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Tough MEAN, MEDIAN question...Please help me

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Tough MEAN, MEDIAN question...Please help me

by nandy1984 » Sun Oct 09, 2011 6:54 pm
The sum of the integers in list S is the same as the sum of the integers in list T. Does S contain more integers
than T?
(1) The average (arithmetic mean) of the integers in S is less than the average of the integers in T.
(2) The median of the integers in S is greater than the median of the integers in T.

Sol:
I HAVE TRIED LIKE THIS.....................
if S = 2, 2, 2 and T = 3, 3, then the sums are both 6, the average of S is less (2 vs. 3), and S has
more integers.
if S = -3, -3 and T = -2, -2, -2, then the sums are both -6, the average of S is less (-3 vs. -2),
and S has fewer integers.
so (a) is insufficient.
if T = 2, 2, 2 and S = 3, 3, then the sums are both 6, the median of S is greater (3 vs. 2), and S
has fewer integers.
if T = -3, -3 and S = -2, -2, -2, then the sums are both -6, the median of S is greater (-2 vs. -3),
and S has more integers.
so (b) is insufficient.

NOW I AM CONFUSED HOW I CAN COMBINE THE TWO STATEMENTS....
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Source: — Data Sufficiency |
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by cans » Sun Oct 09, 2011 9:16 pm
The sum of the integers in list S is the same as the sum of the integers in list T. Does S contain more integers
than T?
(1) The average (arithmetic mean) of the integers in S is less than the average of the integers in T.
(2) The median of the integers in S is greater than the median of the integers in T.
S(s) = S(t). N(s)>N(t)????
A) avg(s) < avg(t)
S(s)/N(s) < S(t) / N(t)
S(s)/N(s) - S(s)/N(t) <0
S(s)* (N(t)- N(s))<0
now if sum >0, N(t) < N(s)
and is sum <0, N(t) > N(s)

insufficient

B) median(s) > median (t) insufficient

A&B) insufficient

IMO E
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by nandy1984 » Mon Oct 10, 2011 6:43 am
cans wrote:
The sum of the integers in list S is the same as the sum of the integers in list T. Does S contain more integers
than T?
(1) The average (arithmetic mean) of the integers in S is less than the average of the integers in T.
(2) The median of the integers in S is greater than the median of the integers in T.
S(s) = S(t). N(s)>N(t)????
A) avg(s) < avg(t)
S(s)/N(s) < S(t) / N(t)
S(s)/N(s) - S(s)/N(t) <0
S(s)* (N(t)- N(s))<0
now if sum >0, N(t) < N(s)
and is sum <0, N(t) > N(s)

insufficient

B) median(s) > median (t) insufficient

A&B) insufficient

IMO E

Takeway:
1) Median can never give us the information about the number of items in the set.
2) whenever we are not able to solve the problem using the plugin numbers immediately switch to the algebra and viceversa.

Here my intution said to me to go to algebra first but i tried the plugin numbers. Do u have any thing to comment on when to go to Algebra and when to go for Plugin numbers..It will help if you can give a small example of these types... Thanks for the answer.tc
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by cans » Tue Oct 11, 2011 9:21 pm
Hey,
I always try to simplify equation a bit with algebra and try to put numbers. Specially in cases such as mean and median...
But if question is like a>b is given, find if b>c or not, I try to pick numbers..
Dunno if this is of much help but always go with your intuition :)
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by AbhiJ » Tue Oct 11, 2011 10:21 pm
Actually this is a statistics result, you can have mean > median for left skewed distribution and vice versa. This is independent of no of terms in distribution.
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by eaakbari » Fri Nov 09, 2012 11:13 am
nandy1984 wrote:The sum of the integers in list S is the same as the sum of the integers in list T. Does S contain more integers
than T?
(1) The average (arithmetic mean) of the integers in S is less than the average of the integers in T.
(2) The median of the integers in S is greater than the median of the integers in T.
The answer here is given as E, but shouldn't it be A.


Soln:

(1) M(S)< M(T)
Mean = SUM / N

Where n is number of values in the set.

Since M (S) < M (T)
and the sum of both sets is the same.

SUM / N(S) < SUM / N(T)
Which implies N(S) > N(T)

Hence S has more integers. Therefore Suff.

(2) Clearly Insuff.

Hence IMO A.

Clarifications please?
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by eaakbari » Fri Nov 09, 2012 11:14 am
nandy1984 wrote:The sum of the integers in list S is the same as the sum of the integers in list T. Does S contain more integers
than T?
(1) The average (arithmetic mean) of the integers in S is less than the average of the integers in T.
(2) The median of the integers in S is greater than the median of the integers in T.
The answer here is given as E, but shouldn't it be A.


Soln:

(1) M(S)< M(T)
Mean = SUM / N

Where n is number of values in the set.

Since M (S) < M (T)
and the sum of both sets is the same.

SUM / N(S) < SUM / N(T)
Which implies N(S) > N(T)

Hence S has more integers. Therefore Suff.

(2) Clearly Insuff.

Hence IMO A.

Clarifications please?
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by GMATGuruNY » Fri Nov 09, 2012 3:38 pm
eaakbari wrote:
nandy1984 wrote:The sum of the integers in list S is the same as the sum of the integers in list T. Does S contain more integers
than T?
(1) The average (arithmetic mean) of the integers in S is less than the average of the integers in T.
(2) The median of the integers in S is greater than the median of the integers in T.
The answer here is given as E, but shouldn't it be A.


Soln:

(1) M(S)< M(T)
Mean = SUM / N

Where n is number of values in the set.

Since M (S) < M (T)
and the sum of both sets is the same.

SUM / N(S) < SUM / N(T)
Which implies N(S) > N(T)

Hence S has more integers. Therefore Suff.

(2) Clearly Insuff.

Hence IMO A.

Clarifications please?

If the problem is restricted to positive integers, then the correct answer is A.
If the problem is not restricted to positive integers, then the correct answer is E.

Consider the following cases:

Case 1:
S = -2, 3, 3 and T = 2, 2.
The sum of each list is 4.
The average of S (4/3) is less than the average of T (2).
The median of S (3) is greater than the median of T (2).
In this case, S has more integers than T.

Case 2:
S = -4, -2 and T = -4, -4, 2.
The sum of each list is -6.
The average of S (-3) is less than the average of T (-2).
The median of S (-3) is greater than the median of T (-4).
In this case, S has fewer integers than T.
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