Question

The average of nine numbers is \(M\), and the average of three of these numbers is \(P\). If the average of the remaining six numbers is \(N\), which of the following must be true?

• A. M = N + P
• B. 2M = N + P
• C. 3M = 2N + P
• D. 3M = 2P + N

Hint:

Convert averages to totals: sum of 9 = 9M, sum of 3 = 3P, so sum of remaining 6 = 9M - 3P, and this divided by 6 equals N.

Answer 1

The Correct Option is C

Step 1: Think of M as a weighted average of the two sub-groups.
The 9 numbers split into a group of 3 (average $P$) and a group of 6 (average $N$). The overall average $M$ is a weighted average of $P$ and $N$, weighted by how many numbers are in each group.

Step 2: Write the weighted average formula.
\[ M = \frac{(3 \times P) + (6 \times N)}{3 + 6} \]

Step 3: Simplify the denominator.
\[ M = \frac{3P + 6N}{9} \]

Step 4: Clear the fraction.
Multiply both sides by 9:
$9M = 3P + 6N$.

Step 5: Divide through to simplify.
Divide every term by 3:
$3M = P + 2N$, which is the same as $3M = 2N + P$.

Step 6: Sanity check with real numbers.
Suppose the 3-number group is $\{3,3,3\}$ so $P=3$, and the 6-number group is $\{6,6,6,6,6,6\}$ so $N=6$. All 9 numbers together average $M = \frac{9+36}{9} = 5$. Check: $3M = 15$ and $2N+P = 12+3=15$. The formula holds.

Final Answer:
The required relation is $3M = 2N + P$. \[ \boxed{3M = 2N + P} \]