Question

A certain sum of money is invested at the rate 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
pls give step by step answer​

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Principal=4,740.45\:rupees}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\underline{ \underline\text{Given : }} \\ \implies Rate = 10\% \\ \\ \implies t_{1} = 1st \: year \\ \\ \implies t_{2} = 3rd \: year \\ \\ \implies Difference = 1,105 \\ \\ \underline{ \underline\text{To \: Find : }} \\ \implies Principal = ?$

• According to given question :

$\text{Let \: Principal \: be \: x \: rupees} \\ \\ \bold{For \: 1st \: yea r :} \\ \implies Amount = Principal(1 + \frac{rate}{100} )^{time} \\ \\ \implies A_{1} = x(1 + \frac{10}{100} )^{1} \\ \\ \implies A_{1} = x \times (1 + 0.1) \\ \\ \implies { A_{1} = 1.1x} \\ \\ \bold{for \: Compound \: Interest :} \\ \implies C.I_{1}= A_{1} - P \\ \\ \implies C.I_{1} = 1.1x - x\\ \\ \implies C.I_{1}= 0.1x \\ \\ \bold{For \: 3rd \: year : } \\ \implies A_{2} = p \times (1 + \frac{r}{100} )^{t} \\ \\ \implies A_{2} = x \times (1 + \frac{10}{100} )^{3} \\ \\ \implies A_{2} = x \times (1.1)^{3} \\ \\ \implies A_{2}= 1.331x \\ \\ \bold{For \: 3rd \: year \: Compound \: Interest: } \\ \implies C.I_{2} = A_{2} - P \\ \\ \implies C.I_{2} = 1.331x - x \\ \\ \implies C.I_{2} = 0.331x$

$\text{For \: difference \: of \: 1st \: and \: 3rd \: year : } \\ \implies C.I_{2} - C.I_{1} = 1,105 \\ \\ \implies 0.331x - 0.1x = 1,105\\ \\ \implies 0.231x = 1105 \\ \\ \implies x = \frac{1,105}{0.231} \\ \\ \bold{\implies x = 4,740.45 \: rupees}$

Answer 2

$\huge\sf{Answer:-}$

Refer the given above attachament.

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