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If sequence Q has four terms: x, 12, y, and 20, is it true that the [#permalink]
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01 Oct 2014, 11:39
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If sequence Q has four terms: x, 12, y, and 20, is it true that the median of sequence Q is 16? (1) The average of sequence Q is 16. (2) x=16
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Re: If sequence Q has four terms: x, 12, y, and 20, is it true that the [#permalink]
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01 Oct 2014, 20:10
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goodyear2013 wrote: If sequence Q has four terms: x, 12, y, and 20, is it true that the median of sequence Q is 16? (1) The average of sequence Q is 16. (2) x=16 As the sequence consists of 4 terms, so if the sequence is rearranged in increasing order the median will be average of the middle 2 terms. Hence the question is reduced to finding the middle 2 terms. (1) \(Average =\frac{(x + 12 + y + 20)}{4} = 16\) => \((x + y) = 32\) There are 3 possible combinations for x and y a) x < 12, then y > 20  Check for possible possible values if x = 1 then y = 33, if x = 11 then y = 21 This means that the middle 2 terms are 12 & 20. And the median = 16 b) x = 12, then y = 20 This also means that the middle 2 terms are 12 & 20. And the median = 16 c) 12 < x < 20, then 12 < y < 20 if x = 13, then y = 19 => \(Median = \frac{(13 + 19)}{2} = 16\) if x = 15, then y = 17 => \(Median = \frac{(15 + 17)}{2} = 16\) Hence Stmt I alone is sufficient (2) x = 16 => Q = 12, 16, y, 20 Now y can be any value and it can change the median. Hence Stmt II is not sufficient.
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Re: If sequence Q has four terms: x, 12, y, and 20, is it true that the [#permalink]
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Re: If sequence Q has four terms: x, 12, y, and 20, is it true that the [#permalink]
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23 Mar 2017, 11:29
prasun9 wrote: goodyear2013 wrote: If sequence Q has four terms: x, 12, y, and 20, is it true that the median of sequence Q is 16? (1) The average of sequence Q is 16. (2) x=16 As the sequence consists of 4 terms, so if the sequence is rearranged in increasing order the median will be average of the middle 2 terms. Hence the question is reduced to finding the middle 2 terms. (1) \(Average =\frac{(x + 12 + y + 20)}{4} = 16\) => \((x + y) = 32\) There are 3 possible combinations for x and y a) x < 12, then y > 20  Check for possible possible values if x = 1 then y = 33, if x = 11 then y = 21 This means that the middle 2 terms are 12 & 20. And the median = 16 b) x = 12, then y = 20 This also means that the middle 2 terms are 12 & 20. And the median = 16 c) 12 < x < 20, then 12 < y < 20 if x = 13, then y = 19 => \(Median = \frac{(13 + 19)}{2} = 16\) if x = 15, then y = 17 => \(Median = \frac{(15 + 17)}{2} = 16\) Hence Stmt I alone is sufficient (2) x = 16 => Q = 12, 16, y, 20 Now y can be any value and it can change the median. Hence Stmt II is not sufficient. Hi Bunuel, In this question, a "term" of the sequence was assumed to be an integer. Why can't a term of the sequence be a noninteger in this case?
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Re: If sequence Q has four terms: x, 12, y, and 20, is it true that the [#permalink]
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23 Mar 2017, 12:10
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[/quote] Hi Bunuel, In this question, a "term" of the sequence was assumed to be an integer. Why can't a term of the sequence be a noninteger in this case?[/quote] duahsolo : I don't think that is an assumption here. Let me try to explain with examples You can try incorporating noninteger values in the 1st and 3rd case mentioned by prasun9 case a : x < 12, then y > 20 x=10.5 (assuming) then y=32x = 21.5 median of Q> 10.5,12,20,21.5 is 16 , since ((12+20)/2))=16 case b : x=12 and y=20 Fixed value median of Q> 12,12,20,20 is 16, since ((12+20)/2))=16 case c : x > 12, then y < 20 x=13.5 (assuming) then y=32x = 18.5 median of Q> 12,13.5,18.5, 20 is 16, since ((13.5+18.5)/2))=16 Looking at these cases, I think we should see that the median will be either : 1) Arithmetic mean of 12 and 20 (i.e. if x and y are not between these two numbers on the number line), or 2) Arithmetic mean of x and y (if x and y are between 12 and 20 on the number lines) Try to note here that it can't be the case that one of x and y lie between 12 and 20, while the other does not For both these cases we would have a value of 16, since 12+20=32/2=16 and x+y=32, so (x+y)/2=16 Hence we know the median of the sequence Q with statement 1 alone

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Re: If sequence Q has four terms: x, 12, y, and 20, is it true that the [#permalink]
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24 Mar 2017, 06:33
himanshuv wrote: Hi Bunuel, In this question, a "term" of the sequence was assumed to be an integer. Why can't a term of the sequence be a noninteger in this case?[/quote] duahsolo : I don't think that is an assumption here. Let me try to explain with examples You can try incorporating noninteger values in the 1st and 3rd case mentioned by prasun9 case a : x < 12, then y > 20 x=10.5 (assuming) then y=32x = 21.5 median of Q> 10.5,12,20,21.5 is 16 , since ((12+20)/2))=16 case b : x=12 and y=20 Fixed value median of Q> 12,12,20,20 is 16, since ((12+20)/2))=16 case c : x > 12, then y < 20 x=13.5 (assuming) then y=32x = 18.5 median of Q> 12,13.5,18.5, 20 is 16, since ((13.5+18.5)/2))=16 Looking at these cases, I think we should see that the median will be either : 1) Arithmetic mean of 12 and 20 (i.e. if x and y are not between these two numbers on the number line), or 2) Arithmetic mean of x and y (if x and y are between 12 and 20 on the number lines) Try to note here that it can't be the case that one of x and y lie between 12 and 20, while the other does not For both these cases we would have a value of 16, since 12+20=32/2=16 and x+y=32, so (x+y)/2=16 Hence we know the median of the sequence Q with statement 1 alone[/quote] Hi himanshuv, Got it, +1 kudos
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