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Set X of has an average of 61. If the largest element is 7 greater tha [#permalink]
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29 Jan 2017, 05:50
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Set \(X\) of has an average of \(61\). If the largest element is 7 greater than 6 times the smallest element, how many values out of \({2,3,5,8,9,11,53,102,123,178,210,267,283,311,376,383,399,401}\) can be a part of Set \(X\)? (A) 18 (B) 13 (C) 9 (D) 7 (E) cannot be determined
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Re: Set X of has an average of 61. If the largest element is 7 greater tha [#permalink]
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29 Jan 2017, 08:53
ziyuenlau wrote: Set \(X\) of has an average of \(61\). If the largest element is 7 greater than 6 times the smallest element, how many values out of \({2,3,5,8,9,11,53,102,123,178,210,267,283,311,376,383,399,401}\) can be a part of Set \(X\)?
(A) 18 (B) 13 (C) 9 (D) 7 (E) cannot be determined I think this Q has already been discussed earlier. I hope you have searched for it before posting. Average is 61..find the max and min values possible, you will get the answer
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Re: Set X of has an average of 61. If the largest element is 7 greater tha [#permalink]
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29 Jan 2017, 10:17
The average is 61 therefore the max value must be higher
We know that the largest value is 6 times the smallest plus 7
=> (617)/6 = 9 => The smallest value must be grater than 9
The max value will be 61*6 + 7 = 373 => The largest value must be less than 373
=> The number in the set must be 9<X<373 => 9 values



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Re: Set X of has an average of 61. If the largest element is 7 greater tha [#permalink]
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29 Jan 2017, 11:38
LHC8717 wrote: The average is 61 therefore the max value must be higher
We know that the largest value is 6 times the smallest plus 7
=> (617)/6 = 9 => The smallest value must be grater than 9
The max value will be 61*6 + 7 = 373 => The largest value must be less than 373
=> The number in the set must be 9<X<373 => 9 values Hi, Let's say smallest is X then, {X+(6X+7)}/2 = 61 7X = 122  7 X = 115/7 Now, pls suggest what am I doing wrong? Thanks,



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Set X of has an average of 61. If the largest element is 7 greater tha [#permalink]
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29 Jan 2017, 13:32
Answer (C) is right.
2,3,5,8, can not be a part of X. Even if one of them is a min element of X then the max element is not greater 61 and the average of X cannot be 61. 376,383,399,401 can not be a part of X. Even if one of them is a max element of X then the min element is not less than 61 and the average of X cannot be 61.



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Set X of has an average of 61. If the largest element is 7 greater tha [#permalink]
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05 Aug 2017, 22:54
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Here is what I did for this one> Smallest value = a(let) Largest value => 6a+7 Now the smallest value of the set must be less than the mean and the largest value must be greater than the mean. Hence a>61 and 6a+7>61 ==> a>9 At a=61 => Largest value => 6a+7= 373 Hence the range of the set would be (9,373) Only 9 elements are in the range. Hence C. NOTE > THIS QUESTION is a part of the MOCK SERIES > https://gmatclub.com/forum/stonecolds ... l#p1676182
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Re: Set X of has an average of 61. If the largest element is 7 greater tha [#permalink]
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10 Aug 2017, 10:44
annusngh wrote: LHC8717 wrote: The average is 61 therefore the max value must be higher
We know that the largest value is 6 times the smallest plus 7
=> (617)/6 = 9 => The smallest value must be grater than 9
The max value will be 61*6 + 7 = 373 => The largest value must be less than 373
=> The number in the set must be 9<X<373 => 9 values Hi, Let's say smallest is X then, {X+(6X+7)}/2 = 61 7X = 122  7 X = 115/7 Now, pls suggest what am I doing wrong? Thanks, You are assuming the numbers in the set to be in AP. Its not given in the problem!



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Set X of has an average of 61. If the largest element is 7 greater tha [#permalink]
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Updated on: 15 May 2018, 21:31
Given average \(61\) and \(max\) = \(min * 6 + 7\), so set must have alteast two different integers (meaning it is not a set of all same integers)
in a set of different integers, min value can't be equal to average, so \(min\) < \(61\) => \(max\) < \(61 * 6 + 7\), so \(max\) < \(373\) similarly, max value can't be equal to average, so \(max\) > \(61\) => \(61\) < \(max\) < \(373\)
now we found, \(max\) > \(61\)=> \(min * 6 + 7\) > \(61\) => \(min\) > \(9\).
all the elements in the set must be between \(9 < S < 373\), only 9 integers of the given falls within the range => Answer (C)



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Re: Set X of has an average of 61. If the largest element is 7 greater tha [#permalink]
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10 Jan 2018, 19:12
stonecold wrote: Here is what I did for this one> Smallest value = a(let) Largest value => 6a+7 Now the smallest value of the set must be less than the mean and the largest value must be greater than the mean. Hence a>61 and 6a+7>61 ==> a>9 At a=61 => Largest value => 6a+7= 373 Hence the range of the set would be (9,373) Only 9 elements are in the range. Hence C. NOTE > THIS QUESTION is a part of the MOCK SERIES > https://gmatclub.com/forum/stonecolds ... l#p1676182I'm confused how you derived the largest value 373 by taking a=61. Can you please explain.




Re: Set X of has an average of 61. If the largest element is 7 greater tha
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