Question

If $x$ is the average (arithmetic mean) of $m$ and $9,y$ is the average of $2m$ and $15$ and $z$ is the average of $3m$ and $18$, what is the average of $x,y$ and $z$ in terms of $m$ ?

• A. $m+6$
• B. $m+7$
• C. $2m+14$
• D. $3m+21$

Answer 1

Answer: (B)

$m+7$

Given, $x=\frac{m+9}{2}$
$y=\frac{2m+15}{2}$
$z=\frac{3m+18}{2}$
Average of $x,y,z=\frac{\frac{m+9}{2}+\frac{2m+15}{2}+\frac{3m+18}{2}}{3}=\frac{\frac{6m+42}{2}}{3}=m+7$