If list S contains nine distinct integers, at least one of w

Question

If list S contains nine distinct integers, at least one of which is negative, is the median of the integers in list S positive?
 
(1) The product of the nine integers in list S is equal to the median of list S.
 
(2) The sum of all nine integers in list S is equal to the median of list S.

mau5 · Accepted Answer

Let the set of 9 distinct integers in increasing order be a_1  ,  a_2  ...  a_9 .

From F.S 1, we know that  a_1*a_2*....a_4*a_5....a_9  =  a_5   	o   a_5(a_1*a_2*...a_4....a_9-1)  = 0.

Thus, either  a_5  = 0 OR  (a_1*a_2*...*a_5....a_9-1)  = 0, the latter is not possible as no product of 8 distinct integers can ever equal 1.

Thus, the median, a_5  = 0 and not positive. Sufficient.

From F. S 2, for -4,-3,-2,-1,0,1,2,3,4,the median is 0 and a NO for the question stem. Again, for the series -10,-3,-1,0,1,2,3,4,5, the median is 1, which is positive, and hence a YES for the question stem. Thus, Insufficient.

A.

Rock750 · Answer

How did u come up with this :  a_1*a_2*....a_4*a_5....a_9  =  a_5   	o   a_5(a_1*a_2*...a_4....a_9-1)  = 0. ?

mau5 · Answer

Could you elaborate on your doubt? The above equation is what is given in the Fact Statement.

Rock750 · Answer

The first equation is given in the fact statement OK But what about the transition from the first to the second ?

I think u factorized by a_5 from both the two parts of the equation which must lead to the following :

a_1*a_2*....a_4*a_5....a_9  =  a_5   	o   (a_1*a_2*...a_4....a_9)  = 1.

Am I missing something here ?

mau5 · Answer

Yes, we cannot cancel  a_5  from both sides because  a_5  can be zero.

hfbamafan · Answer

For statement 1)

Can we deduce that since at least 1 member of the set must be negative, that would give a no answer to the stem of the question.

Ie, since at least one member is negative, the only possible answers we could get are either are 0 or <0.

Which would give a no answer to the question.

Or is this reasoning incorrect for this exam.

Thanks, 
hunter

Nunuboy1994 · Answer

Statement 1:

Two possible numbers to start with are 1 and 0. We cannot have a set of 9 numbers where product of all numbers equals the median if we use 1 (e.x --3 x -2 x -1 x 1 x 2 x 3 x4 ); however, any number times 0 is 0 so if we use 0 as a median we can simply take any product of four negative numbers and multiply them by the product of four positive numbers and then by zero to yield a number equal to the mean or vice versa the product of any four negative numbers times zero times the product of any four positive numbers equals 0 a.k.a the median

Sufficient.

Statement 2

We could have a set of numbers where 0 is the median (1,2,3,4,0,-1,-2,-3,-4) the result of which would be the media- however, we could also have --6,-4,-3,-2,-1,2,3,5,7 which would equal -1 which is also the median of that set and a  negative  number

Insufficient

GMATinsight · Answer

Statement 1: Product if integers = Median
Which is true only if either all terms are 1 or ,-1 or Median is zero

Since integers are distinct so median has to be zero
Sufficient

Statement 2: set may be

(-5,-4,-3,-2,1,2,3,4,5) Or
(-5,-4,-3,-2,0,2,3,4,5)

Hence median may be 1 or zero or likewise 
Not sufficient

Answer Option A

Nunuboy1994 · Answer

Tricky but you just to be mindful of the possibilities -9 integers- all must be different

St 1

The only way this scenario could hold true is if the median were 0

suff

St 2

You could actually have different scenarios as demonstrated below

-5,-4,-3,-2,1,2,3,4,5

-5,-4,-3,-2,-1,1,2,5,6

insuff

A

hellosanthosh2k2 · Answer

Statement 1:
 Prod of all integers is equal median
 Let prod = product of all integers except median.

given,  prod  *  median  =  median 
  prod  *  median  -  median  = 0
  median  * ( prod  - 1) = 0 => median = 0 or prod = 1
 but product of all integers except median cannot be 1, as all are distinct => median = 0 => not positive => sufficient

Statement 2:
 Sum of all nine integers to equal to median.
 case 1: we can have median as positive, sum of left 4 integers and right 4 integers to cancel each other
 case 2: we can have median as negative, sum of left 4 integers and right 4 integers to cancel each other
 case 3: we can have median as 0, sum of left 4 integer and right 4 integers to cancel each other.
 insuff

(A)

standyonda · Answer

-- EASY SOLUTION --

STATEMENT 1
 x1 is the product of the 4 numbers left to the median
M is the median
x2 is the product of the 4 numbers right to the median

We have x1*x2*M=M or M(x1*x2-1)=0
(i) If M=0 it is a NO
(ii) If x1*x2=1 means that either of x1, x2 is 1 or either is -1. This is impossible since all the numbers must be different. So we cancel option (ii).

SUFFICIENT

Bunuel  check

Meghakothari98 · Answer

But if there are even no. of negatives numbers in the , the product of all 9 integers will be positive and the median will be positive
where as if there is an odd no of negative numbers in the set, the product will be negative and hence the mean will also be negative.
how is A sufficient?

Meghakothari98 · Answer

But if there are even no. of negatives numbers in the , the product of all 9 integers will be positive and the median will be positive
where as if there is an odd no of negative numbers in the set, the product will be negative and hence the mean will also be negative.
how is A sufficient?

Thelionking1234 · Answer

Because list S has nine integers (an odd number), the median must be the middle integer.
 
(1) SUFFICIENT: The median is an integer. According to this statement, the nine numbers must collectively multiply to this same integer. None of the numbers is a fraction, and all nine are different numbers. The only way to multiply a bunch of integers together and arrive at the same starting point is to multiply by 0 or 1. In order to use 1, every number on the list except one would have to equal 1; for example, 1 × 1 × 1 × 3 = 3. In order to use 0, though, only one number has to equal zero; for example, 1 × 18 × -3 × 0 = 0.
 
Therefore, the only way in which the product of nine different integers can equal just one of the integers on that list is if the product of the nine integers is zero. (Not convinced? Try to disprove that rule using some real numbers. If the list is -4, -3, -2, -1, 1, 2, 3, 4, 5 then the median is 1 but the product is definitely not 1. If the list is -10, -9, -8, -7, -6, -5, -4, -3, -2, the median is -6 but the product is not -6. If, on the other hand, the list is -4, -3, -2, -1, 0, 1, 2, 3, 4, then the median is 0 and the product is also 0.)
 
Further, since the product of the nine integers equals the median, the median itself must also be zero. The median, then, is definitely not positive; the statement is sufficient to answer the question.
 
(2) INSUFFICIENT: The median of list S is one of the nine integers in list S. Therefore, if the sum of all nine integers equals the median, it follows that the sum of the eight integers on either side of the median must be zero:
Sum of all nine integers = (Median) + (Sum of other eight integers)
Statement 2 says that the Sum of all nine integers equals the Median. Substitute that info into the above equation:
Median = (Median) + (Sum of other eight integers)
0 = Sum of other eight integers
 
This isn’t enough, though, to determine the median; in fact, any number can still be the median of the list, as long as the surrounding numbers sum to zero. For example, the list –10, –9, –8, –7, 5, 7, 8, 9, 10 satisfies the criterion and has a median value of 5; the list –10, –9, –8, –7, -1, 6, 8, 9, 11 also satisfies the criterion and has a median value of -1.
 
 The correct answer is A.

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If list S contains nine distinct integers, at least one of w

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If list S contains nine distinct integers, at least one of w

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