Practice Test on Compound Interest | Compound Interest Questions and Answers

Students can find several questions on Compound Interest. Practice the Objective Questions of Compound Interest Over here and be prepared for the exams. Learn how to solve Compound Interest Problems by checking the Solved Examples. Use the Sample Problems over here covering various questions including the Compound Interest Formula.

Compound Interest Practice Test has questions when the Interest Rate is Compounded Annually, Half-Yearly, Quarterly, Various Rate of Interest, Amount Calculations, etc. Solve the Questions on CI and test your knowledge on the related areas and bridge the gap accordingly.

1. The compound interest on $ 20,000 at 5 % per annum for 3 years, compounded annually is?

Solution:

P = $20,000

R = 5%

n = 3 Years

A = P(1+R/100)n

= 20,000(1+5/100)3

=20,000(105/100)3

= 20,000(1.157)

= $23152

CI = A – P

= $23152 – $20,000

= $3152

2. The simple interest on a sum of money for 2 years at 3 % per annum is $ 6250. What will be the compound interest on the same sum at the same rate for the same period, compounded annually?

Solution:

From given data SI = $6250

T = 2 Years

R = 3%

SI = PTR/100

6250 = P*2*3/100

6250 = 6P/100

6250*100 = 6P

6P = 625000

P = $1,04,166

We know A = P(1+R/100)n

=1,04,166(1+3/100)2

=1,04,166(103/100)2

=1,10,509

Compound Interest = Amount – Principal

= $1,10,509 – $1,04,166

= $6343

3. If a sum of Rs. 10,000 lent for 10% per annum at compound interest then the sum of the amount will be Rs. 14,161 in

Solution:

We know A = P(1+R/100)n

Given P = Rs. 10,000

R = 10%

A = 14,161

n = ?

Substitute the input values in the formula of Amount we have

14161 = 10000(1+10/100)n

14161/10000 = (1+10/100)n

(11/10)4 = (11/10)n

n = 4 Years

4. The population of a city is 1,20,000. It increases by 5% in the first year and increases by 10% in the second year. What is the population of the town at the end of 2 yrs?

Solution:

The population of city = 1,20,000

The population of city after 2 years = P(1+R1/100)(1+R2/100)

= 1,20,000(1+5/100)(1+10/100)

= 1,20,000(1.05)(1.1)

= 1,38,600

Therefore, the Population of the city by the end of 2 years is 1,38,600.

5. The difference between simple interest and compound on Rs. 1500 for one year at 20% per annum reckoned half-yearly is

Solution:

Given Data is Principal = 1500

R = 20%

T = 1 year

SI = PTR/100

= (1500*1*20)/100

= Rs. 300

Amount A = P(1+R/100)n

=1500(1+(10/100))2

= 1500(1+1/10)2

=1500(1.1)2

= Rs. 1815

CI = A – P

= Rs. 1815 – Rs. 1500

= Rs. 315

Difference = CI – SI

= Rs. (315 – 300)

= Rs. 15

Difference of Compound Interest and Simple Interest | How to Calculate Difference Between SI and CI?

The Major Difference Between Simple Interest and Compound Interest is just that Simple Interest is calculated on the Principal whereas Compound Interest is calculated on the Principal Amount along with the Interest accumulated for a certain period of time. Both Simple and Compound Interest are widely used concepts in the majority of financial services. Check out Solved Example Problems for finding the difference between CI and SI in the later sections. Get to know about the concept Difference of Compound Interest and Simple Interest in detail by going through the entire article.

How to find the Difference Between Simple Interest and Compound Interest?

Let us discuss in detail how to find the Difference between Simple Interest and Compound Interest. They are along the lines

Consider the Rate of Interest is the same for both Compound Interest and Simple Interest

Difference = Compound Interest for 2 years – Simple Interest for 2 Years

= P{(1+r/100)2-1} – P*r*2/100

= P*r/100*r/100

=((P*r/100)*r*1)/100

Solved Examples on Difference of Compound Interest and Simple Interest

1. Find the difference of the compound interest and simple interest on $ 10,000 at the same interest rate of 10 % per annum for 2 years?

Solution:

Simple Interest = PTR/100

= 10,000*10*2/100

= $2000

To find the Compound Interest firstly calculate the Amount

Amount A = P(1+R/100)n

A = 10,000(1+10/100)2

= 10,000(110/100)2

= 10,000(1.21)

= $12,100

Compound Interest = Amount – Principal

= 12,100 – 10,000

= $2,100

Difference between CI and SI = CI for 2 Years – SI for 2 Years

= $2100- $2000

= $100

2. What is the sum of money on which the difference between simple and compound interest in 2 years is $ 100 at the interest rate of 5% per annum?

