Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 1912:30 PM EST
-01:30 PM EST
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Nov 2001:30 PM EST
-02:30 PM IST
Learn how Kamakshi achieved a GMAT 675 with an impressive 96th %ile in Data Insights. Discover the unique methods and exam strategies that helped her excel in DI along with other sections for a balanced and high score.
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2407:00 PM PST
-08:00 PM PST
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
Nov 2510:00 AM EST
-11:00 AM EST
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.
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
64% (01:36) correct
36%
(01:32)
wrong
based on 1408
sessions
History
Date
Time
Result
Not Attempted Yet
A certain list consists of five different integers. Is the average (arithmetic mean) of the two greatest integers in the list greater than 70 ?
(1) The median of the integers in the list is 70.
(2) The average of the integers in the list is 70.
(1) The median of the integers in the list is 70.
(2) The average of the integers in the list is 70.
20 Jul 2007, 03:54
Kudos
Bookmarks
i believe it's D
Statement 1: you have 5 numbers _ _ _ _ _
if the median is 70 you have _ _ 70 _ _
the problem states "list consists of five DIFFERENT integers" thus the two greatest must be at LEAST 71 and 72 making their average > 70
Statement 2: start with imagining all 5 numbers as 70.
70 70 70 70 70, this is the only case where the average of all numbers will be 70 with the two greatest NOT being greater than 70.
we know from the problem that all the integers are different. so for each number that gets smaller, another will get bigger. so even in this case
69 70 70 70 71, already the average of the 2 greatest is > 70, but even this case isn't possible.
68 69 70 71 72 is the limit. the average of the 2 greatest will be at least 71.5
it's hard to explain, but to me is the fastest way of logically accepting B
maybe an easier way of explaining statement 2 is:
if the 3 smaller are less than 70, the 2 greater have no choice but to average more than 70... to double check yourself, 69 + 69 + 69 = 207...
then the sum of 2 larger numbers = 143 (divide by 2) the avg is > 70
if 4 of the numbers are smaller than 70. the largest number must be greater than 70 to the point it must make up for all of the numbers less than 70. if the greatest and 2nd greatest (where the 2nd greatest is less than 70) average 70 the other 3 numbers will offset the average to less than 70. thus the greatest number must be bigger taking the average of it plus the 2nd greatest to greater than 70.
--------------a-b-c-d--------------------70-------------------e
a visual way to see it is if d and e are both the same distance from 70 (avg 70) the avg of all 5 will be < 70. e must be farther to the right to offset a, b, and c making d and e's average > 70
all apologies for the abstract and unnecessarily long explanation... but i think getting into the habit of testing limits and thinking like this will help a lot.
Statement 1: you have 5 numbers _ _ _ _ _
if the median is 70 you have _ _ 70 _ _
the problem states "list consists of five DIFFERENT integers" thus the two greatest must be at LEAST 71 and 72 making their average > 70
Statement 2: start with imagining all 5 numbers as 70.
70 70 70 70 70, this is the only case where the average of all numbers will be 70 with the two greatest NOT being greater than 70.
we know from the problem that all the integers are different. so for each number that gets smaller, another will get bigger. so even in this case
69 70 70 70 71, already the average of the 2 greatest is > 70, but even this case isn't possible.
68 69 70 71 72 is the limit. the average of the 2 greatest will be at least 71.5
it's hard to explain, but to me is the fastest way of logically accepting B
maybe an easier way of explaining statement 2 is:
if the 3 smaller are less than 70, the 2 greater have no choice but to average more than 70... to double check yourself, 69 + 69 + 69 = 207...
then the sum of 2 larger numbers = 143 (divide by 2) the avg is > 70
if 4 of the numbers are smaller than 70. the largest number must be greater than 70 to the point it must make up for all of the numbers less than 70. if the greatest and 2nd greatest (where the 2nd greatest is less than 70) average 70 the other 3 numbers will offset the average to less than 70. thus the greatest number must be bigger taking the average of it plus the 2nd greatest to greater than 70.
--------------a-b-c-d--------------------70-------------------e
a visual way to see it is if d and e are both the same distance from 70 (avg 70) the avg of all 5 will be < 70. e must be farther to the right to offset a, b, and c making d and e's average > 70
all apologies for the abstract and unnecessarily long explanation... but i think getting into the habit of testing limits and thinking like this will help a lot.
17 Jun 2018, 09:15
Kudos
Bookmarks
IMO "D"
A certain list consists of five different integers. Is the average (arithmetic mean) of the two greatest integers in the list greater than 70 ?
Let's say these 5 unique integers are A, B, C, D, E (from min to max)
So now we need to find if (D+E)/2 > 70 ---> D+E > 140
(1) The median of the integers in the list is 70.
Tell us that C = 70. To test that D+E > 140, I plug in the lowest possible values which are 71 and 72 (since they must be unique integers)
71 + 72 > 140 ---> sufficient
(2) The average of the integers in the list is 70.
which means (A+B+C+D+E)/5 = 70 ---> A+B+C+D+E = 350
To test whether D+E>140, we need to maximize the values A, B, and C
In this case the greatest possible values of A, B, and C are 68, 69, and 70 given that D and E must be greatest 2 integers.
(Let's say if A, B, C are 69, 70, 71 ---> D+E = 350-69-70-71 = 140 which is not possible that D and E are greatest 2 integers)
So we got the minimum values of D and E are 71 and 72. ---> Sufficient
A certain list consists of five different integers. Is the average (arithmetic mean) of the two greatest integers in the list greater than 70 ?
Let's say these 5 unique integers are A, B, C, D, E (from min to max)
So now we need to find if (D+E)/2 > 70 ---> D+E > 140
(1) The median of the integers in the list is 70.
Tell us that C = 70. To test that D+E > 140, I plug in the lowest possible values which are 71 and 72 (since they must be unique integers)
71 + 72 > 140 ---> sufficient
(2) The average of the integers in the list is 70.
which means (A+B+C+D+E)/5 = 70 ---> A+B+C+D+E = 350
To test whether D+E>140, we need to maximize the values A, B, and C
In this case the greatest possible values of A, B, and C are 68, 69, and 70 given that D and E must be greatest 2 integers.
(Let's say if A, B, C are 69, 70, 71 ---> D+E = 350-69-70-71 = 140 which is not possible that D and E are greatest 2 integers)
So we got the minimum values of D and E are 71 and 72. ---> Sufficient
