Step 1: Calculate S’s final amount.Nbsp;
S invests 22,000 Rs. for 6 years at 4% per annum, compounded half-yearly.Nbsp;
The half-yearly interest rate is:
\[\text{Rate per half-year} = \frac{4}{2} = 2\%.\]
The total number of compounding periods (half-years) is:
\[\text{Number of half-years} = 6 \text{ years} \times 2 \text{ half-years/year} = 12.\]
Using the compound interest formula:
\[A = P \cdot \left(1 + \frac{r}{100}\right)^n\]
where \(P\) is the principal amount, \(r\) is the interest rate per period, and \(n\) is the number of compounding periods.
Substitute the values for S:
\[A_S = 22000 \cdot \left(1 + \frac{2}{100}\right)^{12}.\]
Simplify the expression:
\[A_S = 22000 \cdot (1.02)^{12}.\]
Using an approximate value for \((1.02)^{12}\):
\[(1.02)^{12} \approx 1.26824.\]
Calculate S's final amount:
\[A_S = 22000 \cdot 1.26824 = 27,901.28 \, \text{Rs}.\]
Step 2: Calculate B’s initial investment
Let B’s initial investment be \(P_B\). Nbsp;
B invests for 5 years in the same scheme (compounded half-yearly at 4%) and then reinvests the resulting amount for 1 year at 10% simple interest.
Step 2.1: Calculate B’s amount after 5 years of compound interest.
The number of half-years in 5 years is:
\[\text{Number of half-years} = 5 \text{ years} \times 2 \text{ half-years/year} = 10.\]
Using the compound interest formula for B’s initial investment:
\[A_B = P_B \cdot \left(1 + \frac{2}{100}\right)^{10}.\]
Substitute the simplified rate:
\[A_B = P_B \cdot (1.02)^{10}.\]
Using an approximate value for \((1.02)^{10}\):
\[(1.02)^{10} \approx 1.21900.\]
So, B’s amount after 5 years is:
\[A_B = P_B \cdot 1.21900.\]
Step 2.2: Reinvest B’s amount for 1 year at 10% simple interest.
The simple interest formula for the final amount is:
\[A = P \cdot \left(1 + \frac{r \cdot t}{100}\right).\]
In this case:
The principal \(P\) is B's amount after 5 years, so \(P = A_B = P_B \cdot 1.21900\).
The annual interest rate \(r = 10\%\).
The time period \(t = 1\) year.
Substitute these values into the simple interest formula:
\[A_B^\text{final} = \left(P_B \cdot 1.21900\right) \cdot \left(1 + \frac{10 \cdot 1}{100}\right).\]
Simplify the calculation:
\[A_B^\text{final} = \left(P_B \cdot 1.21900\right) \cdot \left(1 + 0.1\right).\]
\[A_B^\text{final} = \left(P_B \cdot 1.21900\right) \cdot 1.1.\]
Thus, B’s final amount is:
\[A_B^\text{final} = P_B \cdot 1.3409.\]
Step 3: Equating final amounts
The problem states that S’s final amount is equal to B’s final amount:
\[A_S = A_B^\text{final}.\]
Substitute the calculated values:
\[27,901.28 = P_B \cdot 1.3409.\]
Solve for \(P_B\) (B's initial investment):
\[P_B = \frac{27,901.28}{1.3409}.\]
\[P_B \approx 20,800 \, \text{Rs}.\]
Final Answer
B’s initial investment is:
\[\boxed{20,800 \, \text{Rs}}.\]