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
Given:
A sum of Rs 30,000 is invested for 3 years at % 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:
Principal, P = Rs. 30,000
Time, n = 3 years
The rate of interest, R =
(i). Finding the sum at the end of the 2nd year:
Here, n = 2 years
∴
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
∴
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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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.