Seven friends planned a tea party. The expenses per boy in rupees is numerically 1 less than the number of girls and the expenses per girl in rupees is numerically 1 less than the number of boys. If the ratio of the total expenses of the boys and the girls is ` 8 : 9`, then what is the expenditure of each boy?

Options

A. Rs.2
B. Rs.3
C. Rs.1
D. Rs.4

Explanation

To solve the problem, let's denote the following:

- Let \( b \) be the number of boys.
- Let \( g \) be the number of girls.
- Let \( x \) be the expenses per boy.
- Let \( y \) be the expenses per girl.

From the problem, we know:

1. The total number of friends (boys + girls) is 7:
\[ b + g = 7 \quad \text{(1)} \]

2. The expenses per boy is numerically 1 less than the number of girls:
\[ x = g - 1 \quad \text{(2)} \]

3. The expenses per girl is numerically 1 less than the number of boys:
\[ y = b - 1 \quad \text{(3)} \]

4. The ratio of the total expenses of boys to girls is \( 8:9 \):
\[ \frac{b \cdot x}{g \cdot y} = \frac{8}{9} \quad \text{(4)} \]

Now, let's substitute equations (2) and (3) into equation (4).

From (2):
\[ x = g - 1 \]

From (3):
\[ y = b - 1 \]

Substituting these into equation (4):
\[ \frac{b \cdot (g - 1)}{g \cdot (b - 1)} = \frac{8}{9} \]

Cross-multiplying gives:
\[ 9b(g - 1) = 8g(b - 1) \]

Expanding both sides:
\[ 9bg - 9b = 8gb - 8g \]

Rearranging terms:
\[ 9bg - 8gb = 9b - 8g \]
\[ bg = 9b - 8g \]

Now, let's isolate \( g \):
\[ bg + 8g = 9b \]
\[ g(b + 8) = 9b \]
\[ g = \frac{9b}{b + 8} \quad \text{(5)} \]

Now, substitute equation (5) back into equation (1):
\[ b + \frac{9b}{b + 8} = 7 \]

To eliminate the fraction, multiply through by \( b + 8 \):
\[ b(b + 8) + 9b = 7(b + 8) \]
\[ b^2 + 8b + 9b = 7b + 56 \]
\[ b^2 + 10b - 7b - 56 = 0 \]
\[ b^2 + 3b - 56 = 0 \]

Now, we can factor this quadratic equation:
\[ (b + 8)(b - 7) = 0 \]

Thus, \( b = -8 \) (not possible) or \( b = 7 \).

So, \( b = 7 \).

Now substituting \( b = 7 \) back into equation (1):
\[ 7 + g = 7 \]
\[ g = 0 \]

Now, substituting \( b = 7 \) into equation (2) to find \( x \):
\[ x = g - 1 = 0 - 1 = -1 \quad \text{(not possible)} \]

This indicates that we made an error in our assumptions or calculations. Let's go back to our equations and check.

Instead, let's assume \( b = 3 \) and \( g = 4 \) (since they must add up to 7).

Now substituting \( b = 3 \) into equation (2):
\[ x = g - 1 = 4 - 1 = 3 \]

And substituting \( g = 4 \) into equation (3):
\[ y = b - 1 = 3 - 1 = 2 \]

Now we can calculate the total expenses:
- Total expenses for boys: \( 3 \cdot 3 = 9 \)
- Total expenses for girls: \( 4 \cdot 2 = 8 \)

Now checking the ratio:
\[ \frac{9}{8} \text{ (not matching)} \]

So we need to adjust our values.

After checking various combinations, we find:

Let’s try \( b = 4 \) and \( g = 3 \):
- Then \( x = g - 1 = 3 - 1 = 2 \)
- And \( y = b - 1 = 4 - 1 = 3 \)

Now checking the expenses:
- Total expenses for boys: \( 4 \cdot 2 = 8 \)
- Total expenses for girls: \( 3 \cdot 3 = 9 \)

Now checking the ratio:
\[ \frac{8}{9} \text{ (matches)} \]

Thus, the expenditure of each boy is \( 2 \) rupees.

### Final Answer:
The expenditure of each boy is **2 rupees**.