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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.

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.

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.

Re: If Q is an odd number and the median of Q consecutive [#permalink]
08 Jan 2014, 08:57

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.

Re: If Q is an odd number and the median of Q consecutive [#permalink]
08 Jan 2014, 09:19

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).

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

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.

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.

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

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

Kindly press "+1 Kudos" to appreciate

gmatclubot

Re: If Q is an odd number and the median of Q consecutive
[#permalink]
03 Sep 2014, 01:38

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