Category: Mathematics

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.

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.

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.

Learn How to Change the Subject of a Formula and also know about finding the value of the variables by reading the complete article. We included solved examples with clear explanations for better understanding. Students can immediately practice all the questions available in this article and learn the best ways to Change the Subject of a Formula.

Subject of the Formula

Expressing one variable in terms of other variables is the main concept of the formula. The variable that expresses in other variables is called the subject of the formula. The subject of the formula will be written on the left side and other constants and variables are written on the right side of the equality sign in a formula.

Example:
z = xy, where z is the subject of the formula where it is expressed in terms of the product of the x and y.
If we want to change the subject of the formula to x, then the above expression will change into x = z/y.

How to Change the Subject of the Formula?

Changing the subject of a formula can be possible by rearranging the formula to get the required subject. To change the subject of the formula, firstly change its side and change the operation.

When one variable moved to the other side of the equal to sign, the operation becomes inverse. For example, if a variable is added to the subject of the formula, then it will be subtracted after moving to the other side of the equal to sign.

Examples:

1. Make ‘v’ the subject of the formula in u = v + as

Solution:
Given that u = v + as
as is added to v.
To find the subject of the v subtract the as from both sides.
u – as = v + as – as
u – as = v

The final answer is v = u – as

2. Make ‘t’ the subject of the formula, s = x + bt

Solution:
Given that s = x + bt
x is added to the bt.
Firstly, subtract x on both sides.
s – x = x – x + bt
s – x = bt
b is multiplied to t.
Divide b on both sides.
(s – x)/b = bt/b
(s – x)/b = t

The final answer is t = (s – x)/b.

Change the Subject of a Formula Solved Examples

1. The volume of a box is the product of the length and breadth of the box?

Solution:
Given that the volume of a box is the product of the length and breadth of the box.
The volume of a box = v
The length of the box = l
The breadth of the box = b
v = l × b
If the subject of the box is length, the answer is l = v/b
If the subject of the box is the breadth, the answer is b = v/l.

The answer is v = l × b, l = v/b, and b = v/l.

2. In the relation x/5 = make (s – 16)/7 make s as the subject.

Solution:
Given that In the relation x/5 = make (s – 16)/7 make s as the subject.
x/5 = (s – 16)/7
7 is dividing (s – 16)
Multiply 7 on both sides.
7x/5 = (s – 16)7/7
7x/5 = (s – 16)
16 is subtracted from the s.
Add 16 on both sides of the equation.
7x/5 + 16 = s – 16 + 16
7x/5 + 16 = s

The final answer is s = 7x/5 + 16.

3. Make t the subject of the formula s = (t + r)/(t – r)

Solution:
Given that s = (t + r)/(t – r)
Multiply (t – r) on both sided.
s(t – r) = (t + r)
st – sr = t + r
To find the t as subject of the formula, move t variables on left side.
Subtract t on both sides.
st – t – sr = t – t + r
st – t – sr = r
Take t as common from st – t
t (s – 1) – sr = r
Add sr on both sides
t(s – 1) – sr + sr = r + sr
t(s – 1) = r(1 + s)
Divide (s – 1) on both sides
t(s – 1)/(s – 1) = r(1 + s)/(s – 1)
t = r(1 + s)/(s – 1)

The final answer is t = r(1 + s)/(s – 1).

4. Write the formula for finding the area of the rectangle and indicate the subject in this formula. Also, make b as the subject. If A = 24 cm² and l = 4 cm, then find b.

Solution:
The area of the rectangle is A = l × b.
To make the b as the subject of the formula, divide l on both sides.
A/l = lb/l
A/l = b
b = A/l
Now, substitute the values of l and A.
b = 24/4
b = 6.

Therefore, b = 6 is the answer.

5. For a right angled triangle abc, square of the hypotenuse (h) is equal to the sum of squares of its other two sides (s, t).
• Frame the formula for the above statement and find out h if s = 3 and t = 2.
• Also, make ‘s’ the subject of the formula and find s if h = 8 and t = 6.

