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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
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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
