Last visit was: 18 Dec 2024, 09:20 It is currently 18 Dec 2024, 09:20

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.

Close

Request Expert Reply

Confirm Cancel
Verbal Expert
Joined: 18 Apr 2015
Posts: 30355
Own Kudos [?]: 36752 [22]
Given Kudos: 26080
Send PM
Most Helpful Expert Reply
Verbal Expert
Joined: 18 Apr 2015
Posts: 30355
Own Kudos [?]: 36752 [7]
Given Kudos: 26080
Send PM
Most Helpful Community Reply
avatar
Retired Moderator
Joined: 16 Oct 2019
Posts: 63
Own Kudos [?]: 175 [11]
Given Kudos: 21
Send PM
General Discussion
Verbal Expert
Joined: 18 Apr 2015
Posts: 30355
Own Kudos [?]: 36752 [0]
Given Kudos: 26080
Send PM
Re: Of the 150 houses in a certain development, 60 percent have [#permalink]
Expert Reply
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
avatar
Manager
Manager
Joined: 29 Apr 2020
Posts: 67
Own Kudos [?]: 9 [0]
Given Kudos: 0
Send PM
Re: Of the 150 houses in a certain development, 60 percent have [#permalink]
any other explanation
avatar
Intern
Intern
Joined: 17 Aug 2020
Posts: 40
Own Kudos [?]: 31 [0]
Given Kudos: 0
Send PM
Re: Of the 150 houses in a certain development, 60 percent have [#permalink]
1
Thank you for your answers. I wonder something though. How do you reconcile what you found with the formula
P(AUBUC)=P(A)+P(B)+P(C) - P(A^B) - P(B^C) - P(A^C) + P(A^B^C)? where ^ means intersection.
Because if I did that, I would find:
P(AUBUC) = 150 - 5
P(A)=90 P(B)=75 P(C)=45
P(A^B^C)=5
So the sum P(A^B) + P(B^C) + P(A^C) = 90 + 75 + 45 + 6 - (150-5) = 71 which sounds clearly false...
I looked at the topic on overlapping sets and I don't get why this doesn't match the formula given there.
Many thanks for your help!
avatar
Manager
Manager
Joined: 22 Jan 2020
Posts: 120
Own Kudos [?]: 241 [6]
Given Kudos: 10
Send PM
Re: Of the 150 houses in a certain development, 60 percent have [#permalink]
5
1
Bookmarks
fraise wrote:
Thank you for your answers. I wonder something though. How do you reconcile what you found with the formula
P(AUBUC)=P(A)+P(B)+P(C) - P(A^B) - P(B^C) - P(A^C) + P(A^B^C)? where ^ means intersection.
Because if I did that, I would find:
P(AUBUC) = 150 - 5
P(A)=90 P(B)=75 P(C)=45
P(A^B^C)=5
So the sum P(A^B) + P(B^C) + P(A^C) = 90 + 75 + 45 + 6 - (150-5) = 71 which sounds clearly false...
I looked at the topic on overlapping sets and I don't get why this doesn't match the formula given there.
Many thanks for your help!




A= house with an AC
B= house with a Sun porch
C= house with a pool

|A|=150(.6)=90
|B|=150(.4)=75
|C|=150*(.3)=45
|AUBUC|=150-5=145
|A,B,C|=5

145=|AUBUC|=|A|+|B|+|C|-|A,B|-|A,C|-|B,C|+|A,B,C|=90+75+45-|A,B|-|A,C|-|B,C|+5
|A,B|+|A,C|+|B,C|= 90+75+45+5-150= 215-145= 70
This is the number of house with 2 or more of the amenities. So we need to subtract those that have all three.

But the intersection of A,B, and C as shown in the Venn Diagram above is in each of the intersections of two events.
That is we have |A,B,C| in |A,B|, |A,B,C| in |A,C|, and |A,B,C| in |B,C|.

