Answer :
To determine at what rate of percent per annum the principal will be 1/4 of the amount in 5 years, we need to use the compound interest formula and solve for the rate [tex]\( r \)[/tex].
### Step-by-Step Solution
1. Understanding the Problem:
- Let [tex]\( P \)[/tex] be the principal amount.
- We know that the amount [tex]\( A \)[/tex] after 5 years will be:
[tex]\[ A = P + \frac{1}{4}P = P \left(1 + \frac{1}{4}\right) = P \left(\frac{5}{4}\right) \][/tex]
2. Setting Up the Compound Interest Formula:
- The compound interest formula is given by:
[tex]\[ A = P \left(1 + \frac{r}{100}\right)^t \][/tex]
- Here, [tex]\( t = 5 \)[/tex] years, and we are trying to find [tex]\( r \)[/tex] such that:
[tex]\[ P \left(\frac{5}{4}\right) = P \left(1 + \frac{r}{100}\right)^5 \][/tex]
3. Equating and Simplifying:
- Cancel out [tex]\( P \)[/tex] from both sides of the equation:
[tex]\[ \frac{5}{4} = \left(1 + \frac{r}{100}\right)^5 \][/tex]
- We now need to solve for [tex]\( r \)[/tex] from this equation.
4. Solving for [tex]\( r \)[/tex]:
- Take the 5th root of both sides to isolate [tex]\( 1 + \frac{r}{100} \)[/tex]:
[tex]\[ \left(\frac{5}{4}\right)^{1/5} = 1 + \frac{r}{100} \][/tex]
- Subtract 1 from both sides to solve for [tex]\( \frac{r}{100} \)[/tex]:
[tex]\[ \left(\frac{5}{4}\right)^{1/5} - 1 = \frac{r}{100} \][/tex]
5. Calculating the Rate [tex]\( r \)[/tex]:
- Compute [tex]\( \left(\frac{5}{4}\right)^{1/5} \)[/tex]:
[tex]\[ \left(\frac{5}{4}\right)^{1/5} \approx 1.0456 \][/tex]
- Subtract 1 to find:
[tex]\[ 1.0456 - 1 = 0.0456 \][/tex]
- Convert this to a percentage:
[tex]\[ 0.0456 \times 100 = 4.56\% \][/tex]
Therefore, the required rate of interest per annum is approximately [tex]\( 4.56\% \)[/tex].
6. Comparing with Given Options:
- Option (a): [tex]\( \frac{16}{6} \% = 2.67\% \)[/tex]
- Option (b): [tex]\( \frac{20}{6} \% = 3.33\% \)[/tex]
- Option (c): [tex]\( \frac{25}{6} \% = 4.17\% \)[/tex]
- Option (d): [tex]\( \frac{28}{5} \% = 5.6\% \)[/tex]
Thus, the closest match to our calculated rate is Option (c): [tex]\( 4.17\% \)[/tex].
Hence, the correct answer is:
[tex]\[ \boxed{\frac{25}{6}\%} \][/tex]
### Step-by-Step Solution
1. Understanding the Problem:
- Let [tex]\( P \)[/tex] be the principal amount.
- We know that the amount [tex]\( A \)[/tex] after 5 years will be:
[tex]\[ A = P + \frac{1}{4}P = P \left(1 + \frac{1}{4}\right) = P \left(\frac{5}{4}\right) \][/tex]
2. Setting Up the Compound Interest Formula:
- The compound interest formula is given by:
[tex]\[ A = P \left(1 + \frac{r}{100}\right)^t \][/tex]
- Here, [tex]\( t = 5 \)[/tex] years, and we are trying to find [tex]\( r \)[/tex] such that:
[tex]\[ P \left(\frac{5}{4}\right) = P \left(1 + \frac{r}{100}\right)^5 \][/tex]
3. Equating and Simplifying:
- Cancel out [tex]\( P \)[/tex] from both sides of the equation:
[tex]\[ \frac{5}{4} = \left(1 + \frac{r}{100}\right)^5 \][/tex]
- We now need to solve for [tex]\( r \)[/tex] from this equation.
4. Solving for [tex]\( r \)[/tex]:
- Take the 5th root of both sides to isolate [tex]\( 1 + \frac{r}{100} \)[/tex]:
[tex]\[ \left(\frac{5}{4}\right)^{1/5} = 1 + \frac{r}{100} \][/tex]
- Subtract 1 from both sides to solve for [tex]\( \frac{r}{100} \)[/tex]:
[tex]\[ \left(\frac{5}{4}\right)^{1/5} - 1 = \frac{r}{100} \][/tex]
5. Calculating the Rate [tex]\( r \)[/tex]:
- Compute [tex]\( \left(\frac{5}{4}\right)^{1/5} \)[/tex]:
[tex]\[ \left(\frac{5}{4}\right)^{1/5} \approx 1.0456 \][/tex]
- Subtract 1 to find:
[tex]\[ 1.0456 - 1 = 0.0456 \][/tex]
- Convert this to a percentage:
[tex]\[ 0.0456 \times 100 = 4.56\% \][/tex]
Therefore, the required rate of interest per annum is approximately [tex]\( 4.56\% \)[/tex].
6. Comparing with Given Options:
- Option (a): [tex]\( \frac{16}{6} \% = 2.67\% \)[/tex]
- Option (b): [tex]\( \frac{20}{6} \% = 3.33\% \)[/tex]
- Option (c): [tex]\( \frac{25}{6} \% = 4.17\% \)[/tex]
- Option (d): [tex]\( \frac{28}{5} \% = 5.6\% \)[/tex]
Thus, the closest match to our calculated rate is Option (c): [tex]\( 4.17\% \)[/tex].
Hence, the correct answer is:
[tex]\[ \boxed{\frac{25}{6}\%} \][/tex]