If Q is an odd number and the median of Q consecutive integers is 120

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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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

A very fast solution is to see what happens when Q = 1.
This means that there's only ONE integer in the set.
So, if the median of the set is 120, then the set is {120}, which means the greatest value in the set is 120

So the correct answer choice should yield 120 when Q = 1.

a) (1-1)/2 + 120 = 120 PERFECT!
b) 1/2 + 119 = some non-integer
c) 1/2 + 120 = some non-integer
d) (1+119)/2 = 60
e) (1+120)/2 = some non-integer

Since only answer choice A yield the correct output, it is the correct answer.

Cheers,
Brent

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

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.

Posted from my mobile device

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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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}

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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Hi reto,

You'll likely find it best to break Quant prompts down into 'pieces', so that you can deal with each piece one-at-a-time and simplify any complex-looking situations.

Here, the first thing we're told is "Q is an ODD number..." Now what does THAT mean? Here are some examples...

Q COULD be -5, -3, -1, 1, 3, 5, 7, etc.

Next, we're told "the median of Q consecutive integers is 120." 'Q' clearly refers to the 'Q' in the earlier part of the sentence; we're also dealing with the statistical term "median" (which means 'middle number of a group, when the numbers are ordered from least to greatest').

So let's list out some possibilities (it's worth noting that you can't have "negative consecutive integers", so the negative examples that I listed are not possible here):

IF...Q = 1....the median of 1 consecutive number is 120......so the 'group' is {120}
IF...Q = 3....the median of 3 consecutive numbers is 120....so the 'group' is {119, 120, 121}
IF...Q = 5....the median of 5 consecutive numbers is 120....so the 'group' is {118, 119, 120, 121, 122}
Etc.

GMAT assassins aren't born, they're made,
Rich

Really nice method Bunuel!

Dear Bunuel
Is this question properly formulated? "If Q is an odd number and the median of Q consecutive integers is 120, what is the largest of these integers?"
Shouldn't it say "If Q is a set of consecutive numbers and Q is is odd, then wat is the largest of..."?
I find the wording a little confusing.

The key here is that each question on the GMAT has exactly one correct answer.
So, it can't be the case that, when Q = 1, the correct answer is A, but when Q = 3, the correct answer is to something else.

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.

Hi, I know it's given that Q has to be an integer and basis that we eliminate B, C and D. But C gives us 121.5 cant we round this to 122 and this would be greater than A: 121?

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Hi woohoo921,

Yes - when TESTing VALUES, if only one of the 5 answers matches what you are looking for, then that answer is the correct answer. If more than one answer matches, then one of those answers is the correct answer (and the other(s) only sometimes match what you are looking for) - meaning that you would then have to TEST again with a different value (or values) to find the one answer that is ALWAYS a match.

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: Rich.C@empowergmat.com

woohoo921

Maybe more cooks in the kitchen than you need, especially since Rich and Brent have already done a great job answering...great enough that I'll simply add a few words to Rich's post (in red) juuuuuust in case it wasn't already clear.

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

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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