BTGModeratorVI wrote: ↑Sat Aug 08, 2020 7:11 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
Answer:
A
Source: Official guide
Solution:
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
