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# A list of numbers has six positive integers. Three of those integers a

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A list of numbers has six positive integers. Three of those integers a  [#permalink]

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07 Jul 2015, 02:03
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A list of numbers has six positive integers. Three of those integers are known: 4, 5 and 24 and three of those are unknown: x, y and z. The three unknowns are known to be distinct. It is also known that the mean of the list is 10 and the median lies between 7 and 8 (exclusive).

Which of the following CANNOT be the value of any one of the unknowns?

(A) 13
(B) 12
(C) 11
(D) 10
(E) 5

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Re: A list of numbers has six positive integers. Three of those integers a  [#permalink]

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10 Apr 2017, 19:14
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3
Bunuel wrote:
A list of numbers has six positive integers. Three of those integers are known: 4, 5 and 24 and three of those are unknown: x, y and z. The three unknowns are known to be distinct. It is also known that the mean of the list is 10 and the median lies between 7 and 8 (exclusive).

Which of the following CANNOT be the value of any one of the unknowns?

(A) 13

(B) 12

(C) 11

(D) 10

(E) 5

Terms are {4, 5, 24, x, y, z}
Mean = 10 i.e. Sum = 10*6 = 60
i.e.4+5+24+x+y+z = 60
i.e. x+y+z = 27

Media = between 7 and 8
But since number of terms in the set is even so median is the average of two middle terms which can only be 7.5
Hence, Median = 7.5
Media 7.5 is possible when two of the terms in the sets are {7,8} or {6,9} or {5,10} or {4,11} or {3,12} or {2,13} or {1,14}

Lets check options
(A) 13 {x, y, z} may be {2, 13, 12} hence Possible

(B) 12 {x, y, z} may be {5, 10, 12} hence Possible

(C) 11 {x, y, z}hence NOT Possible as the pair needed is {4, 11} but 4 can't be third term ever as sum of other two of x, y, z is 16 which can't be each smaller than 4. CORRECT ANSWER

(D) 10

(E) 5

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Re: A list of numbers has six positive integers. Three of those integers a  [#permalink]

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07 Jul 2015, 04:10
6
Mean = 10
Therefore sum of the 6 digits = 6*10 = 60 -------- (1)
The 3 digits are 4,5 & 24 -------------- (2)

Now, 7 < median < 8
-> 7 < (sum of 2 digits)/2 < 8
-> 14 < (sum of 2 digits) < 16
Therefore the sum of 2 digits = 15.

The 2 digits whose sum is 15 & whose median is b/w 7 & 8 can be :
1. 7, 8,
2. 6, 9,
3. 5, 10, or

But as soon as we take 4, 11 -> the series becomes 4,4,5,11,12,24 &
11 pairs up with 5 to give median as 8.

Hence the number can't be 11 !!
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Re: A list of numbers has six positive integers. Three of those integers a  [#permalink]

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07 Jul 2015, 10:17
3
Bunuel wrote:
A list of numbers has six positive integers. Three of those integers are known: 4, 5 and 24 and three of those are unknown: x, y and z. The three unknowns are known to be distinct. It is also known that the mean of the list is 10 and the median lies between 7 and 8 (exclusive). Which of the following CANNOT be the value of any one of the unknowns?

(A) 13
(B) 12
(C) 11
(D) 10
(E) 5

If the mean is 10, the sum is 60. The three known numbers add to 33.
If the median is between 7 and 8, that means two digits were averaged to the median. (7<(x+y)/2<8 is the equation form).
x+y must be 15 (between 14 and 16). The other number is 12 (to equal 27).

I chose to solve using the answer choices. 10 and 5 add up to 15, so certainly both cannot be the answer.

For 11 to be the answer, the other number is 4.

