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If the average arithmetic mean of the five numbers x, 7, 2, [#permalink]
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If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and 11 is equal to the median of five numbers, what is the value of X (1) 7 < x < 11 (2) x is the median of the five numbers first statement is definitely sufficient, but in the second statement, if X is the median, it can take three numbers, 10. 9, 8, so ...how is it sufficient, i might be missing some basic point here,,,,so need help,,,,as the second statement is also sufficient for this problem
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Re: ds question [#permalink]
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19 Oct 2010, 03:22
satishreddy wrote: if the average arithmetic mean of the five numbers X, 7,2,16.11 is equal to the median of five numbers, what is the value of X
1) 7<X<11 2) X is the median of five numbers
first statement is definitely sufficient, but in the second statement, if X is the median, it can take three numbers, 10. 9, 8, so ...how is it sufficient, i might be missing some basic point here,,,,so need help,,,,as the second statement is also sufficient for this problem For finding out median you have to arrange the numbers in increasing or decresing order, then only you have to take the middle term as median. So now the order will be 2,7,x,11,16 And in that way ,you will have the same median value i.e, 9. So answer is D. Please consider KUDOS if it is helpful to u .



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Re: ds question [#permalink]
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19 Oct 2010, 11:26
satishreddy wrote: if the average arithmetic mean of the five numbers X, 7,2,16.11 is equal to the median of five numbers, what is the value of X
1) 7<X<11 2) X is the median of five numbers
first statement is definitely sufficient, but in the second statement, if X is the median, it can take three numbers, 10. 9, 8, so ...how is it sufficient, i might be missing some basic point here,,,,so need help,,,,as the second statement is also sufficient for this problem Both statements are saying the SAME thing ! (X + 7 + 2 + 16 + 11) / 5 = MEDIAN if it is given that X = Median, then (X + 7 + 2 + 16 + 11) / 5 = X (7 + 2 + 16 + 11) = 4X ....... and so on.
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Re: ds question [#permalink]
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20 Oct 2010, 06:24
satishreddy wrote: if the average arithmetic mean of the five numbers X, 7,2,16.11 is equal to the median of five numbers, what is the value of X
1) 7<X<11 2) X is the median of five numbers
first statement is definitely sufficient, but in the second statement, if X is the median, it can take three numbers, 10. 9, 8, so ...how is it sufficient, i might be missing some basic point here,,,,so need help,,,,as the second statement is also sufficient for this problem I am not convinced with the above answers... from 1 ) 7 < X< 11 === > X is like 7.1, or 7.2 or 7.5 or 8,8.1 or 9 0r 9.5 ... there are so many possibilites so i can not take only 8,9,10 as X so i can not decide from 1 from 2) X is the median of five numbers then === > X, 7,2,16,11 .... ==> how can i arrange if i dont know the value of X in either descending or ascending.. so from the both also it wont be possible to get the ans i believe... So ans E ... Do let me know if i am wrong....



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Re: ds question [#permalink]
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20 Oct 2010, 08:46
vitamingmat wrote: satishreddy wrote: if the average arithmetic mean of the five numbers X, 7,2,16.11 is equal to the median of five numbers, what is the value of X
1) 7<X<11 2) X is the median of five numbers
first statement is definitely sufficient, but in the second statement, if X is the median, it can take three numbers, 10. 9, 8, so ...how is it sufficient, i might be missing some basic point here,,,,so need help,,,,as the second statement is also sufficient for this problem I am not convinced with the above answers... from 1 ) 7 < X< 11 === > X is like 7.1, or 7.2 or 7.5 or 8,8.1 or 9 0r 9.5 ... there are so many possibilites so i can not take only 8,9,10 as X so i can not decide from 1 from 2) X is the median of five numbers then === > X, 7,2,16,11 .... ==> how can i arrange if i dont know the value of X in either descending or ascending.. so from the both also it wont be possible to get the ans i believe... So ans E ... Do let me know if i am wrong.... There is one more thing we know from the stem: "the average arithmetic mean of the five numbers x, 7, 2, 16, 11 is equal to the median of five numbers" > \(\frac{x+2+7+11+16}{5}=median\). As there are odd number of terms then the median is the middle term when arranged in ascending (or descending) order but we don't know which one (median could be x, 7 or 11). (1) 7 < x < 11 > when ordered we'll get 2, 7, x, 11, 16 > \(\frac{x+2+7+11+16}{5}=median=x\) > \(\frac{x+36}{5}=x\) > \(x=9\). Sufficient. (2) x is the median of five numbers > the same info as above: \(\frac{x+2+7+11+16}{5}=median=x\) > \(\frac{x+36}{5}=x\) > \(x=9\). Sufficient. Answer: D.
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Re: If the average arithmetic mean of the five numbers x, 7, 2, [#permalink]
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09 Oct 2013, 11:17
The above set is not evenly spaced. Is it possible to have a nonevenly spaced set where \(mean=median\)?
Thank you for your help.



