Question

in a period of 3 years,if RS 6250 earned compound interest of RS 1623.20,find the rate per annum. the answer will be 8% but some one tell me how​

Answer 1

Answer:-

Given:

Principle (P) = Rs. 6250

Compound interest (CI) = Rs. 1623.20

Time period (n) = 3 years.

Let the rate of interest be r.

We know that,

$\boxed{\sf CI \: = P \bigg(1 + \frac{r}{100} \bigg) ^{n} - P}$

So,

$\implies \sf \: 1623.20 = 6250 \bigg(1 + \frac{r}{100} \bigg) ^{3} - 6250 \\ \\ \implies \sf \: 1623.20 + 6250 = 6250 \bigg( \frac{100 + r}{100} \bigg) ^{3} \\ \\ \implies \sf \:7873.2 \times \frac{1}{6250} \times 100 \times 100 \times 100 = {(100 + r)}^{3} \\ \\ \implies \sf \:1259712 = (100 + r) ^{3} \\ \\ \implies \sf \:(108) ^{3} = {(100 + r)}^{3}$

Powers are equal So bases are also equal.

$\: \implies \sf \:108 = 100 + r \\ \\ \implies \sf \:108 - 100 = r \\ \\ \: \implies \boxed{ \sf \:r = 8 \: \%}$

∴ The rate of interest is 8 % per annum.

Answer 2

$\bf \huge {\underline {\underline \red{AnSwEr}}}$

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Given

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• n = 3 yrs
• p = 6250
• C.I = 1623.80

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To Find

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• Rate per annum

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Solution

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Formula of CI

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$\bf \pink {CI = p( {(1 + \frac{r}{100} )}^{n} - 1)}$

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Answer

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$\bf \implies 1623.20 = 6250((1 + \frac{r}{100} ) {}^{3} - 1)$

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$\bf \implies 1623.20 = 6250(( \frac{100 + r }{100} ) {}^{3} - 1)$

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$\bf \implies \frac{1623.20}{6250} = {( \frac{100 + r}{100}) }^{3} - 1$

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$\bf \implies \frac{1623.20}{6250} + 1 = {( \frac{100 + r}{100}) }^{3}$

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$\bf \implies \frac{1623.20 + 6250}{6250} = ( \frac{100 + r}{100} ) {}^{3}$

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$\bf \implies \frac{7873.2}{6250} = {( \frac{100 + r}{100}) }^{3}$

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$\bf \implies \sqrt[3]{ \frac{7873.2}{6250} } = \frac{100 + r}{100}$

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$\bf \implies \frac{27}{25} = \frac{100 + r}{100}$

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$\bf \implies \frac{27}{25} \times 100 = 100 + r$

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$\bf \implies 27 \times 4 = 100 + r$

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$\bf \implies 100 + r = 27 \times 4$

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$\bf \implies 100 + r = 108$

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$\bf \implies r = 108 - 100$

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$\bf \implies r = 8 \%$

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Therefore, Rate is 8%.