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Difficulty:
65%
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Question Stats:
56% (02:27) correct
44%
(02:08)
wrong
based on 135
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List L consists of k consecutive integers, where k is an odd integer. The median of the integers in L is m. Which of the following statements must be true?
I. The sum of the integers in L is an odd integer.
II. The least integer in L is equal to \(m - \frac{k−1}{2}\)
III. The greatest integer in L is equal to \(m+\frac{k+1}{2}\)
A. I only
B. II only
C. III only
D. I and II only
E. II and III only
I. The sum of the integers in L is an odd integer.
II. The least integer in L is equal to \(m - \frac{k−1}{2}\)
III. The greatest integer in L is equal to \(m+\frac{k+1}{2}\)
A. I only
B. II only
C. III only
D. I and II only
E. II and III only
11 Nov 2022, 13:39
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List L consists of k consecutive integers, where k is an odd integer. The median of the integers in L is m. Which of the following statements must be true?
I. The sum of the integers in L is an odd integer.
II. The least integer in L is equal to \(m - \frac{k−1}{2}\)
III. The greatest integer in L is equal to \(m+\frac{k+1}{2}\)
A. I only
B. II only
C. III only
D. I and II only
E. II and III only
I. The sum of the integers in L is an odd integer.
II. The least integer in L is equal to \(m - \frac{k−1}{2}\)
III. The greatest integer in L is equal to \(m+\frac{k+1}{2}\)
A. I only
B. II only
C. III only
D. I and II only
E. II and III only
Let's see what inferences can we make -
- "List L consists of k consecutive integers" : Inference - The list is evenly spaced; the terms are in arithmetic progression.
- "k is an odd integer." : Inference - The median is a member of the set
- As the list is evenly spaced, the sum of the list = Median * Number of terms
Answer choice elimination
I. The sum of the integers in L is an odd integer.
Sum = Median * number of terms
number of terms is odd. Is the value of median odd ? Well we aren't sure. Eliminate the option as it cannot be MUST be true.
II. The least integer in L is equal to \(m - \frac{k−1}{2}\)
Its given there are k integers in the list. So \(\frac{k-1}{2}\) integers lie to the left of m and \(\frac{k-1}{2}\) lie to the right
----------- \(\frac{k-1}{2}\)----------- m----------- \(\frac{k-1}{2}\)-----------
So the lowest term is \(m - \frac{k−1}{2}\). Will keep
III. The greatest integer in L is equal to \(m+\frac{k+1}{2}\)
With reference to the above explanation, the greatest term is \(m + \frac{k−1}{2}\). Hence Eliminate
Option B
