S.I. on a sum is 9/16 of its principal. Accord ingly if the no. of annual rate of interest and related no. of years is same, then what was the no. of years for which sum is lent?

Options

A. `5(1)/(2)`
B. `6(1)/(2)`
C. `7(1)/(2)`
D. 7

Answer

To solve the problem, we need to find the number of years for which the sum is lent, given that the Simple Interest (SI) is \( \frac{9}{16} \) of the principal (P). We also know that the rate of interest (R) and the number of years (T) are the same.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
We know that Simple Interest (SI) is given by the formula:
\[ SI = \frac{P \times R \times T}{100} \]
According to the problem, \( SI = \frac{9}{16} P \).

2. **Setting Up the Equation**:
Since \( SI = \frac{9}{16} P \), we can substitute this into the SI formula:
\[ \frac{9}{16} P = \frac{P \times R \times T}{100} \]

3. **Cancelling P**:
Assuming \( P \neq 0 \), we can divide both sides by \( P \):
\[ \frac{9}{16} = \frac{R \times T}{100} \]

4. **Using the Given Condition**:
Since \( R = T \) (the rate and time are the same), we can replace \( R \) with \( T \):
\[ \frac{9}{16} = \frac{T \times T}{100} \]
This simplifies to:
\[ \frac{9}{16} = \frac{T^2}{100} \]

5. **Cross-Multiplying**:
Cross-multiply to eliminate the fraction:
\[ 9 \times 100 = 16 \times T^2 \]
This gives us:
\[ 900 = 16 T^2 \]

6. **Solving for T²**:
Divide both sides by 16:
\[ T^2 = \frac{900}{16} = 56.25 \]

7. **Taking the Square Root**:
Now, take the square root of both sides to find T:
\[ T = \sqrt{56.25} = 7.5 \]

8. **Conclusion**:
Therefore, the number of years for which the sum is lent is:
\[ T = 7.5 \text{ years} \]

### Summary:
The number of years for which the sum is lent is **7.5 years**.

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