DS : The sum of the integers in list S

[Rathna]

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.

How to approach this problem?

[gauravbawa+gm]

Is b the answer?

[RonPurewal]

ah, evil.

to get at the heart of this problem, you must realize that "greater" is NOT the same thing as "farther away from zero".
for positive numbers, these two concepts are the same, but they are not so for negative numbers (for which the the greater number is actually closer to zero).
that realization is the crux of this problem.

see, here's the deal:
the SUM is the same for each of the two sets. therefore, since average = (sum) / (# of data points), the average will be CLOSER TO ZERO if there are more data points.

the problem is that this doesn't mean that the average is lower. if the sum is negative, then just the opposite will occur.

examples:
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.

together:
this takes a little more creativity.
if S = -7, 9, 10 and T = 6, 6, then the sums are both 12, the average of S is less (4 vs. 12), the median of S is greater (9 vs. 6), and S has more integers.
if S = -6, -6 and T = -10, -9, 7, then the nums are both -12, the average of S is less (-6 vs. -4), the median of S is greater (-6 vs. -9), and S has fewer integers.
so, insufficient.

answer = (e).

incidentally, if this problem is supposed to say that the integers are positive or non-negative (and the original poster simply forgot this caveat), then the problem is much easier, and the answer is different (at a glance i think it would be (a)).

[Rathna]

Infact Ron, OA is A

[RonPurewal]

Infact Ron, OA is A

if that's the case, then the problem statement must contain a condition stating that the numbers are positive, or at the very least non-negative. if it doesn't, then, as i have demonstrated above, that's simply the wrong answer.

anybody with the official version of this problem out there?
screen shot?

[Rathna]

i have the screen shot...but i am not able to post it :(

[RonPurewal]

i have the screen shot...but i am not able to post it :(

does it mention that the numbers must be positive, or, at the very least, non-negative?

[moty.98]

I think that this is the reason for A being correct:

Average = Sum/(numbers in set)

Sum of numbers in S = Ave(S) * (number of numbers in set S)
Sum of numbers in T = Ave(T) * (number of numbers in set T)

It is given that Sum(S)=Sum(t).

Thus,
Ave(S) * (number of numbers in set S) = Ave(T) * (number of numbers in set T)

If Ave(s)<Ave(t), then
(number of numbers in set S) > (number of numbers in set T).

------------------------------------
Moty Keret

[esledge]

I think that this is the reason for A being correct:

Average = Sum/(numbers in set)

Sum of numbers in S = Ave(S) * (number of numbers in set S)
Sum of numbers in T = Ave(T) * (number of numbers in set T)

It is given that Sum(S)=Sum(t).

Thus,
Ave(S) * (number of numbers in set S) = Ave(T) * (number of numbers in set T)

If Ave(s)<Ave(t), then
(number of numbers in set S) > (number of numbers in set T).

------------------------------------
Moty Keret

That is algebraically correct, but ONLY IF the average of each set is positive.

However, if Ave(s) = 0 and Ave(t) = positive, then according to this relationship,

I Ave(S) * (number of numbers in set S) = Ave(T) * (number of numbers in set T)

the # of terms in set S could be anything, and the # of terms in set T would have to be 0. That's nonsensical, so (1) constitutes a hidden, implied constraint that Ave(s) can't be 0. Ave(t) can't be 0, either, for the same reason.

Also, if Ave(s) and Ave(t) are both negative, then the relationship between the # of terms in the set is the opposite of your algebraic conclusion. In this case, (# of terms in set S) < (# of terms in set T). Algebraically, this is because of the old "flip the sign when multiplying or dividing by a negative" rule. You might think of it this way:

Given: Ave(S) * (# of terms in set S) = Ave(T) * (# of terms in set T)
From (1): Ave(s) < Ave(t)

Without the specification that the averages are positive, one of the cases we must consider for (1) is Ave(s) < Ave(t) < 0.

So, the given becomes:
Ave(S) * (# of terms in set S) = Ave(T) * (# of terms in set T)
(more negative) * (# of terms in set S) = (less negative) * (# of terms in set T)

The # of terms in the sets must be positive, so to make the equation balance:
(more negative) * (less positive) = (less negative) * (more positive)

Therefore, 0 < (# of terms in set S) < (# of terms in set T)

I also like Ron's sample sets as proof:

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.

In short, we really do need clarification of whether the problem included the constraint that the averages are positive. If it did not, the problem is incorrect.

[johnhillescobar]

I have tried to attach a jpg file with a print screen of this question. I think I am not allowed to upload files or attach images from my computer.

Could you people (instructor and staff) give me an e-mail to send you this file?

[johnhillescobar]

OA is A

[RonPurewal]

I have tried to attach a jpg file with a print screen of this question. I think I am not allowed to upload files or attach images from my computer.

Could you people (instructor and staff) give me an e-mail to send you this file?

we generally don't give out private e-mail addresses for forum moderators.
try the following image posting site:

http://postimage.org/

that's a pretty no-frills image hosting site; i know essentially nothing about computers or the internet, and it took me about 10 seconds to learn how to use it.
if your computer barred from uploading images to a site like that, then you would also be barred from attaching those images to an e-mail; the upload process is exactly the same.

once you do upload the image to that site, go ahead and copy the direct link (it should be the last of the 3-4 options that appears upon your successful upload), and post it here. thanks.

[johnhillescobar]

Here we are with the OA



[img=http://s4.postimage.org/6HrCi.jpg]

<a href="http://www.postimage.org/image.php?v=aV6HrCi" target="_blank"><img src="http://s4.postimage.org/6HrCi.jpg" border="0" alt="Free image hosting powered by PostImage.org" /></a>

http://www.postimage.org/image.php?v=aV6HrCi

[akhp77]

This is official problem but why OA is A.

Sum of integers in S = Sum of integers in T = SUM (Either +ve or -ve)
No of integers in S = s
No of integers in T = t

Integers may be -ve or +ve. It is not mentioned.

Statement 1:
Assume SUM = 6
6 / s < 6 / t; s > t

Assume SUM = -6
-6 / s < -6 / t; 6 / s > 6 / t; s < t

Not Sufficient

[mschwrtz]

Consider these positive and negative examples (each consistent with the question and S1), then we'll talk algebra.

Suppose that the sum of the integers in each list is 12, that there are 6 numbers in S, and that there are 4 numbers in T.

This is consistent with S1, since it yields a mean for S of 2 and a mean for T of 3. It also gives us "yes."

Suppose that the sum of the integers in each list is -12, that there are 4 numbers in S, and that there are 6 numbers in T.

This is consistent with S1, since it yields a mean for S of -3 and a mean for T of -2. It also gives us "no."

If we can get both yes and no, it's not sufficient.

OK, here's a simpler approach than Ron offered, but still a sound one. We'll borrow akhp77's variables.

Sum of integers in S = Sum of integers in T = SUM
No of integers in S = s
No of integers in T = t

Question: Is s>t?

S1: SUM/s<SUM/t
cross-multiply (both s and t are pos, so no worries about the inequality): SUMt<SUMs
Divide each side by SUM
If SUM is pos, t<s YES
If SUM is neg, t>s NO
Not sufficient

Now, I don't know why the OA is A....