4,4,5,11,12,27 is the new set. The median is 8. The question said 7 and 8 exclusive. Therefore, 11 is the answer.
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Re: A list of numbers has six positive integers. Three of those integers a  [#permalink]

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07 Jul 2015, 10:45
1
Bunuel wrote:
A list of numbers has six positive integers. Three of those integers are known: 4, 5 and 24 and three of those are unknown: x, y and z. The three unknowns are known to be distinct. It is also known that the mean of the list is 10 and the median lies between 7 and 8 (exclusive). Which of the following CANNOT be the value of any one of the unknowns?

(A) 13
(B) 12
(C) 11
(D) 10
(E) 5

Kudos for a correct solution.

Set of terms is {4, 5, x, y, z, 24}

Since x, y and z are distinct so let's assume Let, x < y < z

Mean = 10 i.e. Sum = Mean*no. of terms = 10*6 = 60

i.e. Sum of {4, 5, 24, x, y, z} = 60
i.e. x+y+z = 27

median = Average of two middle terms (for even no. of terms in set) = between 7 and 8 which can only be 7.5
if x+y = 7.5*2 = 15
i.e. z = 27 - 15 = 12

Case 1: x = 5 and y = 10 and z = 12
Case 2: x = 6 and y = 9 and z = 12
Case 3: x = 7 and y = 8 and z = 12

OR if z = 13 then x+y = 14
i.e. x and y can be 4 and 10 respectively and the set becomes {4, 4, 5, 10, 13, 24} i.e. Mean = 7.5

So x, y and z can be anything but 11

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Re: A list of numbers has six positive integers. Three of those integers a  [#permalink]

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13 Jul 2015, 03:59
Bunuel wrote:
A list of numbers has six positive integers. Three of those integers are known: 4, 5 and 24 and three of those are unknown: x, y and z. The three unknowns are known to be distinct. It is also known that the mean of the list is 10 and the median lies between 7 and 8 (exclusive). Which of the following CANNOT be the value of any one of the unknowns?

(A) 13
(B) 12
(C) 11
(D) 10
(E) 5

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

The question gives us concrete information about mean – it is 10 – but not about median – it is between 7 and 8 (exclusive). What can we say about median from this? That it cannot be 7 or 8 but anything in between. But we know that the list has all integers. When we have even number of integers, we know that the median is the average of the middle two numbers – when all are placed in increasing order. So can the average of the two middle numbers be, say, 7.1? Which two positive integers can average to give 7.1? None! Note that if the average of two integers is a decimal, the decimal must be (some number).5 such as 7.5 or 9.5 or 22.5 etc. This happens in case one number is odd and the other is even. In all other cases, the average would be an integer.

Since the median is given to be between 7 and 8, the median of the list of the six positive integers must be 7.5 only.

Now we know that the mean = 10 and median = 7.5

Method 1: Algebra/Logic

Let’s try to solve the question algebraically/logically first.

There are 6 elements in the list. The average of the list is 10 which means the sum of all 6 elements = 6*10 = 60

4 + 5 + 24 + x + y + z = 60

x + y + z = 27

Median of the list = 7.5

So sum of third and fourth elements must be 7.5 * 2 = 15

There are two cases possible:

Case 1: Two of the three integers x, y and z could be the third and the fourth numbers. In that case, since already 4 and 5 are less than 7.5, one of the unknown number would be less than 7.5 (the third number) and the other two would be greater than 7.5.

The sum of the third and fourth elements of the list is 15 so

15 + z = 27

z = 12

So, two numbers whose sum is 15 such that one is less than 7.5 and the other greater than 7.5 could be

5 and 10

6 and 9

7 and 8

x, y and z could take values 5, 6, 7, 8, 9, 10 and 12.

Case 2: The known 5 could be the third number in which case one of the unknown numbers is less than 5 and two of the unknown numbers would be more than 7.5.

If the third number is 5, the fourth number has to be 10 to get a median of 7.5. Hence, 10 must be one of the unknown numbers.

The sum of the other two unknown numbers would be 27 – 10 = 17.

