How To Solve Compound Interest Questions Quickly

July 15, 2023

Solve Compound Interest Problems Quickly

Go through the page completely to know How To Solve Compound Interest Quickly.

How to solve Compound interest problems quickly

Definition of 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.

Options:         

A. 408

B. 406.04

C. 409.03

D. 405.50

Solution:    According to the question

P =10000, r = 2%, t = 2

We know that,

Amount = P \mathbf{(1+ \frac{{\frac{r}{2}}}{100})^{2t}}

A = 10000 \mathbf{(1+ \frac{{\frac{2}{2}}}{100})^{2*2}}

A = 10000 \mathbf{(1+ \frac{1}{100})^{4}}

A = 10000 (1+ 0.01)4

A = 10000 (1.01)4

A = 10000 (1.04)

A = 10406.04

Now, CI = A – P

CI = 10406.04 – 10000

CI = 406.04

Correct option: B

Question 3. Find the CI on Rs. 2000 in 9 months at 12% p.a. if the interest is calculated quarterly. 

Options:         

A. 251.01

B. 251

C. 304

D.185.45

Solution:     According to the question,

P = 2000, r = 6%, t = 9 months (3 quarter)

We know that,

Amount = P (1+ \frac{{\frac{r}{4}}}{100})^{4t}

A = 2000 (1+ \frac{{\frac{12}{4}}}{100})^{4*\frac{9}{12}}

A = 2000 (1+ \frac{{\frac{12}{4}}}{100})^{4*\frac{3}{4}}

A = 2000 (1+ \frac{3}{100})^{3}

A = 2000 (1+0.03)3

A = 2000 (1.03) 3

A = 2000 (1.09)

A = 2185.45

Now, CI = A – P

CI = 2185.45– 2000 = 185.45

Correct option: D

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Type 2: When Rate of Interest, Time period, Principal are given

Question 1. The CI on Rs. 20,000 at 6% per annum is Rs. 2472. Find the period (in years):

Options:         

A. 2

B. 4

C. 5

D. 3

Solution:    Amount = 20000 + 2472 = Rs. 22472

Let the time = n years

So, 20000 (1+ \frac{6}{100})^{n} = 22472

(\frac{106}{100})^{n} = (\frac{22472}{20000}) = (\frac{11296}{10000}) = (\frac{106}{100})^{2}

Therefore, n = 2 years

Correct option: A

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?

Options:

A. 8%
B. 5%
C. 4%
D. 3%

Solution:    Difference = P × \mathbf{ \frac{(R)^{2}}{(100)^{2}}}

96 =15000 × \frac{(R)^{2}}{10000}

\frac{96 ×10000 }{15000} =R^{2}

R^{2} = 64

R = 8%

Correct option: A

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?

Options:

A. Rs. 7270

B. Rs. 2270

C. Rs. 7720

D. Rs. 7027

Solution:    According to the question,

P = 20000

R = 5%, 10%, and 20%

T= 1, 2, and 3 years

Amount = 20000 (1+\frac{5}{100}) ( 1+\frac{10}{100})(1+\frac{20}{100})

Amount = 20000 (\frac{105}{100}) (\frac{110}{100})(\frac{120}{100})

Amount = (20000) (1.05) (1.1) ( 1.2)

Amount = 27720

We know that, CI = A – P

CI = 27720 – 20000

CI = Rs. 7720

Correct option: C

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})

Amount = 10000 (1+\frac{8}{100}) (1+\frac{10}{100}) (1+\frac{15}{100})

Amount = 10000 (1.08) (1.1) (1.15)

Amount = 13662

Now, CI = amount –principal

CI = 13662 – 10000

CI = 3662

Correct option: D

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}

Amount after two years = (60000)(1.05)(1.15)

Amount after two years = Rs. 72450

Correct Option: A

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