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
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Official Correct Answer: 20808. First, let me parse the problem again to make sure I get all the details right. Anil invests Rs. 22,000 for 6 years in a scheme with 4% annual interest, compounded half-yearly. On the other hand, Sunil invests in the same scheme for 5 years and then takes the entire amount he gets and reinvests it for one more year at 10% simple interest. The key point is that after 6 years, both Anil and Sunil have the same amount. We need to find Sunil's initial investment.
Okay, so We calculate the amount Anil gets after 6 years and set it equal to the amount Sunil gets after his 6-year investment journey (5 years in the first scheme and 1 year at 10% simple interest). Then, solve for Sunil's initial investment.
Let me recall the formulas for compound interest and simple interest.
For compound interest, the formula is:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
where:
- \( A \) is the amount after time \( t \),
- \( P \) is the principal amount,
- \( r \) is the annual interest rate (in decimal),
- \( n \) is the number of times interest is compounded per year,
- \( t \) is the time in years.
For simple interest, the formula is:
\[ A = P(1 + rt) \]
where:
- \( A \) is the amount after time \( t \),
- \( P \) is the principal amount,
- \( r \) is the annual interest rate (in decimal),
- \( t \) is the time in years.
Alright, so Anil's investment is straightforward. He invests Rs. 22,000 at 4% annual interest, compounded half-yearly for 6 years. Since it's compounded half-yearly, \( n = 2 \). So, Computing Anil's amount first.
Anil's amount:
\[ A_{Anil} = 22000 \left(1 + \frac{0.04}{2}\right)^{2 \times 6} \]
Simplify the terms inside the bracket:
\[ 1 + \frac{0.04}{2} = 1 + 0.02 = 1.02 \]
And the exponent:
\[ 2 \times 6 = 12 \]
So,
\[ A_{Anil} = 22000 \times (1.02)^{12} \]
I might need to compute \( (1.02)^{12} \). Let me remember that \( (1.02)^{12} \) is approximately 1.2682, but I should verify this calculation.
Wait, maybe I can compute it step by step:
- \( 1.02^1 = 1.02 \)
- \( 1.02^2 = 1.0404 \)
- \( 1.02^3 = 1.061208 \)
- \( 1.02^4 = 1.082432 \)
- \( 1.02^5 = 1.104089 \)
- \( 1.02^6 = 1.126161 \)
- \( 1.02^7 = 1.148644 \)
- \( 1.02^8 = 1.171897 \)
- \( 1.02^9 = 1.195625 \)
- \( 1.02^{10} = 1.220190 \)
- \( 1.02^{11} = 1.244984 \)
- \( 1.02^{12} = 1.268245 \)
So, approximately 1.268245. Let's use this value for now.
Thus,
\[ A_{Anil} = 22000 \times 1.268245 \]
Calculating that:
22000 * 1.268245. Computing 22000 * 1.268245.
First, 22000 * 1 = 22000
22000 * 0.268245 = ?
Compute 22000 * 0.2 = 4400
22000 * 0.068245 = ?
Compute 22000 * 0.06 = 1320
22000 * 0.008245 = approximately 22000 * 0.008 = 176, and 22000 * 0.000245 ≈ 5.405. So total ≈ 176 + 5.405 ≈ 181.405.
So, 1320 + 181.405 ≈ 1501.405.
Therefore, 22000 * 0.268245 ≈ 4400 + 1501.405 ≈ 5901.405.
So, total amount ≈ 22000 + 5901.405 ≈ 27901.405.
So, Anil's amount after 6 years is approximately Rs. 27,901.41.
Wait, but let me check if I can compute this more accurately. Alternatively, maybe I can use the exact value of (1.02)^12.
Alternatively, perhaps I can use logarithms or another method, but maybe it's faster to just accept that 1.02^12 is approximately 1.268245 and proceed.
So, Anil's amount is approximately 22000 * 1.268245 ≈ 27901.40.
Now, moving on to Sunil's investment. Sunil invests an initial amount, let's call it P, in the same scheme for 5 years, compounded half-yearly at 4% annual interest. Then