Question

For a sum compound interest of fifth year is rs. 22743 and compound interst of fourth year is rs. 19950, then find the compound interst of third year

Answer 1

Answer:

Rs 17500

Step-by-step explanation:

For a sum compound interest of fifth year is rs. 22743 and compound interest of fourth year is rs. 19950, then find the compound interest of third year

Compound interest = P (1 + R/100)ⁿ  - P

P = Sum Invested

R  = Rate of Interest

n = Time in Years

Compound interest in 5 Years = P(1 + R/100)⁵  - P

Compound interest in 4 Years = P(1 + R/100)⁴  - P

Compound interest in 3 Years = P(1 + R/100)³  - P

Compound interest in 5th Year = Compound interest in 5 Years - Compound interest in 4 Years

=> Compound interest in 5th Year = P(1 + R/100)⁵  - P - (P(1 + R/100)⁴  - P)

=> P(1 + R/100)⁴( 1 + R/100 - 1)  = 22743

=> P(1 + R/100)⁴(R/100)  = 22743    - eq 1

Compound interest in 4th Year = Compound interest in 4 Years - Compound interest in 3 Years

=> Compound interest in 4th Year = P(1 + R/100)⁴  - P - (P(1 + R/100)³  - P)

=> P(1 + R/100)³( 1 + R/100 - 1)  = 19950

=> P(1 + R/100)³(R/100)  = 19950   - eq 2

Eq1 / Eq 2

=> 1 + R/100 = 22743/19950

=> R/100 = 2793/19950

=> R = 14

Compound interest for 3rd Year = P(1 + R/100)³  - P - (P(1 + R/100)²  - P)

P (1 + R/100)²(R/100) = X  - eq 3

Eq 3 / Eq 2

P (1 + R/100)²(R/100) / P(1 + R/100)³(R/100) = X/19950

=> 1/(1 + R/100) = X/19950

=> 1/( 1.14) = X/19950

=> X = 19950/1.14

=> X = 17500

compound interest of third year = Rs 17500

Putting in eq 2 R = 14

=> P (1 + 14/100)³(14/100) = 19950

=>  P ( 1.14)³ = 142500

=> P = 96183.44