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by Frankenstein » Sat Jun 04, 2011 12:40 pm
Hi,
mean = (x+y+x+y+x-4y+xy+2y)/6 = x(y+3)/6
Given mean = y+3. So x=6
y>6. So, y>x
Writing the terms in an increasing order: x-4y,x,y,(x+y),2y,xy.
Median = [y+(x+y)]/2 = y+(x/2) = y+3.
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by Stuart@KaplanGMAT » Sat Jun 04, 2011 12:40 pm
Vishnu88 wrote:x, y, x + y, x - 4y, xy, 2y

For the set of terms shown above, if y > 6 and the mean of the set equals y + 3, then the median must be =

A) (x+y)/2

B) y+3

C) y

D) (3y)/2

E) (x/3) + y


[spoiler]OA = B[/spoiler]
Hi,

we can solve this either via algebra or picking numbers. Let's try picking numbers.

All we know is that y>6, so let's pick y=7. We now know that the mean is 7+3=10.

So:

Avg = (sum of terms)/(# of terms)
10 = (x + 7 + x+7 + x-28 + 7x + 14)/6
60 = 10x + 28 - 28
60 = 10x
6 = x

We want the median; since we have an even number of terms, the median is the average of the two middle terms.

Plugging in x=6 and y=7, our terms are:

{6, 7, 13, -22, 42, 14}

arranging them:

{-22, 6, 7, 13, 14, 42}

and our median is (7+13)/2 = 20/2 = 10

Now we sub into the choices:

a) 13/2... nope
b) 7+3... yup!
c) 7... nope
d) 21/2... nope
e) 2 + 7 = 9... nope

Only (B) is a match - choose (B)!

Picking numbers is an incredibly effective method for problem solving questions with unknowns and variables in the choices. A picking numbers expert could solve this question in about 90 seconds (and certainly under 2 minutes).
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by Jayanth2689 » Sat Jun 04, 2011 8:31 pm
6 terms in this set. Hence, Median = Average of the two middle most numbers. Also, Mean = Median in a set of of numbers.

Mean = y+3 (given), Hence, Median = y+3

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by cans » Sat Jun 04, 2011 8:33 pm
x, y, x + y, x - 4y, xy, 2y
y>6
mean=(3x+xy)/6=y+3 -> 3x + xy = 6y +18
best way to solve is put y=7. mean=10. 10x=60 ->x=6
numbers are 6,7,13,-22,42,14
in order:- -22,6,7,13,14,42
median=(7+13)/2 = 10
a)5
b)10
c)7
d)21/2
e)9
Thus IMO B
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by Frankenstein » Sat Jun 04, 2011 8:48 pm
Jayanth2689 wrote:6 terms in this set. Hence, Median = Average of the two middle most numbers. Also, Mean = Median in a set of of numbers.

Mean = y+3 (given), Hence, Median = y+3

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Hi,
So, you mean to say mean is the average of middle most terms too? It is just a coincidence that mean and median are equal in this case.
Consider 1,2,3,6
mean = (1+2+3+6)/4 = 3
but median = (2+3)/2 = 2.5 (not 3)
Mean = median works in an arithmetic series(progression). Not in all sets of numbers.
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by Jayanth2689 » Sat Jun 04, 2011 9:05 pm
Frankenstein wrote:
Jayanth2689 wrote:6 terms in this set. Hence, Median = Average of the two middle most numbers. Also, Mean = Median in a set of of numbers.

Mean = y+3 (given), Hence, Median = y+3

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Hi,
So, you mean to say mean is the average of middle most terms too? It is just a coincidence that mean and median are equal in this case.
Consider 1,2,3,6
mean = (1+2+3+6)/4 = 3
but median = (2+3)/2 = 2.5 (not 3)
Mean = median works in an arithmetic series(progression). Not in all sets of numbers.
Aah! Right! I made a sweeping generalization for this question! Thanks for the example!
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by [email protected] » Thu Jun 09, 2011 7:57 am
But Frankenstein do let me know about the above equation i.e {-22, 6, 7, 13, 14, 42}

The above series given by the expert is also not in AP but still its Mean = its Median


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Hi,
So, you mean to say mean is the average of middle most terms too? It is just a coincidence that mean and median are equal in this case.
Consider 1,2,3,6
mean = (1+2+3+6)/4 = 3
but median = (2+3)/2 = 2.5 (not 3)
Mean = median works in an arithmetic series(progression). Not in all sets of numbers.
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by Stuart@KaplanGMAT » Thu Jun 09, 2011 8:52 am
[email protected] wrote:But Frankenstein do let me know about the above equation i.e {-22, 6, 7, 13, 14, 42}

The above series given by the expert is also not in AP but still its Mean = its Median
Hi,

just because it's true that the mean=median for all arithmetic progressions doesn't mean that it's true that mean won't equal median for all other sets; there's just no fixed rule for other sets.

So, if your set is created by an AP you know that mean=median; if you have another set mean may or may not equal median.
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