# How To Solve Compound Interest Questions Quickly

## Solve Compound Interest Problems Quickly

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