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Compound Interest, in general, is the Interest Calculated on a Principal and the Interest Accumulated over the Previous Period. It is not similar to the Simple Interest where Interest is not added. Get to know the Compound Interest Formula, Procedure on How to find the Compound Interest on a Daily, Monthly, Quarterly, Yearly Basis. Check out the Solved Examples explained step by step for a better understanding of the concept.

Compound Interest Definition

In Simple Words, Compound Interest is nothing but the Interest that adds back to the Principal Sum so that Interest will be earned during the next compounding period.

Formula to Calculate Compound Interest

Compound Interest Formula is given by Compound Interest = Amount – Principal

Amount A = P(1+r/n)^nt

where,

A= Amount

P= Principal

R= Rate of Interest

n = number of times the interest is compounded per year.

Compound Interest When Interest is Compounded Annually

The Amount Formula mentioned above is the general formula for the number of times the principal is compounded in a year. If the Amount is Compounded Yearly or Annually then Amount Formula is given by

A= P(1+R/100)^t

Compound Interest (CI) when Interest is Compounded Half-Yearly

In this case, if  R is the Rate of Interest Per Annum then it is clearly R/2 per half-year.
A = P(1 + (R/2)/100)^2*1

R/2 = R^‘

CI = A – P

= P(1 + (R/2)/100)^2*1 – P

In the Cases, When the Rate is compounded Half-Yearly, we divide the rate by 2 and multiply the time by 2 before using the general formula for the amount in case of the compound interest.

Compound Interest when Interest is Compounded Quarterly

Let us consider the Compound Interest on a Principal P kept for 1 Year and Interest Rate is R %. Since the Interest is Compounded Quarterly Principal Amount will be changed after 3 months. Interest for the Next Three Months will be calculated on the Amount after 3 first months.

In the same way, Interest for Third Quarter will be calculated on the amount left after the first 6 months. Last Quarter will be calculated on the amount remaining after the first 9 months.

A = P(1+ (R/4)/100)^4T

CI = A – P

= P(1+ (R/4)/100)^4T – P

Solved Example Questions on Compound Interest

1. A town had 15,000 residents in 2000. Its population declines at a rate of 10% per annum. What will be its total population in 2004?

Solution:

The population of a town decreases by 10% every year. Thus, the population of a town next year is calculated on the current year. For Decrease, we have the formula

A = P(1-R/100)ⁿ

= 15000(1-10/100)⁴

= 15000(0.9)⁴
= 9841

The population of the town in 2004 is 9841.

2. The price of a radio is Rs 2000 and it depreciates by 5% per month. Find its value after 4 months?

Solution:

For depreciation, the Amount is A = P(1-R/100)ⁿ

= 2000(1-5/100)⁴

= 2000(1-0.05)⁴

=2000(0.95)⁴
= 1629

Price of radio is Rs. 1629 after 4 months.

3. Calculate the compound interest (CI) on Rs. 10000 for 2 years at 5% per annum compounded annually?

Solution:

We know the formula for Compound Interest Annually is

A= P(1+R/100)^t

= 10000(1+5/100)²

= 10000(105/100)²

= 10000(1.1025)

= 11025

CI = A – P

= 11025 – 10000

= Rs. 1025

4. Calculate the compound interest to be paid on a loan of Rs. 5000 for 3/2 years at 10% per annum compounded half-yearly?

Solution:

Rate of Interest when compounded half-yearly we need to divide R by 2 and multiply Time with 2.

A = P(1+R/100)ⁿ

= P(1+R/2*100)^2*n

= 5000(1+10/2*100)^2*3/2

= 5000(1+5/100)³

= 5000(105/100)³

= 5000(1.157)

= 5788

CI = A – P

= 5788 – 5000

= 788

The Compounded Interest to be paid on a loan is Rs.788

Are you looking eagerly everywhere to find a Practice Test on Simple Interest with Solutions? You have come the right way we have curated the Solved Examples explaining how to calculate Simple Interest. We have listed the Formula to Calculate Simple Interest in the coming modules. Check out the Step by Step Solution provided for the Problems finding the Related Terms like Interest, Time Duration, Principal, etc. Assess your preparation standard using the Simple Interest Practice Test available and improve on the areas you are lagging.

