Question

The compound Interest on ₹ 25000 for 3 years at 4% p.a compounded annually is _______​

Answer 1

Answer :-

• The compound interest on Rs 25000 for 3 years at 4% p.a compounded annually is Rs 3121.6.

Given :-

• Principal = Rs 25000.
• Rate = 4% p.a.
• Time = 3 years.

To find :-

• The compound inerest.

Step-by-step explanation :-

• First let's find the amount, then we will use it to find the compound interest!

We know that :-

$\underline{ \boxed{\sf Amount = Principal \Bigg(1 + \dfrac{Rate}{100}\Bigg)^{Time}}}$

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Here,

• Principal = Rs 25000.
• Rate = 4% p.a.
• Time = 3 years.

Hence,

$\rm Amount = 25000 \bigg(1 + \dfrac{4}{100} \bigg)^{3}$

Making 1 a fraction by taking 1 as the denominator,

$\rm Amount = 25000 \bigg( \dfrac{1}{1} + \dfrac{4}{100} \bigg)^{3}$

The LCM of 1 and 100 is 100, so adding the fractions using their denominators,

$\rm Amount = 25000 \bigg( \dfrac{1 \times 100 + 4 \times 1}{100} \bigg)^{3}$

On simplifying,

$\rm Amount = 25000 \bigg( \dfrac{100 + 4}{100} \bigg)^{3}$

Adding 4 to 100,

$\rm Amount = 25000 \bigg( \dfrac{104}{100} \bigg)^{3}$

The power here is 3, so removing the brackets and multiplying 104/100 with itself 3 times,

$\rm Amount = 25000 \times \dfrac{104}{100} \times \dfrac{104}{100} \times \dfrac{104}{100}$

Let's multiply 104/100 with itself 3 times first.

$\rm Amount =25000 \times \dfrac{104 \times 104 \times 104}{100 \times 100 \times 100}$

On multiplying,

$\rm Amount =25000 \times \dfrac{1124864}{1000000}$

Cutting off the zeroes,

$\rm Amount = 25 \times \dfrac{1124864}{1000}$

Reducing the numbers,

$\rm Amount =1 \times \dfrac{1124864}{40}$

Multiplying 1 with 1124864,

$\rm Amount = \dfrac{1124864}{40}$

Dividing 1124864 by 40,

$\overline{ \boxed{\rm Amount = Rs \: 28121.6}}$

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Now, as we know the amount, let's find the compound interest!

We know that :-

$\underline{ \boxed{ \sf CI = Amount - Principal}}$

Here,

• Amount = Rs 28121.6.
• Principal = Rs 25000.

Hence,

$\boxed{ \bf CI =28121.6 - 25000}$

$\overline{ \boxed{ \bf CI =Rs \: 3121.6}}$

• Hence, the compound interest is Rs 3121.6.

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Abbreviations used :-

$\rm CI = Compound \: Interest.$

Answer 2

Given :

• Principal (P) = Rs 25000
• Time (n) = 3 years
• Rate (R) = 4% per annum

To Find :

• Compound interest

Formulas :

$\red \bigstar \: {\underline{\boxed{\mathcal {\pmb{\quad Amount \: = \: Principal \: \times \bigg(\:1 \: + \: \dfrac{Rate}{100}\bigg)^{n}\quad}}}}}$

$\red \bigstar \: {\underline{\boxed{\mathcal {\pmb{\quad Compound \: Interest \: = \: Amount \: - \: Principal \quad}}}}}$

Solution :

${:} \longrightarrow \sf \: Amount \: = \: Principal \: \times \bigg(\:1 \: + \: \dfrac{Rate}{100}\bigg)^{n} \\ \\ {:} \longrightarrow \sf \: Amount \: = \: 25000 \: \times \bigg(\:1 \: + \: \dfrac{4}{100}\bigg)^{3} \\ \\ {:} \longrightarrow \sf \: Amount \: = \: 25000 \: \times \bigg(\dfrac{104}{100}\bigg)^{3} \\ \\ {:} \longrightarrow \sf \: Amount \: = \: 25000\: \times \: \dfrac{104}{100} \: \times \: \dfrac{104}{100} \: \times \: \dfrac{104}{100} \\ \\ {:} \longrightarrow \sf \: Amount \: = \: 25\cancel{000} \: \times \: \dfrac{104}{1 \cancel{00}} \: \times \: \dfrac{104}{10 \cancel{0}} \: \times \: \dfrac{104}{100} \\ \\ {:} \longrightarrow \sf \: Amount \: = \: \dfrac{25 \: \times \: 104 \: \times \: 104 \: \times \: 104}{10 \: \times \: 100} \\ \\ {:} \longrightarrow \sf \: Amount \: = \: \dfrac{2600 \: \times \: 10816}{ 1000} \\ \\ {:} \longrightarrow \sf \: Amount \: = \: \dfrac{2,81,21,600}{ 1000} \\ \\ {:} \longrightarrow \underline{\boxed{\sf \: Amount \: = \: Rs \: 28,121.6}} \: \blue \bigstar$

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$\longmapsto {\pmb{ Compound \: Interest \: = \: Amount \: - \: Principal} } \\ \\ \longmapsto {\sf{ Compound \: Interest \: = \: 28,121.6 \: - \: 25000} } \\ \\ \longmapsto \underline{\boxed{\pmb{ \red{ Compound \: Interest \: = \:Rs \:3,121.6} }}}$

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$\qquad \therefore\: \sf{ Amount \: = \underline {\underline{ Rs \: 28,121.6}}}$

$\qquad \therefore\: \sf{ Compound \: Interest\: = \underline {\underline{ Rs \: 3,121.6}}}$

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