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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Sep 28Live Online Class with Top GRE Expert Mike Butville
08:00 PM EDT
-10:00 PM EDT
Top scores are possible when you enroll in a powerful GRE course, taught live online + 6 months access to TTP GRE OnDemand video courses included! Class starts Mon/Wed Sept. 28, 2026 - Dec. 9, 2026, 8:00pm-10:00pm EST
In a neighborhood consisting of 2,000 homes, 80 percent of
[#permalink]
21 Dec 2017, 16:59
21 Dec 2017, 16:59
Expert Reply
00:00
Question Stats:
57% (01:07) correct
42% (01:04) wrong based on 42 sessions
Hide Show timer Statistics
In a neighborhood consisting of 2,000 homes, 80 percent of the homes are valued at $325,000 or less. Which of the following statements about the values of the homes in the neighborhood must be true?
Indicate \(all\) such statements.
A. The average (arithmetic mean) value is at most $325,000.
B. The median value is at most $325,000.
C. At most 400 homes have values greater than $325,000.
kudo for the right solution and explanation
Indicate \(all\) such statements.
A. The average (arithmetic mean) value is at most $325,000.
B. The median value is at most $325,000.
C. At most 400 homes have values greater than $325,000.
kudo for the right solution and explanation
Re: In a neighborhood consisting of 2,000 homes, 80 percent of
[#permalink]
22 Dec 2017, 00:56
22 Dec 2017, 00:56
1
So 0,8 * 2000 = 1600 values evaluated at at least 325 000
So 400 are greater than 325 000 C must be true
Since we have 1600 houses at 325 000 or less and 400 at more than 325 000
We cannot conclude that the median is 325 000. The 1000th and 1001 th values could be either less than 325 000 or equals 325000. So B is out.
The average is the number of houses * prices divided by the total of houses.
So it could be written as (<= 325 000)(1600) + (400)((>325000) /2000
Therefore A is out
So 400 are greater than 325 000 C must be true
Since we have 1600 houses at 325 000 or less and 400 at more than 325 000
We cannot conclude that the median is 325 000. The 1000th and 1001 th values could be either less than 325 000 or equals 325000. So B is out.
The average is the number of houses * prices divided by the total of houses.
So it could be written as (<= 325 000)(1600) + (400)((>325000) /2000
Therefore A is out
Re: In a neighborhood consisting of 2,000 homes, 80 percent of
[#permalink]
22 Dec 2017, 21:52
22 Dec 2017, 21:52
1
Popo wrote:
So 0,8 * 2000 = 1600 values evaluated at at least 325 000
So 400 are greater than 325 000 C must be true
Since we have 1600 houses at 325 000 or less and 400 at more than 325 000
So 400 are greater than 325 000 C must be true
Since we have 1600 houses at 325 000 or less and 400 at more than 325 000
I donot think your assumption is wright- can you plz explain how you assumed the remaing 400 houses have values more than 325000. As nothing is mentioned about the remaining 400 houses
