Let `(C)/(5)=(F-32)/(9)` If C lies between `10` and `20` then :

Options

A. `50ltFlt78`
B. `50ltFlt68`
C. `49ltFlt68`
D. `49ltFlt78`

Answer

To solve the problem, we need to manipulate the given equation and find the range of values for \( F \) based on the given range for \( C \).

### Step-by-step Solution:

1. **Start with the given equation:**
\[ \frac{C}{5} = \frac{F - 32}{9} \]

2. **Cross-multiply to eliminate the fractions:**
\[ 9C = 5(F - 32) \]

3. **Distribute the 5 on the right side:**
\[ 9C = 5F - 160 \]

4. **Rearrange the equation to solve for \( F \):**
\[ 5F = 9C + 160 \]
\[ F = \frac{9C + 160}{5} \]

5. **Now, substitute the values of \( C \) to find the range of \( F \). Given that \( C \) lies between 10 and 20:**
- For the minimum value of \( C = 10 \):
\[ F_{\text{min}} = \frac{9(10) + 160}{5} = \frac{90 + 160}{5} = \frac{250}{5} = 50 \]

- For the maximum value of \( C = 20 \):
\[ F_{\text{max}} = \frac{9(20) + 160}{5} = \frac{180 + 160}{5} = \frac{340}{5} = 68 \]

6. **Conclusion:**
Therefore, \( F \) lies between 50 and 68:
\[ 50 < F < 68 \]

### Final Answer:
The correct range for \( F \) is \( 50 < F < 68 \).

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