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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
Posts: 173
If Q is an odd number and the median of Q consecutive  [#permalink]

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07 Dec 2012, 04:47
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35% (medium)

Question Stats:

75% (01:49) correct 25% (02:04) wrong based on 2857 sessions

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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: 58445
Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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07 Dec 2012, 04:52
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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

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.

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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, 10:19
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3
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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Joined: 13 Aug 2012
Posts: 401
Concentration: Marketing, Finance
GPA: 3.23
Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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19 Dec 2012, 00:28
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!

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Posts: 695
Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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29 Dec 2012, 20:08
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1
Really nice method Bunuel!
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Joined: 11 Jan 2013
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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, 08:09
1
Bunuel 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.

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: 58445
Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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25 Jun 2013, 10:18
2
1
Revenge2013 wrote:
Bunuel 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.

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, 11:22
Ah thanks - was indeed a stupid question, but got it now.

Thanks a lot!
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Joined: 21 Sep 2013
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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, 09:57
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: 142
Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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13 Jan 2014, 04:41
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, 12:23
2
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: 11
Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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04 Feb 2014, 13:15
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: 58445
Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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05 Feb 2014, 01:55
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, 11:14
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
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Posts: 58445
Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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08 Feb 2014, 11:20
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 52516 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, 21:27
1
Bunuel 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.

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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28 Feb 2014, 00:48
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, 03:35
2
davidfrank wrote:
Bunuel 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.

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, 13:30
[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, 02:38
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
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Re: If Q is an odd number and the median of Q consecutive   [#permalink] 03 Sep 2014, 02:38

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