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30 Sep 2023, 05:15
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E
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Difficulty:
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Question Stats:
54% (02:04) correct
46%
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wrong
based on 168
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If a, b, c, d, e and f are integers, is their median greater than their average (arithmetic mean)?
(1) a < b < c < d < e < f
(2) b - a = d - c = f - e
(1) a < b < c < d < e < f
(2) b - a = d - c = f - e
30 Sep 2023, 07:36
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Bunuel
Statement 1
(1) a < b < c < d < e < f
On a number line, the relative position of \(a, b, c, d, e,\) and\( f\) is as shown below -
Attachment:
Screenshot 2023-09-30 194456.jpg [ 5.6 KiB | Viewed 1843 times ]
This statement only provides the relative position of the numbers w.r.t each other, hence we can have different possible values of mean and median. This flexibility allows us to create different scenarios one in which the mean is equal to the median and the answer to the question "Is their median greater than the average (arithmetic mean)" is No, and the other in which the mean is greater than the median and the the answer to the question "Is their median greater than the average (arithmetic mean)" is Yes.
Therefore the statement alone is not sufficient to answer the question and we can eliminate A and D.
Attachment:
Screenshot 2023-09-30 200239.jpg [ 35.32 KiB | Viewed 1825 times ]
Statement 2
(2) b - a = d - c = f - e
We know that the distance between the given pairs is the same. However, we do not know the relative position of the pairs with respect to each other. Hence, just as in Statement 1, we can create multiple possible combinations, one in which the mean = median and the other in which the mean is greater than the median.
Therefore the statement alone is not sufficient to answer the question and we can eliminate B.
Attachment:
Screenshot 2023-09-30 200346.jpg [ 36.79 KiB | Viewed 1783 times ]
Combined
The statements combined don't help either as we have no information on the distances between b and c, or the distances between d and e. Hence, we can use the examples of statement 2 to prove that the statements combined are insufficient to answer the question.
Option E
01 Oct 2023, 00:37
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If a, b, c, d, e and f are integers, is their median greater than their average (arithmetic mean)?
This tests extreme scenarios and somewhat weighted average concept.
(1) a < b < c < d < e < f
A. If integers are equidistant then NO.
Eg. 1,2,3,4,5,6
Median = (3+4)/2 = 3.5
Average = 21/6 = 3.5
(From weighted average pov the average is exactly between the integers because of uniformity in distance between integers)
B. If integers are not equidistant then YES.
Eg. 2,4,6,7,8,9
Median = (6+7)/2 = 6.5
Average = 36/6 = 6
(From weighted average pov the average shifts slightly towards left hand side even after some form of uniformity in distance between first 3 integers which is not same as that of last 3)
Best would have been to put first one or the last one at extreme ends which would have cause weighted average towards the big value.
INSUFFICIENT.
(2) b - a = d - c = f - e
A. If all are equidistant like we have a case in Statement 1 then NO
Eg. 1,2,3,4,5,6 where 2 - 1 = 4 - 3 = 6 - 5 = 1
Median = (3+4)/2 = 3.5
Average = 21/6 = 3.5
(From weighted average pov the average is exactly between the integers because of uniformity in distance between integers)
B. Another case could be 1,2,7,8,31,32 then NO
2 - 1 = 8 - 7 = 32 - 31
Median = (7+8)/2 = 7.5
Average = 81/6 = 13.5
OR
C. a, b, c, d, e, f are as follows
-2, -3, 1, 0, 2, 1 where
-3-(-2) = 0 - 1 = 1 - 2 = -1 then YES
Median = (0+1)/2 = 0.5
Average = (-1)/6 = -0.166..
INSUFFICIENT.
Together 1 and 2
A. St.1 A satisfies with NO
B. Taking exact opposite of St.1 A but with extreme situation then YES
-26,-25,-4,-3,-2,-1
Median = -3.5
Median = -10.5
Answer E.
This tests extreme scenarios and somewhat weighted average concept.
(1) a < b < c < d < e < f
A. If integers are equidistant then NO.
Eg. 1,2,3,4,5,6
Median = (3+4)/2 = 3.5
Average = 21/6 = 3.5
(From weighted average pov the average is exactly between the integers because of uniformity in distance between integers)
B. If integers are not equidistant then YES.
Eg. 2,4,6,7,8,9
Median = (6+7)/2 = 6.5
Average = 36/6 = 6
(From weighted average pov the average shifts slightly towards left hand side even after some form of uniformity in distance between first 3 integers which is not same as that of last 3)
Best would have been to put first one or the last one at extreme ends which would have cause weighted average towards the big value.
INSUFFICIENT.
(2) b - a = d - c = f - e
A. If all are equidistant like we have a case in Statement 1 then NO
Eg. 1,2,3,4,5,6 where 2 - 1 = 4 - 3 = 6 - 5 = 1
Median = (3+4)/2 = 3.5
Average = 21/6 = 3.5
(From weighted average pov the average is exactly between the integers because of uniformity in distance between integers)
B. Another case could be 1,2,7,8,31,32 then NO
2 - 1 = 8 - 7 = 32 - 31
Median = (7+8)/2 = 7.5
Average = 81/6 = 13.5
OR
C. a, b, c, d, e, f are as follows
-2, -3, 1, 0, 2, 1 where
-3-(-2) = 0 - 1 = 1 - 2 = -1 then YES
Median = (0+1)/2 = 0.5
Average = (-1)/6 = -0.166..
INSUFFICIENT.
Together 1 and 2
A. St.1 A satisfies with NO
B. Taking exact opposite of St.1 A but with extreme situation then YES
-26,-25,-4,-3,-2,-1
Median = -3.5
Median = -10.5
Answer E.
