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

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:
$\boxed{20,800 \, \text{Rs}}.$