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₹8,829 is invested into three different sectors in such a way that their amounts at 4% p.a. S.I. after 5 years; 6 and 8 years are equal. Find each part of the sum.?
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₹8,829 is invested into three different sectors in such a way that the...
Calculation of Investment in Three Different Sectors

- Let the investments in the three sectors be x, y, and z respectively.
Formulas Used

- The formula for calculating simple interest is: \(SI = \frac{P \times R \times T}{100}\)
- Given that the amounts at 4% p.a. after 5, 6, and 8 years are equal, we can set up the following equations:
\[
x \times 1.04^5 = y \times 1.04^6 = z \times 1.04^8
\]
Calculating the Amounts at the End of 5, 6, and 8 Years

- Using the formula for amount: \(A = P + SI\)
- The amounts at the end of 5, 6, and 8 years can be calculated as follows:
- For x: \(x(1 + 0.04 \times 5) = x \times 1.04^5\)
- For y: \(y(1 + 0.04 \times 6) = y \times 1.04^6\)
- For z: \(z(1 + 0.04 \times 8) = z \times 1.04^8\)
Equating the Amounts

- Since the amounts at the end of 5, 6, and 8 years are equal, we have:
- \(x \times 1.04^5 = y \times 1.04^6 = z \times 1.04^8\)
Solving the Equations

- From the above equation, we get:
- \(x = \frac{1.04^8}{1.04^5} \times z\)
- \(y = \frac{1.04^8}{1.04^6} \times z\)
Substitute and Solve

- Substituting the values of x and y in terms of z into the total investment equation:
- \(x + y + z = ₹8,829\)
- \(\frac{1.04^8}{1.04^5} \times z + \frac{1.04^8}{1.04^6} \times z + z = ₹8,829\)
- Solve the above equation to find the individual investments in each sector.
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₹8,829 is invested into three different sectors in such a way that their amounts at 4% p.a. S.I. after 5 years; 6 and 8 years are equal. Find each part of the sum.?
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