Source: Manhattan Prep
Set S includes elements {8, 2, 11, x, 3, y} and has a mean of 7 and a median of 5.5. If x < y, then which of the following is the maximum possible value of x?
A. 0
B. 1
C. 2
D. 3
E. 4
The OA is D
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Set S includes elements {8, 2, 11, x, 3, y} and has a mean of 7 and a median of 5.5. If x < y, then which of the...
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Set S = {2, 3, 8, 11, x, y}
$$Mean\ of\ S=\frac{2+3+8+11+x+y}{6}$$
$$7=\frac{24+x+y}{6}$$
Median = 5.5 and x<y
From the mean of S
$$24+x+y=42$$
$$x+y=18$$
From the median, an average of 2 numbers in the middle = 5.5
$$i.e\ \frac{a+b}{2}=5.5\ and\ a+b=11$$
From the numbers in the set 3+8=11. This means that x and y might not be in the middle and since x<y, $$The\max imum\ possible\ value\ of\ x\ is\le3$$
Answer = option D
$$Mean\ of\ S=\frac{2+3+8+11+x+y}{6}$$
$$7=\frac{24+x+y}{6}$$
Median = 5.5 and x<y
From the mean of S
$$24+x+y=42$$
$$x+y=18$$
From the median, an average of 2 numbers in the middle = 5.5
$$i.e\ \frac{a+b}{2}=5.5\ and\ a+b=11$$
From the numbers in the set 3+8=11. This means that x and y might not be in the middle and since x<y, $$The\max imum\ possible\ value\ of\ x\ is\le3$$
Answer = option D
