Last visit was: 20 Nov 2025, 01:34 It is currently 20 Nov 2025, 01:34
Close
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.
Close
Request Expert Reply
Confirm Cancel
User avatar
helloanupam
Joined: 13 Oct 2010
Last visit: 20 Mar 2012
Posts: 12
Own Kudos:
97
 [71]
Posts: 12
Kudos: 97
 [71]
8
Kudos
Add Kudos
63
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,408
Own Kudos:
Given Kudos: 99,987
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,408
Kudos: 778,430
 [19]
8
Kudos
Add Kudos
11
Bookmarks
Bookmark this Post
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,408
Own Kudos:
778,430
 [8]
Given Kudos: 99,987
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,408
Kudos: 778,430
 [8]
5
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
General Discussion
User avatar
helloanupam
Joined: 13 Oct 2010
Last visit: 20 Mar 2012
Posts: 12
Own Kudos:
Posts: 12
Kudos: 97
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel,
Sorry, I could not understand the language of the question and hence probably the answer is not clear to me. If possible , could you reframe the question or explain as to what the question is asking?
User avatar
LM
Joined: 03 Sep 2006
Last visit: 04 Apr 2015
Posts: 444
Own Kudos:
Given Kudos: 33
Posts: 444
Kudos: 7,673
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Don't we somehow need to use the information 1/10 mile increment signposts in calculating the probability?
Approach mentioned above is logical but there must be simple mathematical solution using the increment signposts?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,408
Own Kudos:
778,430
 [3]
Given Kudos: 99,987
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,408
Kudos: 778,430
 [3]
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
LM
Don't we somehow need to use the information 1/10 mile increment signposts in calculating the probability?
Approach mentioned above is logical but there must be simple mathematical solution using the increment signposts?

Stem gives us the information about the signs only to fix the point of 2.5 miles on the track and thus fix the 1 mile interval the car should end within. 1/10 mile increment is totally irrelevant.

Consider the following: A circular racetrack is 3 miles in length. If a race car starts at a random location on the track and travels exactly Y miles, what is the probability that the car ends within a half mile of some point X on the track?

The answer will be the same: as the interval for the endpoint of the car is 1 mile (from x-0.5 to x+0.5) then the probability will be 1/3.

Note that a car starts at a random location on the track and travels exactly Y miles means that the endpoint of the car will also be at random location on the track (travel part is also to confuse us: random location plus Y miles=random location). So the question basically ask what is the probability that the car ends within the particular 1 mile interval on the track of 3 miles.

Hope it's clear.
User avatar
tfincham86
Joined: 17 Aug 2010
Last visit: 02 Jan 2013
Posts: 322
Own Kudos:
Given Kudos: 46
Status:Bring the Rain
Location: United States (MD)
Concentration: Strategy, Marketing
Schools: Michigan (Ross) - Class of 2014
GMAT 1: 730 Q49 V39
GPA: 3.13
WE:Corporate Finance (Aerospace and Defense)
Schools: Michigan (Ross) - Class of 2014
GMAT 1: 730 Q49 V39
Posts: 322
Kudos: 63
Kudos
Add Kudos
Bookmarks
Bookmark this Post
+1 for C.

Thanks Bunuel
User avatar
helloanupam
Joined: 13 Oct 2010
Last visit: 20 Mar 2012
Posts: 12
Own Kudos:
Posts: 12
Kudos: 97
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Thankyou Bunuel for the extra efforts in explaining. I have now understood your explanation.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,408
Own Kudos:
778,430
 [1]
Given Kudos: 99,987
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,408
Kudos: 778,430
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on probability problems: math-probability-87244.html

All DS probability problems to practice: search.php?search_id=tag&tag_id=33
All PS probability problems to practice: search.php?search_id=tag&tag_id=54

Tough probability questions: hardest-area-questions-probability-and-combinations-101361.html
avatar
lchen
Joined: 09 Apr 2013
Last visit: 20 Apr 2019
Posts: 37
Own Kudos:
176
 [2]
Given Kudos: 2
Location: United States (DC)
Concentration: Strategy, Social Entrepreneurship
GMAT 1: 750 Q50 V41
GPA: 3.55
WE:General Management (Non-Profit and Government)
Products:
GMAT 1: 750 Q50 V41
Posts: 37
Kudos: 176
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Here is a rewording of the original question:
What is the probability that the car starts between the 1.5 mile and 2.5 mile mark on a 3 mi racetrack?

