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)

Dereno

Joined: 22 May 2020

Last visit: 24 Apr 2026

Posts: 1,398

Own Kudos:

1,373

[1]

Given Kudos: 425

Products:

Posts: 1,398

Kudos: 1,373

[1]

17 Dec 2025, 12:40

Kudos

Bookmarks

Bunuel

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

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

flippedeclipse

Joined: 26 Apr 2025

Last visit: 22 Apr 2026

Posts: 105

Own Kudos:

[1]

Given Kudos: 37

GMAT Focus 1: 655 Q80 V87 DI80 Verified score

Products:

GMAT Focus 1: 655 Q80 V87 DI80 Verified score

Posts: 105

Kudos: 73

[1]

17 Dec 2025, 12:43

Kudos

Bookmarks

Bunuel

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

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.

Vamikaaa

Joined: 31 Oct 2025

Last visit: 04 Apr 2026

Posts: 17

Own Kudos:

[1]

Given Kudos: 13

Posts: 17

Kudos: 9

[1]

17 Dec 2025, 13:34

Kudos

Bookmarks

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.

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

adityamntr

Joined: 15 Jul 2023

Last visit: 21 Feb 2026

Posts: 111

Own Kudos:

[1]

Given Kudos: 13

Location: India

Concentration: General Management, Strategy

Schools: Booth '27 Fuqua '27 Kellog '27

Posts: 111

Kudos: 81

[1]

17 Dec 2025, 15:07

Kudos

Bookmarks

Bunuel

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

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

sriharsha4444

Joined: 06 Jun 2018

Last visit: 05 Mar 2026

Posts: 125

Own Kudos:

Given Kudos: 803

Posts: 125

Kudos: 84

17 Dec 2025, 18:37

Kudos

Bookmarks

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

poojaarora1818

Joined: 30 Jul 2019

Last visit: 24 Apr 2026

Posts: 1,624

Own Kudos:

784

Given Kudos: 3,818

Location: India

Concentration: General Management, Economics

Schools: Duke '27 Haas '27 Kelley '27 Kellogg '27 McDonough '27

GPA: 3

WE:Human Resources (Real Estate)

Products:

Schools: Duke '27 Haas '27 Kelley '27 Kellogg '27 McDonough '27

Posts: 1,624

Kudos: 784

17 Dec 2025, 19:18

Kudos

Bookmarks

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

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

bhanu29

Joined: 02 Oct 2024

Last visit: 22 Apr 2026

Posts: 358

Own Kudos:

270

[1]

Given Kudos: 263

Location: India

Schools: Anderson '27 Booth '27 Foster '27 Haas '27 Kellogg '27 Kellogg Mashall '27 McCombs '27 Ross '27 Scheller '27 Tepper '27 Yale '27 Sloan (MIT) Stanford '27

GMAT Focus 1: 675 Q87 V85 DI79 Verified score

GMAT Focus 2: 715 Q87 V84 DI86 Verified score

GPA: 9.11

WE:Engineering (Technology)

Products:

Schools: Anderson '27 Booth '27 Foster '27 Haas '27 Kellogg '27 Kellogg Mashall '27 McCombs '27 Ross '27 Scheller '27 Tepper '27 Yale '27 Sloan (MIT) Stanford '27

GMAT Focus 2: 715 Q87 V84 DI86 Verified score

Posts: 358

Kudos: 270

[1]

17 Dec 2025, 19:47

Kudos

Bookmarks

Bunuel

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

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

obedear

Joined: 05 Sep 2024

Last visit: 23 Apr 2026

Posts: 61

Own Kudos:

[1]

Given Kudos: 11

Products:

Posts: 61

Kudos: 39

[1]

17 Dec 2025, 19:52

Kudos

Bookmarks

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.

AbhishekP220108

Joined: 04 Aug 2024

Last visit: 24 Apr 2026

Posts: 501

Own Kudos:

213

[1]

Given Kudos: 137

GMAT Focus 1: 555 Q81 V78 DI74 Verified score

Products:

GMAT Focus 1: 555 Q81 V78 DI74 Verified score

Posts: 501

Kudos: 213

[1]

17 Dec 2025, 20:14

Kudos

Bookmarks

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

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

arnab24

Joined: 16 Jan 2024

Last visit: 25 Feb 2026

Posts: 96

Own Kudos:

[1]

Given Kudos: 7

Location: India

Schools: ISB '26

GPA: 8.80

Products:

Schools: ISB '26

Posts: 96

Kudos: 81

[1]

17 Dec 2025, 20:28

Kudos

Bookmarks

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.

Bunuel

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

gemministorm

Joined: 26 May 2025

Last visit: 23 Apr 2026

Posts: 143

Own Kudos:

110

[1]

Given Kudos: 57

GMAT Focus 1: 565 Q82 V79 DI73 Verified score

GMAT Focus 2: 605 Q84 V83 DI73 Verified score

Posts: 143

Kudos: 110

[1]

17 Dec 2025, 21:35

Kudos

Bookmarks

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.

prepapr

Joined: 06 Jan 2025

Last visit: 23 Apr 2026

Posts: 92

Own Kudos:

Given Kudos: 5

GMAT Focus 1: 615 Q85 V80 DI77 Verified score

Posts: 92

Kudos: 82

17 Dec 2025, 22:08

Kudos

Bookmarks

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

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

SBN

Joined: 12 Jun 2023

Last visit: 15 Mar 2026

Posts: 22

Own Kudos:

[1]

Given Kudos: 68

Posts: 22

Kudos: 16

[1]

17 Dec 2025, 23:08

Kudos

Bookmarks

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