So we need to subtract it three times.
70-3(5)
=70-15
=55

Final Answer: D
avatar
Intern
Intern
Joined: 23 Jun 2022
Posts: 1
Own Kudos [?]: 0 [0]
Given Kudos: 0
Send PM
Of the 150 houses in a certain development, 60 percent have [#permalink]
chacinluis wrote:
fraise wrote:
Thank you for your answers. I wonder something though. How do you reconcile what you found with the formula
P(AUBUC)=P(A)+P(B)+P(C) - P(A^B) - P(B^C) - P(A^C) + P(A^B^C)? where ^ means intersection.
Because if I did that, I would find:
P(AUBUC) = 150 - 5
P(A)=90 P(B)=75 P(C)=45
P(A^B^C)=5
So the sum P(A^B) + P(B^C) + P(A^C) = 90 + 75 + 45 + 6 - (150-5) = 71 which sounds clearly false...
I looked at the topic on overlapping sets and I don't get why this doesn't match the formula given there.
Many thanks for your help!




A= house with an AC
B= house with a Sun porch
C= house with a pool

|A|=150(.6)=90
|B|=150(.4)=75
|C|=150*(.3)=45
|AUBUC|=150-5=145
|A,B,C|=5

145=|AUBUC|=|A|+|B|+|C|-|A,B|-|A,C|-|B,C|+|A,B,C|=90+75+45-|A,B|-|A,C|-|B,C|+5
|A,B|+|A,C|+|B,C|= 90+75+45+5-150= 215-145= 70
This is the number of house with 2 or more of the amenities. So we need to subtract those that have all three.

But the intersection of A,B, and C as shown in the Venn Diagram above is in each of the intersections of two events.
That is we have |A,B,C| in |A,B|, |A,B,C| in |A,C|, and |A,B,C| in |B,C|.

So we need to subtract it three times.
70-3(5)
=70-15
=55

Final Answer: D



Hi, I wonder why 70 is the number of house with 2 or more of the amenities, isn's it exactly the number of houses with 2 amenities (|A,B|+|A,C|+|B,C|= 70)? We've already subtracted 5 houses that have all 3 amenities.
Retired Moderator
Joined: 02 Dec 2020
Posts: 1831
Own Kudos [?]: 2148 [1]
Given Kudos: 140
GRE 1: Q168 V157

GRE 2: Q167 V161
Send PM
Re: Of the 150 houses in a certain development, 60 percent have [#permalink]
1
The \(5\) houses that were subtracted were those that have zero aminities.

LucaZh00 wrote:
Hi, I wonder why 70 is the number of house with 2 or more of the amenities, isn's it exactly the number of houses with 2 amenities (|A,B|+|A,C|+|B,C|= 70)? We've already subtracted 5 houses that have all 3 amenities.
avatar
Intern
Intern
Joined: 27 Feb 2023
Posts: 2
Own Kudos [?]: 0 [0]
Given Kudos: 0
Send PM
Re: Of the 150 houses in a certain development, 60 percent have [#permalink]
Carcass wrote:
Total houses = number with air conditioning + number with sunporch + number with pool - number with only two of the three things - 2(number with all three things) + number with none of the three things

150 = 0.6(150) + 0.5(150) + 0.3(150) - D - 2(5) + 5

150 = 90 + 75 + 45 - D - 10 + 5

150 = 205 - D

D = 55

Why are you multiplying the number with all three things by 2??
Verbal Expert
Joined: 18 Apr 2015
Posts: 30355
Own Kudos [?]: 36752 [0]
Given Kudos: 26080
Send PM
Re: Of the 150 houses in a certain development, 60 percent have [#permalink]
Expert Reply
It is a formula for the three overlapping set
User avatar
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 5090
Own Kudos [?]: 76 [0]
Given Kudos: 0
Send PM
Re: Of the 150 houses in a certain development, 60 percent have [#permalink]
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
Re: Of the 150 houses in a certain development, 60 percent have [#permalink]
Moderators:
GRE Instructor
88 posts
GRE Forum Moderator
37 posts
Moderator
1115 posts
GRE Instructor
234 posts

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