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List J consists of 65 consecutive integers. If the integer with the le [#permalink]
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13 Feb 2017, 03:09
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List J consists of 65 consecutive integers. If the integer with the le [#permalink]
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13 Feb 2017, 04:08
Bunuel wrote: List J consists of 65 consecutive integers. If the integer with the least value is 12, what is the median of the positive integers in List J?
A. 20 B. 25.5 C. 26 D. 26.5 E. 27 Since least value is 12, the list of 65 consecutive integers goes from 12 to 52 both included Positive consecutive integers from the list goes from 1 to 52 Median of these Positive consecutive integers = (26+27)/2 = 26.5 Hence Option D is correct. Hit Kudos if you liked it



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Re: List J consists of 65 consecutive integers. If the integer with the le [#permalink]
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17 Mar 2017, 11:48
Bunuel wrote: List J consists of 65 consecutive integers. If the integer with the least value is 12, what is the median of the positive integers in List J?
A. 20 B. 25.5 C. 26 D. 26.5 E. 27 total positive numbers = 6512= 53 total number of non negative numbers (including 0 ) = 53+1= 54 thus median = (26+27)/2 =26.5 Ans D



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Re: List J consists of 65 consecutive integers. If the integer with the le [#permalink]
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18 Mar 2017, 04:06
Isnt 0 considered a neutral integer? i.E neither positive nor negative.
I did not consider 0 while calculating the median.



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Re: List J consists of 65 consecutive integers. If the integer with the le [#permalink]
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18 Mar 2017, 04:23



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Re: List J consists of 65 consecutive integers. If the integer with the le [#permalink]
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18 Mar 2017, 08:06
0akshay0 wrote: Bunuel wrote: List J consists of 65 consecutive integers. If the integer with the least value is 12, what is the median of the positive integers in List J?
A. 20 B. 25.5 C. 26 D. 26.5 E. 27 Since least value is 12, the list of 65 consecutive integers goes from 12 to 52 both included Positive consecutive integers from the list goes from 1 to 52 Median of these Positive consecutive integers = (26+27)/2 = 26.5 Hence Option D is correct. Hit Kudos if you liked it I considered the median of 25 and 26 In larger numbers, how do you ensure that you are taking the right median value? I got confused between taking 26 and 27 and 25 and 26.



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Re: List J consists of 65 consecutive integers. If the integer with the le [#permalink]
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19 Mar 2017, 11:33
Hi Bhavanahiremath, By definition, when you have the median of a group of terms, there will be the same number of terms "to the left" of the median as "to the right" of the median. Here, you're dealing with the integers 1 to 52, inclusive. Since we have an EVEN number of terms, the median will be the average of the two "middle terms." Is that the average of 25 and 26? There are 24 terms to the left of 25. How many are to the right of 26??? Counting 25 and 26 and the 24 terms to the left of 25, that's 26 terms  meaning that there are 26 terms to the right of 26. 24 to the left and 26 to the right are NOT equal, so 25 and 26 CAN'T be the two 'middle terms.' How about 26 and 27? There are 25 terms to the left of 26. How many are to the right of 27??? Counting 26 and 27 and the 25 terms to the left of 25, that's 27 terms  meaning that there are 25 terms to the right of 26. 25 to the left and 25 to the right ARE equal, so 26 and 27 ARE the two 'middle terms.' GMAT assassins aren't born, they're made, Rich
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Re: List J consists of 65 consecutive integers. If the integer with the le [#permalink]
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19 Mar 2017, 20:47
EMPOWERgmatRichC wrote: Hi Bhavanahiremath,
By definition, when you have the median of a group of terms, there will be the same number of terms "to the left" of the median as "to the right" of the median.
Here, you're dealing with the integers 1 to 52, inclusive. Since we have an EVEN number of terms, the median will be the average of the two "middle terms."
Is that the average of 25 and 26?
There are 24 terms to the left of 25. How many are to the right of 26??? Counting 25 and 26 and the 24 terms to the left of 25, that's 26 terms  meaning that there are 26 terms to the right of 26.
24 to the left and 26 to the right are NOT equal, so 25 and 26 CAN'T be the two 'middle terms.'
How about 26 and 27?
There are 25 terms to the left of 26. How many are to the right of 27??? Counting 26 and 27 and the 25 terms to the left of 25, that's 27 terms  meaning that there are 25 terms to the right of 26.
25 to the left and 25 to the right ARE equal, so 26 and 27 ARE the two 'middle terms.'
GMAT assassins aren't born, they're made, Rich Thanks for the explanation!



