Read the given statements carefully and state T for true and F for false.
(i) The value of k for which the equations `x+2y=5 and 3x + ky-15=0` has no solution, is 6.
(ii) The system of equations `3x - 5y = 20 and 6x – 10y = 40` has infinitely many solutions.
(iii) If the sum of the digits of a two digit number is 8 and difference between the number and the number formed by reversing the digits is 18, then the number is 34.

Options

A. `{:(i,ii,iii),(T,F,T):}`
B. `{:(i,ii,iii),(F,F,T):}`
C. `{:(i,ii,iii),(T,T,F):}`
D. `{:(i,ii,iii),(F,T,F):}`

Answer

To solve the problem, we need to evaluate each of the three statements provided and determine whether they are true (T) or false (F). Let's go through each statement step by step.

### Step 1: Evaluate the First Statement
**Statement (i):** The value of k for which the equations \(x + 2y = 5\) and \(3x + ky - 15 = 0\) has no solution is 6.

1. **Identify the equations:**
- First equation: \(x + 2y - 5 = 0\) (let's call this Equation 1)
- Second equation: \(3x + ky - 15 = 0\) (let's call this Equation 2)

2. **Condition for no solution:**
- For the system of equations to have no solution, the ratios of the coefficients must be equal while the ratio of the constants must be different:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]

3. **Coefficients from the equations:**
- From Equation 1: \(a_1 = 1\), \(b_1 = 2\), \(c_1 = -5\)
- From Equation 2: \(a_2 = 3\), \(b_2 = k\), \(c_2 = -15\)

4. **Set up the ratios:**
\[ \frac{1}{3} = \frac{2}{k} \quad \text{and} \quad \frac{-5}{-15} = \frac{1}{3} \]

5. **Solving for k:**
- From \(\frac{1}{3} = \frac{2}{k}\):
\[ k = 6 \]
- Now check the third ratio:
\[ \frac{1}{3} = \frac{1}{3} \quad \text{(which is true)} \]

6. **Conclusion for Statement (i):**
- Since the ratios are equal, the first statement is **False (F)**.

### Step 2: Evaluate the Second Statement
**Statement (ii):** The system of equations \(3x - 5y = 20\) and \(6x - 10y = 40\) has infinitely many solutions.

1. **Identify the equations:**
- First equation: \(3x - 5y - 20 = 0\) (let's call this Equation 3)
- Second equation: \(6x - 10y - 40 = 0\) (let's call this Equation 4)

2. **Condition for infinitely many solutions:**
- For the system to have infinitely many solutions, the ratios of the coefficients must be equal:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

3. **Coefficients from the equations:**
- From Equation 3: \(a_1 = 3\), \(b_1 = -5\), \(c_1 = -20\)
- From Equation 4: \(a_2 = 6\), \(b_2 = -10\), \(c_2 = -40\)

4. **Set up the ratios:**
\[ \frac{3}{6} = \frac{-5}{-10} = \frac{-20}{-40} \]

5. **Simplifying the ratios:**
\[ \frac{3}{6} = \frac{1}{2}, \quad \frac{-5}{-10} = \frac{1}{2}, \quad \frac{-20}{-40} = \frac{1}{2} \]

6. **Conclusion for Statement (ii):**
- All ratios are equal, hence the second statement is **True (T)**.

### Step 3: Evaluate the Third Statement
**Statement (iii):** If the sum of the digits of a two-digit number is 8 and the difference between the number and the number formed by reversing the digits is 18, then the number is 34.

1. **Let the two-digit number be represented as \(10a + b\)** where \(a\) is the tens digit and \(b\) is the units digit.

2. **Set up the equations based on the statement:**
- Sum of the digits: \(a + b = 8\) (Equation 5)
- Difference between the number and its reverse:
\[ (10a + b) - (10b + a) = 18 \implies 9a - 9b = 18 \implies a - b = 2 \quad \text{(Equation 6)} \]

3. **Solve the system of equations (5) and (6):**
- From Equation 6: \(a = b + 2\)
- Substitute into Equation 5:
\[ (b + 2) + b = 8 \implies 2b + 2 = 8 \implies 2b = 6 \implies b = 3 \]
- Substitute \(b\) back to find \(a\):
\[ a = 3 + 2 = 5 \]

4. **The number is:**
\[ 10a + b = 10(5) + 3 = 53 \]

5. **Conclusion for Statement (iii):**
- The number is 53, not 34. Therefore, the third statement is **False (F)**.

### Final Summary of Statements
- (i) F
- (ii) T
- (iii) F

### Final Answer
- (i) F
- (ii) T
- (iii) F