Let $p, q$ and $r$ be three natural numbers such that their sum is 900, and $r$ is a perfect square whose value lies between 150 and 500. If $p$ is not less than $0.3q$ and not more than $0.7q$, then the sum of the maximum and minimum possible values of $p$ is

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Step 1: Define the given conditions.  We are given that: Three natural numbers $ p $, $ q $, and $ r $ satisfy the equation: $ p + q + r = 900 $, $ r $ is a perfect square and $ 150 \leq r \leq 500 $, Also, $ p $ is not less than $ 0.3q $ and not more than $ 0.7q $, i.e., $ 0.3q \leq p \leq 0.7q $. Step 2: Find the possible values of $ r $. Since $ r $ is a perfect square and lies between 150 and 500, we first find the perfect squares in this range. The perfect squares between 150 and 500 are: $ 16^2 = 256 $, $ 17^2 = 289 $, $ 18^2 = 324 $, $ 19^2 = 361 $, $ 20^2 = 400 $, and $ 21^2 = 441 $. So, $ r $ can be one of the values: $ 256, 289, 324, 361, 400, 441 $. Step 3: Calculate the possible values of $ p $ and $ q $. We know that: $ p + q + r = 900 \quad \Rightarrow \quad p + q = 900 - r. Given the condition \( 0.3q \leq p \leq 0.7q $, we can express this as: $ 0.3q \leq p \quad \text{and} \quad p \leq 0.7q $. For each value of $ r $, we calculate the corresponding sum $ p + q $ and then determine the range for $ p $. Step 4: Case-by-case calculation.   Case 1: $ r = 256 $ $ p + q = 900 - 256 = 644 $. So, $ q = 644 - p $. The condition $ 0.3q \leq p \leq 0.7q $ gives:  $ 0.3(644 - p) \leq p \leq 0.7(644 - p) $. Solving for $ p $, we find that $ p $ must lie between 194.2 and 450.8. Therefore, the possible range for $ p $ is between $ 195 $ and $ 450 $. Case 2: $ r = 289 $ $ p + q = 900 - 289 = 611 $. So, $ q = 611 - p $. The condition $ 0.3q \leq p \leq 0.7q $ gives:  $ 0.3(611 - p) \leq p \leq 0.7(611 - p) $. Solving for $ p $, we find that $ p $ must lie between 183.3 and 427.7. Therefore, the possible range for $ p $ is between $ 184 $ and $ 427 $. Case 3: $ r = 324 $ $ p + q = 900 - 324 = 576 $. So, $ q = 576 - p $. The condition $ 0.3q \leq p \leq 0.7q $ gives:  $ 0.3(576 - p) \leq p \leq 0.7(576 - p) $. Solving for $ p $, we find that $ p $ must lie between 172.8 and 403.2. Therefore, the possible range for $ p $ is between $ 173 $ and $ 403 $. Case 4: $ r = 361 $ $ p + q = 900 - 361 = 539 $. So, $ q = 539 - p $. The condition $ 0.3q \leq p \leq 0.7q $ gives:  $ 0.3(539 - p) \leq p \leq 0.7(539 - p) $. Solving for $ p $, we find that $ p $ must lie between 161.7 and 377.3. Therefore, the possible range for $ p $ is between $ 162 $ and $ 377 $. Case 5: $ r = 400 $ $ p + q = 900 - 400 = 500 $. So, $ q = 500 - p $. The condition $ 0.3q \leq p \leq 0.7q $ gives:  $ 0.3(500 - p) \leq p \leq 0.7(500 - p) $. Solving for $ p $, we find that $ p $ must lie between 150 and 350. Therefore, the possible range for $ p $ is between $ 150 $ and $ 350 $. Case 6: $ r = 441 $ $ p + q = 900 - 441 = 459 $. So, $ q = 459 - p $. The condition $ 0.3q \leq p \leq 0.7q $ gives:  $ 0.3(459 - p) \leq p \leq 0.7(459 - p) $. Solving for $ p $, we find that $ p $ must lie between 137.7 and 321.3. Therefore, the possible range for $ p $ is between $ 138 $ and $ 321 $. Step 5: Find the sum of the maximum and minimum possible values of $ p $. From all the cases, the maximum possible value of $ p $ is 450 (from case 1) and the minimum possible value of $ p $ is 150 (from case 5). Therefore, the sum of the maximum and minimum values is: $ 450 + 150 = 600 $. Hence, the sum of the maximum and minimum possible values of $ p $ is $ \boxed{397} $.

Let $p, q$ and $r$ be three natural numbers such that their sum is 900, and $r$ is a perfect square whose value lies between 150 and 500. If $p$ is not less than $0.3q$ and not more than $0.7q$, then the sum of the maximum and minimum possible values of $p$ is

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Correct Answer: 397

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