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If x/3 is a positive odd integer and x ≤ 15
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13 Aug 2020, 01:43
13 Aug 2020, 01:43
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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 \(1, 3, 5, 7, 9, 11, 13, 15, 17, x, \frac{x}{3}\) ?
Indicate all such values.
A. 5
B. 6
C. 7
D. 9
E. 11
KUDOS for the right solution and explanation
Indicate all such values.
A. 5
B. 6
C. 7
D. 9
E. 11
KUDOS for the right solution and explanation
Re: If x/3 is a positive odd integer and x ≤ 15
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15 Aug 2020, 18:28
15 Aug 2020, 18:28
1
The only difficulty here is keeping the details straight and a little bit of repetition. First we find the one or more x's that fit the criteria. x must be a multiple of 3 to produce an integer.
If x = 3, x/3 = 1, which is a positive odd integer, so that is one possiblity.
If x = 6, x/3 = 2, no good.
If x = 9, x/3 = 3, good.
If x = 12, x/3 = 4, no good.
If x = 15, x/4 = 5, good. x cannot exceed 15, so we're done finding our x's. They are 3, 9, and 15. x/3's are 1, 3, and 5.
We take each possiblity and insert our (x, x/3) pairs into the list, one pair at a time. Then we find the sixth element, because that is where the median is found in a list of 11 elements.
1, 1, 3, 3, 5, 7, stop. We found our first median, 7.
1, 3, 3, 5, 7, 9, stop. We found our second median, 9.
1, 3, 5, 5, 7, 9, stop We've already found 9.
The answers are C and D, 7 and 9.
If x = 3, x/3 = 1, which is a positive odd integer, so that is one possiblity.
If x = 6, x/3 = 2, no good.
If x = 9, x/3 = 3, good.
If x = 12, x/3 = 4, no good.
If x = 15, x/4 = 5, good. x cannot exceed 15, so we're done finding our x's. They are 3, 9, and 15. x/3's are 1, 3, and 5.
We take each possiblity and insert our (x, x/3) pairs into the list, one pair at a time. Then we find the sixth element, because that is where the median is found in a list of 11 elements.
1, 1, 3, 3, 5, 7, stop. We found our first median, 7.
1, 3, 3, 5, 7, 9, stop. We found our second median, 9.
1, 3, 5, 5, 7, 9, stop We've already found 9.
The answers are C and D, 7 and 9.
Carcass wrote:
If \(\frac{x}{3}\) is a positive odd integer and \(x ≤ 15\), which values could be the median of the list of numbers \(1, 3, 5, 7, 9, 11, 13, 15, 17, x, \frac{x}{3}\) ?
Indicate all such values.
A. 5
B. 6
C. 7
D. 9
E. 11
KUDOS for the right solution and explanation
Indicate all such values.
A. 5
B. 6
C. 7
D. 9
E. 11
KUDOS for the right solution and explanation
gmatclubot
Re: If x/3 is a positive odd integer and x ≤ 15 [#permalink]
15 Aug 2020, 18:28
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