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# If Q is an odd number and the median of Q consecutive

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Manager
Joined: 02 Dec 2012
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If Q is an odd number and the median of Q consecutive  [#permalink]

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07 Dec 2012, 03:47
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If Q is an odd number and the median of Q consecutive integers is 120, what is the largest of these integers?

(A) (Q - 1)/2 + 120
(B) Q/2 + 119
(C) Q/2 + 120
(D) (Q + 119)/2
(E) (Q + 120)/2
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Joined: 02 Sep 2009
Posts: 65785
Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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07 Dec 2012, 03:52
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Walkabout wrote:
If Q is an odd number and the median of Q consecutive integers is 120, what is the largest of these integers?

(A) (Q - 1)/2 + 120
(B) Q/2 + 119
(C) Q/2 + 120
(D) (Q + 119)/2
(E) (Q + 120)/2

Consider the easiest case, say Q=3, then;
Set = {119, 120, 121};
The largest integer = 121.

Now, plug Q=3 into the answers to see which yields 121. Only answer choice A works. Notice that we don't really need to plug for B, C, or E, since these options do not yield an integer value for any odd value of Q.

Answer: A.
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Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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08 Jan 2014, 09:19
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If 120 is the middle number,then there will be Q-1/2 numbers both before and after 120.Therefore,largest number will be Q-1/2+120.
Smallest number will be 120-(Q-1/2).

Posted from my mobile device
##### General Discussion
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Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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18 Dec 2012, 23:28
1
Walkabout wrote:
If Q is an odd number and the median of Q consecutive integers is 120, what is the largest of these integers?

(A) (Q - 1)/2 + 120
(B) Q/2 + 119
(C) Q/2 + 120
(D) (Q + 119)/2
(E) (Q + 120)/2

Let Q = 3
Set = {119,120,121} which makes 121 the largest integer

(A) (3-1)/2 + 120 = 121
(B) and (C) yields a decimal. ELIMATE!
(D) 122/2 = 61 ELIMINATE!
(E) yields a decimal. ELIMINATE!

Answer: A
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Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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29 Dec 2012, 19:08
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Really nice method Bunuel!
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Joined: 11 Jan 2013
Posts: 14
Location: United States
Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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25 Jun 2013, 07:09
1
Bunuel wrote:
Walkabout wrote:
If Q is an odd number and the median of Q consecutive integers is 120, what is the largest of these integers?

(A) (Q - 1)/2 + 120
(B) Q/2 + 119
(C) Q/2 + 120
(D) (Q + 119)/2
(E) (Q + 120)/2

Consider the easiest case, say Q=3, then;
Set = {119, 120, 121};
The largest integer = 121.

Now, plug Q=3 into the answers to see which yields 121. Only answer choice A works. Notice that we don't really need to plug for B, C, or E, since these options do not yield an integer value for any odd value of Q.

Answer: A.

Sorry, if this is a really stupid question. I am probably missing something, but I do not understand the logic of this approach:

If 120 is the median (i.e. in my mind the middle number of the set) how can 121 then be the largest number? I am asking, since this does not make sense to me, yet and thus I would have never come up with such a good shortcut.

Many thanks
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Posts: 65785
Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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25 Jun 2013, 09:18
2
1
Revenge2013 wrote:
Bunuel wrote:
Walkabout wrote:
If Q is an odd number and the median of Q consecutive integers is 120, what is the largest of these integers?

(A) (Q - 1)/2 + 120
(B) Q/2 + 119
(C) Q/2 + 120
(D) (Q + 119)/2
(E) (Q + 120)/2

Consider the easiest case, say Q=3, then;
Set = {119, 120, 121};
The largest integer = 121.

Now, plug Q=3 into the answers to see which yields 121. Only answer choice A works. Notice that we don't really need to plug for B, C, or E, since these options do not yield an integer value for any odd value of Q.

Answer: A.

