i) SUFF 3...11 (9*(3+11)*2=63)
ii) SUFF 6..14 (9*(4+14)*2=81)
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goyalsau
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GHong14 wrote:Please explain a little more
Q) - AM of 11 consecutive integers
1) - Average of first 9 integers is 7 ,
Sum of first 9 integers will be 7 * 9 = 63
There is a formula of sum of consecutive integers N ( F + L ) / 2
N - Number of terms
F - First term in the series
L - Last term in the series.
Let the first term be N , Last term must be N + 8
Take for example First term is 1 , last term is 9 , Or it can be written as 1 + 8
63 = 9 { N + N + 8 } / 2
on solving N = 3
First term is 3 , 9th term must be 11 ( 3 + 8 ) , 11th term must be 3 + 10 = 13
So its sufficient.
Same way we can do it with last terms,
It will be good if you try by your self and if you have problem Do let me know i will try to put in best possible efforts
Saurabh Goyal
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Geva@EconomistGMAT
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the concept tested by this question is as follows: in any set of consecutive terms, the average is also the median.GHong14 wrote:Can anyone explain why the selected answer is wrong
If the number of terms in the set is odd (such as 7 or 9), the median is simply the number in the middle when the set is ordered in ascending order.
Thus, if we;'re looking for the average of 11 cons. integers, we need the 6th integer.
Stat. (1), using the same concept, tells us that the median of the first 9 integers is 7. In other words, then 5th integer (the number in the middle of the first 9) is 7, so the 6th integer we need to find the average of the entire set is 8. Sufficient.
Stat. (2) same idea: the median of the last 9 terms is 9. the median of the last 9 terms will be 5 away from the top: in a set of 11 integers, 5 away from the top is the 7th term (you can list them out backwards: A11, A10, A9, A8, A7 is the 5th term). So stat. (2) tells you that the 7th term is 9, making the required 6th term 8. Sufficient.
