Bunuel
Let \(x_1, \ x_2, \ … , \ x_9, \ y_1, \ y_2, \ … , \ y_9\) are all distinct positive integers less than 100 such that \(x_1 < x_2 < x_3 < x_4 < … < x_9\) and \(y_1 < y_2 < y_3 < y_4 < … < y_9\). Suppose arithmetic mean of x1 and x2 is less than arithmetic mean of \(x_3, \ x_4, \ x_5, \ …, \ x_9\) by 9. Also, arithmetic mean of \(y_1\) and \(y_2\) is less than arithmetic mean of \(y_3, \ y_4, \ y_5, \ …, \ y_9\) by 9. If the difference between arithmetic mean of (\(x_1 \ and \ x_2\)) and (\(y_1 \ and \ y_2\)) is maximum possible, which of the following is arithmetic mean of \(x_1, \ x2, \ y_1, \ y_2\)?
A. 48.5
B. 46.5
C. 44.5
D. 42.5
E. 40.5
Are You Up For the Challenge: 700 Level Questionsx's and y's need to be as far apart as possible so let's put x's close to 0 and y's close to 100.
Mean of x1 and x2 is 9 less than mean of other 7 x's. Since all are distinct positive integers, mean of other 7's will be the middle number, an integer.
So x1 and x2, in the least case, will be 1 and 3 with mean 2 (an integer). They cannot be 1 and 2 because mean in that case will be 1.5.
Mean of y1 and y2 must be as great at possible and 9 less than mean of other 7 y's. So other 7 y's must be 93 to 99 with mean of 96 (max possible) which means mean of y1 and y2 will be 87. So y1 and y2 must be 86 and 88.
Mean of 2 and 87 is 44.5 so that will be the mean of 1, 3, 86, 88 (mean can replace each number in the list so 1 and 3 can be replaced by 2 and 2 and 86 and 88 can be replaced by 87 and 87. So mean of 2 an d87 will give the mean of 1, 3, 86, 88)
Answer (C)