A sum of money is rounded off to the nearest rupee, if `((m)/(n))^2` be the probability that the round off error is atleast ten prizes, where m and n are positive relative primes, then value of (n-m) is

Answer

To solve the problem, we need to determine the probability that the round-off error is at least 10 paisa when a sum of money is rounded to the nearest rupee. We will express this probability in the form \((\frac{m}{n})^2\) where \(m\) and \(n\) are coprime integers, and then find the value of \(n - m\).

### Step-by-Step Solution:

1. **Understanding Round-off Error**:
- When rounding to the nearest rupee, the round-off error can be at least 10 paisa if the value is at least 0.10 rupees away from the nearest whole number. This means that if the amount is between \(x.00\) and \(x.09\), it rounds down to \(x\), and if it is between \(x.10\) and \(x.99\), it rounds up to \(x+1\).

2. **Identifying the Range for Round-off Error**:
- The round-off error is at least 10 paisa when the amount is in the ranges:
- From \(x.00\) to \(x.09\) (rounds down)
- From \(x.10\) to \(x.99\) (rounds up)
- The critical points for rounding are at \(x + 0.05\). Thus, the ranges can be defined as:
- Rounds down: \(x.00\) to \(x.04\) (error = -0.04)
- Rounds up: \(x.06\) to \(x.99\) (error = +0.06)

3. **Calculating the Favorable Outcomes**:
- The total range of values from \(0.00\) to \(0.99\) is 100 paisa (or 100 values).
- The favorable outcomes for rounding down (0.00 to 0.04) is 5 paisa (5 values).
- The favorable outcomes for rounding up (0.06 to 0.99) is 94 paisa (94 values).
- Therefore, the total favorable outcomes where the round-off error is at least 10 paisa is \(5 + 94 = 99\).

4. **Calculating the Total Outcomes**:
- The total possible outcomes when rounding to the nearest rupee is 100 (from 0.00 to 0.99).

5. **Calculating the Probability**:
- The probability \(P\) that the round-off error is at least 10 paisa is given by:
\[ P = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{99}{100} \]

6. **Expressing the Probability in Required Form**:
- We can express this probability as:
\[ P = \left(\frac{99}{100}\right)^2 \]
- Here, \(m = 99\) and \(n = 100\).

7. **Finding \(n - m\)**:
- Now, we need to find \(n - m\):
\[ n - m = 100 - 99 = 1 \]

### Final Answer:
The value of \(n - m\) is \(1\).