Sorry, if this is a really stupid question. I am probably missing something, but I do not understand the logic of this approach:

If 120 is the median (i.e. in my mind the middle number of the set) how can 121 then be the largest number? I am asking, since this does not make sense to me, yet and thus I would have never come up with such a good shortcut.

Many thanks

We are told that there are Q consecutive integers in a set and Q is odd. We are also told that the median of the set is 120.

Now, say Q=3=odd. So, we have that the median of 3 consecutive integers is 120.
Question: what is the largest of these 3 integers? The set in this case must be {119, 120, 121} (3 consecutive integers with median of 120), so the largest of these 3 integers is 121.

Hope it's clear.
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Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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25 Jun 2013, 10:22
Ah thanks - was indeed a stupid question, but got it now.

Thanks a lot!
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Joined: 21 Sep 2013
Posts: 19
Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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08 Jan 2014, 08:57
3
I came across an alternate method.
Since Q is odd, therefore Q/2 is a fraction this options B and C are eliminated.( questions talks about integers only).E option is nullified since Q+120 is odd and the result will be a fraction on division with respect to 2. Left A and D just put Q =1,3 or any odd number D gives a value less than 120 therefore it cannot be the largest.

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Joined: 12 Jan 2013
Posts: 137
Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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13 Jan 2014, 03:41
1
Walkabout wrote:
If Q is an odd number and the median of Q consecutive integers is 120, what is the largest of these integers?

(A) (Q - 1)/2 + 120
(B) Q/2 + 119
(C) Q/2 + 120
(D) (Q + 119)/2
(E) (Q + 120)/2

Pick numbers:

Q = 5, thus we have the numbers 118, 119, 120, 121, 122..

We need to pick an answer that yields 122... A: $$\frac{(5-1)}{2} + 120 = 122$$, so A is correct
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Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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28 Jan 2014, 11:23
2
Walkabout wrote:
If Q is an odd number and the median of Q consecutive integers is 120, what is the largest of these integers?

(A) (Q - 1)/2 + 120
(B) Q/2 + 119
(C) Q/2 + 120
(D) (Q + 119)/2
(E) (Q + 120)/2

Average of Q consecutive ints in a list = average of first and the last ints in the list
also for consecutive int mean = median
F = First Number
l = Last Number

avg = \frac{(F + L)}{2}

now L = F + (Q-1)

\frac{(F + F +(Q-1))}{2} = 120

F = 120 - \frac{(Q-1)}{2}

L = 120 - \frac{(Q-1)}{2} + (Q-1) = 120 + \frac{(Q-1)}{2}
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Joined: 04 Feb 2014
Posts: 10
Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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04 Feb 2014, 12:15
I understand the question if you put it in parenthesis but how would you know which answers need parenthesis and which ones dont
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Posts: 65785
Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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05 Feb 2014, 00:55
kedusei wrote:
I understand the question if you put it in parenthesis but how would you know which answers need parenthesis and which ones dont

Not sure I follow... The answer choices are given, you don't need to put/add anything there...
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Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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08 Feb 2014, 10:14
Bunuel wrote:
kedusei wrote:
I understand the question if you put it in parenthesis but how would you know which answers need parenthesis and which ones dont

Not sure I follow... The answer choices are given, you don't need to put/add anything there...

These choices have parenthesis but the choices in the book dont
Math Expert
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Posts: 65785
Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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08 Feb 2014, 10:20
kedusei wrote:
Bunuel wrote:
kedusei wrote:
I understand the question if you put it in parenthesis but how would you know which answers need parenthesis and which ones dont

Not sure I follow... The answer choices are given, you don't need to put/add anything there...

