Last visit was: 07 Apr 2025, 10:01 It is currently 07 Apr 2025, 10:01
Decision Tracker
My Rewards
New comers' posts
New posts
Unanswered

Close

GRE Prep 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 GRE 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
Close

Request Expert Reply

Confirm Cancel
PREVIOUS TOPIC
NEXT TOPIC

The seating chart of an airplane shown 30 rows of seats

Tags :
Find Similar Topics  About  Add a Tag

Show Tags
Hide Tags

avatar
sarahl
Intern
Intern
Joined: 23 Apr 2018
Posts: 26
Own Kudos [?]: 38 [3]
Given Kudos: 0
Send PM
The seating chart of an airplane shown 30 rows of seats [#permalink] New post   Updated on: 21 Oct 2018, 07:01
1
2
Bookmarks
00:00
A
B
C
D
E

Question Stats:

62% (01:52) correct 37% (01:26) wrong based on 29 sessions

Show timer Statistics

The seating chart of an airplane shown 30 rows of seats. Each row has 3 seats on each side of the center aisle, and one of the seats on each side is a window seat. The view from the window seats in 5 of the rows is obscured by the wings of the airplane. If the first person to be assigned a seat is assigned a window seat and the window seat is assigned randomly, what is the probability that the person will get a seat with an unobscured view?
(A)16
(B)13
(C)23
(D)56
(E)1718
ShowHide Answer
Official Answer
D

Originally posted by sarahl on 20 Oct 2018, 19:15.
Last edited by chetan2u on 21 Oct 2018, 07:01, edited 1 time in total.
updated the OA
avatar
pranab223
Retired Moderator
Joined: 20 Apr 2016
Posts: 1307
Own Kudos [?]: 2294 [0]
Given Kudos: 251
WE:Engineering (Energy and Utilities)
Send PM

Top Member of the Month
Re: The seating chart of an airplane shown 30 rows of seats [#permalink] New post   20 Oct 2018, 19:49
1
sarahl wrote:
The seating chart of an airplane shown 30 rows of seats. Each row has 3 seats on each side of the center aisle, and one of the seats on each side is a window seat. The view from the window seats in 5 of the rows is obscured by the wings of the airplane. If the first person to be assigned a seat is assigned a window seat and the window seat is assigned randomly, what is the probability that the person will get a seat with an unobscured view?
(A)16
(B)13
(C)23
(D)56
(E)1718



Total seats = 30 *3 = 90

and total seat with obscured view = 5 * 3 = 15

Now total seat with unobscured view = 90 - 15 = 75

Therefore the probability that the person will get a seat with an unobscured view = 7590=56.

*** Please at the end of your question do mention the source of the question***
avatar
chetan2u
Supreme Moderator
Joined: 01 Nov 2017
Posts: 371
Own Kudos [?]: 477 [0]
Given Kudos: 0
Send PM
Re: The seating chart of an airplane shown 30 rows of seats [#permalink] New post   21 Oct 2018, 07:00
Expert Reply
sarahl wrote:
The seating chart of an airplane shown 30 rows of seats. Each row has 3 seats on each side of the center aisle, and one of the seats on each side is a window seat. The view from the window seats in 5 of the rows is obscured by the wings of the airplane. If the first person to be assigned a seat is assigned a window seat and the window seat is assigned randomly, what is the probability that the person will get a seat with an unobscured view?
(A)16
(B)13
(C)23
(D)56
(E)1718


hi, OA marked is wrong ...
It will be 56....

There are 30 rows and each row has 6 seats, 3 on each side of centre aisle.. so total seats = 6*30 = 180
there are 2 window seats in each row, one on each side of aisle, so total = 30∗2=60
seats with obscured view is 5∗2=10..

So, the probability that the person will get a seat with an unobscured view = 60−1060=56
avatar
sarahl
Intern
Intern
Joined: 23 Apr 2018
Posts: 26
Own Kudos [?]: 38 [0]
Given Kudos: 0
Send PM
Re: The seating chart of an airplane shown 30 rows of seats [#permalink] New post   21 Oct 2018, 17:28
chetan2u wrote:
sarahl wrote:
The seating chart of an airplane shown 30 rows of seats. Each row has 3 seats on each side of the center aisle, and one of the seats on each side is a window seat. The view from the window seats in 5 of the rows is obscured by the wings of the airplane. If the first person to be assigned a seat is assigned a window seat and the window seat is assigned randomly, what is the probability that the person will get a seat with an unobscured view?
(A)16
(B)13
(C)23
(D)56
(E)1718


hi, OA marked is wrong ...
It will be 56....

There are 30 rows and each row has 6 seats, 3 on each side of centre aisle.. so total seats = 6*30 = 180
there are 2 window seats in each row, one on each side of aisle, so total = 30∗2=60
seats with obscured view is 5∗2=10..

