Question

4. At what rate will 1,000 amount to 1,331 in 3 years if compounded annually?​

Answer 1

$\huge\mathfrak\red{answer \: \: \: \: \: \: \: \: ✎}$

Given: Principal = P = 1000 ₹

period = N = 3 years

Amount = A = 1331 ₹

To find : Rate = R

Solution: Amount =

$p[1 + [{ \frac{r}{100} }]^{3}$

$: \: 1331 = 1000 \: [1 + { \frac{r}{100}] }^{3}$

$\frac{1331}{1000} = [1 + { \frac{r}{100} ]}^{3}$

Taking value root on both sides,

$\frac{11}{10} = 1 + \frac{r}{100} ⇒ \frac{r}{100} = { \frac{11}{10} }^{ - 1}$

$\frac{r}{100} = \frac{11 - 10}{10} ⇒r = \frac{1}{10} \times 100$

$: \: r = 10\%$

Answer: Hence the rate of the intersast is

10% P.a.

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$\sf \colorbox{gold} {\red(ANSWER ᵇʸ ⁿᵃʷᵃᵇ⁰⁰⁰⁸}$