Simple Interest Formula Questions | Word Problems on Simple Interest

1. Find the Principal when

(a) S.I. = 100 Rate = 5% per annum Time = 2 years

(b) S.I = 1200 Rate = 2% per annum Time = 10 Months

Solution:

(a) Formula to Calculate the Simple Interest SI = PTR/100

Rearranging the basic formula we get P = SI*100/T*R

= 100*100/2*5

= 1000

Prinicipal = Rs. 1000

(b) S.I = 1200 Rate = 2% per annum Time = 10 Months

Simple Interest in Months = (P × n × R)/ (12 ×100)

Rearranging we have the P as under

1200 = (P*10*2)/(12*100)

(1200*1200) = 20P

P = 1440000/20

= Rs. 72, 000

2. Shubash deposited Rs. 10000 at simple interest for 2 years. Had the interest been 2%, how much amount he would have got?

Solution:

P = 10000

R = 2%

T = 2 Years

Simple Interest = PTR/100

= 10000*2*2/100

= 400

Amount = Principal + Interest

= 10000+400

= 10400

Shubash gains an amount of 10400 after 2 years.

3. A sum fetched a total simple interest of Rs. 4000 at the rate of 10 %.p.a. in 5 years. What is the sum?

Solution:

Formula for SI = PTR/100

Rearranging we have P = SI*100/(TR)

= 4000*100/(10*5)

= 40000/5

= 8000

Sum taken is Rs. 8000

4. At what rate percent annum will a sum of money double itself in 5 years?

Solution:

We know A = P(1+TR/100)

From given data Amount doubles in 5 years

2P = P(1+R*5/100)

2-1= 5R/100

100 = 5R

R = 20%

Sum of the amount doubles in 5 years at a rate of 20%.

5. What will be the ratio of simple interest earned by a certain amount at the same rate of interest for 5 years and that for 10 years?

Solution:

Let the Principal in both the cases be P and Rate of Interest be R %

From given data

Ratio of Simple Interest = (P*5*R)/100 : (P*10*R)/100

= 5:10

= 1:2

6. The simple interest on a certain sum for 6 months at 8% per annum is Rs. 150 less than the simple interest on the same sum for 15 months at 5% per annum. The sum is?

Solution:

Consider the sum to be x

From the given info we have the equation as under

(x*5*15/(100*12)) – (x*8*6/(100*12)) = 150

(75x-48x)/1200 = 150

27x = 1200*150

x = (1200*150)/27

= Rs. 6666.667

Therefore, the Sum is Rs. 6666.667

7. A sum of money at simple interest amounts to Rs. 875 in 3 years and to Rs. 900 in 4 years. The sum is?

Solution:

From given data Simple Interest per 1 Year = 900 – 875

= Rs. 25

SI for 3 Years = (25*3) = 75

Principal = Amount – Interest

= 875 – 75

= Rs. 800

Therefore the Sum is Rs. 800

Simple Interest is an easy method of finding Interest Over a Loan/ Principal Amount. It is necessary to recall the concept of Interest and Ways to Calculate it. The Concept of Simple Interest is Applicable in Many Sectors such as Automobile, Finance, Banking, etc. Whenever you borrow money from a bank or someone you need to repay along with an extra amount. Get to know the Simple Interest Definition, Formula, and How to find the Simple Interest in the later modules.

What is Simple Interest?

Simple Interest is the method of finding the Interest Over a Certain Amount. When you borrow from a bank you need to return back some extra money while paying. While returning back the amount it depends on the rate of interest as well as the time for which you borrow.

How to find the Simple Interest?

Formula to Calculate the Simple Interest will give you the Interest Amount if Principal, Rate of Interest, Time Duration are Given.

SI = (P*T*R)/100

Where, P is the Principal Amount

T is the Time Duration

R is the Rate of Interest

Amount(A) = Principal(P)+Interest(I)

The formula above is to calculate the Simple Interest on a yearly basis. To Calculate the Simple Interest when time duration is months the formula is as such

Simple Interest in Months = (P × n × R)/ (12 ×100)

Solved Examples of Calculating Simple Interest

1. Calculate the Simple Interest if the principal amount is Rs. 4000, the time period is 2 years and the rate is  5%. Also, calculate the total amount at the end of 2 years?

