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

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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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New post 29 Jan 2017, 04:50
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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
[Reveal] Spoiler: OA

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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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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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New post 29 Jan 2017, 09:17
The average is 61 therefore the max value must be higher

We know that the largest value is 6 times the smallest plus 7

=> (61-7)/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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New post 29 Jan 2017, 10:38
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

=> (61-7)/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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New post 29 Jan 2017, 12:32
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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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/stonecold-s- ... 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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New post 10 Aug 2017, 09:44
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

=> (61-7)/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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New post 03 Jan 2018, 00:42
Given average \(61\) and \(max\) = \(min * 6 + 7\), so set must have 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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New post 10 Jan 2018, 18:12
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/stonecold-s- ... l#p1676182



I'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   [#permalink] 10 Jan 2018, 18:12
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Set X of has an average of 61. If the largest element is 7 greater tha

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