Question

4000 for 2 years at 4% p.a. compound annually?​

Answer 1

Appropriate Question :-

• 4000 for 2 years at 4% p.a. compound annually. Find the amount and Compound Interest.  

Given :-  

• 4000 for 2 years at 4% p.a. compound annually

To Find :-  

• The amount and the compound interest

Solution :-

~Here, we’re given the principal , rate of interest and time for which the money is invested and we need to find the Compound Interest ( CI ) after the given time. We can easily find the amount by putting the values in it’s formula and then Compound Interest.  

 _____________

Here,  

• The principal  ( P ) is Rs. 4,000

• The rate ( R ) is 4 %  

• The time ( n )  is 2 years  

• The amount will be ( A )  

• Compound Interest be ( CI )  

 _____________

As we know that ,  

$\underline{\boxed{\sf \maltese \;\; A = P \bigg\{ 1 + \dfrac{R}{100} \bigg\} ^{n}}}$

$\underline{\boxed{\sf \maltese \;\; CI = A - P}}$

 _____________

Finding the amount ::  

$\sf \implies A = 4000 \bigg\{ 1 + \dfrac{4 }{100} \bigg\} ^{2}$

$\sf \implies A = 4000 \bigg\{ \dfrac{104}{100} \bigg\} ^{2}$

$\sf \implies A = 4000 \times \dfrac{104}{100} \times \dfrac{104}{100}$

$\sf \implies A = 4 \times 10.4 \times 104$

$\sf \leadsto A = 4326.4$

Finding the Compound Interest ::  

$\sf \implies CI = 4326.4 -4000$

$\sf \leadsto CI = 326.4$

  _____________

Hence,  

• The amount is Rs. 4326.4 and the compound interest is 326.4  

 _____________

Answer 2

Given :-

• Principal (P) = Rs 4000
• Time (n) = 2 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 :-

${:} \longrightarrow \sf \: Amount \: = \: Principal \: \times \bigg(\:1 \: + \: \dfrac{Rate}{100}\bigg)^{n} \\ \\ {:} \longrightarrow \sf \: Amount \: = \: 4000 \: \times \bigg(\:1 \: + \: \dfrac{4}{100}\bigg)^{2} \\ \\ {:} \longrightarrow \sf \: Amount \: = \: 4000 \: \times \bigg(\dfrac{104}{100}\bigg)^{2} \\ \\ {:} \longrightarrow \sf \: Amount \: = \: 4000\: \times \: \dfrac{104}{100} \: \times \: \dfrac{104}{100} \\ \\ {:} \longrightarrow \sf \: Amount \: = \: 4\cancel{000} \: \times \: \dfrac{104}{1 \cancel{00}} \: \times \: \dfrac{104}{10 \cancel{0}} \\ \\ {:} \longrightarrow \sf \: Amount \: = \: \dfrac{4 \: \times \: 104 \: \times \: 104}{10}\\ \\ {:} \longrightarrow \sf \: Amount \: = \: \dfrac{416 \: \times \: 104}{10}\\ \\{:} \longrightarrow \sf \: Amount \: = \: \dfrac{ 43264 }{10}\\ \\ {:} \longrightarrow \underline{\boxed{\sf \: Amount \: = \: Rs \: 4326.4}} \: \blue \bigstar$

━━━━━━━━━━

$\longmapsto {\pmb{ Compound \: Interest \: = \: Amount \: - \: Principal} } \\ \\ \longmapsto {\sf{ Compound \: Interest \: = \: 4326.4 \: - \: 4000} } \\ \\ \longmapsto \underline{\boxed{\pmb{ \red{ Compound \: Interest \: = \:Rs \:326.4} }}}$

━━━━━━━━━━

$\qquad \therefore\: \sf{ Amount \: = \underline {\underline{ Rs \: 4326.4}}}$

$\qquad \therefore\: \sf{ Compound \: Interest\: = \underline {\underline{ Rs \: 326.4}}}$

━━━━━━━━━━