Solution:

Principal = Rs. 4000

Time Period= 2 Years

Rate of Interest = 5%

Simple Interest = (P*T*R)/100

= (4000*2*5)/100

= 400

Amount to be returned = Principal + Interest

= 4000+400

= 4400

2. Manoj Pays Rs. 8000 as an amount on the Sum of Rs. 5000 that he had borrowed for 2 years. Find the rate of interest?

Solution:

From the given data

Amount = Rs 8000

Principal = Rs 5000

SI = A – P = 8000 – 5000 = Rs 3000

T = 2 years

We need to find the Value of R

SI = (P × R ×T) / 100

R = (SI × 100) /(P× T)

R = (3000 × 100 /5000 × 2) =

Thus, R = 30%

Manoj Pays Rs. 8000 as an amount on the Sum of Rs. 5000 at a rate of 30% for 2 Years.

Algebraic fractions are nothing but fractions where its numerator and denominator are algebraic polynomial expressions. These fractions are subject to the laws of arithmetic expressions. Reducing algebraic fractions to their lowest term means the fraction has no common factor other than 1. Get the definition, step by step process to reduce any type of algebraic fractions to its lowest term easily, few example questions and answers in the following sections.

Reduce Algebraic Fractions to its Lowest Term

Reducing algebraic fractions to its lowest term and simplifaction of algebraic fractions are one and the same. Here, we are changing the numerator, denominator of the given fraction so that it should not have a common factor between the numerator and denominator. Actually, the reduced form of the fraction is always equal to the original fraction. While we reduce an algebraic fraction to its lowest term we need to remember one important thing i.e if the numerator and denominator of the fractions are multiplied or divided by the same quantity, then the fraction value remains unchanged.

How to Reduce Algebraic Fractions to Lowest Terms?

Go through the below-mentioned steps to get help with reducing a fraction to its lowest term manually. You will find the procedure quite easy with the detailed explanation provided. Remember that the simplified algebraic fraction numerator, denominator H.C.F is always 1.

• Let us take any polynomial algebraic fraction having numerator and denominator.
• Find the factors of numerator and denominator separately.
• Cancel the common factors in both numerator and denominator.
• Reduce the algebraic fraction to the lowest term.

Solved Examples on Simplifying Algebraic Fractions

Example 1.

Simplify (x² + 5x) / (x² – 5).

Solution:

Given algebraic fraction is (x² + 5x) / (x² – 5).

We see that the numerator and denominator of the given algebraic fraction is polynomial, which can be factorized.

= [x (x + 5)] / [(x – 5) (x + 5)]

Cancel the common factor (x + 5) in the numerator, denominator.

= x / (x – 5)

Example 2.

Reduce the algebraic fraction (x² + 15x + 56) / (x² + 5x – 24) to its lowest term.

Solution:

Given algebraic fraction is (x² + 15x + 56) / (x² + 5x – 24)

Both numerator and denominator of the fraction are polynomials, which can be factorized.

= (x² + 8x + 7x + 56) / (x² + 8x – 3x – 24)

= (x(x + 8) + 7 (x + 8)) / (x(x + 8) – 3 (x + 8))

= ((x + 8) (x + 7)) / ((x + 8) (x – 3))

Cancel the common term (x + 8) in the numerator, denominator.

= (x + 7) / (x – 3)

(x² + 15x + 56) / (x² + 5x – 24) = (x + 7) / (x – 3).

Example 3.

Reduce the algebraic fraction (2x⁵ – 2x⁴ – 4x³) / (x⁴ – 1) to the lowest term.

Solution:

Given algebraic fraction is (2x⁵ – 2x⁴ – 4x³) / (x⁴ – 1)

Both numerator and denominator are polynomials, factorize them.

= (2x³ (x² – x – 2)) / ((x²)² – 1²)

= (2x³ (x² – 2x + x – 2)) / ((x² – 1) (x² + 1))

= (2x³ (x (x – 2) + 1 (x – 2)) / ((x² – 1²) (x² + 1))

= (2x³ (x + 1) (x – 2)) / ((x – 1) (x + 1) (x² + 1))

cancel the common term (x + 1) in the numerator, denominator.

= (2x³ (x – 2)) / ((x – 1) (x² + 1))

(2x⁵ – 2x⁴ – 4x³) / (x⁴ – 1) = (2x³ (x – 2)) / ((x – 1) (x² + 1))

Learn about the relation between the highest common factor and least common factor of two polynomials on this page. First of all, know the definitions of H.C.F, L.C.M, and the relationship between them. Check the solved example questions in the below sections. We have listed the step-by-step solutions for all the example problems provided making it easy for you to understand the concepts.

