jaspreetsra wrote:If Q is an odd number and the median of Q consecutive integers is 120, what is the largest of these integers?
1) (Q-1)/2 + 120
2) Q/2 + 119
3) Q/2 + 120
4) Q+119/2
5) Q+120/2

Let Q=3, implying that there are 3 consecutive integers.
Since the median is 120, the 3 integers are {119, 120, 121}.
The question stem asks for the largest integer: 121.
This is our target.
Now plug Q=3 into the answers to see which yields our target of 121.
Only A works:
(Q-1)/2 + 120 = (3-1)/2 + 120 = 121.

The correct answer is A.

jaspreetsra wrote:If Q is an odd number and the median of Q consecutive integers is 120, what is the largest of these integers?
1) (Q-1)/2 + 120
2) Q/2 + 119
3) Q/2 + 120
4) Q+119/2
5) Q+120/2

Didn't get this question. Help plz.

Another way to do it is by seeing the logic of it.

First you can look at the answer choices to see what they are looking for and see that they all involve using Q along with some number related to the median, 120.

Then figure out what makes sense, which is this.

We have Q consecutive integers. Since they are consecutive, they are all evenly spaced 1 apart from each other.

120 is the median and the number of integers, Q, is odd. So 120 is in the exact middle of the set, and the rest of the set is composed of an even number of integers, Q - 1, half of which are higher than 120 and the other half of which are lower than 120.

So in the set, other than 120 there are Q - 1 integers, and half of them are greater than 120. This means there are (Q - 1)/2 integers that are greater than 120.

This is already looking like one of the answer choices, but just to be sure that's the right choice let's finish the math.

How high is the highest one? Well we know that they are evenly spaced 1 apart, so to get to the highest one we can start at 120 and add the number of higher integers to get (Q - 1)2 + 120, and that exactly matches one of the answer choices.

Choose 1.

Holy moly Brent. That's genius.

jaspreetsra wrote:If Q is an odd number and the median of Q consecutive integers is 120, what is the largest of these integers?
1) (Q-1)/2 + 120
2) Q/2 + 119
3) Q/2 + 120
4) Q+119/2
5) Q+120/2

Didn't get this question. Help plz.

Let, Q=3

Median of 3 consecutive Integer{a, a+1, a+2} = a+2
i.e. a+1 = 120
i.e. a = 119
i.e. Largest = a+2 = 121

Substitute Q=3 in options,

Answer: Option A

Let's do something other than substitution, for a change of pace. (The other three answers are all substitution, which is fine but also kinda fast food math: instantly gratifying, but not satisfying or particularly good for you. )

Since Q is an odd number, we can write Q = 2*(some integer) + 1. It doesn't matter what that integer is, so let's call it k, and say that Q = 2k + 1.

Since Q = 2k + 1, we have 2k + 1 integers in our set. There will be k integers greater than the median, k integers less than the median, and 1 integer that's the median itself.

Since the median is 120, we know that the GREATEST integer in the set is k integers greater than 120, or 120 + k. The answers are in terms of Q, so let's fix this. We know 2k + 1 = Q, so k = (Q - 1)/2.

Hence the greatest integer is 120 + k, or 120 + (Q-1)/2. Voila!