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Means, Medians & Numbers

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Means, Medians & Numbers

by theCodeToGMAT » Sun Sep 22, 2013 11:28 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.

OA after some replies
[E]
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by vinay1983 » Mon Sep 23, 2013 1:40 am
To say whether the numerator (average) of Set S is greater than the average of set T, we need to know the denominator i.e is the no.of integers

Statement 1

Average of set S is greater than Set T

Suppose sum of integers of Set S and of Set T is 20 each

S= 0,1,1,6,6,6 T = 2,3,4,5,6

Then average of S = 20/6 = 3.66 and average of T is 20/5= 4

This means that the denominator is inversely proportional to average if the numerator or sum of integers is kept constant.....Sufficient

STatement 2

S = 1,1,6,6,6 T = 2,3,4,5,6

Median of S = 6 and Median of T = 4 Does this tell us something about the no.of integers in Set S or Set T?

Not sufficient

So OA for me is A
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by theCodeToGMAT » Mon Sep 23, 2013 1:52 am
vinay1983 wrote:To say whether the numerator (average) of Set S is greater than the average of set T, we need to know the denominator i.e is the no.of integers

Statement 1

Average of set S is greater than Set T

Suppose sum of integers of Set S and of Set T is 20 each

S= 0,1,1,6,6,6 T = 2,3,4,5,6

Then average of S = 20/6 = 3.66 and average of T is 20/5= 4

This means that the denominator is inversely proportional to average if the numerator or sum of integers is kept constant.....Sufficient

STatement 2

S = 1,1,6,6,6 T = 2,3,4,5,6

Median of S = 6 and Median of T = 4 Does this tell us something about the no.of integers in Set S or Set T?

Not sufficient

So OA for me is A
The answer is NOT {A} .. you need to correct/complete the assumptions you made....
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by vipulgoyal » Mon Sep 23, 2013 2:14 am
OA E,
this is somewhat brute force method, got it in old blogs
any quick solution is highly appreciated,
1.
sum/s < sum/t, if sum is positive then s < t, if sum is negative then s > t(reverse the eneaulity),Insuffi
2.
The median of the integers in S is greater than the median of the integers in T. If S={1, 1} and T={0, 0, 2} then the median of S (1) is greater than the median of T (0) and S contains less elements than T but if S={-1, -1, -1} and T={-3, 0} then the median of S (-1) is greater than the median of T (-1.5) and S contains more elements than T. Not sufficient.

(1)+(2):
If S={-1, 2, 2} and T={1, 2} then the sum is equal (3), the average of S (1) is less than the average of T (1.5), the median of S (2) is greater than the median of T (1.5) and S contains more elements than T.

If S={-2, -1} and T={-2, -2, 1} then the sum is equal (-3), the average of S (-1.5) is less than the average of T (-1), the median of S (-1.5) is greater than the median of T (-2) and S contains less elements than T.

Not sufficient.

Answer: E.
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by theCodeToGMAT » Mon Sep 23, 2013 2:20 am
Yes Vipul, the Answer is indeed [spoiler] {E}[/spoiler] ...

the problem with Vinay's assumption was that he only considered Positive numbers and then decided the result using algebra..

The problem Stem:
"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?"

It doesn't mention that integers are positive..

On the first attempt I also got {A}, then after spending some time on question I found my mistake..

Thought that this question needs to be shared... so posted it here :)
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by vipulgoyal » Mon Sep 23, 2013 2:27 am
indeed its worth of posting, but my concern is, this brute force solution is very time consuming, is there any conceptual solution rather then plugging numbers ??
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by theCodeToGMAT » Mon Sep 23, 2013 2:35 am
vipulgoyal wrote:indeed its worth of posting, but my concern is, this brute force solution is very time consuming, is there any conceptual solution rather then plugging numbers ??
Not sure.. Experts can definitely provide some... let's wait for their comments..
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by vinay1983 » Mon Sep 23, 2013 2:48 am
theCodeToGMAT wrote:Yes Vipul, the Answer is indeed [spoiler] {E}[/spoiler] ...

the problem with Vinay's assumption was that he only considered Positive numbers and then decided the result using algebra..

The problem Stem:
"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?"

It doesn't mention that integers are positive..

On the first attempt I also got {A}, then after spending some time on question I found my mistake..

Thought that this question needs to be shared... so posted it here :)

Whoa...never thought of it...I think we need experts here
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by lunarpower » Mon Sep 23, 2013 3:21 am
I received a private message about this thread.
vipulgoyal wrote:this brute force solution is very time consuming
It's not very time-consuming.
If you understand why you're choosing numbers -- i.e., you know exactly what you want the numbers to do, or not to do -- then it's a pretty efficient process. You'll have to explore a bit, but it's not a time-suck.

A bit of introspection is in order here.
99.9% of the time when people say "I think this will take too long", that's not the real thought process. The real issue is that people don't want to do number plugging, and they're just making up cheap justifications. In almost all of these cases, the number plugging really doesn't take very long at all, as long as you have a purpose in mind -- it's just a false rationale to cover up "I don't want to do it."

If this is what's going on in your head (which it almost certainly is, at least in part), then try to figure out where your opposition to number plugging comes from.
For many students, it comes down to "I feel dumb if I can't solve this problem with theory", which, if you think about it, is a non-reason (and is pretty silly, too). Anything that solves the problem is just as good as anything else that solves the problem.
Ron has been teaching various standardized tests for 20 years.

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by rakeshd347 » Wed Sep 25, 2013 3:41 am
I totally agree with RON on this.
Half of the time I give up on awkward looking or wordy PS or DS not because it is hard or time consuming but because it boring and I am lazy. However, i have changed that habit now...if you start with what you know about the relevant topic whether by picking numbers or assigning variables questions look easy enough to solve within time limit.

lunarpower wrote:I received a private message about this thread.
vipulgoyal wrote:this brute force solution is very time consuming
It's not very time-consuming.
If you understand why you're choosing numbers -- i.e., you know exactly what you want the numbers to do, or not to do -- then it's a pretty efficient process. You'll have to explore a bit, but it's not a time-suck.

A bit of introspection is in order here.
99.9% of the time when people say "I think this will take too long", that's not the real thought process. The real issue is that people don't want to do number plugging, and they're just making up cheap justifications. In almost all of these cases, the number plugging really doesn't take very long at all, as long as you have a purpose in mind -- it's just a false rationale to cover up "I don't want to do it."

If this is what's going on in your head (which it almost certainly is, at least in part), then try to figure out where your opposition to number plugging comes from.
For many students, it comes down to "I feel dumb if I can't solve this problem with theory", which, if you think about it, is a non-reason (and is pretty silly, too). Anything that solves the problem is just as good as anything else that solves the problem.
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