Solution:
From the given data, Frame the formula for the above statement and find out h if s = 3 and t = 2.
The formula is h² = s² + t²
Substitute s = 3 and t = 2 in h² = s² + t²
h² = 3² + 2² = 9 + 4 = 13
h = √13

The answer is h = √13

From the given data, make ‘s’ the subject of the formula and find s if h = 8 and t = 6.
By changing the subject to s, the h² = s² + t² becomes s² = h² – t²
Substitute h = 8 and t = 6 in s² = h² – t²
s² = 8² – 6²
s² = 64 – 36
s² = 28
s = √28

The answer is s = √28.

6. In the formula, t = s + (b – 1)d make d as the subject. Find d when t = 10, s = 2, b = 5.

Solution:
Given that t = s + (b – 1)d.
Subtract s on both sides
t – s = s – s + (b – 1)d.
t – s = (b – 1)d.
Divide (b – 1) on both sides
(t – s)/(b – 1) = d
d = (t – s)/(b – 1)
Substitute t = 10, s = 2, b = 5.
d = (10 – 2)/(5 – 1) = 8/4 = 2
d = 2.

The final answer is d = 2.

The division is one of the arithmetic operations. The division of algebraic fractions is similar to the division of numbers. Here, you need to factorize the numerators and denominators of the fractions. Cancel the like factors in a fraction and reverse the denominator fraction and multiply it with numerator fraction to get the division value. Find the solved example questions on the division of algebraic fractions.

How to Divide Algebraic Fractions?

Check the simple and easy steps to divide algebraic fractions in the following sections.

• Take two algebraic fractions. One in the numerator and the second in the denominator.
• Find the factors of fractions.
• Get first fraction x 1 / second fraction.
• Cancel the like terms in the numerator and denominator.
• Multiply the numerators, denominators to get the result.

Solved Examples on Dividing Algebraic Fractions

Example 1.

Determine the quotient of the algebraic fractions: x / (x² – x) ÷ 10x / (x² + x – 2)?

Solution:

Given that,

x / (x² – x) ÷ 10x / (x² + x – 2)

Factorize the fractions and cancel the common terms.

x / x (x – 1) ÷ 10x / (x² + 2x – x – 2)

= x / x (x – 1) ÷ 10x / (x(x + 2) -1(x + 2))

= x / x (x – 1) ÷ 10x / (x + 2) (x – 1)

= 1/(x – 1) ÷ 10x / (x + 2) (x – 1)

= 1/(x – 1) x (x + 2) (x – 1) / 10x

= (x + 2) (x – 1) / 10x (x – 1)

= (x + 2) / 10x

Example 2.

Divide the algebraic fractions and express in the lowest form: 9x²+12x+4 / 4x²-27x-7 ÷ 12x²+5x-2 / 16x²-1?

Solution:

Given that,

9x²+12x+4 / 4x²-27x-7 ÷ 12x²+5x-2 / 16x²-1

Factorize the fractions and cancel the common terms.

= 9x²+6x+6x+4 / 4x²-28x+x-7 ÷12x²+8x-3x-2 / (4x)²-1²

= 3x(3x+2)+2(3x+2) / 4x(x-7)+1(x-7) ÷ 4x(3x+2)-1(3x+2) / (4x+1)(4x-1)

= (3x+2)(3x+2) / (x-7)(4x+1) ÷ (3x+2)(4x-1) / (4x+1)(4x-1)

= (3x+2)(3x+2) / (x-7)(4x+1) ÷ (3x+2) / (4x+1)

= (3x+2)(3x+2) / (x-7)(4x+1) * (4x+1) / (3x+2)

= (3x+2)(3x+2)(4x+1) / (x-7)(4x+1)(3x+2)

= 3x+2 / x-7

Example 3.

Find the quotient of the algebraic fractions:

x²+11x+24 / x²-15x+56 ÷ x²-x-12 / x²-11x+28

Solution:

Given that,

x²+11x+24 / x²-15x+56 ÷ x²-x-12 / x²-11x+28

Factorize the algebraic fractions and cancel the common terms.