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Re: If the average arithmetic mean of the five numbers x, 7, 2, [#permalink]
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10 Oct 2013, 02:43



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Re: ds question [#permalink]
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13 Jun 2014, 20:19
Bunuel wrote: vitamingmat wrote: satishreddy wrote: if the average arithmetic mean of the five numbers X, 7,2,16.11 is equal to the median of five numbers, what is the value of X
1) 7<X<11 2) X is the median of five numbers
first statement is definitely sufficient, but in the second statement, if X is the median, it can take three numbers, 10. 9, 8, so ...how is it sufficient, i might be missing some basic point here,,,,so need help,,,,as the second statement is also sufficient for this problem I am not convinced with the above answers... from 1 ) 7 < X< 11 === > X is like 7.1, or 7.2 or 7.5 or 8,8.1 or 9 0r 9.5 ... there are so many possibilites so i can not take only 8,9,10 as X so i can not decide from 1 from 2) X is the median of five numbers then === > X, 7,2,16,11 .... ==> how can i arrange if i dont know the value of X in either descending or ascending.. so from the both also it wont be possible to get the ans i believe... So ans E ... Do let me know if i am wrong.... There is one more thing we know from the stem: "the average arithmetic mean of the five numbers x, 7, 2, 16, 11 is equal to the median of five numbers" > \(\frac{x+2+7+11+16}{5}=median\). As there are odd number of terms then the median is the middle term when arranged in ascending (or descending) order but we don't know which one (median could be x, 7or 16). (1) 7 < x < 11 > when ordered we'll get 2, 7, x, 11, 16 > \(\frac{x+2+7+11+16}{5}=median=x\) > \(\frac{x+36}{5}=x\) > \(x=9\). Sufficient. (2) x is the median of five numbers > the same info as above: \(\frac{x+2+7+11+16}{5}=median=x\) > \(\frac{x+36}{5}=x\) > \(x=9\). Sufficient. Answer: D. Hi Bunuel, Two questions: 1) Why do you say that the median could be x, 7 or 16  couldn't it also be 11? 2) Isn't statement 1 and 2 saying the same thing or am I missing something?



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Re: If the average arithmetic mean of the five numbers x, 7, 2, [#permalink]
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13 Jun 2014, 20:48
russ9
1) I believe that it can also be 11 since if x was 11 the median in fact would be 11 (but the median could never be 16 since if x=16, then you'd have 2, 7, 11, 16, 16 and the median would be 11 not 16).
2) Statement 1 and 2 are essentially saying the say thing, since the constraint "the mean is equal to the median" and that is the only value [9] that satisfies that constraint  therefore it is D.
Please Kudos if this helped!



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Re: If the average arithmetic mean of the five numbers x, 7, 2, [#permalink]
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14 Jun 2014, 01:39



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Re: If the average arithmetic mean of the five numbers x, 7, 2, [#permalink]
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29 Mar 2016, 12:07
Dear Bunuel , I have a doubt on statement B
Statement B says  x is the median of the 5 no's. So x can be 7,8,9,10,11 . Then how can we determine the value of x using B alone ?



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Re: If the average arithmetic mean of the five numbers x, 7, 2, [#permalink]
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29 Mar 2016, 12:19
ThePlayer wrote: Dear Bunuel , I have a doubt on statement B
Statement B says  x is the median of the 5 no's. So x can be 7,8,9,10,11 . Then how can we determine the value of x using B alone ? This is explained above: iftheaveragearithmeticmeanofthefivenumbersx103134.html#p803613There is one more thing we know from the stem: "the average arithmetic mean of the five numbers x, 7, 2, 16, 11 is equal to the median of five numbers" > \(\frac{x+2+7+11+16}{5}=median\). As there are odd number of terms then the median is the middle term when arranged in ascending (or descending) order but we don't know which one (median could be x, 7 or 11).(1) 7 < x < 11 > when ordered we'll get 2, 7, x, 11, 16 > \(\frac{x+2+7+11+16}{5}=median=x\) > \(\frac{x+36}{5}=x\) > \(x=9\). Sufficient. (2) x is the median of five numbers > the same info as above: \(\frac{x+2+7+11+16}{5}=median=x\) > \(\frac{x+36}{5}=x\) > \(x=9\). Sufficient. Answer: D.
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Re: If the average arithmetic mean of the five numbers x, 7, 2, [#permalink]
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09 Apr 2016, 13:54
oh man..I just had this question in my GMATPrep test, and I panicked and picked A... quick question to Bunuel, can we consider only INTEGER values, knowing that the question stem tells that we have NUMBERS (aka any numbers)? 1. x has values 8, 9, 10. we know that (36+x)/5 = x 36+x=5x x=9 so only value if x=9 2. x is the median of the five 36+x=5x x=9 yes, 9 is the median. D I dismissed B because I considered noninteger values for x...but still..illogical..because if we apply what we are given..we always get x=9...



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Re: If the average arithmetic mean of the five numbers x, 7, 2, [#permalink]
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Updated on: 02 Aug 2017, 06:13
On Generalizing the question the equation we got x+36=5M so if x is the median then M=x then statement 2 will become sufficient Sent from my HTC Desire 816G dual sim using GMAT Club Forum mobile app
Originally posted by arif24 on 02 Aug 2017, 05:52.
Last edited by arif24 on 02 Aug 2017, 06:13, edited 1 time in total.



