Anil invests Rs. 22000 for 6 years in a certain scheme with 4% interest per annum, compounded half-yearly. Sunil invests in the same scheme for 5 years, and then reinvests the entire amount received at the end of 5 years for one year at 10% simple interest. If the amounts received by both at the end of 6 years are same, then the initial investment made by Sunil, in rupees, is
Question:
Anil invests Rs. 22000 for 6 years in a certain scheme with 4% interest per annum, compounded half-yearly. Sunil invests in the same scheme for 5 years, and then reinvests the entire amount received at the end of 5 years for one year at 10% simple interest. If the amounts received by both at the end of 6 years are same, then the initial investment made by Sunil, in rupees, is
Anil invests Rs. 22000 for 6 years in a certain scheme with 4% interest per annum, compounded half-yearly. Sunil invests in the same scheme for 5 years, and then reinvests the entire amount received at the end of 5 years for one year at 10% simple interest. If the amounts received by both at the end of 6 years are same, then the initial investment made by Sunil, in rupees, is
Options
Answer: 20808
Explanation:
Anil: Principal = 22000 Rate = 4% per annum, compounded half-yearly → 2% per half-year Number of half-years in 6 years = 12 Amount after 6 years = 22000 × (1.02)^12 Sunil: Let initial investment = S Same scheme for 5 years → 10 half-years Amount after 5 years = S × (1.02)^10 Reinvested for 1 year at 10% simple interest Final amount = S × (1.02)^10 × 1.10 Given both final amounts are equal: S × (1.02)^10 × 1.10 = 22000 × (1.02)^12 Solve for S: S = 22000 × (1.02^12) / (1.10 × (1.02^10)) S = 22000 × (1.02^2) / 1.10 S = 22000 × 1.0404 / 1.10 S = 22000 × 0.94581818... S ≈ 20807.09
Explanation:
Anil: Principal = 22000 Rate = 4% per annum, compounded half-yearly → 2% per half-year Number of half-years in 6 years = 12 Amount after 6 years = 22000 × (1.02)^12 Sunil: Let initial investment = S Same scheme for 5 years → 10 half-years Amount after 5 years = S × (1.02)^10 Reinvested for 1 year at 10% simple interest Final amount = S × (1.02)^10 × 1.10 Given both final amounts are equal: S × (1.02)^10 × 1.10 = 22000 × (1.02)^12 Solve for S: S = 22000 × (1.02^12) / (1.10 × (1.02^10)) S = 22000 × (1.02^2) / 1.10 S = 22000 × 1.0404 / 1.10 S = 22000 × 0.94581818... S ≈ 20807.09
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