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If x is an integer, is the median of the 5 numbers shown gre
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17 Jun 2007, 19:05
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x,3,1,12,8 If x is an integer, is the median of the 5 numbers shown greater than the average (arithmetic mean ) of the 5 numbers ? (1) x > 6 (2) x is greater than the median of the 5 numbers.
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If x is an integer, is the median of the 5 numbers shown gre
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02 Nov 2010, 09:09
metallicafan wrote: \(x, 3, 1, 12, 8\)
If x is an integer, is the median of the 5 numbers shown greater than the average (arithmetic mean) of the 5 numbers?
(1) \(x>6\) (2) x is greater than the median of the 5 numbers We have a set: {1, 3, 8, 12, x} Question: is \(median>mean=\frac{x+1+3+8+12}{5}=\frac{x+24}{5}\)? Note that as we have odd (5) # of terms in the set then the median will be the middle term when arranged in ascending (or descending) order. So: if \(x\leq{3}\): {1, x, 3, 8, 12} then \(median=3\); if \(3<x\leq{8}\): {1, 3, x, 8, 12} then \(median=x\); if \(x\geq{8}\): {1, 3, 8, x, 12} then \(median=8\). (1) \(x>6\). If \(x=7\) then the median will be 7 as well: {1, 3, 7, 8, 12} and mean will be \(mean=\frac{7+24}{5}=6.2\), so \(median=7>mean=6.2\) and the answer is YES BUT if \(x\) is very large number then the median will be 8: {1, 3, 8, 12, x=very large number} and mean will be more than median (for example if \(x=26\) then \(mean=\frac{26+24}{5}=10\), so \(median=8<10=mean\)) and the answer will be NO. Not sufficient. (2) x is greater than the median of the 5 numbers > so \(median=8\): now, if \(x=11\) then \(mean=\frac{11+24}{5}=7\), so \(median=8>7=mean\) and the answer is YES. Again it's easy to get answer NO with very large \(x\). Not sufficient. (1)+(2) Again, x=11 and x=very large number give two diffrent answers to the question. Not sufficeint. Answer: E.
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Both are sufficient..
put the nos in ascending seq. 1,3,8,12 and x. We don't no where x is positioned.
X > 6 could mean that X is the median or 8 is the median.
if X =7 then mean = 6.2 and X > mean
if x= 11 then mean = 7 and X > mean
X greater than median of 5 nos. Which means that median HAS to be 8.
x> 8
Again satisfies the question for any value of x.
D



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ArvGMAT wrote: Both are sufficient.. put the nos in ascending seq. 1,3,8,12 and x. We don't no where x is positioned.
X > 6 could mean that X is the median or 8 is the median. if X =7 then mean = 6.2 and X > mean if x= 11 then mean = 7 and X > mean
X greater than median of 5 nos. Which means that median HAS to be 8. x> 8 Again satisfies the question for any value of x.
D
what if x is 16 then mean = 8 = median which is not greater so answer is E



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Re: DS Mean, Median
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19 Jun 2007, 15:35
LM wrote: x,3,1,12,8
If x is an integer, is the median of the 5 numbers shown greater than the average (arithmetic mean ) of the 5 numbers ?
(1) X > 6
(2) X is greater than the median of the 5 numbes.
Got E.
(1) plug in x = 7, the lowest avg value is (7+3+1+12+8) / 5 = 31/5 =~ 6
This is lower than the median, which is 7 in this case. However, if x is really large, the avg will shoot through the roof, but the median will remain at 8. INSUFFICIENT.
(2) This tells us that x>8, so plug in x=9. We get (9+3+1+12+8) / 5 = 33/5 =~ 6.6. This still less than 8 ,which is the median. Same reason as above, INSUFFICIENT.
Together, we get x>8, INSUFFICIENT.



