In an examination, the average marks of 4 girls and 6 boys is 24. Each of the girls has the same marks while each of the boys has the same marks. If the marks of any girl are at most double the marks of any boy, but not less than the marks of any boy, then the number of possible distinct integer values of the total marks of 2 girls and 6 boys is:

Question

Accepted Answer

Let the marks of each girl be $$ g $$ and the marks of each boy be $$ b $$.

#### Step 1: Total Marks Relation  
The average marks of 4 girls and 6 boys is given as 24:  
$$
\frac{4g + 6b}{10} = 24
$$  
Simplify:  
$$
4g + 6b = 240
$$  
Divide by 2:  
$$
2g + 3b = 120 \quad 	ext{(1)}.
$$

#### Step 2: Constraints on $$ g $$ and $$ b $$  
It is given that the marks of any girl are at most double the marks of any boy, but not less than the marks of any boy:  
$$
b \leq g \leq 2b.
$$

#### Step 3: Express Total Marks for Two Girls and Six Boys  
The total marks for two girls and six boys is:  
$$
2g + 6b = (2g + 3b) + (3b).
$$  
From equation (1):  
$$
2g + 6b = 120 + 3b.
$$

#### Step 4: Integer Values for $$ b $$  
For $$ g $$ to be an integer, $$ b $$ must satisfy equation (1). Rearrange equation (1):  
$$
g = \frac{120 - 3b}{2}.
$$  
For $$ g $$ to be an integer, $$ (120 - 3b) $$ must be even. This means $$ b $$ must be even.

#### Step 5: Range of Values for $$ b $$ and Constraints on $$ g $$  
From the constraint $$ g \leq 2b $$, substitute for $$ g $$:  
$$
\frac{120 - 3b}{2} \leq 2b.
$$  
Simplify:  
$$
120 - 3b \leq 4b.
$$  
$$
120 \leq 7b.
$$  
$$
b \geq \frac{120}{7} \approx 17.14.
$$

From the constraint $$ b \leq g $$, substitute for $$ g $$:  
$$
\frac{120 - 3b}{2} \geq b.
$$  
Simplify:  
$$
120 - 3b \geq 2b.
$$  
$$
120 \geq 5b.
$$  
$$
b \leq \frac{120}{5} = 24.
$$

Thus, possible values for $$ b $$ are even integers from **18 to 24** inclusive.

#### Step 6: Calculate Distinct Values for Total Marks  
For each valid value of $$ b $$, calculate total marks:  
$$
2g + 6b = (120) + (3b).
$$

Since there are **7 distinct values** for $$ b $$(18 to 24), there are **21 distinct values** for total marks when considering all combinations.

Thus, the number of possible distinct integer values is **21**.

Answer

Answer

Answer