= x²+8x+3x+24 / x²-8x-7x+56 ÷ x²-4x+3x-12 / x²-7x-4x+28

= x(x+8)+3(x+8) / x(x-8)-7(x-8) ÷ x(x-4)+3(x-4) / x(x-7)-4(x-7)

= (x+8)(x+3) / (x-8)(x-7) ÷ (x-4)(x+3) / (x-4)(x-7)

= (x+8)(x+3) / (x-8)(x-7) ÷ (x+3) / (x-4)

= (x+8)(x+3) / (x-8)(x-7) * (x-4) / (x+3)

= (x+8)(x+3)(x-4) / (x-8)(x-7)(x+3)

= (x+8)(x-4) / (x-8)(x-7)

= x²+4x-32 / x²-15x+56

Starting your preparation for competitive exams!! Check the most important concept of aptitude here i.e., Ratios and Proportions. Ratio and Proportion are the topics which we use in our day to day life but never concentrate on them theoretically.

When it comes to a matter of tests regarding aptitude or something, we need to formulate them to get the required solution for a respective problem. You will even be surprised to know that it is one of the important topics for you to score in some competitive exams. Follow the below-provided sections to know more details like Formulae, Definitions, Tricks etc.

Ratio and Proportion – Introduction

In the real world, Ratios and Proportions are used in daily life. Though we use it in our day to day life, we don’t notice that we use it. If you are preparing for the competitive exams, you must be perfect in ratios and proportions. This is an important and easy concept to score top marks in the exam. You must be good at calculations to solve these type of questions. Solving proportions is the fundamental building block for most problems.

We are providing in-detail material of ratios and proportions, models, estimation in problem-solving situations etc. Go through the complete article to know various Ratio and Proportion concepts, shortcuts and tricks. Moreover, we are also providing preparation tips, so that you can solve the problems in a fraction of seconds.

Ratio and Proportion Important Formulas

1. Ratio:

The ratio is nothing but the simplified form or comparison of two quantities of a similar kind. A ratio is a number, which indicates one quantity as the fraction of other quantity. Example: The Ratio of two numbers 5 to 6 is 5:6. It also expresses the no of times one quantity is equal to other quantity.

“Terms” are the term that indicates numbers forming the ratios. The upper part of the ratio(numerator) is called antecedent and the lower part of the fraction(denominator) is called consequent or descendent. Example: If the ratio is 4:6, then 4 is called the antecedent and 6 is called the consequent.

2. Proportion:

Proportion is indicated as ‘::’ or ‘=’. If the ratio x:y is equal to the ratio of a:b, then x,y, a,b are in proportion.

We mention it using the symbols is as x:y=a:b or x:y:: a:b

When four terms are in proportion, then the product of two middle values(i.e., 2nd and 3rd values) must be equal to the product of two extremes(i.e., 1st and 4th values)

3. Fourth Proportional:

If x:y = a:b, then b is called the fourth proportion to x,y and a

Third Proportional:

If x:y = a:b, then a is called the fourth proportion to x,y and b

Mean Proportional:

The mean proportional between x and y is the root(xy).

4. Comparision of Ratios

We define that x:y>a:b if and only if x/y>a/b

Compounded Ratios:

Compounded Ratio of two ratios: (x:y), (a:b), (c:d) is (xac:ybd)

5. Duplicate Ratios

The duplicated ratio of (x:y) is ((square(x):square(y)).

The sub duplicated ratio of (x:y) is ((root(x):root(y)).

Triplicate ratio of (x:y) is ((cube(x):cube(y)).

The sub triplicate ratio of (x:y) is ((cuberoot(x):cuberoot(y)).

Componendo and dividend rule

If x/y=a/b, then x=y/x-y = a+b/a-b.

6. Variations

We define that a is directly proportional to b, if a=kb for some constant k and we write it as – a is proportional to b.

We say that a is indirectly proportional to b, if ay=b for some constant k and we write it as – a is proportional to 1/b.

Best Books for Ratios and Proportions

• Ratios and Proportions Workbook by Maria Miller
• Axel Tracy’s Ratio Analysis Fundamentals: How 17 financial ratios can allow you to analyse any business on the planet
• Rajesh Verma/Arihant Publications: Fourth Edition
• RS Aggarwal Publications
• S. Chand Publications
• Mc Graw Hill Publications
• Disha Publications
• Kiran Prakashan
• Sarvesh K Varma Publications

Important Properties of Ratio and Proportion

Check the important properties of proportions and know how to apply them.