Solution:

Simple Interest = PTR/100

Principal = P

T = 2 years

Substituting the given data in the formula for the simple interest we have

SI = (P*2*5)/100

To find the Compound Interest firstly, find out the Amount

Amount A = P(1+R/100)n

= P(1+5/100)2

CI = Amount – Principal

= P(1+5/100)2 – P

= P((1+5/100)2 -1)

Given the difference between CI and SI = $100

P((1+5/100)2 -1) – (P*2*5)/100 = $100

P((105/100)2 -1)-10P/100 = $100

P(1.1025-1)-10P/100 = $100

100P(0.1025)-10P =$10000

110.25P-10P = $10000

100.25P = $10000

P = $10000/100.25

= $99.75

Therefore, the Sum of Money is $99.75

Variable Rate of Compound Interest | Compound Interest Formula with Successive Rate of Interest

Let us discuss how to find the Compound Interest when a Variable Rate is given. Check out the Solved Examples on finding the Compound Interest When Rate of Successive Years is Different. We tried explaining each and every step for all the Problems provided here. Use the Problems over here and learn the concept behind them in no time. After going through this article, you will learn the concept of Variable Rate of Compound Interest quite easily.

How to find Compound Interest When Successive Years Rate of Interest is Different?

Get to know in detail how to find the Compound Interest when Consecutive/ Successive Years Rate of Interest is Different from the below sections.

Let us consider the amount be A and Principal be P,

Rate of Compound Interest for Successive Years is different i.e. r1%, r2%, r3%, r4%, …… then the Formula to calculate amount is given by

A = P(1+r1/100)(1+r2/100)(1+r3/100)(1+r4/100)……

Where A = Amount

P = Principal

r1%, r2%, r3%, r4%, ……  are the rate of successive years

Solved Problems on Variable Rate of Compound Interest

1. Find the compound interest accrued by Amar from a bank on $ 12000 in 3 years, when the rates of interest for successive years are 8%, 10%, and 12% respectively?

Solution:

Formula for Amount A = P(1+r1/100)(1+r2/100)(1+r3/100)(1+r4/100)……

From given data P = $12,000

n = 3 years

r1 = 8% r2= 10% r3 = 12%

A = 12,000(1+8/100)(1+10/100)(1+12/100)

= 12,000(1+0.08)(1+0.1)(1+0.12)

= 12,000(1.08)(1.1)(1.12)

= $15966

CI = A – P

= 15966 – 12000

= $3966

2. A company offers the following growing rates of compound interest annually to the investors on successive years of investment 5%, 6% and 7%

(i) A man invests $ 30,000 for 2 years. What amount will he receive after 2 years?

(ii) A man invests $ 20,000 for 3 years. What amount he will receive after 3 years?

Solution:

Formula to Calculate the Amount is A = P(1+r1/100)(1+r2/100)(1+r3/100)(1+r4/100)……

(i) Principal = $30,000

n = 2 years

r1 = 5%, r2 = 6%

Substitute the input values in the formula we have the equation as under

A = 30,000(1+5/100)(1+6/100)

= 30,000(1.05)(1.06)

= $ 33,390

Therefore, the Man receives $33,390 by the end of 2 years.

(ii) From the given data

Principal = $20,000

n = 3 years

r1 = 5%, r2 = 6%, r3 = 7%

Substitute the input values in the formula we have the equation as under

A = 20,000(1+5/100) (1+6/100)(1+7/100)

= 20,000(1.05)(1.06)(1.07)

= $23,818

Therefore, the Man receives $23,818 by the end of 3 years.

Problems on Compound Interest | Compound Interest Problems with Solutions

Compound Interest is the Interest calculated on the principal and accumulated interest on the previous period’s loan. Get to know the formula and steps to calculate Compound Interest from here. Check out the Solved Problems on finding the Compound Interest. Try Practicing the Examples over here and refer to them as a quick guide to solve problems on Compound Interest.

Compound Interest Questions and Answers

1. Find the amount if Rs. 10,000 is invested at 10% p.a. for 2 years when compounded annually?

Solution:

We know A = P(1+R/100)n

From given data P = 10,000

R = 10%

n = 2 years

Substituting the input values we have the equation as under

A = 10,000(1+10/100)2

= 10,000(1+0.1)2

= 10,000(1.1)2

= 10,000(1.21)

= Rs. 12,100

2. Find the CI, if Rs 5000 was invested for 2 years at 10% p.a. compounded half-yearly?

Solution:

We know A = P(1+R/100)n

From given data P = 5,000

R = 10%

n = 2 years

Substituting the input values we have the equation as under

A = 5000(1+10/100)2

= 5000(1+0.1)2

=5000(1.1)2

= 5000(1.21)

= Rs. 6050

CI = A – P

= 6050 – 5000

= Rs. 1050

3. The CI on a sum of Rs 1000 in 2 years is Rs 440. Find the rate of interest?

Solution:

Given P = 1000

n = 2 years

CI = 150

We know CI = A – P

440 = A – 1000

A = 1440

R = ?