One of them must be less than 5 and the other greater than 10. So possible options are

4 and 13

3 and 14

2 and 15

1 and 16

x, y and z could take various values but none of them could be 11

Method 2: Process of Elimination

Let’s now try to look at the process of elimination here and see if we can find an easier way.

The three unknowns need to add up to 10*6 – 4 – 5 – 24 = 27.

Two of the given options are 5 and 10. They have a median of 7.5 so lets assume that two of the unknown numbers are 5 and 10 (5 can be one of the unknowns since we are not given that all six integers need to be distinct). If two unknowns make up third and fourth numbers in the list and have a median of 7.5, their sum would be 15 and the third unknown will be 12 (to get the mean of 10). This case (5, 10, 12) satisfies all conditions so options (B), (D) and (E) are out of play.

Now we are left with two options 13 and 11. Check any one of them and you will know which one is not possible. Let’s check 13.

From the given options, any number greater than 7.5 must be either the fourth number or the fifth number. 13 cannot be the fourth number since the third number would need to be 2 in that case to get median 7.5. But we have 4 and 5 more than 2 so it cannot be the third number. So 13 must be the fifth number of the list. We saw in the case above that if two unknowns are third and fourth numbers then the fifth number HAS TO BE 12. So the already present 5 must be the third number and the fourth number must be 10. In that case, the leftover unknown would be 4 (to get a sum of 27). So the three unknowns would be 4, 10 and 13. This satisfies all conditions and is possible. Hence answer must be (C). 11 will not be possible.

Let’s see what would have happened had you picked 11 to try out. If 11 were the fourth number, to get a median of 7.5, we would need 4 as the third number. That is not possible since we already have a 5 given. So 11 must have been the fifth number. This would mean that the already present 5 and one unknown 10 would make the median of 7.5. So the third unknown in this case would be 6 (to get a sum of 27). But 6 would be the third number and the median in this case would be (6 + 10)/2 = 8. So one of the numbers cannot be 11.

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Re: A list of numbers has six positive integers. Three of those integers a  [#permalink]

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07 Mar 2016, 12:55
The 6 positive integers can be 4, 5, x, y, z, 24.
Mean = 10 --> 4 + 5 + x + y + z + 24 = 60 --> x + y + z = 27
Median lies between 7 and 8. Since all the numbers are integers, median = (x + y)/2 --> x + y = 2 * Median

x + y = integer value between 14 and 16 --> x + y = 15. So value of z = 12

We now have 4, 5, x, y, 12, 24. We can now substitute for the options for x and y and check for the median value.

If one of the unknowns is 13 then the values take 2, 4, 5, 12, 13, 24 and Median = 17/2 > 8.

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Re: A list of numbers has six positive integers. Three of those integers a  [#permalink]

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05 Feb 2017, 00:13
1
4
Bunuel wrote:
A list of numbers has six positive integers. Three of those integers are known: 4, 5 and 24 and three of those are unknown: x, y and z. The three unknowns are known to be distinct. It is also known that the mean of the list is 10 and the median lies between 7 and 8 (exclusive).

Which of the following CANNOT be the value of any one of the unknowns?

(A) 13

(B) 12

(C) 11

(D) 10

(E) 5

Hi all those looking for a short method..

points to NOTE..
1) the median of 6 integers will be the centre value of two MIDDLE integers, here it will be 3rd and 4th in ascending/descending order.
2) when you are looking for centre of two integers, the centre value will be either an integer or a number with .5 in decimals. Here it is given between 7 and 8, the value is 7.5
3) sum of two of the three unknowns is 7.5*2=15

Let's work further..
Total of unknowns=10*6-(24+5+4)=27..
The third unknown =27-15=12..

for other two totalling 15, the lower one cannot be less than 5, as the two lowest are 4 and 5..
So values can be 5 and 10,... or 6 and 9... or 7 and 8...
So A and C are not possible...
Bunuel , we may have to change the median from ' between 7 and 8(exclusive) to from 7 to 8(exclusive)..