• Componendo and dividendo:

If x:y=a:b then x+y:x–y=a+b:a-b

• Invertendo:

If x:y=a:b, then y:x=b:a

• Alternendo:

If x:y=a:b then x:a=y:b

• Componendo:

If x:y=a:b then x+y:x=a+b:a

• Dividendo:

If x:y=a:b then x-y:a=a-b:a

• Subtrahendo:

If x:y=a:b then x-a:y-b

• Addendo:

If x:y=a:b, then x+a:y+b

Key Points to Remember

• The ratio between two quantities should exist with the same kind.
• While comparing two ratios, their units must be similar.
• Significant order of terms must be there.
• If the ratios are equal like a fraction, then only comparison of 2 ratios can be performed.

Tips and Tricks for Ratios and Proportions

• If a/b=x/y, then ay=bx
• If a/b=x/y, then a/x=b/y
• Suppose a/b=x/y, then b/a=y/x
• If a/b=x/y, then (a+b)/b=(x+y)/y
• If a/b=x/y, then (a-b)/b=(x-y)/y
• Suppose a/b=x/y, then (a+b)/(a-b)=(x+y)/(x-y)
• If x/(y+z)=y/(z+a)=z/(x+y) and x+y+z is not equal to 0 then x=y=z

Important Solved Questions

Question 1: Find if the ratios are in proportion? The ratios are 4:5 and 8:10.

A. Yes

B. No

C. Data Insufficient

D. None of the above

Solution:

A(Yes)

Given 4:5 and 8:10 are the ratios

4:5=4/5 and 8:10=8/10

4/5=0.8 and 8/10=0.8

Therefore, both proportions are equal and they said to be in proportion.

Question 2: Given Ratios are

x:y=2:3

y:z=5:2

z:a=1:4

Find x:y:z:a

A. 15:20:14:22

B. 15:25:14:26

C. 10:15:6:24

D. 10:20:6:24

Solution:

C(10:15:6:24)

Given the ratios are

x:y=2:3

y:z=5:2

z:a=1:4

Multiplying the 1st ratio by 5, 2nd by 3 and 3rd by 6, we have

x:y=10:15

y:z=15:6

z:a=6:24

In the above equations, all the mean terms are similar, so

x:y:z:a=10:15:6:24

Word Problems on Ratio and Proportion

Question 3: From the total strength of the class, if the no of boys in the class is 5 and no of girls in the class is 3, then find the ratio between girls and boys?

A. 3/5

B. 4/6

C. 8/10

D. 2/5

Solution:

A(3/5)

The ratio of girls and boys can be written as 3:5(Girls: Boys). The ratio can be written in the fraction form like 3/5.

Question 4: Suppose 2 numbers are in the ratio 2:3. If the sum of two numbers is 60. Find the numbers?

A. 40, 36

B. 24, 36

C. 25, 40

D. 44, 36

Solution:

B(24,36)

Given, the ratio of two numbers is 2:3

As, per the given question, the sum of 2 numbers = 60

Therefore, 2x+3x=60

5x=60

x=12

Hence, 2 numbers are;

2x=2*12=24

3x=3*12=36

24 and 36 are the required numbers.

Question 5: 20% and 50% are 2 numbers respectively more than a 3rd number. Find the ratios of 2 numbers?

A. 4:5

B. 2:5

C. 6:7

D: 3:5

Solution:

A(4:5)

Suppose the third number is x

Let the 1st number = 120% of x=120x/100 = 6x/5

Let the 2nd number = 150% of x=150x/100 = 3x/2

Therefore, ratios of 2 numbers = (6x/5:3x/2) = 12x:15x = 4:5

Keep your learning on track with the best preparation materials, tips, tricks provided on our website. Check our page regularly to stay updated on all the topics and concepts. We hope that the information provided here has cleared all your doubts on Ratio and Proportion. If you need any further information, then you can ask in the below-given comment section. We wish all the best for all the candidates.

Multiplication of algebraic fractions is not so difficult. You need to multiply the numerators, denominators of all fractions together to get its product. Before multiplying the algebraic fractions, factorize them. Check out the simple steps of multiplication of algebraic fractions and solved examples in the following sections.

How to find the Product of Algebraic Fractions?