2.5 - 1.5 = 1
that is 1/3 of the racetrack

Answer is C

Everything else in this question is irrelevant.
Although, if you really wanted you could rephrase the question to account for the 1/10th mile sign posts:
What is the probability that the car stars between the 15th and 25th sign posts if there are 30 sign posts total.
But that is unnecessary extra work.
User avatar
cumulonimbus
Joined: 14 Nov 2011
Last visit: 10 Feb 2023
Posts: 97
Own Kudos:
Given Kudos: 102
Location: United States
Concentration: General Management, Entrepreneurship
GPA: 3.61
WE:Consulting (Manufacturing)
Posts: 97
Kudos: 62
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
helloanupam
Bunuel,
Sorry, I could not understand the language of the question and hence probably the answer is not clear to me. If possible , could you reframe the question or explain as to what the question is asking?

Look at the diagram:
Attachment:
untitled.PNG
The car ends within a half mile of the sign indicating 2.5 miles means that the car should end in one mile interval, between the signs indicating 2 (2.5-0.5=2) and 3 miles (2+0.5=3), so within the red segment on the diagram.

Now if the cars appears somewhere between the blue dots, between 1.5 and 2 miles signs then after traveling 0.5 miles the car will be in the red segment. So in order after traveling 0.5 miles the car to end within the red segment it should appear between 1.5 and 2.5 miles, so within 1 mile interval, as the circumference of the track is 3 miles then the probability of that will be P=favorable/total=1/3. As I mentioned in my previous post actually it doesn't matter where the car appears or what distance it travel, as long as favorable interval in the end is 1 mile and total interval is 3 miles then the probability will be 1/3 miles.

Hope it's clear.

Hi Bunnel,

What about if car starts between 0.5 and 2.5 miles?

So this will make probability = 2*1/3 ?
User avatar
ScottTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 14 Oct 2015
Last visit: 19 Nov 2025
Posts: 21,717
Own Kudos:
26,998
 [2]
Given Kudos: 300
Status:Founder & CEO
Affiliations: Target Test Prep
Location: United States (CA)
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 21,717
Kudos: 26,998
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
helloanupam
A circular racetrack is 3 miles in length and has signs posted to indicate each 1/10 mile increment. If a race car starts at a random location on the track and travels exactly one half mile, what is the probability that the car ends within a half mile of the sign indicating 2 1/2 miles?

A. 1/6
B. 3/10
C. 1/3
D. 1/2
E. 2/3

We are given that a circular racetrack is 3 miles in length and has signs posted to indicate each 1/10 mile increment. We are also given that a race car starts at a random location on the track and travels exactly one half mile, and we need to determine the probability that the car ends within a half mile of the sign indicating 2 1/2 (or 2.5) miles.

If the car ends within a half mile of the 2.5-mile sign, that means the car can end as far as the 2.0-mile sign or the 3.0-mile sign. However, since the car travels exactly one half mile, the starting point of the car can be anywhere from the 1.5 mile sign to the 2.5-mile sign. In other words, the car can be anywhere in this 2.5 - 1.5 = 1 mile stretch. Since the racetrack is 3 miles long, the probability that the car is in this 1 mile stretch is ⅓.

Answer: C
User avatar
ShashankDave
Joined: 03 Apr 2013
Last visit: 26 Jan 2020
Posts: 218
Own Kudos:
288
 [1]
Given Kudos: 872
Location: India
Concentration: Marketing, Finance
GMAT 1: 740 Q50 V41
GPA: 3
GMAT 1: 740 Q50 V41
Posts: 218
Kudos: 288
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
helloanupam
A circular racetrack is 3 miles in length and has signs posted to indicate each 1/10 mile increment. If a race car starts at a random location on the track and travels exactly one half mile,what is the probability that the car ends within a half mile of the sign indicating 2 1/2 miles?
(A) 1/6
(B) 3/10
(C) 1/3
(D) 1/2
(E) 2/3


Answer: C

Could somebody explain how to work this out?

The car ends within a half mile of the sign indicating 2 1/2 miles means that the car will end in one mile interval, between the signs indicating 2 and 3 miles.

