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The median of the list of positive integers above is 5.5 [#permalink]
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25 Mar 2012, 18:46
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8, 5, x, 6 The median of the list of positive integers above is 5.5. Which of the following could be the average (arithmetic mean) of the list? A. 3 B. 5.5 C. 6.25 D. 7 E. 7.5
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Re: Mediating the mean [#permalink]
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25 Mar 2012, 18:57
Given "The median of the list of positive integers above is 5.5"
So arranging the numbers from lowest to highest, avg of 2 middle terms needs to be 5.5
so the sequence will be x 5 6 8
Lets say x = 4, which gives us mean = sum/4 = 5.5
B is correct



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The median of the list of positive integers above is 5.5 [#permalink]
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26 Mar 2012, 01:05
piyatiwari wrote: Given "The median of the list of positive integers above is 5.5"
So arranging the numbers from lowest to highest, avg of 2 middle terms needs to be 5.5
so the sequence will be x 5 6 8
Lets say x = 4, which gives us mean = sum/4 = 5.5
B is correct 8, 5, x, 6
The median of the list of positive integers above is 5.5. Which of the following could be the average (arithmetic mean) of the list?A. 3 B. 5.5 C. 6.25 D. 7 E. 7.5 Notice that we are told that x is a positive integer.The median of a set with even number of terms is the average of two middle terms when arranged in ascending/descending order. Since given that the median is 5.5 (the average of 5 and 6) then integer x must be less than or equal to 5, in order 5 and 6 to be two middle numbers. So our list in ascending order is: {x, 5, 6, 8}. Now, since \(x\leq{5}\) then the maximum sum of the numbers is 5+5+6+8=24, so the maximum average is 24/4=6 (options C, D, and E are out). 3 cannot be the average because in this case the sum of the numbers must be 4*3=12, which is less than the sum of just two numbers 6 and 8 (remember here that since x is a positive integer it cannot decrease the sum to 12). So we are left only with option B (for x=3 the average is (3+5+6+8)/4=22/4=5.5). Answer: B.
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Re: The median of the list of positive integers above is 5.5 [#permalink]
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20 Oct 2012, 20:58
This is amazing soln I solved by plugging in.. (8+5+x+6)/4 . . plug nos from options 1 by 1.. when ans 5.5 is plugged in,you get x=3. . so you have 3,5,6,8 as the order which confirms with the given data
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Re: 8, 5, x, 6 The median of the list of positive integers [#permalink]
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27 Mar 2013, 03:40
8, 5, x, 6 The median of the list of positive integers above is 5.5. Which of the following could be the average (arithmetic mean) of the list? If the median is 5.5, then x cannot be >8 ( 5,6,8,9 median 7>5,5). It cannot be > 6 (5,6,7,8 median 6.5>5.5). So we have found that \(x<=5\) \(\frac{x+5+6+8}{4}=mean\) \(x=mean*419\) x is an integer so 4*mean  19 must be an integer, option (C)6.25 is out. (D)7 (E)7.5 these options return values of x > 9, so they cannot be right. Option (A)3 returns a negative integer, but x must be positive. Option (B)5.5 is correct, and the value of x is 3, which is possible given the initial condition of "The median of the list of positive integers above is 5.5"
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Re: The median of the list of positive integers above is 5.5 [#permalink]
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27 Mar 2013, 04:26
The median will be the average of the two elements in the middle as the list contains even number of elements. So,the sum of the two numbers in the middle will be 5.5 * 2 = 11. With the above data,we can write the list as follows : x 5 6 8 where 1 < x < 6. Avg = (x+5+6+8)/4 = (19+x)/4. So, 5< Avg <6 Hence, B will be the answer. ** Alternatively, we can work out this problem taking options one by one.
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Re: 8, 5, x, 6 The median of the list of positive integers [#permalink]
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03 Aug 2013, 05:43
Zarrolou wrote: n  19 must be an integer, option (C)6.25 is out.
The reason c is out is because if the average was 6.25, x = 6 and the media would become 6 (6.25 x 4) = 25 x = 25  19
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Re: The median of the list of positive integers above is 5.5 [#permalink]
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20 Aug 2015, 22:46
8, 5, x, 6
The median of the list of positive integers above is 5.5. Which of the following could be the average (arithmetic mean) of the list? A. 3 B. 5.5 C. 6.25 D. 7 E. 7.5
Notice that we are told that x is a positive integer.
Experts your view please??
In my opinion this q is flawed unless it should explicitly mentioned "list of different positive no"
list could be 5,6,6,8 and in that case we have mean 6.25



