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Compound InterestCompound interest is the multiplication of the initial principal amount and one plus the annual interest rate raised to the number of compound periods minus one i.e Compound Interest
Compound interest is the interest calculated on the original principal and on the accumulated past interest of a deposit or loan.
Compound interest calculated by multiplying the original principal amount one plus the annual interest rate raised to the number of compound periods minus one.
Basic Formula for the Compound Interest ,
A= P(1 +\mathbf{\frac{r}{n}})nt
Here, A = Amount
P = Principal
r = Interest rate(decimal)
n = number of times interest is compounded per unit ‘t’
t = total time
Type 1: Problems on Compound Interest (Yearly, Quarterly, and Half-yearly)
Question 1 . If a sum of Rs.100 is invested for 10% p.a at CI then the sum of the amount will be Rs.121 in
Options:
A. 2 years
B.1 year
C.1.5 years
D. 3 years
Solution: We know that, Amount =P(1+ \mathbf{\frac{r}{100}})^{t}
121 = 100 (1 + \frac{10}{100})^t
(\frac{121}{100}) = (\frac{11}{10})^t
(\frac{11}{10})^2= (\frac{11}{10})^t
t = 2 years
Correct option: A
Question 2. Find the CI on Rs. 10,000 in 2 years at 2 % per annum, the interest being compounded half-yearly.
Question 2. The principal amount is put on CI for two years at 40%. It gets 964 more if the interest is payable half yearly. Calculate the sum.
Options:
A. Rs 8485 B. Rs 8485.91 C. Rs 8480 D. Rs 8455.91
Solution: Let us assume the Principal as Rs. 100
When compounded annually
A = 100 (1+ \frac{40}{100})^{2}
A= 196
When compounded half yearly
A = 100 (1+ \frac{\frac{40}{2}}{100})^{4}
A = 100 (1+ \frac{20}{100})^{4}
A = 207.36
Difference, 207.36 – 196 = 11.36
If difference is 11.36, then Principal = Rs 100
If difference is 964, then Principal = ( \frac{100}{11.36}) × 964
P = Rs 8485.91
Correct option: B
Question 3. In what time the CI on Rs 800 at 30% pa will amount to Rs.1352 if calculated annually?
Options:
A. 3 years B. 1.6 years C. 2 years D. 5 years
Solution: We know that,
Amount = P (1+ \frac{r}{100})^{t}
1352 = 800 (1+ \frac{30}{100})^{t}
(\frac{1352}{800})= (1.3)^{t}
1.69 = (1.3)^{t}
1.69 = (1.3)^{2}
Therefore, t = 2 years
Correct option: C
Type 3: When difference of compound and simple interest is given
Question 1. If the difference between compound interest and simple interest on a certain principal amount is Rs 500 at 10% p.a. for 2 years. Then calculate the sum.
Options:
A. 45000 B. 50000 C. 55000 D. 5000
Solution: Given the difference between SI and CI = 500
Time = 2 years
When the difference between SI and CI is of two years then,
Difference = P \mathbf{ \frac{(R)^{2}}{(100)^{2}}}
Where P = principal amount, R = rate of interest
According to the question:
500 = P \frac{(10)^{2}}{(100)^{2}}
P = 50,000
Correct option: B
Question 2. The difference between CI and SI on an amount of Rs. 15,000 for 2 years is Rs. 96. Find the rate of interest?
Question 3. The difference between CI and SI calculated annually on a certain amount of money for two years at 4% pa is Re. 1. Find the sum:
Options:
A. 600 B. 700 C. 650 D. 625
Solution: When the difference between SI and CI is of two years then,
Difference =P × \mathbf{ \frac{(R)^{2}}{(100)^2}}
Where P = principal amount, R = rate of interest
1 = P × \frac{(4)^{2}}{(100)^2}
P = 625
Correct option: D
Type 4: Problem related to sum of money becomes x time in ‘a’ years and y times in ‘b’ years.
Question 1. A sum of money borrowed under CI gets double in 5 years. When will it become eight times of itself if the rate of interest remains same?
Options:
A. 15 years B. 20 years C. 10 years D. 5 years
Solution: We know that,
(x)^{\frac{1}{a}} =(y)^{\frac{1}{b}}
(2)^{\frac{1}{5}} =(8)^{\frac{1}{b}}
(2)^{\frac{1}{5}} =(2)^{\frac{3}{x}}
\frac{1}{5} =\frac{3}{x}
x = 15 years
Correct option: A
Question 2. If a certain amount of sum becomes 9 times in 2 years at compound interest, then find out the rate of interest?
Options:
A. 250% B. 100% C. 200% D. 20%
Solution: We know that,
r = 100 ((\frac{A}{P})^\frac{1}{t} – 1)
Therefore,
r = 100((\frac{9P}{P})^\frac{1}{2} – 1)
r =100 (9^\frac{1}{2} – 1)
r = 100 (3-1)= 100*2
r = 200%
Correct option: C
Question 3. At what rate percentage will certain amount of money become 8 times in three years?
Options:
A. 113. 79% B. 100% C. 110. 79% D. 130%
Solution: We know that
r = 100 ((\frac{A}{P})^\frac{1}{t} – 1)
r = 100 ((8)^\frac{1}{3} – 1)
r = 100 * (2-1)
r = 100%
Correct option: B
Type 5: How To Solve Compound Interest Quickly (When rates are different for different years)
Question 1. Ravi took an amount of Rs.20000 as loan at CI charging 5%, 10% and 20% for the 1st year, 2nd year, and 3rd year respectively. Find out the total interest to be paid by Ravi at the end of the 3rd year?
Question 2. If the rate of CI for the 1st year is 8%, 2nd year is 10%, and for 3rd year is 15% then find the amount and the CI on Rs 10000 in three years.
Options:
A. 3626
B. 6236
C. 3666
D. 3662
Solution: We know that, Amount = P (1 + \frac{r_{1}}{100}) (1 + \frac{r_{2}}{100}) (1 + \frac{r_{3}}{100})
Question 3. Karan bought a bike of Rs. 60000 by paying cash. He borrowed the cash from his friend at rate of interest 5% for the 1st year and 15% for the 2nd year. Find out the total amount he has to pay after 2 years to his friend.
Options:
A. Rs. 72450
B. Rs. 74250
C. Rs. 72540
D. Rs. 75420
Solution We know that, Amount = P (1+\frac{r_{1}}{100})(1+\frac{r_{2}}{100})
Amount after two years = 60000 (1+\frac{5}{100}(1+\frac{15}{100}
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