Question

S invested 22,000 for 6 years at a 4% rate compounded half-yearly. B also invested a certain amount of money for 5 years in the same scheme and then reinvested it for a year at a 10% rate of simple interest. If their total amounts become the same after 6 years, what was B’s initial investment?

Answer 1

Correct Answer: 20 - 800

Step 1: Calculate S’s final amount
S invests 22,000 Rs. for 6 years at 4% per annum, compounded half-yearly. 
The half-yearly rate is:
\[\text{Rate per half-year} = \frac{4}{2} = 2\%.\]
The total number of half-years is:
\[\text{Number of half-years} = 6 \cdot 2 = 12.\]

Using the compound interest formula:
\[A = P \cdot \left(1 + \frac{r}{100}\right)^n\]
where \(P\) is the principal, \(r\) is the rate, and \(n\) is the number of compounding periods.

Substitute the values:
\[A_S = 22000 \cdot \left(1 + \frac{2}{100}\right)^{12}.\]

Simplify:
\[A_S = 22000 \cdot (1.02)^{12}.\]

Using approximation:
\[(1.02)^{12} \approx 1.26824.\]

Thus:
\[A_S = 22000 \cdot 1.26824 = 27,901.28 \, \text{Rs}.\]

Step 2: Calculate B’s initial investment
Let B’s initial investment be \(P_B\).  
B invests for 5 years in the same scheme (compounded half-yearly at 4%) and then reinvests the amount for 1 year at 10% simple interest.

Step 2.1: Calculate B’s amount after 5 years
The number of half-years is:
\[\text{Number of half-years} = 5 \cdot 2 = 10.\]
Using the compound interest formula:
\[A_B = P_B \cdot \left(1 + \frac{2}{100}\right)^{10}.\]

Substitute:
\[A_B = P_B \cdot (1.02)^{10}.\]

Using approximation:
\[(1.02)^{10} \approx 1.21900.\]

Thus:
\[A_B = P_B \cdot 1.21900.\]

Step 2.2: Reinvest B’s amount for 1 year at 10% simple interest
The simple interest formula is:
\[A = P \cdot \left(1 + \frac{r \cdot t}{100}\right).\]

Here:
\[P = A_B = P_B \cdot 1.21900, \quad r = 10\%, \quad t = 1.\]

Substitute into the formula:
\[A_B^\text{final} = \left(P_B \cdot 1.21900\right) \cdot \left(1 + \frac{10}{100}\right).\]

Simplify:
\[A_B^\text{final} = \left(P_B \cdot 1.21900\right) \cdot 1.1.\]Thus:
\[A_B^\text{final} = P_B \cdot 1.3409.\]

Step 3: Equating final amounts
The final amounts of S and B are equal:
\[A_S = A_B^\text{final}.\]

Substitute:
\[27,901.28 = P_B \cdot 1.3409.\]

Solve for \(P_B\):
\[P_B = \frac{27,901.28}{1.3409} \approx 20,800 \, \text{Rs}.\]

Final Answer
B’s initial investment is:
\[\boxed{20,800 \, \text{Rs}}.\]