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The median of the list of positive integers above is 5.5 [#permalink]
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26 Dec 2015, 11:17
the question directly says that the median is 5.5. this can be only if 5 and 6 are the middle numbers. thus, x must be <=5. now, let's take x=5 = the sum of the numbers is 24, and the average is 6. since the maximum value of x can be 5, it must be that the maximum value of the mean can't be greater than 6. thus, we can eliminate C, D, and E. now, we have the sum = 24. we have in the answer choices 3 and 5.5 since these numbers should be the mean, it means that 12 or 22 is the sum of the integers. Now, we have 5,6, and 8. the sum of these numbers is way above 12. We are told that we have only positive numbers. thus, 12 can't be the sum and 3 can't be the average. A is out, and B must be the possible average. to test, let's say that x=3. now, we have 3+5+6+8=22. 22/4 = 5.5. all good. looks like I got to the right answer the way bunuel did



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Re: The median of the list of positive integers above is 5.5 [#permalink]
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15 Dec 2016, 16:26
Excellent Question this one. Here is my solution >
Notice that the number of terms here are even => The median of a set with even number of terms is the average of two middle terms when arranged in either increasing or decreasing order.
Median =5.5
Hence the sum of two middle terms must be 11
For that allowed values of positive integer x => 1,2,3,4,5
Hence the mean => 19+x/4 = Value x=4*Value 19
This value is nothing but the option.
Lets put in each value and see which on of them gives us x as {1,2,3,4,5) A)Value=3 => x=1219=7 => Rejected B)Value=>5.5=> x=2219=3=> Allowed C)Value=6.25=>x=2519=6=> Rejected
D)Value=7=>x=2819=9=> Rejected E)Rejected
Hence B
Method 2=> Back solving
Let Mean = 3 Hence x=7 => Rejected Mean = 5.5 => x=3 => Median of set = 5.5 => Accepted.
Hence B
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Re: The median of the list of positive integers above is 5.5 [#permalink]
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05 Jan 2018, 06:42
calreg11 wrote: 8, 5, x, 6
The median of the list of positive integers above is 5.5. Which of the following could be the average (arithmetic mean) of the list?
A. 3 B. 5.5 C. 6.25 D. 7 E. 7.5 Median = 5.5 = Average of two middle numbers in case the number of terms in teh set is even. i.e. 5 and 6 are the two middle terms of the set i.e. x must be smaller than 5 for it to be the second term when terms are arranged in the ascending order so the sum of the terms in the set = x+5+6+8 = x+19 average = (x+19)/4 = (x/4)+4.75 and since x is smaller than or equal to 5 average ≤ (5/4)+4.75 i.e. average ≤ (1.25)+4.75 average ≤ 6 since x is positive hence average > 4.75 ANswer: Option B
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Re: The median of the list of positive integers above is 5.5 [#permalink]
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05 Jan 2018, 09:13
The median is above 5.5, shouldn't we consider options like x,5,6,8 & 5,6,8,x & 5,6,x,8 ?
What Am I missing?



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Re: The median of the list of positive integers above is 5.5 [#permalink]
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05 Jan 2018, 09:22
siddharthselvamohan wrote: The median is above 5.5, shouldn't we consider options like x,5,6,8 & 5,6,8,x & 5,6,x,8 ?
What Am I missing? The median of a set with even number of terms is the average of two middle terms when arranged in ascending/descending order. So, if the set in ascending order is {5, 6, x, 8} (meaning if \(6\leq x \leq 8\)), then the median would be the average of 6 and x, and it cannot be 5.5, it could be at least 6, for x = 6. Similarly if if the set in ascending order is {5, 6, 8, x} (meaning if \(x \geq 8\)), then the median would be the average of 6 and 8, so 7, not 5.5. Thus, since given that the median is 5.5 (the average of 5 and 6) then integer x must be less than or equal to 5, in order 5 and 6 to be two middle numbers. So our list in ascending order is: {x, 5, 6, 8}. Check complete solution here: https://gmatclub.com/forum/themediano ... l#p1064955Hope it helps.
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Re: The median of the list of positive integers above is 5.5
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