Question

Savita invested 1,000 in a finance company and received Rs 1331 after 3 years, find the rate of interest per cent per annum compounded annually.
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Answer 1

as we know,

Amount = p₁ (1 + r/100)ⁿ

Given :

Amount = 1331 rs.

principal = 1000 rs.

Time = 3 yrs.

Solution :

p₁ (1 + r/100)ⁿ = amount

(by putting all the values here)

1000(1 + r/100)³ = 1331

(1 + r/100)³ = 1331/1000 or (11/10)³

(cube in RHS cancelled by cube in LHS)

1 + r/100 = 11/10

100 + r/100 = 11/10

(cross multiplication)

(100 + r)10 = (11)100

1000 + 10r = 1100

r = 10%

Answer 2

$\bf \red{ \underline{ \underline{given}}}$

• Principal ,P = Rs 1000
• Amount, A = Rs 1331
• Time, n = 3 years

$\bf \red{ \underline{ \underline{to \: find \: out}}}$

find the rate of interest per cent per annum compounded annually.

find the rate of interest per cent per annum compounded annually.

$\bf \red{ \underline{ \underline{formula \: used}}}$

$\blue{A = p(1 + \frac{r}{100} ) {}^{n} }$

$\bf \red{ \underline{ \underline{solution}}}$

$A = p(1 + \frac{r}{100} ) {}^{n}$

$\star \: putting \: the \: value$

$= > 1331 = 1000(1 + \frac{r}{100}) {}^{3}$

$= > \frac{1331}{1000} = (1 + \frac{r}{100} ) {}^{3}$

$= >( \frac{11}{10} ) {}^{3} = ( 1 + \frac{r}{100} ) {}^{3}$

$= > \frac{11}{10} = 1 + \frac{r}{100}$

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

$= > \frac{r}{100} = \frac{1}{10}$

$= > r = \frac{1}{10} \times 100$

$= > \purple{r = 10}$

$\therefore \: required \: rate \: of \: interest \: is \: 10\%$