These choices have parenthesis but the choices in the book dont

Below is a screenshot of this question:
Attachment:

Untitled.png [ 14.82 KiB | Viewed 65145 times ]
Which is the same as the options in original post:
(A) (Q - 1)/2 + 120
(B) Q/2 + 119
(C) Q/2 + 120
(D) (Q + 119)/2
(E) (Q + 120)/2

Can you please tell me what is confusing?
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Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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27 Feb 2014, 20:27
1
Bunuel wrote:
Walkabout wrote:
If Q is an odd number and the median of Q consecutive integers is 120, what is the largest of these integers?

(A) (Q - 1)/2 + 120
(B) Q/2 + 119
(C) Q/2 + 120
(D) (Q + 119)/2
(E) (Q + 120)/2

Consider the easiest case, say Q=3, then;
Set = {119, 120, 121};
The largest integer = 121.

Now, plug Q=3 into the answers to see which yields 121. Only answer choice A works. Notice that we don't really need to plug for B, C, or E, since these options do not yield an integer value for any odd value of Q.

Answer: A.

Little eager....might be stupid too but can you assume q=1. This brings me to an important question can you have a median in a set that has only one element. I know this might sound stupid but is this possible. Further if that being so my largest number through option 1 becomes 120. But that is wrong. So am I missing something or the question has been set up assuming odd number starts with 3.

Bunuel request you to please clarify this.
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Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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27 Feb 2014, 23:48
Walkabout wrote:
If Q is an odd number and the median of Q consecutive integers is 120, what is the largest of these integers?

(A) (Q - 1)/2 + 120
(B) Q/2 + 119
(C) Q/2 + 120
(D) (Q + 119)/2
(E) (Q + 120)/2

Let us say that the numbers are {119, 120, 121}
Q = 3

(A) (3 - 1)/2 + 120 = 121
(B) 119 + 1.5
(C) 1.5 + 120
(D) (122)/2
(E) (123)/2

Hence option A is the answer
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Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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28 Feb 2014, 02:35
2
davidfrank wrote:
Bunuel wrote:
Walkabout wrote:
If Q is an odd number and the median of Q consecutive integers is 120, what is the largest of these integers?

(A) (Q - 1)/2 + 120
(B) Q/2 + 119
(C) Q/2 + 120
(D) (Q + 119)/2
(E) (Q + 120)/2

Consider the easiest case, say Q=3, then;
Set = {119, 120, 121};
The largest integer = 121.

Now, plug Q=3 into the answers to see which yields 121. Only answer choice A works. Notice that we don't really need to plug for B, C, or E, since these options do not yield an integer value for any odd value of Q.

Answer: A.

Little eager....might be stupid too but can you assume q=1. This brings me to an important question can you have a median in a set that has only one element. I know this might sound stupid but is this possible. Further if that being so my largest number through option 1 becomes 120. But that is wrong. So am I missing something or the question has been set up assuming odd number starts with 3.

Bunuel request you to please clarify this.

The median of a single element set is that number itself. For example, the median of {-11} is -11.

Next, you can consider Q to be 1, in this case the set is {120} and the largest integer is 120. Substituting Q=1 into the options gives A as the answer.

Hope it's clear.
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Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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01 Sep 2014, 12:30
[quote="AKG1593"]If 120 is the middle number,then there will be Q-1/2 numbers both before and after 120.Therefore,largest number will be Q-1/2+120.
Smallest number will be 120-(Q-1/2).

Posted from my mobile device [/quote

Can you please elaborate on this? How did you get Q-1/2? Versus Q*1/2?
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Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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03 Sep 2014, 01:38
Chin926926 wrote:
AKG1593 wrote:
If 120 is the middle number,then there will be Q-1/2 numbers both before and after 120.Therefore,largest number will be Q-1/2+120.
Smallest number will be 120-(Q-1/2).

Posted from my mobile device [/quote

Can you please elaborate on this? How did you get Q-1/2? Versus Q*1/2?

Just refer to Bunuel's method at the top. Plugging in the numbers of example 119, 120, 121 should give the perfect result
Re: If Q is an odd number and the median of Q consecutive   [#permalink] 03 Sep 2014, 01:38

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