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02 Nov 2010, 08:41
\(x, 3, 1, 12, 8\) If x is an integer, is the median of the 5 numbers shown greater than the average (arithmetic mean) of the 5 numbers? (1) \(x>6\) (2) x is greater than the median of the 5 numbers
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Re: Average question
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02 Nov 2010, 10:33
Thanks Bunuel! Kudos for you
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If x is an integer, is the median of the 5 numbers shown gre
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16 Feb 2015, 06:03
That moment when you think ans. is D i.e both statements even alone are sufficient,but it turns out that both of them,even together are insufficient ie ans E.
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If x is an integer, is the median of the 5 numbers shown gre
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18 Dec 2015, 08:37
Bunuel wrote: metallicafan wrote: \(x, 3, 1, 12, 8\)
If x is an integer, is the median of the 5 numbers shown greater than the average (arithmetic mean) of the 5 numbers?
(1) \(x>6\) (2) x is greater than the median of the 5 numbers We have a set: {1, 3, 8, 12, x} Question: is \(median>mean=\frac{x+1+3+8+12}{5}=\frac{x+24}{5}\)? Note that as we have odd (5) # of terms in the set then the median will be the middle term when arranged in ascending (or descending) order. So, if \(x\leq{3}\): {1, x, 3, 8, 12} then \(median=3\), if \(3<x\leq{8}\): {1, 3, x, 8, 12} then \(median=x\) and if \(x\geq{8}\): {1, 3, 8, x, 12} then \(median=8\). (1) \(x>6\). If \(x=7\) then the median will be 7 as well: {1, 3, 7, 8, 12} and mean will be \(mean=\frac{7+24}{5}=6.2\), so \(median=7>mean=6.2\) and the answer is YES BUT if \(x\) is very large number then the median will be 8: {1, 3, 8, 12, x=very large number} and mean will be more than median (for example if \(x=26\) then \(mean=\frac{26+24}{5}=10\), so \(median=8<10=mean\)) and the answer will be NO. Not sufficient. (2) x is greater than the median of the 5 numbers > so \(median=8\): now, if \(x=11\) then \(mean=\frac{11+24}{5}=7\), so \(median=8>7=mean\) and the answer is YES. Again it's easy to get answer NO with very large \(x\). Not sufficient. (1)+(2) Again, x=11 and x=very large number give two diffrent answers to the question. Not sufficeint. Answer: E Just wanted to clarify that I have solved this question using random values. Your approach is pretty precise. Is using random values always a disadvantage on GMAT, given the time constraint? It took me arnd 3 mins to do this question , using random values. Given that it's a 700 level question, taking 3 mins on such questions is justified??? Thanks in advance!!



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Re: If x is an integer, is the median of the 5 numbers shown gre
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19 Dec 2015, 05:19
x,3,1,12,8 If x is an integer, is the median of the 5 numbers shown greater than the average (arithmetic mean ) of the 5 numbers ? (1) x > 6 (2) x is greater than the median of the 5 numbers. In the original condition, there is 1 variable(x), which should match with the number of equation. So you need 1 more equation. For 1), 1 equation, for 2) 1 equation, which is likely to make D the answer. In 1) & 2), for 1), when x>0, 1,3,x,8,12/1,3,8,x,12/1,3,8,12,x. mean=(1+3+8+12+x)/5=(24+x)/5 and median=x,8. So, (24+x)/5>x? or (24+x)/5>8? is unknown, which is not sufficient. For 2), in the above, median is 3, x, 8 and from x>3, x>x(impossible), x>8, it is x>8>3. So, although x>8, (24+x)/5>8?, which is x>16?, is not sufficient. Even if 1) & 2), when x>8, you cannot find out (24+x)/5>8?, x>16? from x>8>6>3. Therefore it is not sufficient and the answer is E. > For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: If x is an integer , is the median of 5 numbers shown greater than
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01 Dec 2016, 19:56
Statement 1: For X>6, Avg<median for X=7,8,9,10,11,12,13,14,15, Avg=median for X= 16 Avg> median for X>16 Statement 2: X>median of 5 number => X> 8 Again u can't decide if Median > Avg Both combined X>8 Can't deduct if median> avg So.. Option E Sent from my C6902 using GMAT Club Forum mobile app



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Re: If x is an integer, is the median of the 5 numbers shown gre
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02 Dec 2016, 00:51
Required to put actual values to solve this question. Option 1: considered value of x as 7 and 50. With 7 median > mean. With value of x as 50 median < mean. Not sufficient. Option 2: Same values can be used to eliminate this answer. With 50 median <mean and with value of x as 13, medium > mean. Hence, not sufficient. Considering both the equation also we are not able to solve the question.
My answer is E.



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Re: If x is an integer, is the median of the 5 numbers shown gre
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14 Feb 2019, 19:32
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