Math, asked by yadavsubhash66banka, 10 months ago

3. A sum of Rs 30,000 is invested for 3 years at 71% pa. compound interest
(i) What is the sum at the end of the second year?
(ii) What is the sum at the end of the 3 years?
(iii) What is the total interest earned in 3 years?

Answers

Answered by bhagyashreechowdhury
24

Given:

A sum of Rs 30,000 is invested for 3 years at 7\frac{1}{2}% pa. compound interest

To find:

(i) What is the sum at the end of the second year?

(ii) What is the sum at the end of the 3 years?

(iii) What is the total interest earned in 3 years?

Solution:

Formula to be used:

\boxed{\bold{A = P[1 + \frac{R}{100} ]^n}}\\\\\boxed{\bold{C.I. = A - P}}

Principal, P = Rs. 30,000

Time, n = 3 years

The rate of interest, R = 7\frac{1}{2} \% = \frac{15}{2}\%

(i). Finding the sum at the end of the 2nd year:

Here, n = 2 years

A = 30000 [1 + \frac{\frac{15}{2} }{100} ]^2

\implies A = 30000 [1 + \frac{15}{200} ]^2

\implies A = 30000 [ \frac{215}{200} ]^2

\implies A = 30000 \times \frac{215}{200} \times \frac{215}{200}

\implies A = 0.75 \times 215 \times 215

\implies \bold{A = Rs. \:34668.75}

Thus, the sum at the end of the second year is Rs. 34668.75.

(ii). Finding the sum at the end of the 3 years:

Here, n = 3 years

A = 30000 [1 + \frac{\frac{15}{2} }{100} ]^3

\implies A = 30000 [1 + \frac{15}{200} ]^3

\implies A = 30000 [ \frac{215}{200} ]^3

\implies A = 30000 \times \frac{215}{200} \times \frac{215}{200}\times \frac{215}{200}

\implies \bold{A = Rs. \:37268.90}

Thus, the sum at the end of the second year is Rs. 37268.90.

(iii). Finding the total interest earned in 3 years:

The amount at the end of 3 years = Rs. 37268.90

∴ C.I. = Rs. 37268.90 - Rs. 30000 = Rs. 7268.90

Thus, the total interest earned in 3 years is Rs. 7268.90.

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Answered by gusangha22
0

Solution:

Formula to be used:

\begin{gathered}\boxed{\bold{A = P[1 + \frac{R}{100} ]^n}}\\\\\boxed{\bold{C.I. = A - P}}\end{gathered}

A=P[1+

100

R

]

n

C.I.=A−P

Principal, P = Rs. 30,000

Time, n = 3 years

The rate of interest, R = 7\frac{1}{2} \% = \frac{15}{2}\%7

2

1

%=

2

15

%

(i). Finding the sum at the end of the 2nd year:

Here, n = 2 years

∴ A = 30000 [1 + \frac{\frac{15}{2} }{100} ]^2A=30000[1+

100

2

15

]

2

\implies A = 30000 [1 + \frac{15}{200} ]^2⟹A=30000[1+

200

15

]

2

\implies A = 30000 [ \frac{215}{200} ]^2⟹A=30000[

200

215

]

2

\implies A = 30000 \times \frac{215}{200} \times \frac{215}{200}⟹A=30000×

200

215

×

200

215

\implies A = 0.75 \times 215 \times 215⟹A=0.75×215×215

\implies \bold{A = Rs. \:34668.75}⟹A=Rs.34668.75

Thus, the sum at the end of the second year is Rs. 34668.75.

(ii). Finding the sum at the end of the 3 years:

Here, n = 3 years

∴ A = 30000 [1 + \frac{\frac{15}{2} }{100} ]^3A=30000[1+

100

2

15

]

3

\implies A = 30000 [1 + \frac{15}{200} ]^3⟹A=30000[1+

200

15

]

3

\implies A = 30000 [ \frac{215}{200} ]^3⟹A=30000[

200

215

]

3

\implies A = 30000 \times \frac{215}{200} \times \frac{215}{200}\times \frac{215}{200}⟹A=30000×

200

215

×

200

215

×

200

215

\implies \bold{A = Rs. \:37268.90}⟹A=Rs.37268.90

Thus, the sum at the end of the second year is Rs. 37268.90.

(iii). Finding the total interest earned in 3 years:

The amount at the end of 3 years = Rs. 37268.90

∴ C.I. = Rs. 37268.90 - Rs. 30000 = Rs. 7268.90

Thus, the total interest earned in 3 years is Rs. 7268.90.

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