We know the formula for Amount A = P(1+R/100)n

Substitute the given values in the above formula

1440 = 1000(1+R/100)2

1440/1000 = (1+R/100)2

12/10 = 1+R/100

12/10 -1 = R/100

(12-10)/10 = R/100

2/10 = R/100

Rearranging we get the Rate of Interest as 20%.

4. The difference between SI and CI for 2 years at 10% per annum is Rs 15. What is the principal?

Solution:

We know the formula Difference =  P (R/100)2

15 = P(10/100)2

15 = P(100/10000)

15 = P/100

Therefore, Principal = Rs 1500

5. A certain sum amounts to $ 7200 in 2 years at 6% per annum compound interest, compounded annually. Find the sum?

Solution:

Given Data A = $7200

n = 2 years

R = 6%

Formula to Calculate the Amount A = P(1+R/100)n

7200 = P(1+6/100)2

7200 = P(106/100)2

7200 = P(1.1236)

P = 7200/1.1236

= $ 6407

Therefore, the Sum is $6407.

6. A man deposited $100000 in a bank. In return, he got $133100. Bank gave interest 10% per annum. How long did he kept the money in the bank?

Solution:

Principal = $100000

A = $133100

R = 10%

n = ?

Formula to Calculate the Amount is A = P(1+R/100)n

Substitute the input values in the above formula and rearrange it to obtain the value of n

133100 = 100000(1+10/100)n

133100/100000 = (1+10/100)n

(11/10)3 = (110/100)n

(11/10)3 = (11/10)n

n= 3

Therefore, the man kept his money in the bank for 3 years.

Changing the Subject in an Equation or Formula | How to Change the Subject in an Equation or Formula?

Are you searching for the Changing the Subject in an Equation or Formula Question and Answers? Then, you are at the correct place and you can get them here. All the Problems on Changing the Subject in an Equation or Formula are explained with detailed explanations in this article. Have a look at every problem and completely understand the Changing the Subject in an Equation or Formula concept.

Procedure for Changing Subject in an Equation or Formula?

Changing the Subject in an Equation or Formula is the way of changing the subject of formula and method of substitution to find the value of one variable. The method of substitution is the process of substituting the given values in the place of variables to find the value of the algebraic expression.

  • Find the required variable as the subject.
  • Note down the given values and substitutes them in the concern variables.
  • Finally, simplify the expressions and find the value of the subject.

Solved Examples on Changing the Subject in an Equation or Formula

1. Make s the subject of the below Adding and Subtracting formula
(i) t = s + r
(ii) t = s – r

Solution:
(i) Given that t = s + r
Subtract r on both sides
t – r = s+r-r

t-r = s
The final answer is s = t – r
(ii) Given that t = s – r
Add r on both sides
t + r = s-r+r
The final answer is s = t +r

2. (i) Given x = ty Make y as the subject.
(ii) Given p = rq Make r as the subject.

Solution:
(i)Given that x = ty
t is multiplying y.
Divide t on both sides to get y as the subject.
x/t = y

The final answer is y = x/t

(ii) Given that p = rq
q is multiplying r.
Divide q on both sides to get r as the subject.
p/q = r

The final answer is r = p/q

3. Given B = Q {1 + (t/100)}ⁿ make t as the subject.
Given B = 1102.50 Q = 1000 n = 2, find t.

Solution:
Given that B = Q {1 + (t/100)}ⁿ
Divide Q on both sides
B/Q = Q/Q {1 + (t/100)}ⁿ
B/Q = {1 + (t/100)}ⁿ
Apply the power 1/n on both sides
(B/Q)^1/n = {1 + (t/100)}
Subtract 1 on both sides
(B/Q)1/n – 1 = 1 – 1 + (t/100)
(B/Q)1/n – 1 = (t/100)
Multiply 100 on both sides
100{(B/Q)1/n – 1} = t
The answer is t = 100{(B/Q)1/n – 1}
Substitute the given values B = 1102.50 Q = 1000 n = 2, to find t.
t = 100{(1102.50/1000)1/2 – 1}
t = 100{(441/400)1/2 – 1}
t = 100 [{(21/20)2}1/2 – 1]
t = 100 [(21/20) – 1]
t = 100/20
t = 5

The final answer is t = 5.

4. Make L the subject of the following formula:
(i) s = mL + r
(ii) r = mnL + t

Solution:
(i) Given that s = mL + r
Sutract r on both sides
s – r = mL
Divide m on both sides
(s – r)/m = L

The final answer is L = (s – r)/m

(ii) Given that r = mnL + t
Sutract t on both sides
r – t = mnL
Divide mn on both sides
(r – t)/mn = L

The final answer is (r – t)/mn.