Question

If Q is an odd number and the median of Q consecutive integers is 120, what is the largest of these integers?

• A. $\frac{Q-1}{2}+120$
• B. $\frac{Q}{2}+119$
• C. $\frac{Q}{2}+120$
• D. $\frac{Q+119}{2}$
• E. $\frac{Q+120}{2}$

Answer 1

Answer: (A)

$\frac{Q-1}{2}+120$

For example, if $Q=3$, we would have three integers$:119,120,121.$

The largest integer would be $121.$

We would then plug $Q=3$ into each answer choice, to see which answer choice gives us a result of $121.$

Option $A:\frac{Q-1}{2}+120$

$\Rightarrow$ $\frac{3-1}{2}+120$

$\Rightarrow$ $\frac{2}{2}+120$

$\Rightarrow$ $1+120$

$\Rightarrow$ $121$

$\therefore$ Correct answer is optiom $A$