4. List S consists of 25 different integers. The least and greatest integers in S are 0 and 50, respectively. The integers in list T are obtained by adding 50 to each integer in S. List R consists of the 50 values in S and T combined. Quantity A: The median of the integers in R. Quantity B: 2 × the median of the integers in S.
Answer: C
The two quantities are equal.
Both quantities A and B can be shown to be equal due to the specific construction of lists S and T. Since T is formed by adding 50 to each integer in S, the medians of the two combined lists will align appropriately.
A) Quantity A is greater.
This option is incorrect because the median of list R, which includes all integers from both S and T, will not exceed twice the median of S. In fact, the medians of the two lists are specifically linked due to the addition of a constant.
B) Quantity B is greater.
This option is also incorrect. The relationship between the medians of S and R ensures that Quantity B, which is twice the median of S, cannot be greater than Quantity A. The transformation of S into T maintains the overall balance of the data.
C) The two quantities are equal.
This option is correct because the median of list R, which combines S and T, will equal twice the median of S. The integers in T are symmetrically distributed about the median of S, which results in the combined median being precisely double that of S.
D) The relationship cannot be determined from the information given.
This option is incorrect as the relationship can indeed be determined. Given the systematic addition of 50 to each integer in S to form T, we can definitively calculate the medians for both lists, confirming the equality.
Conclusion
The conclusion that the two quantities are equal is supported by the arithmetic and structural relationship between lists S and T. The median of R, which includes both lists, reflects the relationship created by the constant addition to each integer in S, confirming that Quantity A and Quantity B are equivalent.