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A set S contains five numbers k, k + 1, 2k - 1, 5k + 7 and 6k + 8. If

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itzmyzone911
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A set S contains five numbers k, k + 1, 2k - 1, 5k + 7 and 6k + 8. If [#permalink] New post   03 May 2015, 22:42
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A set S contains five numbers k, k + 1, 2k - 1, 5k + 7 and 6k + 8. If k is a positive integer, determine its value.

(1) The median of the elements of S is k + 1.

(2) The average of the elements of S is less than 10.


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Given OA is A.

My doubt: A is definitely true for k=1. However for k=2, k+1=2k-1, which is also compatible with stmt. 1. This means we have 2 values for k..1 and 2. Then how can A alone be sufficient?


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Re: A set S contains five numbers k, k + 1, 2k - 1, 5k + 7 and 6k + 8. If [#permalink] New post   03 May 2015, 23:41
I got this wrong, eager to see the solution. I worked it out likes this:

Statement 1: insuff

Because k can have several values to meet those conditions (I think?)

if k = 1
k---k+1---2k-1---5k+7---6k+8
1 2 1 12 14

Median = 2


if k = 2
k---k+1---2k-1---5k+7---6k+8
2 3 3 17 20

Median = 3

So K could be 2 or 3, and probably many more numbers (??)

Statement 2: average of set is less than 10

Using the same examples from above the averages are 6 and 8, both under 10. ---> insuff


Not sure how combining these would help either, but there may very well be a way!

Don't understand how the OA is A...

I would have gone for E on an exam, maybe C.
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Re: A set S contains five numbers k, k + 1, 2k - 1, 5k + 7 and 6k + 8. If [#permalink] New post   04 May 2015, 01:12
It's given that the average is less than 10 , so
as per condition 2 :

(k+ k+1 + 2k-1 +5k+7 + 6k+8) / 5 < 10
(15k + 15) /5 <10
3k + 3 < 10
3k < 7
and k is integer
so k = 1, 2

For k=1, s={1 , 2, 1, 12, 14} and Median= 2 (condition 1 satisfied)

for k=2, s={2, 3, 3, 17, 20} and Median =3 (condition 1 satisfied)

So K=1 and 2
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Re: A set S contains five numbers k, k + 1, 2k - 1, 5k + 7 and 6k + 8. If [#permalink] New post   04 May 2015, 01:16
Moumila wrote:
It's given that the average is less than 10 , so
as per condition 2 :

(k+ k+1 + 2k-1 +5k+7 + 6k+8) / 5 < 10
(15k + 15) /5 <10
3k + 3 < 10
3k < 7
and k is integer
so k = 1, 2

For k=1, s={1 , 2, 1, 12, 14} and Median= 2 (condition 1 satisfied)

for k=2, s={2, 3, 3, 17, 20} and Median =3 (condition 1 satisfied)

So K=1 and 2



Yeah but the average being less than 10 is given in statement nr. 2, so how can the answer be "A) Statement 1 alone is sufficient?"

Is there something in the prompt that I'm missing?
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Re: A set S contains five numbers k, k + 1, 2k - 1, 5k + 7 and 6k + 8. If [#permalink] New post   04 May 2015, 01:28
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Re-attempting this after some thought:

Placing all values in a number line, it is unclear where k+1 and 2k-1 go, since if k is given different values it is clear that they will shift depending on values for K

itzmyzone911 wrote:
A set S contains five numbers k, k + 1, 2k - 1, 5k + 7 and 6k + 8. If k is a positive integer, determine its value.

(1) The median of the elements of S is k + 1.


When given this information, we know that this means that k+1 > 2k-1, it is further to the right on the number line

Using this inequality we get that k<2 .... Which means that k MUST be 1 as it is a positive integer.



(2) The average of the elements of S is less than 10.
Given that statement 1 alone is sufficient, this one needs to be either insuff or suff on its own. It is not suff on it's own so that means that the answer is A.

Show Spoiler
Given OA is A.

