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A list of numbers has six positive integers. Three of those integers
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07 Mar 2016, 11:20
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A list of numbers has six positive integers. Three of those integers
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05 Feb 2017, 00:13
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=15Let's work further.. Total of unknowns=10*6(24+5+4)=27.. The third unknown =2715=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 2714=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
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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.
Answer: A



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Re: A list of numbers has six positive integers. Three of those integers
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04 Feb 2017, 09:34
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 How can this be solved in 2 mins?



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Re: A list of numbers has six positive integers. Three of those integers
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04 Feb 2017, 22:24
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 it took me more than 5 min to try each n every possible combination ...how to tackle it within 2 min.. In actual exam ,i would have skipped it :p



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Re: A list of numbers has six positive integers. Three of those integers
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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=15Let's work further.. Total of unknowns=10*6(24+5+4)=27.. The third unknown =2715=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 2714=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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A list of numbers has six positive integers. Three of those integers
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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=15Let's work further.. Total of unknowns=10*6(24+5+4)=27.. The third unknown =2715=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 2714=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
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10 Apr 2017, 19:14
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 Answer: Option C
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Re: A list of numbers has six positive integers. Three of those integers
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13 Jun 2017, 17:54
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=15Let's work further.. Total of unknowns=10*6(24+5+4)=27.. The third unknown =2715=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 2714=13.... And other two values can be 5 and 9...... Or 6 and 8..... Now only 11 is not possible.. Ans C this is not fair, from 7 to 8 (exclusive), 8 is not counted while 7 is included.



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A list of numbers has six positive integers. Three of those integers
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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
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27 Jul 2018, 21:23
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 Answer: Option C If we consider option A, including 13 would make the sequence {4,5,12,13,24} with space to accomodate another integer which is 12. but all numbers are supposed to be distinct. How is it possible?




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