Which of the following statement (s) is/are true ?
I. `(1+(2)/(3)) (1+(2)/(5))(1+(2)/(7))....(1+(2)/(997)) lt 334`
II `15(5)/(8)+13+(3)/(8)-10 (3)/(5) lt 21(2)/(3)+3(1)/(3)-6(3)/(5)`

Options

A. Only I
B. Only II
C. Both I and II
D. Neither I nor II

Answer

To determine which of the statements is true, we will analyze each statement step by step.

### Statement I:
\[ (1 + \frac{2}{3})(1 + \frac{2}{5})(1 + \frac{2}{7}) \ldots (1 + \frac{2}{997}) < 334 \]

1. **Rewrite the Terms**:
Each term can be rewritten as:
\[ 1 + \frac{2}{n} = \frac{n + 2}{n} \]
Therefore, the product becomes:
\[ \prod_{n=3, 5, 7, \ldots, 997} \frac{n + 2}{n} \]

2. **Identify the Terms**:
The numbers \(n\) are all odd numbers from 3 to 997. The total count of these numbers can be found using the formula for the nth odd number:
- The first odd number is 3, and the last is 997.
- The sequence of odd numbers can be expressed as \(3, 5, 7, \ldots, 997\). This is an arithmetic sequence where:
- First term \(a = 3\)
- Common difference \(d = 2\)
- Last term \(l = 997\)

The number of terms \(n\) can be calculated as:
\[ n = \frac{l - a}{d} + 1 = \frac{997 - 3}{2} + 1 = 498 \]

3. **Calculate the Product**:
The product can be simplified:
\[ \prod_{n=3, 5, 7, \ldots, 997} \frac{n + 2}{n} = \frac{(5)(7)(9)\ldots(999)}{(3)(5)(7)\ldots(997)} \]
The numerator is the product of all odd numbers from 5 to 999, and the denominator is the product of all odd numbers from 3 to 997.

4. **Simplify Further**:
Notice that most terms will cancel out:
\[ = \frac{999}{3} = 333 \]

5. **Final Comparison**:
Since \(333 < 334\), Statement I is **true**.

### Statement II:
\[ 15 \frac{5}{8} + 13 + \frac{3}{8} - 10 \frac{3}{5} < 21 \frac{2}{3} + 3 \frac{1}{3} - 6 \frac{3}{5} \]

1. **Convert Mixed Numbers to Improper Fractions**:
- Left Hand Side (LHS):
\[ 15 \frac{5}{8} = \frac{120 + 5}{8} = \frac{125}{8}, \quad 10 \frac{3}{5} = \frac{50 + 3}{5} = \frac{53}{5} \]
Thus:
\[ LHS = \frac{125}{8} + 13 + \frac{3}{8} - \frac{53}{5} \]
Convert \(13\) to eighths:
\[ 13 = \frac{104}{8} \]
Therefore:
\[ LHS = \frac{125 + 104 + 3}{8} - \frac{53}{5} = \frac{232}{8} - \frac{53}{5} \]

2. **Finding a Common Denominator**:
The common denominator for \(8\) and \(5\) is \(40\):
\[ LHS = \frac{232 \times 5}{40} - \frac{53 \times 8}{40} = \frac{1160 - 424}{40} = \frac{736}{40} = 18.4 \]

3. **Right Hand Side (RHS)**:
- Convert:
\[ 21 \frac{2}{3} = \frac{63 + 2}{3} = \frac{65}{3}, \quad 3 \frac{1}{3} = \frac{10}{3}, \quad 6 \frac{3}{5} = \frac{30 + 3}{5} = \frac{33}{5} \]
Thus:
\[ RHS = \frac{65}{3} + \frac{10}{3} - \frac{33}{5} \]
Combine:
\[ RHS = \frac{75}{3} - \frac{33}{5} = 25 - \frac{33}{5} \]

4. **Finding a Common Denominator**:
The common denominator for \(1\) and \(5\) is \(5\):
\[ RHS = \frac{125}{5} - \frac{33}{5} = \frac{92}{5} = 18.4 \]

5. **Final Comparison**:
Since \(LHS = 18.4\) and \(RHS = 18.4\), the statement \(LHS < RHS\) is **false**.

### Conclusion:
- **Statement I is true**.
- **Statement II is false**.