Question

If the average of p numbers is q² and that of q numbers is p², then the average of (p+q) numbers is:

• A. \(\frac{p}{q}\)​
• B. \(pq\)
• C. \(p+q\)
• D. \(p−q\)

Answer 1

The Correct Option is B

Given that the average of \( p \) numbers is \( q^2 \), we can express this as:

\[\frac{\text{Sum of } p \text{ numbers}}{p} = q^2\]

This implies:

\[\text{Sum of } p \text{ numbers} = p \cdot q^2\]

Similarly, if the average of \( q \) numbers is \( p^2 \), we have:

\[\frac{\text{Sum of } q \text{ numbers}}{q} = p^2\]

This implies:

\[\text{Sum of } q \text{ numbers} = q \cdot p^2\]

To find the average of \( p + q \) numbers, calculate the total sum of all numbers:

\[ \text{Total sum of } (p+q) \text{ numbers} = p \cdot q^2 + q \cdot p^2 \]

The average is then given by:

\[\frac{p \cdot q^2 + q \cdot p^2}{p+q}\]

Simplify the numerator:

\[ p \cdot q^2 + q \cdot p^2 = pq(q + p) \]

The expression for the average becomes:

\[\frac{pq(q + p)}{p+q} = pq\]

Since \( p+q \neq 0 \), canceling \( p+q \) from numerator and denominator successfully yields:

\[ \frac{pq(q + p)}{p+q} = pq\]

Thus, the average of \( (p+q) \) numbers is: \(pq\)