My doubt: A is definitely true for k=1. However for k=2, k+1=2k-1, which is also compatible with stmt. 1. This means we have 2 values for k..1 and 2. Then how can A alone be sufficient?
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Re: A set S contains five numbers k, k + 1, 2k - 1, 5k + 7 and 6k + 8. If [#permalink] New post   05 May 2015, 10:13
sabineodf wrote:
Re-attempting this after some thought:

Placing all values in a number line, it is unclear where k+1 and 2k-1 go, since if k is given different values it is clear that they will shift depending on values for K

itzmyzone911 wrote:
A set S contains five numbers k, k + 1, 2k - 1, 5k + 7 and 6k + 8. If k is a positive integer, determine its value.

(1) The median of the elements of S is k + 1.


When given this information, we know that this means that k+1 > 2k-1, it is further to the right on the number line

Using this inequality we get that k<2 .... Which means that k MUST be 1 as it is a positive integer.



(2) The average of the elements of S is less than 10.
Given that statement 1 alone is sufficient, this one needs to be either insuff or suff on its own. It is not suff on it's own so that means that the answer is A.


sabineodf, for statement 1, why don't you consider the equation as \(k+1>=2k-1\)
In this case, you would get \(k<=2\), which means k=1,2.
Thus, insufficient.

Please clarify.
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Re: A set S contains five numbers k, k + 1, 2k - 1, 5k + 7 and 6k + 8. If [#permalink] New post   05 May 2015, 18:32
akhilbajaj

I tested the values.

If k = 2, then as you said 2+1=2*2−1 (3=3). But statement 1) tells us that k+1 is the median, and I think that this statement means that it is the only median, which is wouldn't be if K=2 because then we have a shared median, which leaves that k=1 since it's the only number that fits the parameters, and gives a unique value for the median.

I'm not 100% sure on this.
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Re: A set S contains five numbers k, k + 1, 2k - 1, 5k + 7 and 6k + 8. If [#permalink] New post   05 May 2015, 23:16
sabineodf wrote:
akhilbajaj

I tested the values.

If k = 2, then as you said 2+1=2*2−1 (3=3). But statement 1) tells us that k+1 is the median, and I think that this statement means that it is the only median, which is wouldn't be if K=2 because then we have a shared median, which leaves that k=1 since it's the only number that fits the parameters, and gives a unique value for the median.

I'm not 100% sure on this.



In this question, it is said that k+1 is the median, so you should not assume any extra information.
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Re: A set S contains five numbers k, k + 1, 2k - 1, 5k + 7 and 6k + 8. If [#permalink] New post   05 May 2015, 23:26
PrepTap wrote:
sabineodf wrote:
akhilbajaj

I tested the values.

If k = 2, then as you said 2+1=2*2−1 (3=3). But statement 1) tells us that k+1 is the median, and I think that this statement means that it is the only median, which is wouldn't be if K=2 because then we have a shared median, which leaves that k=1 since it's the only number that fits the parameters, and gives a unique value for the median.

I'm not 100% sure on this.



In this question, it is said that k+1 is the median, so you should not assume any extra information.


K+1 = median

7 values. There must be 3 values on either side of k+1. If k+1 = 2k-1 then they are interchangeable and k+1 is not 100 % the median. But we are given that it IS the median, so that means there must be 2 values either side. I don't think I've made any extreme assumptions/
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Re: A set S contains five numbers k, k + 1, 2k - 1, 5k + 7 and 6k + 8. If [#permalink] New post   07 May 2015, 08:07
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Option A :The median of the elements of S is k + 1 .There are 5 elements so two elements on the left and two elements on the right of k+1
so at most k+1>=2k-1 or k+1<=5k+7 .Solving first equation gives us : k<=2 .Second gives k>=6/4 so k=2

Option B : (k +k + 1+2k - 1 +5k + 7 + 6k + 8)/5 < 10 .Solving 3k+3<10 so k<7/3 .k can have values 1,2 So B is not correct .

Hence A is the correct option.

Press Kudos if you like the solution.

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Re: A set S contains five numbers k, k + 1, 2k - 1, 5k + 7 and 6k + 8. If [#permalink] New post   18 Jun 2015, 23:57
I think the solution A would be correct if we changed 5k + 7 to 5k - 7:
set S contains five numbers k, k + 1, 2k - 1, 5k - 7 and 6k + 8. If k is a positive integer, determine its value.

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Hi there,
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Re: A set S contains five numbers k, k + 1, 2k - 1, 5k + 7 and 6k + 8. If [#permalink]
18 Jun 2015, 23:57
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