So now the Q says that the median could be 7 also,
total of two is 2*7=14 and third unknown becomes 27-14=13....
And other two values can be 5 and 9...... Or 6 and 8.....

Now only 11 is not possible..
Ans C
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Re: A list of numbers has six positive integers. Three of those integers a  [#permalink]

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10 Apr 2017, 15:23
chetan2u wrote:
Bunuel wrote:
A list of numbers has six positive integers. Three of those integers are known: 4, 5 and 24 and three of those are unknown: x, y and z. The three unknowns are known to be distinct. It is also known that the mean of the list is 10 and the median lies between 7 and 8 (exclusive).

Which of the following CANNOT be the value of any one of the unknowns?

(A) 13

(B) 12

(C) 11

(D) 10

(E) 5

Hi all those looking for a short method..

points to NOTE..
1) the median of 6 integers will be the centre value of two MIDDLE integers, here it will be 3rd and 4th in ascending/descending order.
2) when you are looking for centre of two integers, the centre value will be either an integer or a number with .5 in decimals. Here it is given between 7 and 8, the value is 7.5
3) sum of two of the three unknowns is 7.5*2=15

Let's work further..
Total of unknowns=10*6-(24+5+4)=27..
The third unknown =27-15=12..

for other two totalling 15, the lower one cannot be less than 5, as the two lowest are 4 and 5..
So values can be 5 and 10,... or 6 and 9... or 7 and 8...
So A and C are not possible...
Bunuel , we may have to change the median from ' between 7 and 8(exclusive) to from 7 to 8(exclusive)..

So now the Q says that the median could be 7 also,
total of two is 2*7=14 and third unknown becomes 27-14=13....
And other two values can be 5 and 9...... Or 6 and 8.....

Now only 11 is not possible..
Ans C

How did you know that two lowest are 4 and 5 ? ...one of the possibility can be 3,4,5,10,14,24
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Re: A list of numbers has six positive integers. Three of those integers a  [#permalink]

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10 Apr 2017, 18:48
manishcmu wrote:
chetan2u wrote:
Bunuel wrote:
A list of numbers has six positive integers. Three of those integers are known: 4, 5 and 24 and three of those are unknown: x, y and z. The three unknowns are known to be distinct. It is also known that the mean of the list is 10 and the median lies between 7 and 8 (exclusive).

Which of the following CANNOT be the value of any one of the unknowns?

(A) 13

(B) 12

(C) 11

(D) 10

(E) 5

Hi all those looking for a short method..

points to NOTE..
1) the median of 6 integers will be the centre value of two MIDDLE integers, here it will be 3rd and 4th in ascending/descending order.
2) when you are looking for centre of two integers, the centre value will be either an integer or a number with .5 in decimals. Here it is given between 7 and 8, the value is 7.5
3) sum of two of the three unknowns is 7.5*2=15

Let's work further..
Total of unknowns=10*6-(24+5+4)=27..
The third unknown =27-15=12..

for other two totalling 15, the lower one cannot be less than 5, as the two lowest are 4 and 5..
So values can be 5 and 10,... or 6 and 9... or 7 and 8...
So A and C are not possible...
Bunuel , we may have to change the median from ' between 7 and 8(exclusive) to from 7 to 8(exclusive)..

So now the Q says that the median could be 7 also,
total of two is 2*7=14 and third unknown becomes 27-14=13....
And other two values can be 5 and 9...... Or 6 and 8.....

Now only 11 is not possible..
Ans C

How did you know that two lowest are 4 and 5 ? ...one of the possibility can be 3,4,5,10,14,24

Yes , you r correct

only we must look for defined median , mean and the three unknown to be distinct
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Re: A list of numbers has six positive integers. Three of those integers a  [#permalink]

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23 Jul 2018, 03:08
Bunuel wrote:
A list of numbers has six positive integers. Three of those integers are known: 4, 5 and 24 and three of those are unknown: x, y and z. The three unknowns are known to be distinct. It is also known that the mean of the list is 10 and the median lies between 7 and 8 (exclusive).