Now, it doesn't matter where the car starts or what distance it travels, the probability will be P=(favorable outcome)/(total # of outcomes)=1/3 (as the car starts at random point end travels some distance afterwards we can consider its end point as the point where he randomly appeared, so the probability that the car appeared within 1 mile interval out of total 3 miles will be 1/3).

Answer: C.

Hope it's clear.
Bunuel
Shouldn't the answer be 2/3?

The question doesn't specify in which direction the car has to travel, so it could be in either direction. Or may be I'm overthinking :?
User avatar
tejyr
Joined: 26 Dec 2017
Last visit: 24 Dec 2021
Posts: 113
Own Kudos:
90
 [1]
Given Kudos: 109
Products:
Posts: 113
Kudos: 90
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bunuel
helloanupam
Bunuel,
Sorry, I could not understand the language of the question and hence probably the answer is not clear to me. If possible , could you reframe the question or explain as to what the question is asking?

Look at the diagram:
Attachment:
untitled.PNG
The car ends within a half mile of the sign indicating 2.5 miles means that the car should end in one mile interval, between the signs indicating 2 (2.5-0.5=2) and 3 miles (2+0.5=3), so within the red segment on the diagram.

Now if the car appears somewhere between the blue dots, between 1.5 and 2 miles signs then after traveling 0.5 miles the car will be in the red segment. So in order after traveling 0.5 miles the car to end within the red segment it should appear between 1.5 and 2.5 miles, so within 1 mile interval, as the circumference of the track is 3 miles then the probability of that will be P=favorable/total=1/3. As I mentioned in my previous post actually it doesn't matter where the car appears or what distance it travel, as long as favorable interval in the end is 1 mile and total interval is 3 miles then the probability will be 1/3 miles.

Hope it's clear.
Bunuel
A small doubt here to be precise the correct ranges would be 2.1,2.2,2.3,2.4,2.5,2.6,2.7,2.8,2.9 as 2.5 miles -/+ 0.4 miles (within 0.5 miles i.e, <=0.5-> <0.4) is between 2.1 and 2.9
probability would be 9/31. (31- 0,0.1,....3.0)
I know the above is wrong but I am having confusion with distance vs exact point transition.(In other words I am calculating exact point like 2.1,2.2,... as in question given 2.5 (within half mile) but unable to know how 2-3-> as OS is capturing the exact condition of question)
Pls help me to figure out where I faltered.
avatar
outofpocket
Joined: 10 Jul 2018
Last visit: 01 Feb 2019
Posts: 3
Given Kudos: 34
GMAT 1: 600 Q47 V26
GMAT 2: 630 Q48 V28
GMAT 2: 630 Q48 V28
Posts: 3
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Shouldn't the question say "one half miles" instead of "one half mile"? Because "one half mile" can be read as one of 1/2 mile.
User avatar
Hero8888
Joined: 29 Dec 2017
Last visit: 14 Apr 2019
Posts: 300
Own Kudos:
Given Kudos: 273
Location: United States
Concentration: Marketing, Technology
GMAT 1: 630 Q44 V33
GMAT 2: 690 Q47 V37
GMAT 3: 710 Q50 V37
GPA: 3.25
WE:Marketing (Telecommunications)
GMAT 3: 710 Q50 V37
Posts: 300
Kudos: 344
Kudos
Add Kudos
Bookmarks
Bookmark this Post
outofpocket
Shouldn't the question say "one half miles" instead of "one half mile"? Because "one half mile" can be read as one of 1/2 mile.

The question exactly says: 1/2 miles.
avatar
SabnamStha
Joined: 22 Jan 2018
Last visit: 25 Nov 2019
Posts: 1
Own Kudos:
Given Kudos: 57
Concentration: Sustainability, Statistics
GMAT 1: 530 Q42 V22
GPA: 3.69
GMAT 1: 530 Q42 V22
Posts: 1
Kudos: 5
Kudos
Add Kudos
Bookmarks
Bookmark this Post
So, I solved it like this, How many half miles (0.50) miles in in 3 miles, 6 so the probability that car ends within a half mile is 1/6, but, the car could have moved in any direction; clockwise or counter-clockwise so multiply the probability of 1/6 by 2 hence, 1/3.
User avatar
pratik2018
Joined: 09 Apr 2017
Last visit: 23 Jun 2020
Posts: 37
Own Kudos:
Given Kudos: 188
Location: Nepal
Concentration: Finance, Entrepreneurship
GMAT 1: 580 Q47 V22
GMAT 2: 640 Q48 V29
GMAT 3: 690 Q48 V36
WE:Information Technology (Computer Software)
Products:
GMAT 3: 690 Q48 V36
Posts: 37
Kudos: 186
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
helloanupam
A circular racetrack is 3 miles in length and has signs posted to indicate each 1/10 mile increment. If a race car starts at a random location on the track and travels exactly one half mile, what is the probability that the car ends within a half mile of the sign indicating 2 1/2 miles?