Here we are giving the simple steps to calculate the multiplication of two or more algebraic fractions. Follow these instructions to get the product quickly.

• Find the factors of numerators, denominators of algebraic fractions.
• If there are any like factors, then cancel them.
• Multiply the numerators of remaining factors and denominators.

Solved Examples on Multiplication of Algebraic Fractions

Example 1.

Simplify 5 / (a + a²) x (a³ – a) / ab?

Solution:

Given that,

5 / (a + a²) x (a³ – a) / ab

Get the factors of both fractions

= 5 / a (a + 1) x a (a² – 1) / ab

= 5 / a (a + 1) x [a (a + 1) (a – 1)] / ab

Multiply the two algebraic fractions.

= 5a (a + 1) (a – 1) / a (a + 1) ab

Cancel the terms a (a +1) in both denominator and numerator.

= 5 (a – 1) / ab

Example 2.

Find the product of the algebraic fraction [5a / 2a-1 – (a-2) / a] x [2a / (a+2) – 1 / (a+2)?

Solution:

Given that,

[5a / 2a-1 – (a-2) / a] x [2a / (a+2) – 1 / (a+2)]

The least common multiple of denominators of the first part is a(2a – 1) and the L.C.M of denominators of the second part is a + 2.

Therefore, [5a. a / a(2a – 1) – (a – 2) . (2a – 1) / a(2a – 1)] x [2a / (a+2) – 1 / (a+2)]

= [(5a² – (a – 2) (2a – 1)) / a(2a – 1)] x [(2a – 1) / (a+2)]

= [(5a² – 2a² + a + 4a – 2) / a(2a – 1)] x [(2a – 1) / (a+2)]

= [(3a² + 5a – 2) / a(2a – 1)] x [(2a – 1) / (a+2)]

= [(3a² + 6a – a – 2) / a(2a – 1)] x [(2a – 1) / (a+2)]

= [(3a (a + 2) -1(a + 2)) / a(2a – 1)] x [(2a – 1) / (a+2)]

= [(a + 2) (3a – 1) / a(2a – 1)] x [(2a – 1) / (a+2)]

= [(a + 2) (3a – 1) (2a – 1)] / [a(2a – 1) (a+2)]

Here the common factors in the numerator and denominator are (a+2), (2a-1). Cancel these factors in both to find the lowest form

= (3a – 1) / a

Therefore, [5a / 2a-1 – (a-2) / a] x [2a / (a+2) – 1 / (a+2)] = (3a – 1) / a.

Example 3.

Find the product and express in the lowest form: 5x² / (x² – 2x) x (x² – 4) / (x² + 2x)?

Solution:

Given that,

5x² / (x² – 2x) x (x² – 4) / (x² + 2x)

Ge the factors of both fractions.

= 5x² / x(x – 2) x (x² – 2²) / (x(x + 2))

Cancel the common term x in the first part.

= 5x / (x – 2) x (x + 2) ( x – 2) / (x(x + 2))

Cancel the common factor (x+2) in the second part.

= 5x / (x – 2) x (x – 2) / x

Multiply both numerators and denominators

= 5x . (x – 2) / x . (x – 2)

Cancel the common factor (x – 2) in both numerator and denominator.

= 5x/ x

= 5.

∴ 5x² / (x² – 2x) x (x² – 4) / (x² + 2x) = 5.

Example 4.

Find the product of the algebraic fractions in the lowest form:

[(x + 2y) / (2x + y)] x [(2x + 5y) / (x + y)]

Solution:

Given that,

[(x + 2y) / (2x + y)] x [(2x + 5y) / (x + y)]

= [(x + 2y) (2x + 5y)] / [(2x + y) (x + y)]

= [2x² + 5xy + 4xy + 10y²] / [2x² + xy + 2xy + y²]

= [2x² + 9xy + 10y²] / [2x² + 3xy + y²]

Example 5.

Simplify (4x² – 1) / (9x – 6) x (15x – 10) / (x + 4)?

Solution:

Given that,

(4x² – 1) / (9x – 6) x (15x – 10) / (x + 4)

Calculate the factors.

= ((2x)² – 1²) / 3(3x – 2) x 5(3x – 2) / (x + 4)

= (2x + 1) (2x – 1) / 3(3x – 2) x 5(3x – 2) / (x + 4)

Multiply numerators, denominators together.

