Paras takes a loan of 20,000 at a compound interest rate of 5% per annum (p.a.)
(i) Find the compound interest after one year.
(ii) Find the compound interest for two years.
(iii) Find the sum of money required to clean the debt at the end of two years.
(iv) Find the difference between the compound interest and the simple interest
same rate for two years.
Answers
Answer:
i) P = Rs.20000
R = 5%
T = 1 year
\sf{A = P\bigg(1+\dfrac{r}{100}\bigg)^t}A=P(1+
100
r
)
t
= \sf{A = 20000\bigg(1+\dfrac{5}{100}\bigg)^1}A=20000(1+
100
5
)
1
= \sf{A = 20000\bigg(\dfrac{100+5}{100}\bigg)^1}A=20000(
100
100+5
)
1
= \sf{A = 20000\bigg(\dfrac{105}{100}\bigg)^1}A=20000(
100
105
)
1
= \sf{A = 20000\times\dfrac{105}{100}}A=20000×
100
105
= \sf{A = 21000}A=21000
\sf{CI = A - P}CI=A−P
= \sf{CI = 21000 - 20000}CI=21000−20000
= \sf{CI = 1000}CI=1000
Therefore CI after 1 year will be Rs.1000.
ii) P = Rs.20000
R = 5%
T = 2 years
\sf{A = 20000\bigg(1+\dfrac{5}{100}\bigg)^2}A=20000(1+
100
5
)
2
= \sf{A = 20000\bigg(\dfrac{100+5}{100}\bigg)^2}A=20000(
100
100+5
)
2
= \sf{A = 20000\bigg(\dfrac{105}{100}\bigg)^2}A=20000(
100
105
)
2
= \sf{A = 20000\times\dfrac{105}{100}\times\dfrac{105}{100}}A=20000×
100
105
×
100
105
= \sf{A = 22050}A=22050
= \sf{CI = 22050 - 20000}CI=22050−20000
= \sf{CI = Rs.2050}CI=Rs.2050
Therefore CI after 2 years will be Rs.2050
iii) Sum of money at the end of 2 year to clear the debt = (CI after 1 year) + (CI after 2 years)
\sf{Sum\:of\:money = 2050 + 1000}Sumofmoney=2050+1000
\sf{Sum\:of\:money = 3050}Sumofmoney=3050
Therefore the sum of money required to clear the debt at the end of 2 years is Rs.3050
iv) Difference between SI and CI after two years:-
P = 20000
R = 5%
T = 2 years
\sf{SI = \dfrac{P\times R\times T}{100}}SI=
100
P×R×T
= \sf{SI =\dfrac{20000\times 5\times 2}{100}}SI=
100
20000×5×2
= \sf{SI = Rs.2000}SI=Rs.2000
CI after 2 years = Rs.2050
Difference between CI and SI
= \sf{2050-2000}2050−2000
= \sf{50}50
Therefore the difference between CI and SI after 2 years will be 50.
Answer:
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