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

Given: $p + q + r = 900$, where $p, q, r$ are natural numbers, $r$ is a perfect square with $150 < r < 500$, and $0.3q \leq p \leq 0.7q$.

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Perfect squares between 150 and 500:

$13^2 = 169$, $14^2 = 196$, ..., $22^2 = 484$

Smallest $r = 169$ and Largest $r = 484$

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Since $p + q + r = 900$, we get $q = 900 - r - p$.

From $p \leq 0.7q$:

$p \leq 0.7(900 - r - p)$

$p \leq 630 - 0.7r - 0.7p$

$1.7p \leq 630 - 0.7r$

$p \leq \dfrac{7(900 - r)}{17}$

From $p \geq 0.3q$:

$p \geq 0.3(900 - r - p)$

$p \geq 270 - 0.3r - 0.3p$

$1.3p \geq 270 - 0.3r$

$p \geq \dfrac{3(900 - r)}{13}$

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To maximize $p$, use $p \leq \dfrac{7(900 - r)}{17}$ with the smallest $r = 169$:

$p_{max} = \dfrac{7 \times 731}{17}$

$= \dfrac{5117}{17}$

$= 301$

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To minimize $p$, use $p \geq \dfrac{3(900 - r)}{13}$ with the largest $r = 484$:

$p_{min} = \dfrac{3 \times 416}{13}$

$= \dfrac{1248}{13}$

$= 96$

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$p_{max} + p_{min} = 301 + 96 = 397$