Question

A certain sum of money is invested at the rate of 10% per annum compound
interest, the interest compounded annually. If the difference between the
interests of third year and first year is 1,105, find the sum invested.​

Answer 1

||✪✪ QUESTION ✪✪||

A certain sum of money is invested at the rate of 10% per annum compound interest, the interest compounded annually. If the difference between the interests of third year and first year is 1,105, find the sum invested. ?

|| ★★ FORMULA USED ★★ ||

To Calculate the interest: ---

➳ P[ (1+r/100)^n - 1 ] Where P = Principle(Sum invested), r = rate of interest and n = Time period.

|| ✰✰ ANSWER ✰✰ ||

⟿ Let Our sum invested = P

⟿ Rate = 10%

⟿ T1 = 3rd year

⟿ T2 = 1st year

⟿ Difference between the interests of third year and first year = 1105.

Putting all values we get,

➺ P[(1+10/100)^3-1] - P[(1+10/100)^1-1] = 1105

➺ P[(11/10)^3-1] - P[(11/10)^1-1] = 1105

➺ P(1331/1000-1) - P(11/10-1) = 1105

➺ P(331/1000) - P(1/10) = 1105

Taking P common now,

➺ P[ (331/1000) - (1/10) ] = 1105

➺ P(231/1000) = 1105

➺ P = (1105*1000)/231

☛ P = ₹ 4783.55 (Ans.)

Hence, The sum invested was ₹ 4783.55 .

Answer 2

Question :-

→ A certain sum of money is invested at the rate of 10% per annum compound interest, the interest compounded annually. If the difference between the interests of third year and first year is 1,105, find the sum invested.

Answer :-

→ Invested amount is Rs. 4783.55

To Find :-

→ Invested Sum

Solution :-

Let ,invested Sum is "x" ,

$</p><p> \pink{ \mapsto\mathfrak{ Given}\begin{cases}\sf{ \green{ \star \: Rate (R) = 10\% \: }}\\\sf{ \star \blue{Third \: year (T_1) }}\\ \sf{ \star \: \purple{first \: year (T_2) }}\\ \sf{ \star \: difference \: between \: interest \: of \: } \\ \sf{(T_1) \: and \:(T_2) = 1105 }\end{cases}} \:$

$Hence, \\ \mapsto \sf \red{ interest\: for \:} ( {3}^{rd} \: year \: - {1}^{st} \: year) = 1105 \\ \\ \to \sf x \red{ \bigg[{ \bigg(1 + \frac{10}{100} \bigg)}^{3} - 1 \bigg]} - x \green{ \bigg[ \bigg(1 + \frac{10}{100} \bigg)^{1} - 1 \bigg] }= 1105 \\ \\ \to \sf x \bigg \{ { \bigg(\frac{11}{10} \bigg) }^{3} - 1 \bigg \} \purple{- x \bigg \{ \frac{11}{10} - 1 \bigg \} }= 1105 \\ \\ \to \sf x \bigg \{ \frac{1331}{1000} - 1 \bigg \} - \frac{x}{10} = 1105 \\ \\ \to \sf \red{\frac{331 \: x}{1000 } } - \frac{x}{10} = \pink{ 1105 }\\ \\ \to \sf \blue{ \frac{331x - 100x}{1000}} = 1105 \\ \\ \to \sf \green{ \frac{231x}{1000} }= {1105} \\ \\ \sf \to x = \frac{ \red{1105 } \pink{\: \times 1000}}{231} \\ \\ \to \boxed{ \sf \: x = Rs. \: 4783.55}$

Hence invested amount is Rs. 4783.55