Question

find the compound interest if P=4000rs , N= 3 years , R=5% p.c.p.a
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Answer 1

Before, finding the answer. Let's find out how we can find this answer.

• To find the Compound Interest, we must first find the Amount. To find the Amount we use the formula of :

$\sf Amount = P\Bigg[1 + \dfrac{r}{100}\Bigg]^t$

• Then, we must Subtract the Amount and Principal to find the Compound Interest.

_______________

Given :

• Principal = Rs. 4000
• N (Time) = 3 years
• Rate = 5%

To find :

• compound interest

Solution :

$\sf Amount = P\Bigg[1 + \dfrac{r}{100}\Bigg]^t$

            $\sf = 4000 \Bigg[1 + \dfrac{5}{100}\Bigg]^3$

            $\sf = 4000 \Bigg[1 + \dfrac{1}{20}\Bigg]^3$

            $\sf = 4000 \Bigg[1 + \dfrac{ (1 \times 20 + 1 \times 1) }{20}\Bigg]^3 [L.C.M \: of \: 20 \: and \: 1 = 20]$

            $\sf = 4000 \Bigg[1 + \dfrac{ (20 + 1) }{20}\Bigg]^3$

            $\sf = 4000\Bigg[ \dfrac{ 21}{20}\Bigg]^3$

            $\sf = 4000 \times \dfrac{ 21}{20} \times\dfrac{ 21}{20} \times \dfrac{ 21}{20}$

            $\sf = 4000 \times \dfrac{ 21 \times 21 \times 21 }{20 \times 20 \times 20}$

            $\sf = 4000 \times \dfrac{9261 }{8000}$

            $\sf = 40 \not0 \not0 \times \dfrac{9261 }{80 \not 0 \not0}$

            $\sf = 40 \times \dfrac{9261 }{80 }$

            $\sf = \dfrac{40 \times 9261 }{80 }$

            $\sf = \dfrac{370440 }{80 }$

            $\sf = Rs. 4630.5$

Therefore, the Amount is Rs.4630.5.

Compound Interest = Amount - Principal

                                 = 4630.5 - 4000

                                 = Rs. 630.5

Therefore, the Compound Interest is Rs. 630.5.

           

           

Answer 2

Answer:

Amount=P[1+100r]t

Then, we must Subtract the Amount and Principal to find the Compound Interest.

_______________

Given :

Principal = Rs. 4000

N (Time) = 3 years

Rate = 5%

To find :

compound interest

Solution :

\sf Amount = P\Bigg[1 + \dfrac{r}{100}\Bigg]^tAmount=P[1+100r]t

            \sf = 4000 \Bigg[1 + \dfrac{5}{100}\Bigg]^3=4000[1+1005]3

            \sf = 4000 \Bigg[1 + \dfrac{1}{20}\Bigg]^3=4000[1+201]3

            \sf = 4000 \Bigg[1 + \dfrac{ (1 \times 20 + 1 \times 1) }{20}\Bigg]^3 [L.C.M \: of \: 20 \: and \: 1 = 20]=4000[1+20(1×20+1×1)]3[L.C.Mof20and1=20]

            \sf = 4000 \Bigg[1 + \dfrac{ (20 + 1) }{20}\Bigg]^3=4000[1+20(20+1)]3

            \sf = 4000\Bigg[ \dfrac{ 21}{20}\Bigg]^3=4000[2021]3

            \sf = 4000 \times \dfrac{ 21}{20} \times\dfrac{ 21}{20} \times \dfrac{ 21}{20}=4000×2021×2021×2021

            \sf = 4000 \times \dfrac{ 21 \times 21 \times 21 }{20 \times 20 \times 20}=4000×20×20×2021×21×21

            \sf = 4000 \times \dfrac{9261 }{8000}=4000×80009261

            \sf = 40 \not0 \not0 \times \dfrac{9261 }{80 \not 0 \not0}=4000×80009261

            \sf = 40 \times \dfrac{9261 }{80 }=40×809261

            \sf = \dfrac{40 \times 9261 }{80 }=8040×9261

            \sf = \dfrac{370440 }{80 }=80370440

            \sf = Rs. 4630.5=Rs.4630.5

Therefore, the Amount is Rs.4630.5.

Compound Interest = Amount - Principal

                                 = 4630.5 - 4000

                                 = Rs. 630.5

Therefore, the Compound Interest is Rs. 630.5.