Question

At what rate per cent per annum will ₹ 4500 amount to ₹5715 in 3 years?

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Answer 1

ANSWER: 9 %

step by step explanation:

$interest = 5715 - 4500 \\ = 1215rs \\ 1215 = \frac{4500 \times 3 \times r}{100 } \\ \\ = > \frac{1215}{45 \times 3 } = r \\ \\ r = 9\%$

Answer 2

ANSWER :

• [i] When interest is calculated in Simple Interest, Rs. 4500 will be Rs. 5715 in 3 years at the rate of interest 9% p.a.

• [ii] When interest is calculated in Compound Interest, Rs. 4500 will be Rs. 5715 in 3 years at the rate of interest 8.2% p.a.

$\underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }$

SOLUTION :

❒ Given :-

• Principal, P = Rs. 4500

• No. of time, n = 3 years

• Amount, A = Rs. 5715

❒ To Find :-

• Rate of Interest, r = ?

❒ Note :-

★ Interest can be calculated in two ways :

• ➭ Simple Interest

• ➭ Compound Interest

❒ Required formulas :-

$\: \: \: \: \: \bigstar \: \: \boxed{ \bf{Interest = Amount - Principle }}$

$\: \: \: \: \: \bigstar \: \: \boxed{ \bf{S.I = \dfrac{P \times r \times n}{100}}}$

$\: \: \: \: \: \bigstar \: \: \boxed{ \bf{C.I. = A - P}} \\ \\ \: \: \: \: \mapsto \: \: \boxed{ \sf{A = P \: (1 + \dfrac{r}{100} ) {}^{n} }}$

$\underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }$

❖ Calculation of Rate of Interest when interest is calculated in Simple Interest :-

We have,

• A = Rs. 5715

• P = Rs. 4500

$\sf{ \therefore \: \: S.I. = A - P} \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = Rs. \: (5715 - 4500) } \\ \\ \: \: \sf{= Rs. \: 1215}$

Now,

$\: \: \bigstar \: \: \sf{\underline{S.I = \dfrac{ P \times r \times n}{100}}} \\ \\ \sf{ \implies \: 1215 = \frac{45 \cancel{0} \cancel{0} \times r \times 3}{1 \cancel{0} \cancel{0}} } \\ \\ \sf{ \implies \: 1215 = 45 \times r \times 3} \: \: \: \: \: \\ \\ \sf{ \implies \: 1215 = 135 \: r} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf{ \implies \: r = \dfrac{1215}{135}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf{ \large{ \therefore \: \underline{ \: r= 9 \: }}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

• Hence, the rate of interest,r = 9%

$\underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }$

❖ Calculation of Rate of Interest when interest is calculated in Compound Interest :-

We have,

• P = Rs. 4500

• A = Rs. 5715

• n = 3 years

$\bigstar \: \: \sf{\underline{A = P \: (1 + \dfrac{r}{100}) {}^{n}}} \\ \\ \sf{ \implies \: 5715 = 4500(1 + \frac{r}{100}) {}^{3} } \\ \\ \sf{ \implies \: \frac{5715}{4500} = (1 + \frac{r}{100} ) {}^{3}} \: \: \: \: \: \: \\ \\ \sf{ \implies \: 1.27 = (1 + \frac{r}{100} ) {}^{3}} \: \: \: \: \: \: \: \: \\ \\\sf{ \implies \: 1 + \frac{r}{100} = \ \sqrt[3]{1.27}} \: \: \: \: \: \: \: \: \\ \\ \sf{ \implies \: 1 + \frac{r}{100} = 1.082} \: \: \: \: \: \: \: \: \: \: \\ \\ \sf{ \implies \: \frac{r}{100} = 1.082 - 1} \: \: \: \: \: \: \: \: \: \: \\ \\ \sf{ \implies \: \frac{r}{100} = 0.082} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf{ \implies \: r = 0.082 \times 100} \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf{ \large{ \therefore \: \underline{ \: r = 8.2 \: }}} \: \: \: \: \: \: \: \: \: \: \:$

• Hence, the rate of interest, r = 8.2%