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Re: If the average arithmetic mean of the five numbers x, 7, 2, [#permalink]
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02 Aug 2017, 05:58
Q stem states that (2+7+11+16+x)/5 = median
Statement 1) only x=9 in the given range satisfies the condition  suff
Stmnt 2) (x+36)/5 = x > x=9  suff
Answer D



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Re: If the average arithmetic mean of the five numbers x, 7, 2, [#permalink]
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16 Sep 2017, 16:48
If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and 11 is equal to the median of five numbers, what is the value of X
(1) 7 < x < 11 (2) x is the median of the five numbers
Important to note: This is a purely CONCEPTUAL problem. Meaning, there's HARDLY ANY MATH REQUIRED/INVOLVED. You should be able to eye this and take 3045 seconds MAX. I'll show you how.
1 set up #s in increasing order (common step whenever you see the word "median"): 2, 7, 11, 16, x > quickly see 2+7=9...9+11=20, 20+16=36. 2 average = \(\frac{sum of #s in a set}{total numbers in a set}\) > \(\frac{36+x}{5}\) 3 set up question: \(\frac{36+x}{5}\) = median. > key here is to find out what x is (or what the median is)!
1) TELLS US X=MEDIAN! How do you know this? Well, there are 5 terms in the set. > 2 terms are BELOW 7. > 2 terms are ABOVE 11.  Since there are only 5 terms in a set, the 3rd term (middle) = median. Bingo. Eliminate B, C, E.
2) SAME AS 1!  Elim A.
Ans: D



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Re: If the average arithmetic mean of the five numbers x, 7, 2, [#permalink]
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18 Oct 2017, 01:37
Bunuel wrote: andrewwal wrote: russ9
1) I believe that it can also be 11 since if x was 11 the median in fact would be 11 (but the median could never be 16 since if x=16, then you'd have 2, 7, 11, 16, 16 and the median would be 11 not 16).
2) Statement 1 and 2 are essentially saying the say thing, since the constraint "the mean is equal to the median" and that is the only value [9] that satisfies that constraint  therefore it is D.
Please Kudos if this helped! Correct. 15 was a typo there. Edited. BunuelHi Bunuel, I understood the solution however, Elementary doubt here. Since Mean=Median, Shouldn't the numbers in the set be consecutive, However I cannot seem understand the consecutive nature of the above mentioned set. Thanks



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Re: If the average arithmetic mean of the five numbers x, 7, 2, [#permalink]
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18 Oct 2017, 01:48
neilphilip10 wrote: BunuelHi Bunuel, I understood the solution however, Elementary doubt here. Since Mean=Median, Shouldn't the numbers in the set be consecutive, However I cannot seem understand the consecutive nature of the above mentioned set. Thanks For an evenly spaced set (arithmetic progression), the median equals to the mean. Though the reverse is not necessarily true. Consider {0, 1, 1, 2} > median = mean = 1 but the set is not evenly spaced.
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Re: If the average arithmetic mean of the five numbers x, 7, 2, [#permalink]
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22 May 2018, 12:05
Hi, I just found a mistake in a question and would like to let you know
You can answer the question even without looking at the statements 1 and 2  the question stem itself gives sufficient information to find the value of x, which is 9
My reasoning:
if we order the numbers in an increasing order we get 2, 7, 11, 16. we also have x. Regardless of the value of x, the median will be a number that is 7≤X ≤11
The question also says that median equals the mean. (36+x)/5 is the mean, which is also median. (36+x)/5 this tells us that x cannot be 7, because 36/5 is already greater than 7. If x were 7 the median would not be equal to the mean.
If x were 8, the median would be 8, but the mean would be greater than mean. You can check all numbers that satisfies 7≤X ≤11. The only number that satisfies 7≤X ≤11 and also makes median equal to the mean is 9
You can answer the question without looking at the statements



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Re: If the average arithmetic mean of the five numbers x, 7, 2, [#permalink]
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23 May 2018, 05:58
kablayi wrote: Hi, I just found a mistake in a question and would like to let you know
You can answer the question even without looking at the statements 1 and 2  the question stem itself gives sufficient information to find the value of x, which is 9
My reasoning:
if we order the numbers in an increasing order we get 2, 7, 11, 16. we also have x. Regardless of the value of x, the median will be a number that is 7≤X ≤11
The question also says that median equals the mean. (36+x)/5 is the mean, which is also median. (36+x)/5 this tells us that x cannot be 7, because 36/5 is already greater than 7. If x were 7 the median would not be equal to the mean.
If x were 8, the median would be 8, but the mean would be greater than mean. You can check all numbers that satisfies 7≤X ≤11. The only number that satisfies 7≤X ≤11 and also makes median equal to the mean is 9
You can answer the question without looking at the statements That's not correct. From the stem, x can be 1, 9 or 19. If x = 1, then the set is {1, 2, 7, 11, 16} > median = mean = 7. If x = 9, then the set is {2, 7, 9, 11, 16} > median = mean = 9. If x = 19, then the set is {2, 7, 11, 16, 19} > median = mean = 11.
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