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Set X consists of 9 positive elements. Set Y consists of the [#permalink]
12 Jul 2006, 15:37
00:00
A
B
C
D
E
Difficulty:
95% (hard)
Question Stats:
30% (02:48) correct
70% (01:37) wrong based on 37 sessions
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
Set X consists of 9 positive elements. Set Y consists of the squares of the elements of X. Is the average of the elements in Y greater than that of set X?
(1) The median of X is greater than 1. (2) The range of X is 2.
Re: DS: Comparing Means [#permalink]
12 Jul 2006, 17:35
kevincan wrote:
Set X consists of 9 positive elements. Set Y consists of the squares of the elements of X. Is the average of the elements in Y greater than that of set X?
(1) The median of X is greater than 1. (2) The range of X is 2.
first there is no saying if the elements are decimal or integers also it does not mention if the elements are consecutive...
st.1
median = 1, which means that the 5th term once the numbers are put in sequence order, must be 1, but does not limit how big the number can get. depending on the sequence number, we can either answer a yes or no... Not SUFF
st.2
Range=Largest-smallest number in the sequence = 2.
Which means that we can anwer a yes or a no depending on the number sequence... Not SUFF.
Hence lets say the first integer are 1 (remember we are talking about positive integers) the fifth integer is greater than 1 hence lets say its 2 and the lets say the other 4 (has to be >= 2) is 2
the average of these numbers is obviously less than average of square of these numbers.
2) Range is 2
Lets say eight integers are 1 and the last integer has to be 3 to have a range of 2
Hence the average of these numbers is obviously less than average of square of these numbers.
Hence lets say the first integer are 1 (remember we are talking about positive integers) the fifth integer is greater than 1 hence lets say its 2 and the lets say the other 4 (has to be >= 2) is 2
the average of these numbers is obviously less than average of square of these numbers.
2) Range is 2 Lets say eight integers are 1 and the last integer has to be 3 to have a range of 2 Hence the average of these numbers is obviously less than average of square of these numbers.
Hence D
how are you assuming the values to be integers?
in the question, it does not mention integers...
shahnandan,
you said: integers and fractions will yield the same result.
for example: if x=0.5, 0.5
y=0.25, 0.25 since y=x^2, the average of y can not be greater than average of x.
If the values of x are integers, then I agree, but not when they are fractions.
I still don't buy the integer part of the explanation. Just like x^2 = 4, therefore x=2. Wrong x=2 or x=-2... we are not supposed to assume too much in DS...
Re: DS: Comparing Means [#permalink]
13 Jul 2006, 20:34
kevincan wrote:
Set X consists of 9 positive elements. Set Y consists of the squares of the elements of X. Is the average of the elements in Y greater than that of set X?
(1) The median of X is greater than 1. (2) The range of X is 2.
go with E. since we donot know how many elements of x are there in y.
St1: In X if 5 numbers are 1.000000001 and 4 are 0.000000001 then X will be greater. If all elements of X are greater than 1 then average of Y is greater : INSUFF
St2: If range is 2 then average of Y will always be greater than that of X. Lets take teh worst case scenario. When 8 elements of X are 0.0000000001 and one element is 2.0000000001 then sum of elements will be approx 2. But sum of elements of Y will be close to 4. If all elements of X are greater than 1 then also average of Y will be greater.: SUFF _________________
If the median is greater than 1, then for all the numbers above 1, the average will be greater in Y. And in the worst case scenario, the first four numbers are between 0 and 1, but they will have very limited effect on the average.
If the median is greater than 1, then for all the numbers above 1, the average will be greater in Y. And in the worst case scenario, the first four numbers are between 0 and 1, but they will have very limited effect on the average.
B is inconclusive.
So I would go with A.
I will go with A too. since the median has got to be atleast two, the rest of the elements in X has to be atleast 2.01 (or 3 even if they are integers). When squared the increase in value of the last 4 will easily outweigh the decrease in the value of the first four decimals being squared (if they are decimals). So A is suff.
Kevin...your thoughts!!!
Re: DS: Comparing Means [#permalink]
20 Jul 2006, 15:33
It should be B.
stmt1.
counter example
set X : 0.1, 0.1, 0.1, 0.1, 1.01, 1.01, 1.01, 1.01, 1.01
set Y : 0.01, 0.01, 0.01, 0.01, 1.0201, 1.0201, 1.0201, 1.0201, 1.0201
SumX : 1*5 + 0.1*5 + 0.01*5
SumY : 1*5 + 0.01*4+0.0201*5
SumX > SumY
insuff.
Stmt2.
Max (SetX) > 2
Max (SetY) > 4.
Thus
SumY - SumX > 1 (I can prove this using derivative, but not using it, this just become a guess)
Suff. _________________
For example a) x=(0.498, 0.499, 0.5, 0.501, 1, 1.001, 1002, 1003, 1004). y corresponding;
[the elements in x could be the same]
but:
x=(0.1, 0.2, 0.3, 0.4, 1, 10, 100, 1000, 1000000000 [just to be sure])
(Note: x^2-x takes a minimum at x=0.5. The decrease from x to x^2 is 0.25 at x=0.5)
Statement (2) is sufficient.
It is easy to find an example where the average of the elements of y is greater than the average of the elements in x).
Could the average of x be greater than the average of the elements in y? No.
the increase from x=0 to x^2=0 is 0; the increase from x=2 to x^2=2 is 2. Hence the total increase is 2, but the decrease from values around 0.5 is approximately 0.25; therefore the maximum total decrease 0.25*7=1.75 is smaller than the total increase. Hence the average of the elements in y must be larger than the average of the elements in x).
[The argument is based on the fact that (x^2-x) + (x+2)^2-(x+2) takes an extremum at x=-0.5 and is strictly increasing for x>-0.5; hence we have a border solution for x=0].
Another interesting question would arise if the set x could consist of negative numbers as well.