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Re: List J consists of 65 consecutive integers. If the integer with the le [#permalink]
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20 Mar 2017, 01:10
Bunuel wrote: List J consists of 65 consecutive integers. If the integer with the least value is 12, what is the median of the positive integers in List J?
A. 20 B. 25.5 C. 26 D. 26.5 E. 27 since it is the set of odd no. of numbers, and it is progressive, the median will lie at (n+1)/2 = (65+1)/2 = 33 T33 = 12 +(331)*1 = 12 +32 = 20 Option A



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List J consists of 65 consecutive integers. If the integer with the le [#permalink]
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29 Mar 2017, 15:53
List J consists of 65 consecutive integers. If the integer with the least value is 12, what is the median of the positive integers in List J?
A 20 B 25.5 C 26 D 26.5 E 27



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List J consists of 65 consecutive integers. If the integer with the le [#permalink]
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29 Mar 2017, 16:27
jrpf08 wrote: List J consists of 65 consecutive integers. If the integer with the least value is 12, what is the median of the positive integers in List J?
A 20 B 25.5 C 26 D 26.5 E 27 This is another great Veritas question! I'm happy to respond. See: Common GMAT Topic: Descriptive StatisticsFirst of all, we are absolutely guaranteed that if we have an odd number of entries, the median is a member of the set; if we have an even number of members, the median is the average of the middle two members. Here, we know we have 65 members of the set, an odd number, so we are guaranteed that the median is an integer: we can eliminate (B) & (D) right away. If n is an odd integer, and a set has n members, the median is the (n + 1)/2 member of the set. Here, it would be the 33rd member of the set. We have twelve negative integers, then zero. That's 13 before we get to the positive numbers. Starting with the positive numbers, the median is the 20th positive integerin other words, 20. If use inclusive counting, then we see: from 12 to +19, there are 32 integers 20 is the median. From 21 to 52, there are 32 integers. That's 65 members all together. Thus, 20 is the median. Here's a completely different way to think about it. Suppose we start with the list {1, 2, 3, ..., 63, 64, 65}. That has 65 members, all positive. The median of that list is 33. Now, subtract 13 from every number on that listthe minimum will go down from +1 to 12; the maximum will go down from 65 to 52; and the median will go down from 33 to 20. That's the median of the whole list. Does all this make sense? Mike
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Re: List J consists of 65 consecutive integers. If the integer with the le [#permalink]
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29 Mar 2017, 16:43
mikemcgarry wrote: jrpf08 wrote: List J consists of 65 consecutive integers. If the integer with the least value is 12, what is the median of the positive integers in List J?
A 20 B 25.5 C 26 D 26.5 E 27 The question asks for the median of the positive integers only, and not the median of the entire sequence. So, the correct answer should be 26.5 then, right?



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Re: List J consists of 65 consecutive integers. If the integer with the le [#permalink]
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29 Mar 2017, 17:05
llamsivel wrote: jrpf08 wrote: List J consists of 65 consecutive integers. If the integer with the least value is 12, what is the median of the positive integers in List J?
A 20 B 25.5 C 26 D 26.5 E 27 The question asks for the median of the positive integers only, and not the median of the entire sequence. So, the correct answer should be 26.5 then, right? Dear llamsivel, Good eye, my friend! That's my mistake. The positive integers in J go from 1 to 52. The two middle numbers are 26 & 27, so the median is 26.5, (D). Thanks for catching that. Mike
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Re: List J consists of 65 consecutive integers. If the integer with the le [#permalink]
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29 Mar 2017, 18:26
jrpf08 wrote: List J consists of 65 consecutive integers. If the integer with the least value is 12, what is the median of the positive integers in List J?
A 20 B 25.5 C 26 D 26.5 E 27 Responding to a pm: Quote: According to Veritas, the right answer is 26.5 but I think it should be 20. Note that the question asks for the median of the "positive integers". The list would be: 12, 11, 10,... 0, 1, 2, 3, ... 51, 52 The positive integers are: 1, 2, 3, 4, ... 51, 52 The median will be the average of middle two numbers 26 and 27 i.e. the average will be 26.5.
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Re: List J consists of 65 consecutive integers. If the integer with the le [#permalink]
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29 Mar 2017, 21:27



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Re: List J consists of 65 consecutive integers. If the integer with the le [#permalink]
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04 Apr 2017, 16:00
Bunuel wrote: List J consists of 65 consecutive integers. If the integer with the least value is 12, what is the median of the positive integers in List J?
A. 20 B. 25.5 C. 26 D. 26.5 E. 27 If the least value of the integers in list J is 12 and there are 65 consecutive integers, the greatest integer in the list is 12 + 65  1 = 52. Notice that we are asked to determine the median of the positive integers in list J (not all the integers in list J), so we are actually being asked to determine the median of the integers from 1 to 52 inclusive. Since the integers from 1 to 52 inclusive are an evenly spaced set of integers, we can determine the median as follows: (least integer in the set + greatest integer in the set)/2 = (1 + 52)/2 = 53/2 = 26.5 Answer: D
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