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Consecutive integers - complicated

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Consecutive integers - complicated

by AndreiGMAT » Sun May 01, 2016 5:38 pm
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

I have no clue what this question is stating or asking and how to interpret the information.

Please, help
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by GMATGuruNY » Sun May 01, 2016 7:02 pm
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.
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by [email protected] » Sun May 01, 2016 11:03 pm
Hi AndreiGMAT,

Mitch's approach (TESTing VALUES) is a great way to answer this question; it's exactly how I would have approached this prompt.

This prompt is built around an interesting Number Property, which you could have used to answer the question without doing much math at all.

We're told that Q is an ODD number, we're dealing with CONSECUTIVE INTEGERS and we're asked for the LARGEST INTEGER..

Take a look at answer B. Since Q is ODD, will answer B EVER be an integer???
The answer is NO (we'd end up with a fraction, so B can't be the answer.

This same rule holds true for answer C and E. So, we've eliminated B, C and E.

Now look at D; we're taking (Q+119) and cutting it IN HALF. That math doesn't seem like it would produce the LARGEST INTEGER that we're looking for. Eliminate D.

The answer must be A. It would certainly give us an integer: (ODD - 1)/2 = INTEGER. Add THAT integer to 120 and you either get 120 or a bigger integer.

Final Answer: A

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by Brent@GMATPrepNow » Mon May 02, 2016 6:39 am
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
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 answer.

Cheers,
Brent
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by Jeff@TargetTestPrep » Mon Apr 09, 2018 3:44 pm
AndreiGMAT 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
We are given that Q is an ODD NUMBER and that the median of Q CONSECUTIVE INTEGERS is 120. Let's choose a convenient number for Q, such as 3. We can now say:

The median of 3 consecutive integers is 120. Since 120 is the MEDIAN, or middle number of these integers, our 3 integers are the following:

119, 120, 121

Let's now test each answer choice to see which gives us 121:

A) (Q-1)/2 + 120

(3-1)/2 + 120 = 1 + 120 = 121

This IS equal to 121.

We have found our answer.

Alternate solution:

Since Q is an odd number, we see that none of the choices B, C and E will yield an integer; so we can reject those answer choices. This leaves us with either choice A or D as the correct answer. By picking a small number for Q, e.g., Q = 3, we see that choice A will yield (3-1)/2 + 120 = 1 + 120 = 121 and choice D will yield (3+119)/2 = 122/2 = 61. Of course, the largest integer in the set has to be greater than the median of 120, so only choice A can be the right answer.

Answer: A

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