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

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BrentGMATPrepNow
This is very helpful! How do we know not to test another number (e.g., when Q=3)? Overall, I am confused on the rules pertaining to how to know if you need to test more smart numbers, or do you only need to choose one set of smart numbers if they satisfy all the conditions? Thank you in advance

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.

Testing cases is the best option here, but here is the algebra for those who are interested:
S = smallest integer, L = largest integer, Q=number of terms in a set (and we're given that it's odd).

In any set of consecutive integers, the mean is equal to the median and either can be found by finding the average of the smallest and the largest integer. Since median is 120, it follows that:
\(\frac{(S+L)}{2} = 120\)
\(S+L = 240\)
\(S = 240-L\)

The number of terms (inclusive) in a set in which the difference between each term is constant is given by:
\(\frac{(L-S)}{(Difference)} + 1\)
Since we have a set of consecutive integers, the difference between each term is 1 and so:

\(Q=(L-S)+1 \)
\(L = Q-1+S \) (Plug in for S using above)
\(L=Q-1+240-L\)
\(2L=Q-1+240\)
\(L=\frac{(Q-1)}{2} + 120\)

Thank you! To confirm, the only time you would need to choose another smart number or set of smart number(s) to test is when you have more than one answer choice that has the same value? Otherwise, if you get a unique value, you can just move on?

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.

If we were to derive it.
nth term in Arithmetic Progression is

\(t_n = a + (n-1)d\)

where a is first term, n is number of the term in the sequence, d is difference between two successive terms and \(t_n\) is the actual value of nth term

120 is the median and Q is given as odd. So it is the \(\frac{(Q+1)}{2}\) th term. Here d is 1 since numbers are consecutive.

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

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

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

We will use this value of a in the next step.

Now the Qth term which will be the largest term in the sequence

\(= a + (Q-1) 1\)

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

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

Ans: A