Question

A certain sum of money is invested at the rate of 10 parentage per annum compound interest, if the different between the interest of third year and first year r.s1105, the sum invested

Answer 1

Let P is the sum of money .

rate of interest , r = 10%

a/c to question,

compound interest of third year - compound interest in first year = Rs. 1105

compound interest formula, C.I = P(1 + r/100)ⁿ - P

compound interest of 1st year = P(1 + 10/100)¹ - P
= P(1.1) - P = 0.1P

compound interest in two years = P(1 + 10/100)² - P
= P(1.21) - P = (0.21)P

compound interest in three years = P(1 + 10/100)³ - P
= (1.331)P - P = (0.331)P

now, compound in 3rd year = compound interest in three years - compound interest in two years
= (0.331)P - (0.21)P
= (0.121)P

compound interest in 3rd year - compound interest in first year = 1105

or, (0.121)P - (0.1)P = 1105

or, 0.021P = 1105

P = 1105 × 1000/21 = 52619 Rs.

Answer 2

Answer:

Rs 52,619

Step-by-step explanation:

At the end of 1 years:

Amount = P(1 + 0.1)

Amount = 1.1P

At the end of 2 years:

Amount = P(1 + 0.1)²

Amount = 1.21P

At the end of 3 years:

Amount = P(1 + 0.1)³

Amount = 1.331P

Solve P:

The difference between the interest of 1st year and 3rd year is Rs 1105

(1.331P - 1.21P) - (1.1P - P) = 1105

0.121P - 0.1P = 1105

0.021P = 1105

P = Rs 52,619

Answer: The sum invested is Rs 52,619