Cool question: It's difficult to really get what's going on until you reach the second statement.

Stat. (1): since the two sets have different numbers of consecutive integers, how could their sum be the same? One way is hinted in stat. (2): if the sets are both symmetrical around the median of zero, with each positive term canceled out by the equivalent negative term, their sum will be zero. In this case, set X is -3, -2, -1, 0, 1, 2, 3
And
set Y is -4, -3, -2, -1, 0, 1, 2, 3, 4.

The sums are zero, and the medians are equal, and the answer seems to be yes.
But can we find another example, where the two sets have equal sums but different medians?

Let x be the median of set X, and y be the median of set Y. It is now possible to model the terms of the sets as a relationship to the medians:
set X is {x-3, x-2, x-1, x, x+1, x+2, x+3}
set Y is {y-4, y-3, y-2, y-1, y, y+1, y+2, y+3, y+4}

the sums of the terms in set X will be 7x , and 9y for set Y - the numbers in each set cancel each other out.
According to stat. (1), the sums are equal, so we get that 7x=9y. This happens when x and y are both zero (previous scenario), but also when x=9 and y=7, since 7*9 is equal to 9*7 = 63.
Thus, we have two cases, one where the medians are equal (and equal to zero), the other when they're not equal. Stat. (1) is insufficient.

Stat. (2): alone this tells us nothing about set X, so Xs median can be greater, lower or equal to Ys median. Insufficient

Combined: stat. (1) says 7x=9y.
Stat. (2) says y=0. It follows that 7x=9*0, so x must equal zero as well, and the medians are equal. Sufficient. Answer is C.

Geva,

I had similar logic. Please let me know if I did anything wrong:

1) Consecutive 7 number and Consecutive 9 numbers with equal sum. Sum of consecutive of 7 is a multiple of 7 and Sum of consecutive of 9 is multiple of 9. Since they are equal the set has to be multiples of both/ 63. Since X has less number its average has to be higher. Also since median=mean in evenly-spaced set median has to be higher. Here you have Median= Zero or Medians equal multiple of 9s/7s. Thus insufficient.

2) Eliminates the case where the two sets are multiples of 63s thus when combined with 1 is sufficient.

Geva@MasterGMAT wrote:Cool question: It's difficult to really get what's going on until you reach the second statement.

Stat. (1): since the two sets have different numbers of consecutive integers, how could their sum be the same? One way is hinted in stat. (2): if the sets are both symmetrical around the median of zero, with each positive term canceled out by the equivalent negative term, their sum will be zero. In this case, set X is -3, -2, -1, 0, 1, 2, 3
And
set Y is -4, -3, -2, -1, 0, 1, 2, 3, 4.

The sums are zero, and the medians are equal, and the answer seems to be yes.
But can we find another example, where the two sets have equal sums but different medians?

Let x be the median of set X, and y be the median of set Y. It is now possible to model the terms of the sets as a relationship to the medians:
set X is {x-3, x-2, x-1, x, x+1, x+2, x+3}
set Y is {y-4, y-3, y-2, y-1, y, y+1, y+2, y+3, y+4}

the sums of the terms in set X will be 7x , and 9y for set Y - the numbers in each set cancel each other out.
According to stat. (1), the sums are equal, so we get that 7x=9y. This happens when x and y are both zero (previous scenario), but also when x=9 and y=7, since 7*9 is equal to 9*7 = 63.
Thus, we have two cases, one where the medians are equal (and equal to zero), the other when they're not equal. Stat. (1) is insufficient.

Stat. (2): alone this tells us nothing about set X, so Xs median can be greater, lower or equal to Ys median. Insufficient

Combined: stat. (1) says 7x=9y.
Stat. (2) says y=0. It follows that 7x=9*0, so x must equal zero as well, and the medians are equal. Sufficient. Answer is C.

gmatusa2010 wrote:Geva,

I had similar logic. Please let me know if I did anything wrong:

1) Consecutive 7 number and Consecutive 9 numbers with equal sum. Sum of consecutive of 7 is a multiple of 7 and Sum of consecutive of 9 is multiple of 9. Since they are equal the set has to be multiples of both/ 63. Since X has less number its average has to be higher. Also since median=mean in evenly-spaced set median has to be higher. Here you have Median= Zero or Medians equal multiple of 9s/7s. Thus insufficient.

2) Eliminates the case where the two sets are multiples of 63s thus when combined with 1 is sufficient.

Sounds about right.

Geva@MasterGMAT wrote:

gmatusa2010 wrote:Geva,

I had similar logic. Please let me know if I did anything wrong:

1) Consecutive 7 number and Consecutive 9 numbers with equal sum. Sum of consecutive of 7 is a multiple of 7 and Sum of consecutive of 9 is multiple of 9. Since they are equal the set has to be multiples of both/ 63. Since X has less number its average has to be higher. Also since median=mean in evenly-spaced set median has to be higher. Here you have Median= Zero or Medians equal multiple of 9s/7s. Thus insufficient.

2) Eliminates the case where the two sets are multiples of 63s thus when combined with 1 is sufficient.

Sounds about right.

Though if you didn't have stat. (2) to steer you there, getting fixated on that 63 may have taken your eyes of the fact that zero is also a multiple of 7 and 9. Then again, without stat. (2) this question is a lot more difficult than it already is.

We know that sum of numbers from Set X is equal to sum of numbers in set Y.
Also, X contains 7 numbers whereas Y contains 9 numbers.

Also, all numbers are consecutive in both the sets. In that case,

Shouldn't the median of X be less than median of Y ?
If so, shouldn't the answer be A ?

gmatusa2010 wrote:Set X consists of seven consecutive integers, and set Y consists
of nine consecutive integers. Is the median of the numbers in
set X equal to the median of the numbers in set Y ?

(1) The sum of the numbers in set X is equal to the sum of
the numbers in set Y.
(2) The median of the numbers in set Y is 0.

Since both sets contain only consecutive integers, with Set X and Set Y having 7 and 9; respectively.

If (1) is true,the median is the mean so 7x=9y.

Can 7x=9y? This is only true when both x=y=0, and x=9 y=7.
Not Enough Info

If (2) is true, Y's median is 0. This tells nothing about X's median.

With both information together, 7x=9y can be 7x=9(0), therefore, 7x=0.