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Manager  Joined: 08 Aug 2005
Posts: 249
If x is an integer, is the median of 5 numbers shown greater than the  [#permalink]

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43 00:00

Difficulty:   65% (hard)

Question Stats: 64% (02:27) correct 36% (02:10) wrong based on 496 sessions

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x, 3, 1, 12, 8

If x is an integer, is the median of 5 numbers shown greater than the average of 5 numbers?

(1) x > 6
(2) x is greater than median of 5 numbers
Math Expert V
Joined: 02 Sep 2009
Posts: 59125
Re: If x is an integer, is the median of 5 numbers shown greater than the  [#permalink]

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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.

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Manager  Joined: 20 Mar 2005
Posts: 165
Location: Colombia, South America
Re: If x is an integer, is the median of 5 numbers shown greater than the  [#permalink]

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getzgetzu wrote:
x, 3, 1, 12, 8

If x is an integer, is the median of 5 numbers shown greater than the average of 5 numbers?

1) x>6
2) x is greater than median of 5 numbers

(1) insuff.
for instance x can be 7 so median is 7 itself and mean around 5 so mean < median, but x can be 1000 in that case median is 8 but mean is way larger so is not enough

(2) that means that x>8 same problem as statement (1)

together nothing

so I would go with E in this one
Manager  Joined: 14 Mar 2006
Posts: 144
Re: If x is an integer, is the median of 5 numbers shown greater than the  [#permalink]

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getzgetzu wrote:
x, 3, 1, 12, 8

If x is an integer, is the median of 5 numbers shown greater than the average of 5 numbers?

1) x>6
2) x is greater than median of 5 numbers

I go for E.

1) x>6, so x could be lets say 9, then median is 8, and the avg is 6.6, however if x is lets say 100, the mean is greater then median, so not suff.

2) similarly since x is greater then the median, 8 is the median, from there its similar to 1). not suff

therefore, E is correct.
Senior Manager  Joined: 05 Jan 2006
Posts: 316
Re: If x is an integer, is the median of 5 numbers shown greater than the  [#permalink]

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1
1,3,8,12 and Now we need to plug in X

1) x>6...7 to inf
X mean med
7 31/5 7
8 32/5 8
9 33/5 8
100 124/5 8

In suffi

2) x is greater than med

1,3,8,x,12
or
1,3,8,12,x

Again pluging number insufficient

Togather no furthere info

hence E
VP  Joined: 10 Jun 2007
Posts: 1061
Re: If x is an integer, is the median of 5 numbers shown greater than the  [#permalink]

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1
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.
Intern  Joined: 03 Nov 2011
Posts: 7
Re: If x is an integer, is the median of 5 numbers shown greater than the  [#permalink]

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1
1
we can arrange this question like this
1,3,x,8,12 or 1,3,8,x,12 or 1,3,8,12,x
so st 1 (24+x)/5< x,or 8 or 12 it give three different solution i.e 6<x,x<16,x<36 so seems to not sff

st 2 xis greater > median by arrange he series we get 1 3 8 x 12 or 1 3 8 12 x where 8 is median so
24+x<8 ====>x <16 so foe any value of x it gives different result.

and combining 1 n 2 x>6 and x<16 take any value so
it seems to me asn sd be E

if any err let me know plz....
Math Expert V
Joined: 02 Sep 2009
Posts: 59125
Re: If x is an integer, is the median of 5 numbers shown greater than the  [#permalink]

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getzgetzu wrote:
x, 3, 1, 12, 8

If x is an integer, is the median of 5 numbers shown greater than the average of 5 numbers?

(1) x>6
(2) x is greater than median of 5 numbers

Given set {1, 3, 8, 12, x}

The questions asks whether median>average, or whether median>(24+x)/5.

(1) x>6 --> if x=11 then median=8 (the middle number) and average=(24+x)/5=7, so median>average but if x=16 then median=8 and average=(24+x)/5=8, so median=average. Not sufficient.

(2) x is greater than median of 5 numbers --> median=8. Not sufficient.

(1)+(2) Examples from (1) are still valid so we still have two different answers. Not sufficient.

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Re: If x is an integer, is the median of 5 numbers shown greater than the  [#permalink]

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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.

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 the  [#permalink]

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1
Mean has a greater flexibility in terms of how big it can get. So for the median to be greater than the mean we need to limit the upward movement of the mean by restricting the value of x. Thus we need a constraint such as x<(certain number) and since neither of the choices does so answer is E.

For instance, even if we were given x>0
then say x=1 {1 1 3 8 12} 3>25/5? NO
now say x=8 {1 3 7 8 12} 8>(24+8)/5? Yes

on the other hand, say if we were given x<6, the answer would be a definitive No

would you pls confirm if that logic looks good, Bunuel?
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