H.C.F and L.C.M Definition

The highest common factor (H.C.F) is defined as the largest term that divides evenly into all the terms from a group.

Example: H.C.F of 12 and 15 is 3. Because multiples of 12 are 2, 2, 3, and multiples of 15 are 5, 3.

The Lowest Common Multiple (L.C.M) is defined as the smallest term that is a multiple of all numbers from a group.

Example: L.C.M of 12, 15 is 60. Because multiples of 12 are 2, 2, 3, multiples of 15 are 3, 5. And the L.C.M is the multiplication of the smallest common number and remaining numbers from the group.

Relation Between H.C.F. and L.C.M. of Two Polynomials

The relation between L.C.M and H.C.F of polynomials is the product of polynomials is equal to the product of its H.C.F and L.C.M. This relationship can be expressed as follows.

p(x) * q(x) = {L.C.M of p(x) and q(x)} * {H.C.F of p(x) and q(x)]}.

So, find the highest common factor, least common factor of polynomials by using the factorization method. And multiply them to find the product of two polynomial expressions. Using this formula, you can easily find one polynomial when L.C.M, H.C.F of both polynomials, and other polynomial is given.

Solved Examples on Relationship Between H.C.F. and L.C.M. of Two Polynomials

Example 1.

Find H.C.F and L.C.M of two polynomials 2x² – x – 1 and 4x² + 8x + 3 and prove that the product of polynomials is the product of their L.C.M and H.C.F?

Solution:

Given two polynomials are 2x² – x – 1 and 4x² + 8x + 3.

By factoring 2x² – x – 1, we get

= 2x² – 2x + x -1

= 2x(x – 1) + 1(x – 1)

= (x – 1) ( 2x + 1)

By factoring 4x² + 8x + 3, we get

= 4x² + 6x + 2x + 3

= 2x (2x + 3) + 1(2x + 3)

= (2x + 3) (2x + 1)

The common factor in both polynomials is (2x + 1).

Therefore, H.C.F = common factor = (2x + 1)

L.C.M = common factor * remaining factors

= (2x + 1) * (2x + 3) (x – 1) = (2x + 1) (2x + 3) (x – 1)

The relation between H.C.F. and L.C.M. of Two Polynomials is Product of Two Polynomials = Product of Polynomials L.C.M and H.C.F

(2x² – x – 1) (4x² + 8x + 3) = (2x + 1) (2x + 3) (x – 1) (2x + 1)

(x – 1) ( 2x + 1) (2x + 3) (2x + 1) = (2x + 1)²(2x + 3) (x – 1)

Hence proved.

Example 2.

Find the G.C.F of polynomials x³ + y³ and x⁴ + x²y² + y⁴ whose L.C.M is (x³ + y³) (x² + xy + y²)?

Solution:

Given two polynomials are x³ + y³ and x⁴ + x²y² + y⁴

L.C.M of polynomials is (x³ + y³) (x² + xy + y²)

By factoring x⁴ + x²y² + y⁴, we get

= (x² + y²)² – (xy)²

= (x² + xy + y²)² (x² – xy + y²)²

Product of Two Polynomials = Product of Polynomials L.C.M and H.C.F

(x³ + y³) (x⁴ + x²y² + y⁴) = (x³ + y³) (x² + xy + y²) * H.C.F

Cancel the common term (x³ + y³)

(x⁴ + x²y² + y⁴) = (x² + xy + y²) * H.C.F

(x² + xy + y²)² (x² – xy + y²)² = (x² + xy + y²) * H.C.F

Therefore, H.C.F = (x² – xy + y²)²

Example 3.

L.C.M and G.C.F of two polynomials are x³ – 10x² + 11x + 70 and x – 7. And one of the polynomial is x² – 12x + 35, find the other polynomial?

Solution:

Given that,

L.C.M, H.C.F of two polynomials is x³ – 10x² + 11x + 70, x – 7

First polynomial = x² – 12x + 35

Product of Two Polynomials = Product of Polynomials L.C.M and H.C.F

x² – 12x + 35 * Second polynomial = (x³ – 10x² + 11x + 70) * (x – 7)

Second polynomial = (x³ – 10x² + 11x + 70) * (x – 7) / (x² – 12x + 35)

So, second polynomial = (x + 2) (x – 7)

= x² – 7x + 2x -14 = x² -5x – 14.