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.

Step-by-step Explanation:

We are given that $p$, $q$, and $r$ are natural numbers such that:

$p+q+r=900$

From this equation, we can express $q$ in terms of $p$ and $r$:

$q=900-r-p$

We are given the inequality for $p$:

$0.3q\le p\le 0.7q$

1. Finding the Upper Bound for $p$:

Using the right-hand inequality $p\le 0.7q$ and substituting $q=900-r-p$:

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

$p\le 630-0.7r-0.7p$

$1.7p\le 630-0.7r$

Multiplying both sides by 10 gives:

$17p\le 6300-7r$

$p\le \frac{6300-7r}{17}$

To maximize $p$, we must minimize $r$. We are given that $r$ is a perfect square lying between 150 and 500. The smallest perfect square in this range is $r={13}^2=169$.

Substituting $r=169$:

$p_{\text{max}}\le \frac{6300-7\left(169\right)}{17}=\frac{6300-1183}{17}=\frac{5117}{17}=301$

Thus, the maximum possible value of $p$ is $p_{\text{max}}=301$.

2. Finding the Lower Bound for $p$:

Using the left-hand inequality $0.3q\le p$ and substituting $q=900-r-p$:

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

$270-0.3r-0.3p\le p$

$270-0.3r\le 1.3p$

Multiplying both sides by 10 gives:

$2700-3r\le 13p$

$p\ge \frac{2700-3r}{13}$

To minimize $p$, we must maximize $r$. The largest perfect square strictly less than 500 is $r={22}^2=484$.

Substituting $r=484$:

$p_{\text{min}}\ge \frac{2700-3\left(484\right)}{13}=\frac{2700-1452}{13}=\frac{1248}{13}=96$

Thus, the minimum possible value of $p$ is $p_{\text{min}}=96$.

3. Sum of Maximum and Minimum Values:

$\text{Sum}=p_{\text{max}}+p_{\text{min}}=301+96=397$