lheiannie07 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.
What's the best way to determine whether statement 1 is sufficient?
OA B
(1) The median of X is greater than 1.
Case 1: Say X: {2, 2, 2, 2, 2, 2, 2, 2, 2}; thus, Y: {4, 4, 4, 4, 4, 4, 4, 4, 4}; Average of Y (= 4) > Average of X (= 2). The answer is Yes.
Case 2: Say X: {0.1, 0.1, 0.1, 0.1,
1.01, 1.01, 1.01, 1.01, 1.01}; thus, Y: {0.01, 0.01, 0.01, 0.01, 1.02, 1.02, 1.02, 1.02, 1.02};
We see that the first four elements of X are greater than those of Y, whereas the last five elements of Y are greater than those of X. Whether the average of Y is greater than that of X would be determined by differences of the two sums is greater than the other: the difference of the sum of the first four elements of X and those of Y OR the difference of the sum of the last five elements of Y and those of X .
The difference of the sum of the first four elements of X and those of Y = Sum of the first four terms of X - Sum of the first four terms of Y
= 4*0.1 - 4*0.01 = 4*0.09 = 0.36
The difference of the sum of the last five elements of Y and those of X = Sum of the last five terms of Y - Sum of the last five terms of X
= 5*1.02 - 5*1.01 = 5*0.01 = 0.05
Since 0.05 < 0.36, thus, Average of X > Average of Y. The answer is No. No unique answer. Insufficient
(2) The range of X is 2.
Case 1: Say X: {0.1, 2, 2, 2, 2, 2, 2, 2, 2}; thus, Y: {0.01, 4, 4, 4, 4, 4, 4, 4, 4}; Average of Y > Average of X. The answer is Yes.
Since for Case 2, we must have a set of elements such that we are able to prove that average of X > average of Y. To do so, we must take the first eight elements of X as small as possible so that after their squares, the respective elements in Y would be significantly smaller than their counterparts in X.
Case 2: Say X: {0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 20.1, 0.1, ~2}; thus, Y: {0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01, ~4}
We see that the difference of the ninth element of Y (= ~4) and X (= ~2) is significantly far greater (=~ 2); thus, the elements that are greater in X than those in Y will not matter. Thus, we cannot have a set of elements in which we can prove that the average of X > Average of Y.
The correct answer:
B
Hope this helps!
-Jay
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