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# If sequence Q has four terms: x, 12, y, and 20, is it true that the

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Senior Manager
Joined: 21 Oct 2013
Posts: 435
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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85% (hard)

Question Stats:

45% (02:19) correct 55% (01:14) wrong based on 92 sessions

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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
1
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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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 non-integer 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
1
[/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 non-integer 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 non-integer values in the 1st and 3rd case mentioned by prasun9

case a : x < 12, then y > 20
x=10.5 (assuming)
then y=32-x = 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=32-x = 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 non-integer 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 non-integer values in the 1st and 3rd case mentioned by prasun9

case a : x < 12, then y > 20
x=10.5 (assuming)
then y=32-x = 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=32-x = 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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Re: If sequence Q has four terms: x, 12, y, and 20, is it true that the   [#permalink] 24 Mar 2017, 06:33
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