Which of the following CANNOT be the value of any one of the unknowns?

(A) 13

(B) 12

(C) 11

(D) 10

(E) 5

. if u take 10 the order is (3,4,5,10,14,24)
median will stay between 7 and 8.
If u take 13...median won't stay in range
if u take 11 median won't stay in range.
the question needs to introspect and evaluate itself ..... because its wrong in the first place and has no right to challenge us
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Re: A list of numbers has six positive integers. Three of those integers a  [#permalink]

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09 Dec 2018, 21:09
Ok i understand it is C because...
We know that the sum of the 6 digits is 60 for its mean is 6, and its median ranges from 7 to 7.99. Number 8 cannot be..the exercise itself says so "8(exclusive)".
Then we can calculate x+y+z with the set's help (4,5,x,y,z,24)
* 4+5+24+x+y+z=60
* x+y+z=27

Option A) 13
If z= 13 then x+y=14
We can find that x=6;y=8 works fine since the set would be:(4,5,6,8,13,24) median 7 it works
FOR (6+8)/2=7
Option B) 12
If Z=12 then x+y=15
We can find that x=5;y=10 works fine, and we get the set:

(4,5,5,10,12,24) with median 7.5 which still works.
FOR (5+10)/2=7.5

Option C) 11
If Z=11 then x+y= 16. Here the median has to be formed by summing either x or y with 11, if we do it with (x + y) /2 = 16/2 the median will be 8 which cannot be since it's exclusive.
So we need to rearrange the order like this:
X=4; y=11;z=12
So that the sum x+y+z=27
And (11+4)/2=7.5

Buuuut our set's order would change. Remember that the median is calculated with the values on the middle, so with 6 digits we would need the 3th and 4th value, and our set would end like this:
(4,4,5,11,12,24) this median is 8 since (11+5)/2=8
Therefore, there is no way to make it work since the median is out of range and the ANSWER IS C.

Points D and E are resolved with the same set made in point b, so they also work.
Hope it helps

Posted from my mobile device
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Re: A list of numbers has six positive integers. Three of those integers a  [#permalink]

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13 Jul 2019, 14:39
GMATinsight wrote:
Bunuel wrote:
A list of numbers has six positive integers. Three of those integers are known: 4, 5 and 24 and three of those are unknown: x, y and z. The three unknowns are known to be distinct. It is also known that the mean of the list is 10 and the median lies between 7 and 8 (exclusive).

Which of the following CANNOT be the value of any one of the unknowns?

(A) 13

(B) 12

(C) 11

(D) 10

(E) 5

Terms are {4, 5, 24, x, y, z}
Mean = 10 i.e. Sum = 10*6 = 60
i.e.4+5+24+x+y+z = 60
i.e. x+y+z = 27

Media = between 7 and 8
But since number of terms in the set is even so median is the average of two middle terms which can only be 7.5
Hence, Median = 7.5
Media 7.5 is possible when two of the terms in the sets are {7,8} or {6,9} or {5,10} or {4,11} or {3,12} or {2,13} or {1,14}

Lets check options
(A) 13 {x, y, z} may be {2, 13, 12} hence Possible

(B) 12 {x, y, z} may be {5, 10, 12} hence Possible

(C) 11 {x, y, z}hence NOT Possible as the pair needed is {4, 11} but 4 can't be third term ever as sum of other two of x, y, z is 16 which can't be each smaller than 4. CORRECT ANSWER

(D) 10

(E) 5

Thank you. Works best!
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Re: A list of numbers has six positive integers. Three of those integers a   [#permalink] 13 Jul 2019, 14:39
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