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

The correct answer is 397.

We are given that $p$, $q$, and $r$ are three natural numbers such that:
$$p+q+r=900$$
From this, we can express $q$ in terms of $p$ and $r$:
$$q=900-p-r$$

We are also given the constraint on $p$ in terms of $q$:
$$0.3q\le p\le 0.7q$$

Let us analyze the lower bound of this inequality:
$$p\ge 0.3q$$
Substituting $q=900-p-r$ into the inequality:
$$p\ge 0.3(900-p-r)$$
Multiplying both sides by 10 to clear decimals:
$$10p\ge 3(900-p-r)$$
$$10p\ge 2700-3p-3r$$
$$13p\ge 2700-3r$$
$$p\ge \frac{2700-3r}{13}$$

Now let us analyze the upper bound of the inequality:
$$p\le 0.7q$$
Substituting $q=900-p-r$ into the inequality:
$$p\le 0.7(900-p-r)$$
Multiplying both sides by 10:
$$10p\le 7(900-p-r)$$
$$10p\le 6300-7p-7r$$
$$17p\le 6300-7r$$
$$p\le \frac{6300-7r}{17}$$

We are given that $r$ is a perfect square whose value lies between 150 and 500.
The perfect squares in this range are:
$${13}^2=169,{14}^2=196,\dots ,{22}^2=484$$

1. Finding the Minimum Value of $p$:
To find the minimum possible value of $p$, we want to minimize its lower bound: $\frac{2700-3r}{13}$.
This occurs when $r$ is as large as possible. The largest perfect square in the given range is $r=484$.
Substituting $r=484$ into the lower bound equation:
$$p\ge \frac{2700-3(484)}{13}=\frac{2700-1452}{13}=\frac{1248}{13}=96$$
Thus, the minimum possible value of $p$ is $p_{\text{min}}=96$.
(We can verify that if $p=96$ and $r=484$, then $q=900-96-484=320$, which is a natural number and satisfies all given conditions).

2. Finding the Maximum Value of $p$:
To find the maximum possible value of $p$, we want to maximize its upper bound: $\frac{6300-7r}{17}$.
This occurs when $r$ is as small as possible. The smallest perfect square in the given range is $r=169$.
Substituting $r=169$ into the upper bound equation:
$$p\le \frac{6300-7(169)}{17}=\frac{6300-1183}{17}=\frac{5117}{17}\approx 301.0$$
Since $p$ must be a natural number, the maximum possible value is $p_{\text{max}}=301$.
(We can verify that if $p=301$ and $r=169$, then $q=900-301-169=430$, which is a natural number and satisfies all given conditions).

Conclusion:
The sum of the maximum and minimum possible values of $p$ is:
$$p_{\text{max}}+p_{\text{min}}=301+96=397$$