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Bunuel
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If x/3 is a positive odd integer and x 15, which values could be the 12 Aug 2026, 19:00
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75% (02:15) correct 25% (02:15) wrong based on 91 sessions

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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 \(1, 3, 5, 7, 9, 11, 13, 15, 17, x, \frac{x}{3}\) ?

I. 6
II. 7
III. 9

A. I only
B. III only
C. I and II only
D. II and III only
E. I, II and III


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dhart
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If x/3 is a positive odd integer and x 15, which values could be the 12 Aug 2026, 23:09
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looks like 2 and 3 option D
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onlyPlanA
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If x/3 is a positive odd integer and x 15, which values could be the 13 Aug 2026, 01:46
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x can be 3,6,9,12,15

Using all these values, the median can only be 7 and 9.
Option D.
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gautamK0206
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If x/3 is a positive odd integer and x 15, which values could be the 13 Aug 2026, 05:05
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With the given conditions that x<=15, and x/3 = odd positive integer, we can infer that x/3<=5.
Therefore, x can be anything from = [3,9,15], and x/3 from. = [1,3,5].
Now, with this information the possible sets are:
[1,1,3,3,5,7,9,11,13,15,17] or [1,3,3,5,7,9,9,11,13,15,17] or [1,3,5,5,7,9,11,13,15,15,17]
We can see that each possible set has 11 terms, and the median for 11 terms would be (11+1)/2 th term= 6th terms.
So, the possible values are = 7, 9, 9.
Hence, D. II and III only is the correct answer.
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