Question

A certain sum amounts to Rs. 4,000 at the end of 5 years at 12% p.a. interest. Find the sum

Answer 1

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

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• Amount = Rs. 4000
• Time = 5 years
• Rate = 12%

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$\sf{\bold{\green{\underline{\underline{To\:Find}}}}}$

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• Principle = ??

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$\sf{\bold{\green{\underline{\underline{Solution}}}}}$

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$\sf{\red{\boxed{\bold{Amount = Principle + \dfrac{Principle \times Rate\times Time}{100}}}}}$

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• Let principle be p

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$\sf :\implies\: {\bold{4000= p + \dfrac{p\times 12\times 5 }{100} }}$

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$\sf :\implies\: {\bold{4000= p + \dfrac{p\times 12 }{20} }}$

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$\sf :\implies\: {\bold{4000= p + \dfrac{p\times 6 }{10} }}$

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$\sf :\implies\: {\bold{4000= p + \dfrac{6p }{10} }}$

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$\sf :\implies\: {\bold{4000= \dfrac{10p + 6p }{10} }}$

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$\sf :\implies\: {\bold{4000= \dfrac{16p}{10}}}$

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$\sf :\implies\: {\bold{ p = \dfrac{4000\times 10}{16}}}$

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$\sf :\implies\: {\bold{ p = \dfrac{2000\times 10}{8}}}$

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$\sf :\implies\: {\bold{ p = \dfrac{1000\times 10}{4}}}$

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$\sf :\implies\: {\bold{ p = \dfrac{500\times 10}{2}}}$

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$\sf :\implies\: {\bold{ p = 250\times 10}}$

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$\sf :\implies\: {\bold{ p = Rs. 2500}}$

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$\sf{\bold{\green{\underline{\underline{Answer}}}}}$

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• Rs. 2500 will amount to Rs. 4,000 in 5 years at 12% per annum

Answer 2

$\huge\bold{\mathbb{QUESTION}}$

A certain sum amounts to $Rs.\:4000$ at the end of $5$ years at $12 \%$ p.a. interest. Find the sum.

$\huge\bold{\mathbb{GIVEN}}$

• Amount $= Rs.\:4000$

• Time $=5$ yrs

• Rate of interest $=12 \%$

$\huge\bold{\mathbb{TO\:FIND}}$

The sum.

$\huge\bold{\mathbb{SOLUTION}}$

Let the principle be $p.$

We know that:

$\red{\boxed{\Large{a=p+{\frac{p\times r\times t}{100}}}}}$

Where,

• $a$ is the amount.

• $p$ is the principle.

• $r$ is the rate.

• $t$ is the time.

Here,

• $a=4000$

• $r=12$

• $t=5$

Let's find out the principle.

$\therefore 4000=p+{\dfrac{p\times 12\times 5}{100}}$

$\implies 4000=p+{\dfrac{60p}{100}}$

$\implies 4000={\dfrac{(100p+60p)}{100}}$

$\implies 4000={\dfrac{160p}{100}}$

$\implies 4000={\dfrac{16\cancel{0}p}{10\cancel{0}}}$

$\implies 4000={\dfrac{16p}{10}}$

By cross multiplication.

$\implies (4000\times 10) = 16p$

$\implies 40000 = 16p$

$\implies {\dfrac{40000}{16}} = p$

$\implies \cancel{{\dfrac{40000}{16}}} = p$

$\implies 2500 = p$

$\implies p = 2500$

$\huge\bold{\mathbb{HENCE}}$

$p = 2500$

Principle $=p = 2500.$

$\huge\bold{\mathbb{THEREFORE}}$

The sum is $Rs.\: 2500.$

$\huge\bold{\mathbb{DONE}}$