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?
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?
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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.
Updated On: Jul 14, 2026
- M = N + P
- 2M = N + P
- 3M = 2N + P
- 3M = 2P + N
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The Correct Option is C
Solution and Explanation
Step 1: Recall the link between average and total.
For any group of numbers, sum \(=\) average \(\times\) count of numbers. This lets us convert averages into totals that can be added or subtracted.
Step 2: Find the total of all nine numbers.
\[ \text{Sum of all 9 numbers} = 9M \]
Step 3: Find the total of the three numbers with average P.
\[ \text{Sum of these 3 numbers} = 3P \]
Step 4: Find the total of the remaining six numbers.
The remaining 6 numbers' sum is what is left after removing the 3-number group from the full 9-number group.
\[ \text{Sum of remaining 6} = 9M - 3P \]
Step 5: Use that this remaining sum's average is N.
\[ N = \frac{9M - 3P}{6} \]
Step 6: Solve for the relationship between M, N and P.
Multiply both sides by 6: \( 6N = 9M - 3P \).
Rearrange: \( 9M = 6N + 3P \).
Divide everything by 3: \( 3M = 2N + P \).
Step 7: Why the other options are wrong.
Option A and B use the wrong weighting between the two subgroups (3 numbers vs 6 numbers), since a straight sum or average of N and P ignores that the groups are of different sizes.
Option D swaps the coefficients of N and P; since the remaining group of 6 numbers (associated with N) is twice the size of the 3-number group (associated with P), it is N that should carry the larger coefficient, not P.
Final Answer:
The correct relation is \(3M = 2N + P\).
\[ \boxed{3M = 2N + P} \]
For any group of numbers, sum \(=\) average \(\times\) count of numbers. This lets us convert averages into totals that can be added or subtracted.
Step 2: Find the total of all nine numbers.
\[ \text{Sum of all 9 numbers} = 9M \]
Step 3: Find the total of the three numbers with average P.
\[ \text{Sum of these 3 numbers} = 3P \]
Step 4: Find the total of the remaining six numbers.
The remaining 6 numbers' sum is what is left after removing the 3-number group from the full 9-number group.
\[ \text{Sum of remaining 6} = 9M - 3P \]
Step 5: Use that this remaining sum's average is N.
\[ N = \frac{9M - 3P}{6} \]
Step 6: Solve for the relationship between M, N and P.
Multiply both sides by 6: \( 6N = 9M - 3P \).
Rearrange: \( 9M = 6N + 3P \).
Divide everything by 3: \( 3M = 2N + P \).
Step 7: Why the other options are wrong.
Option A and B use the wrong weighting between the two subgroups (3 numbers vs 6 numbers), since a straight sum or average of N and P ignores that the groups are of different sizes.
Option D swaps the coefficients of N and P; since the remaining group of 6 numbers (associated with N) is twice the size of the 3-number group (associated with P), it is N that should carry the larger coefficient, not P.
Final Answer:
The correct relation is \(3M = 2N + P\).
\[ \boxed{3M = 2N + P} \]
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