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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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29 Jan 2017, 05:50
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95% (hard)

Question Stats:

35% (02:33) correct 65% (02:33) wrong based on 202 sessions

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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

=> (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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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

=> (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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29 Jan 2017, 13:32

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/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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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

=> (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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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)

Originally posted by hellosanthosh2k2 on 03 Jan 2018, 01:42.
Last edited by hellosanthosh2k2 on 15 May 2018, 21:31, edited 1 time in total.
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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/stonecold-s- ... l#p1676182

I'm confused how you derived the largest value 373 by taking a=61.

Re: Set X of has an average of 61. If the largest element is 7 greater tha   [#permalink] 10 Jan 2018, 19:12
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