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20 Jul 2020, 02:56
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63% (01:49) correct
37%
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For S, a list of five integers, is the mean of the list an integer in S?
(1) The difference between the smallest integer in S and the median of S is the same as the difference between the median and the highest integer in S.
(2) The difference between the second smallest integer in S and the median of S is the same as the difference between the median and the second highest integer in S.
(1) The difference between the smallest integer in S and the median of S is the same as the difference between the median and the highest integer in S.
(2) The difference between the second smallest integer in S and the median of S is the same as the difference between the median and the second highest integer in S.
yashikaaggarwal
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20 Jul 2020, 03:21
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Let the five number be a1, a2, a3, a4, and a5
statement 1 = a3 - a1 = a5 - a3
=> 2a3 = a1+a5
so, a1, a3, and a5 are at consecutive distance
case 1 : a1, a2, a3, a4, and a5 = 1,2,3,4,5
mean = 3
case 2: a1, a2, a3, a4, and a5 = 4,6,12,16,20
mean = 58/5 = 11.6 (not in s list)
(insufficient)
statement 2 = a2 - a1 = a4 - a3
=> 2a3 = a2+a4
so, a2, a3, and a4 are at consecutive distance
case 1 : a1, a2, a3, a4, and a5 = 1,2,3,4,5
mean = 3
case 2: a1, a2, a3, a4, and a5 = 2,6,12,18,21
mean = 59/5 = 11.8 (not in s list)
(insufficient)
Statement 1&2 together: a1, a2, a3, a4, and a5 are at consecutive distance
so their mean will always be equal to the middle term that is a3
(sufficient)
Answer should be C
statement 1 = a3 - a1 = a5 - a3
=> 2a3 = a1+a5
so, a1, a3, and a5 are at consecutive distance
case 1 : a1, a2, a3, a4, and a5 = 1,2,3,4,5
mean = 3
case 2: a1, a2, a3, a4, and a5 = 4,6,12,16,20
mean = 58/5 = 11.6 (not in s list)
(insufficient)
statement 2 = a2 - a1 = a4 - a3
=> 2a3 = a2+a4
so, a2, a3, and a4 are at consecutive distance
case 1 : a1, a2, a3, a4, and a5 = 1,2,3,4,5
mean = 3
case 2: a1, a2, a3, a4, and a5 = 2,6,12,18,21
mean = 59/5 = 11.8 (not in s list)
(insufficient)
Statement 1&2 together: a1, a2, a3, a4, and a5 are at consecutive distance
so their mean will always be equal to the middle term that is a3
(sufficient)
Answer should be C
20 Jul 2020, 09:52
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Neither statement could be sufficient alone, because using either we can freely determine two values in the set, and can easily get a non-integer mean.
Both Statements together tell you the set is "symmetric" (if you have a value above the median, there is a corresponding value equally far below the median). For any symmetric set, the mean and median are equal, so the two Statements together are sufficient.
If you weren't familiar with symmetric sets, you could do the problem algebraically. If m is the median, we know our smallest and largest values are m - d and m + d (from Statement 1), and our second-largest and second-smallest values are m - c and m + c (from Statement 2). So our list is
m-d, m-c, m, m+c, m+d
and summing this list and dividing by 5, we find the mean is equal to m, the median, which is an integer in our list.
Both Statements together tell you the set is "symmetric" (if you have a value above the median, there is a corresponding value equally far below the median). For any symmetric set, the mean and median are equal, so the two Statements together are sufficient.
If you weren't familiar with symmetric sets, you could do the problem algebraically. If m is the median, we know our smallest and largest values are m - d and m + d (from Statement 1), and our second-largest and second-smallest values are m - c and m + c (from Statement 2). So our list is
m-d, m-c, m, m+c, m+d
and summing this list and dividing by 5, we find the mean is equal to m, the median, which is an integer in our list.