A. 1/6
B. 3/10
C. 1/3
D. 1/2
E. 2/3

Bunuel,
Sorry, I could not understand the language of the question and hence probably the answer is not clear to me. If possible , could you reframe the question or explain as to what the question is asking?

Look at the diagram:



The car ends within a half mile of the sign indicating 2.5 miles means that the car should end in one mile interval, between the signs indicating 2 (2.5-0.5=2) and 3 miles (2+0.5=3), so within the red segment on the diagram.

Now if the car appears somewhere between the blue dots, between 1.5 and 2 miles signs then after traveling 0.5 miles the car will be in the red segment. So in order after traveling 0.5 miles the car to end within the red segment it should appear between 1.5 and 2.5 miles, so within 1 mile interval, as the circumference of the track is 3 miles then the probability of that will be P=favorable/total=1/3. As I mentioned in my previous post actually it doesn't matter where the car appears or what distance it travel, as long as favorable interval in the end is 1 mile and total interval is 3 miles then the probability will be 1/3 miles.

Hope it's clear.


Attachment:
untitled.PNG

One doubt about the question. It is only given that a car starts in the circular race track. Doesnt direction come into play in such scenarios? Shouldn't it be mentioned that what direction the car is travelling in. If we imagine it can be either way the answer changes to the question. Just a thought. Looking at the question, I wasnt sure if we are supposed to take into consideration that the car could have travelled both ways.
User avatar
LMP
Joined: 29 Jun 2018
Last visit: 15 Mar 2022
Posts: 49
Own Kudos:
233
 [1]
Given Kudos: 62
Location: India
GPA: 4
Posts: 49
Kudos: 233
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
helloanupam
A circular racetrack is 3 miles in length and has signs posted to indicate each 1/10 mile increment. If a race car starts at a random location on the track and travels exactly one half mile, what is the probability that the car ends within a half mile of the sign indicating 2 1/2 miles?

A. 1/6
B. 3/10
C. 1/3
D. 1/2
E. 2/3
Ihave solved it this way.
If the vehicle starts at a random location and travels 1/2 mile, it is still at some arbit location.
Hence , I would try to find out the range of desired outcomes.
If the vehicle is between the mileposts indicating 2 miles and 3 miles, it is within a half-mile of the 2 and a half milepost. That's the desired range of 1 mile, out of a possible range of 3 miles length of the track. The probability, then, is 1 / 3.
The answer is C
avatar
ctmerrill50
Joined: 16 Apr 2019
Last visit: 24 Jan 2020
Posts: 4
Own Kudos:
Given Kudos: 5
Posts: 4
Kudos: 1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
helloanupam
A circular racetrack is 3 miles in length and has signs posted to indicate each 1/10 mile increment. If a race car starts at a random location on the track and travels exactly one half mile,what is the probability that the car ends within a half mile of the sign indicating 2 1/2 miles?
(A) 1/6
(B) 3/10
(C) 1/3
(D) 1/2
(E) 2/3


Answer: C

Could somebody explain how to work this out?

The car ends within a half mile of the sign indicating 2 1/2 miles means that the car will end in one mile interval, between the signs indicating 2 and 3 miles.

Now, it doesn't matter where the car starts or what distance it travels, the probability will be P=(favorable outcome)/(total # of outcomes)=1/3 (as the car starts at random point end travels some distance afterwards we can consider its end point as the point where he randomly appeared, so the probability that the car appeared within 1 mile interval out of total 3 miles will be 1/3).

Answer: C.

Hope it's clear.

I got a little mixed up on the wording of question. I interpreted "Within" a half mile to mean "less than a half a mile" not "A half mile or less." Could you rephrase the question.
 1   2   
Moderators:
Math Expert
105408 posts
Tuck School Moderator
805 posts