Each equation below contains a statement followed by Quantity I and Quantity II. You have to study the information along with question and compare the value derived from Quantity I and Quantity II, give the answer.

Given that $p + q + r = 15$ and $p
eq q
eq r$, compare the following quantities:

**Quantity I:** If $p$, $q$, and $r$ are positive integers, find the maximum value of $pqr$.

**Quantity II:** If $p$, $q$, and $r \geq -10$, find the maximum value of $pqr$.

Choose the correct option:

Question

Each equation below contains a statement followed by Quantity I and Quantity II. You have to study the information along with question and compare the value derived from Quantity I and Quantity II, give the answer.

Given that $p + q + r = 15$ and $p 
eq q 
eq r$, compare the following quantities:

**Quantity I:** If $p$, $q$, and $r$ are positive integers, find the maximum value of $pqr$.

**Quantity II:** If $p$, $q$, and $r \geq -10$, find the maximum value of $pqr$.

Choose the correct option:

Accepted Answer

**Solution:**

1. Given the equation:
   $$p + q + r = 15$$
   with $p 
eq q 
eq r$.

2. **Quantity I:** $p, q, r$ are positive integers.

- To maximize the product $pqr$ under the sum constraint, the numbers should be as close as possible.
   - Since $p, q, r$ are distinct positive integers, try values near $5$ (since $15/3 = 5$).
   - Possible distinct triples near 5 are $(4, 5, 6)$.
   - Calculate product: $4 	imes 5 	imes 6 = 120$.
   - Check other combinations:
     - $(3, 5, 7) = 105$ (less)
     - $(3, 6, 6)$ not allowed since $p 
eq q 
eq r$.
   - So, maximum product for Quantity I is $120$.

3. **Quantity II:** $p, q, r \geq -10$ (no restriction on integers or distinctness).

- Since $p, q, r$ can be negative or zero, consider maximizing product.
   - The sum is fixed at 15.
   - To maximize product, consider two negative numbers and one positive number (since product of two negatives is positive).
   - Try $p = -10$, $q = -10$, then $r = 15 - (-10) - (-10) = 15 + 20 = 35$.
   - Product: $(-10) 	imes (-10) 	imes 35 = 100 	imes 35 = 3500$.
   - This is much larger than 120.

4. Therefore,
   $$	ext{Quantity I} = 120 < 3500 = 	ext{Quantity II}$$

**Answer:** (b) Quantity I < Quantity II

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