Averages and Inequality Bounds

Let $g$ be the marks of a girl and $b$ be the marks of a boy. Total marks: $4g + 6b = 10 \times 24 = 240 \implies 2g + 3b = 120$.

Given condition: $b \le g \le 2b$.

Total marks of 2 girls and 6 boys, $T = 2g + 6b = (2g + 3b) + 3b = 120 + 3b$.

Using $2g = 120 - 3b$:

1. $g \ge b \implies \frac{120 - 3b}{2} \ge b \implies 120 - 3b \ge 2b \implies 5b \le 120 \implies b \le 24$.
2. $g \le 2b \implies \frac{120 - 3b}{2} \le 2b \implies 120 - 3b \le 4b \implies 7b \ge 120 \implies b \ge \frac{120}{7} \approx 17.14$.

Now, range of $T = 120 + 3b$:

• Minimum $T = 120 + 3\left(\frac{120}{7}\right) = 120 + 51.42 = 171.42$
• Maximum $T = 120 + 3(24) = 192$

Since $T$ takes integer values, the possible integer values are from $172$ to $192$ inclusive. Number of integer values $= 192 - 172 + 1 = 21$.