j_shreyans wrote:List P contains m numbers; list Q contains n numbers. If the two lists are combined to produce list R, containing m + n numbers, is the median of list R greater than the median of list P ?
(1) The smallest number in list Q is greater than the largest number in list P.
(2) m = n
Statement 1: The smallest number in list Q is greater than the largest number in list P.
Test an EASY CASE that satisfies BOTH statements.
Case 1: P = {0} and Q = {1}, implying that R = {0, 1}.
Median of P = 0.
Median of R = 1/2.
In this case, median of R > median of P, so the answer to the question stem is YES.
Test an easy case that satisfies ONLY STATEMENT 1.
Case 2: P = {0, 0} and Q = {1}, implying that R = {0, 0, 1}.
Median of P = 0.
Median of R = 0.
In this case, median of R = median of P, so the answer to the question stem is NO.
Since the answer is YES in Case 1 but NO in Case 2, INSUFFICIENT.
Statement 2: m=n
Case 1 satisfies statement 2.
In Case 1, the answer to the question stem is YES.
Test a case that satisfies ONLY STATEMENT 2.
Case 3: P = {0} and Q = {0}, implying that R = {0, 0}.
Median of P = 0.
Median of R = 0.
In this case, median of R = median of P, so the answer to the question stem is NO.
Since the answer is YES in Case 1 but NO in Case 3, INSUFFICIENT.
Statements combined:
Let P = {a, b, c}, where a≤b≤c.
Let Q = {d, e, f}, where d≤e≤f.
Statement 1 indicates that d > c.
Thus:
R = {a, b, c, d, e, f}, where a ≤ b ≤ c < d ≤ e ≤ f}.
Adding together c≥b and d>b, we get:
c+d > 2b
(c+d)/2 > b.
Since the median of R = (c+d)/2, and the median of P = b, the resulting inequality implies that the median of R > median of P.
SUFFICIENT.
The correct answer is
C.
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