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03 Dec 2017, 06:14
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D
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Difficulty:
25%
(medium)
Question Stats:
73% (01:06) correct
27%
(01:39)
wrong
based on 158
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What is the average (arithmetic mean) of 12 consecutive even integers ?
(1) The average (arithmetic mean) of the first eight integers is 13.
(2) The average (arithmetic mean) of the last eight integers is 21.
(1) The average (arithmetic mean) of the first eight integers is 13.
(2) The average (arithmetic mean) of the last eight integers is 21.
03 Dec 2017, 12:01
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Consecutive even integers will be in Arithmetic Progression (even spacing between every two terms), and so their Average or Arithmetic Mean = their Median
(1) AM of first 8 integers = 13
So their median = 13, now median for these 8 integers would be the average of 4th and 5th integer.. that means 4th and 5th integers are 12 and 14 respectively. Since these are consecutive even integers, if we can find 4th integer, we can also find the 1st integer, thus we have all 12 integers, and we can find their average also. (No need to calculate the same since its data sufficiency). Sufficient.
(2) AM of last eight integers (i.e., 5th to 12th integers) = 21
Applying the same logic as we did in statement 1, we can find the middle two terms (9th and 10th integers) here, thus we can have all the integers. Sufficient.
(Basically since this is an AP series with a common difference of 2, if we have any numbered term say 5th or 9th term, we can definitely find first term)
Hence D answer
(1) AM of first 8 integers = 13
So their median = 13, now median for these 8 integers would be the average of 4th and 5th integer.. that means 4th and 5th integers are 12 and 14 respectively. Since these are consecutive even integers, if we can find 4th integer, we can also find the 1st integer, thus we have all 12 integers, and we can find their average also. (No need to calculate the same since its data sufficiency). Sufficient.
(2) AM of last eight integers (i.e., 5th to 12th integers) = 21
Applying the same logic as we did in statement 1, we can find the middle two terms (9th and 10th integers) here, thus we can have all the integers. Sufficient.
(Basically since this is an AP series with a common difference of 2, if we have any numbered term say 5th or 9th term, we can definitely find first term)
Hence D answer
18 Nov 2019, 07:05
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The question tells us that we have 12 consecutive even terms.
This can be represented as:
x, x+2, x+4.... and so on.
A tells us that the average of the first eight terms = 13, or (x + x+2 + x+4 ...)/8=13. We don't need to do the math here, but can conclude that we can solve for the value of x, and therefore calculate all values in the set.
We can conclude the same information from B based on the reasoning above.
This can be represented as:
x, x+2, x+4.... and so on.
A tells us that the average of the first eight terms = 13, or (x + x+2 + x+4 ...)/8=13. We don't need to do the math here, but can conclude that we can solve for the value of x, and therefore calculate all values in the set.
We can conclude the same information from B based on the reasoning above.