So, the probability that the person will get a seat with an unobscured view = 60−1060=56


My problem is that the OA is C!! But I think the correct answer should be "D". The explanation is in Chinese language and I don't know Chinese. Please look at the attached file. according to it the correct answer should be "C".
Attachments

gremath set8-1.pdf [24.96 KiB]
Downloaded 309 times

avatar
sarahl
Intern
Intern
Joined: 23 Apr 2018
Posts: 26
Own Kudos [?]: 38 [0]
Given Kudos: 0
Send PM
Re: The seating chart of an airplane shown 30 rows of seats [#permalink] New post   21 Oct 2018, 17:41
pranab01 wrote:
sarahl wrote:
The seating chart of an airplane shown 30 rows of seats. Each row has 3 seats on each side of the center aisle, and one of the seats on each side is a window seat. The view from the window seats in 5 of the rows is obscured by the wings of the airplane. If the first person to be assigned a seat is assigned a window seat and the window seat is assigned randomly, what is the probability that the person will get a seat with an unobscured view?
(A)16
(B)13
(C)23
(D)56
(E)1718



Total seats = 30 *3 = 90

and total seat with obscured view = 5 * 3 = 15

Now total seat with unobscured view = 90 - 15 = 75

Therefore the probability that the person will get a seat with an unobscured view = 7590=56.

*** Please at the end of your question do mention the source of the question***


"Each row has 3 seats on each side of the center aisle" so the total seats would be 180.

About the source...Well I downloaded a zip file in which there were several PDF files including Manhattan books and GRE Math questions. The question belongs to one of these GRE math PDFs: GRE Math **** Set 8-1
The answer explanation is in Chinese.
P.s. The red line is in Chinese but in this site, Chinese characters are not allowed.
avatar
chetan2u
Supreme Moderator
Joined: 01 Nov 2017
Posts: 371
Own Kudos [?]: 477 [0]
Given Kudos: 0
Send PM
Re: The seating chart of an airplane shown 30 rows of seats [#permalink] New post   21 Oct 2018, 17:47
Expert Reply
sarahl wrote:
chetan2u wrote:
sarahl wrote:
The seating chart of an airplane shown 30 rows of seats. Each row has 3 seats on each side of the center aisle, and one of the seats on each side is a window seat. The view from the window seats in 5 of the rows is obscured by the wings of the airplane. If the first person to be assigned a seat is assigned a window seat and the window seat is assigned randomly, what is the probability that the person will get a seat with an unobscured view?
(A)16
(B)13
(C)23
(D)56
(E)1718


hi, OA marked is wrong ...
It will be 56....

There are 30 rows and each row has 6 seats, 3 on each side of centre aisle.. so total seats = 6*30 = 180
there are 2 window seats in each row, one on each side of aisle, so total = 30∗2=60
seats with obscured view is 5∗2=10..

So, the probability that the person will get a seat with an unobscured view = 60−1060=56


My problem is that the OA is C!! But I think the correct answer should be "D". The explanation is in Chinese language and I don't know Chinese. Please look at the attached file. according to it the correct answer should be "C".



Hi..
He mentions in the end 150/180, which I presume is the answer.
And 150/180=5/6 so there must be a typo thereafter.
And yes answer has to be 5/6.
If the source means 2/3, then there must be an error in question or their solution
avatar
pranab223
Retired Moderator
Joined: 20 Apr 2016
Posts: 1307
Own Kudos [?]: 2294 [0]
Given Kudos: 251
WE:Engineering (Energy and Utilities)
Send PM

Top Member of the Month
Re: The seating chart of an airplane shown 30 rows of seats [#permalink] New post   21 Oct 2018, 18:28
sarahl wrote:
pranab01 wrote:
sarahl wrote:
The seating chart of an airplane shown 30 rows of seats. Each row has 3 seats on each side of the center aisle, and one of the seats on each side is a window seat. The view from the window seats in 5 of the rows is obscured by the wings of the airplane. If the first person to be assigned a seat is assigned a window seat and the window seat is assigned randomly, what is the probability that the person will get a seat with an unobscured view?
(A)16
(B)13
(C)23
(D)56
(E)1718



Total seats = 30 *3 = 90

and total seat with obscured view = 5 * 3 = 15

Now total seat with unobscured view = 90 - 15 = 75

Therefore the probability that the person will get a seat with an unobscured view = 7590=56.

*** Please at the end of your question do mention the source of the question***


"Each row has 3 seats on each side of the center aisle" so the total seats would be 180.

About the source...Well I downloaded a zip file in which there were several PDF files including Manhattan books and GRE Math questions. The question belongs to one of these GRE math PDFs: GRE Math **** Set 8-1
The answer explanation is in Chinese.
P.s. The red line is in Chinese but in this site, Chinese characters are not allowed.


Hi, if you can upload the zip file in separate topic

As for the ques, yes u r correct but no need to calculate how many seats are there, as the ques is only interested with window seat

SO, 30 rows and each row has 2 window seat = 30 * 2 = 60

Total seat with obscured view = 5 row * 2 window seat on each side = 10

unobscured seat = 60 - 10 = 50

Probability = 5060=56
User avatar
bumpbot
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 5232
Own Kudos [?]: 77 [0]
Given Kudos: 0
Send PM
Re: The seating chart of an airplane shown 30 rows of seats [#permalink] New post   21 Jul 2022, 02:46
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Prep Club for GRE Bot
gmatclubot
Re: The seating chart of an airplane shown 30 rows of seats [#permalink]
21 Jul 2022, 02:46
Moderators:
adewale223
GRE Instructor
108 posts
bronaugust
GRE Forum Moderator
37 posts
BrushMyQuant
Moderator
1131 posts
VibhuAnurag
GRE Instructor
234 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Main navigation

Partners

GRE Resources

www.ets.org/gre | About | Terms and Conditions and Privacy Policy | Prep Club for GRE Rules | Contact