Question

A certain sum of money is invested at the rate of 10% per annum compound interest, the interest compounded annually.
if the difference between the interests of third year is 1,105,
find the invested?
​

Answer 1

$\star\;{\underline{\rm{\red{GivEn:-}}}}$

• Rate of interest.

• Difference b/w first and third year = Rs. 1105

$\star\;{\underline{\rm{\blue{To\;Find:-}}}}$

• Sum invested

$\star\;{\underline{\rm{\green{SoluTion:-}}}}$

Lets the sum invested be P.

As we know that,

$\dag\;{\underline{\boxed{\bf{\pink{Simple\; Interest\;(SI) = \dfrac{ P \times R \times T}}}}}}$

$\therefore\;\sf Interest\;for\;first\;year\;= \dfrac{P \times 10 \times 1}{100}\\\\\\ :\implies\sf \dfrac{P}{10}$

And

$\therefore\;\sf Amount\;for\;first\;year\;= P + \dfrac{P}{10}\\\\\\ :\implies \dfrac{11P}{10}$

$\therefore\;\sf Principal\;for\;2nd\year = \dfrac{11P}{10}\\\\\\ \sf Now \\\\\\ \sf Interest\;for\;second\;year\;= \dfrac{11P \times \cancel{10} \times 1}{100 \times \cancel{10}}\\\\\\ :\implies\sf \dfrac{11P}{100}\\\\\\ \sf Amount\;for\;second\;year = \dfrac{11P}{100} + \dfrac{11P}{10}\\\\\\ :\implies\sf \dfrac{121P}{100}\\\\\\ \therefore\sf Principal\;for\;third\;year = \dfrac{121P}{100}\\\\\\ \sf Now \\\\\\ \sf Interest\;for\;third\;year\;= \dfrac{121P \times \cancel{10} \times 1}{100 \times \cancel{100}}\\\\\\ :\implies\sf \dfrac{121P}{1000}$

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Hence, Difference b/w interest of 3rd year and first year is,

Here, difference is given by 1105.

$:\implies\sf \dfrac{121P}{1000} - \dfrac{P}{10} = 1105$

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$:\implies\sf \dfrac{121P - 100P}{1000}= 1105$

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$:\implies\sf 121P - 100P = 1105 \times 1000$

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$:\implies\sf 21P = 1105 \times 1000$

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$:\implies\sf P = \dfrac{1105 \times 1000}{21}$

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$:\implies\sf{\underiine{\boxed{\bf{\purple{52619.04}}}}}$

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$\therefore\;\sf \underline{The\;Sum\; invested\;is\; \bf{52619.04.}}$

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Answer 2

sis here

not bro!

sis is more powerful than bro.

careful.