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07 Dec 2023, 02:30
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
67% (02:20) correct
33%
(02:15)
wrong
based on 101
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If \(\frac{x}{3}\) is a positive odd integer and \(x ≤ 15\), which values could be the median of the list of numbers \(1, 3, 5, 7, 9, 11, 13, 15, 17, x, \frac{x}{3}\) ?
B. 6
C. 7
D. 9
A. I only
B. III only
C. I and II only
D. II and III only
E. I, II and III
B. 6
C. 7
D. 9
A. I only
B. III only
C. I and II only
D. II and III only
E. I, II and III
Bunuel
If \(\frac{x}{3}\) is a positive odd integer and \(x ≤ 15\), which values could be the median of the list of numbers \(1, 3, 5, 7, 9, 11, 13, 15, 17, x, \frac{x}{3}\) ?
B. 6
C. 7
D. 9
A. I only
B. III only
C. I and II only
D. II and III only
E. I, II and III
From the stem, we can infer that x is odd and it is a multiple of 3. Therefore, the possible values of X is 3,9,15 and for \(\frac{x}{3}\) it is 1,3,5.B. 6
C. 7
D. 9
A. I only
B. III only
C. I and II only
D. II and III only
E. I, II and III
On rearranging the number in ascending order, the possible values of 6th term is 7 or 9.
Thus Option D is the correct answer.
07 Dec 2023, 05:24
Kudos
Bookmarks
D IMO
Since the no. of terms in the list are 11 i.e odd, the median would be the 6th term in the list.
Possible values of x=3, 9, or 15
If x=3, then \(\frac{x}{3}=1\). Thus, the list becomes: {1,1,3,3,5,7,9,11,13,15,17}. Median is 7.
If x=9, then \(\frac{x}{3}=3\). Thus, the list becomes: {1,3,3,5,7,9,9,11,13,15,17}. Median is 9.
If x=15, then \(\frac{x}{3}=5\). Thus, the list becomes: {1,3,5,5,7,9,11,13,15,15,17}. Median is 9.
Therefore, D is correct.
Since the no. of terms in the list are 11 i.e odd, the median would be the 6th term in the list.
Possible values of x=3, 9, or 15
If x=3, then \(\frac{x}{3}=1\). Thus, the list becomes: {1,1,3,3,5,7,9,11,13,15,17}. Median is 7.
If x=9, then \(\frac{x}{3}=3\). Thus, the list becomes: {1,3,3,5,7,9,9,11,13,15,17}. Median is 9.
If x=15, then \(\frac{x}{3}=5\). Thus, the list becomes: {1,3,5,5,7,9,11,13,15,15,17}. Median is 9.
Therefore, D is correct.
