If the ratio of boys to girls in a class is B and the ratio of girls to boys is G, then 3 (B+ G) is
equal to 3
(b) less than 3
(c ) more than 3
(d) less than ` (1)/(3)`

Answer

To solve the problem, we need to analyze the given ratios of boys to girls (B) and girls to boys (G). Let's break it down step by step.

### Step 1: Define the Ratios
Let the number of boys be \( X \) and the number of girls be \( Y \).
- The ratio of boys to girls is given by \( B = \frac{X}{Y} \).
- The ratio of girls to boys is given by \( G = \frac{Y}{X} \).

### Step 2: Express \( B + G \)
We need to find \( B + G \):
\[ B + G = \frac{X}{Y} + \frac{Y}{X} \]

### Step 3: Find a Common Denominator
To add the two fractions, we need a common denominator, which is \( XY \):
\[ B + G = \frac{X^2}{XY} + \frac{Y^2}{XY} = \frac{X^2 + Y^2}{XY} \]

### Step 4: Multiply by 3
Now we need to calculate \( 3(B + G) \):
\[ 3(B + G) = 3 \left(\frac{X^2 + Y^2}{XY}\right) = \frac{3(X^2 + Y^2)}{XY} \]

### Step 5: Analyze the Expression
Since \( X \) (number of boys) and \( Y \) (number of girls) are both positive integers, \( X^2 + Y^2 \) will always be greater than or equal to \( 2XY \) (by the AM-GM inequality). Therefore:
\[ X^2 + Y^2 \geq 2XY \]
This implies:
\[ 3(X^2 + Y^2) \geq 6XY \]
Thus:
\[ \frac{3(X^2 + Y^2)}{XY} \geq 6 \]

### Step 6: Conclusion
Since \( \frac{3(X^2 + Y^2)}{XY} \) is always greater than or equal to 6, we can conclude:
- \( 3(B + G) \) is always more than 3.

### Final Answer
The correct option is (c) more than 3.