The 15 Hardest SAT Math Questions Ever - Free Printable

Problem Analysis:

We are given the following information:
1. \( x \) is the average (arithmetic mean) of \( m \) and 9.
2. \( y \) is the average of \( 2m \) and 15.
3. \( z \) is the average of \( 3m \) and 18.
4. We need to find the average of \( x \), \( y \), and \( z \) in terms of \( m \).

Step-by-Step Solution:

#### Step 1: Express \( x \) in terms of \( m \)
The average of two numbers \( a \) and \( b \) is given by:
\[
\text{Average} = \frac{a + b}{2}
\]
For \( x \), the average of \( m \) and 9 is:
\[
x = \frac{m + 9}{2}
\]

#### Step 2: Express \( y \) in terms of \( m \)
For \( y \), the average of \( 2m \) and 15 is:
\[
y = \frac{2m + 15}{2}
\]

#### Step 3: Express \( z \) in terms of \( m \)
For \( z \), the average of \( 3m \) and 18 is:
\[
z = \frac{3m + 18}{2}
\]

#### Step 4: Find the average of \( x \), \( y \), and \( z \)
The average of three numbers \( x \), \( y \), and \( z \) is given by:
\[
\text{Average} = \frac{x + y + z}{3}
\]
Substitute the expressions for \( x \), \( y \), and \( z \):
\[
x + y + z = \frac{m + 9}{2} + \frac{2m + 15}{2} + \frac{3m + 18}{2}
\]

#### Step 5: Combine the fractions
Since all the terms have a common denominator of 2, we can combine them:
\[
x + y + z = \frac{(m + 9) + (2m + 15) + (3m + 18)}{2}
\]
Simplify the numerator:
\[
(m + 9) + (2m + 15) + (3m + 18) = m + 2m + 3m + 9 + 15 + 18 = 6m + 42
\]
Thus:
\[
x + y + z = \frac{6m + 42}{2}
\]

#### Step 6: Simplify the expression
Divide both terms in the numerator by 2:
\[
x + y + z = \frac{6m}{2} + \frac{42}{2} = 3m + 21
\]

#### Step 7: Find the average of \( x \), \( y \), and \( z \)
The average of \( x \), \( y \), and \( z \) is:
\[
\frac{x + y + z}{3} = \frac{3m + 21}{3}
\]
Simplify the fraction:
\[
\frac{3m + 21}{3} = m + 7
\]

Final Answer:

The average of \( x \), \( y \), and \( z \) in terms of \( m \) is:
\[
\boxed{B}
\]