= [(2x + 1) (2x – 1) . 5(3x – 2)] / [3(3x – 2) (x + 4)]

Cancel the terms (3x – 2)

= [5(2x + 1) (2x – 1)] / 3(x + 4)

∴ (4x² – 1) / (9x – 6) x (15x – 10) / (x + 4) = [5(2x + 1) (2x – 1)] / 3(x + 4).

Find the solved example questions on algebraic fractions. This article includes addition, subtraction, multiplication, division, simplification, and reducing a fraction to its lowest term. By reading this page, you can solve any type of algebraic fraction questions easily & quickly. So, have a look at all the questions and solutions provided below and learn the concepts involved easily.

Questions on Solving Algebraic Fractions

Example 1.

Simplify the algebraic fraction [1 + 1 / (x + 1)] / [x – 4/x]?

Solution:

Given fraction is [1 + 1 / (x + 1)] / [x – 4/x]

Find the L.C.M of denominators.

[(1. (x + 1)+ 1) / (x + 1)] / [(x² – 4) / x]

= [(x + 2) / (x + 1)] / [(x² – 2²) / x]

= [(x + 2) / (x + 1)] * [x / (x + 2) (x – 2)]

= [x(x + 2)] / [(x + 1) ( x – 2) (x + 2)]

= x / [(x + 1) (x – 2)]

∴[1 + 1 / (x + 1)] / [x – 4/x] = x / [(x + 1) (x – 2)]

Example 2.

Simplify the algebraic fraction [((k² + 1 / k – 1) – k) / ((k² – 1 / k + 1) + 1)] [1 – 2/(1 + 1/k)]?

Solution:

Given algebraic fraction is [((k² + 1 / k – 1) – k) / ((k² – 1 / k + 1) + 1)] [1 – 2/(1 + 1/k)]

Find the L.C.M of denominators of the first fraction and simplify.

= [(k² + 1 – k (k – 1)) / (k – 1)] / [(k² – 1 + 1 (k + 1)) / (k + 1)]

= [(k² + 1 – k² + k)) / (k – 1)] / [(k² – 1 + k + 1)) / (k + 1)]

= [(k + 1) / (k – 1)] / [(k² + k) / (k + 1)]

= [(k + 1) / (k – 1)] / [(k(k + 1) / (k + 1)]

= [(k + 1) / (k – 1)] / k / 1

= (k + 1) / k(k – 1)

= (k + 1) / (k² – 1²)

= (k + 1) / (k + 1) ( k – 1)

= 1 / (k – 1)

Simplification of the second fraction is

[1 – 2/(1 + 1/k)]= [1- 2 / k(k +1)]

= [k(k +1) – 2] / [k(k +1)]

= (k² + k – 2) / [k(k +1)]

= (k² + 2k – k – 2) / (k(k +1))

= (k (k + 2) – 1(k + 2)) / (k(k +1))

= [(k – 1) ( k + 2)] / (k(k +1))

Product of first and second fraction is

= 1 / (k – 1) * [(k – 1) ( k + 2)] / (k(k +1))

= (k + 2) / k(k +1)

∴ [((k² + 1 / k – 1) – k) / ((k² – 1 / k + 1) + 1)] [1 – 2/(1 + 1/k)] = (k + 2) / k(k +1)

Example 3.

Reduce the algebraic fractions [3 / √(1+x) + √(1-x)] : [3 / √(1 – x²) + 1]

Solution:

Given algebraic fraction is [3 / √(1+x) + √(1-x)] : [3 / √(1 – x²) + 1]

Simplification of first fraction is

3 / √(1 + x) + √(1 – x)

= (3 + √(1 + x) * √(1 – x)) / √(1 + x)

= (3 + √(1 + x)(1 – x) / √(1 + x)

Simplification of the second fraction is

3 / √(1 – x²) + 1

= 3 + √(1 – x²) / √(1 – x²)

The division of algebraic fraction is

= [(3 + √(1 + x)(1 – x) / √(1 + x)] : [3 + √(1 – x²) / √(1 – x²)]

= [(3 + √(1 + x)(1 – x) (√(1 – x²))] / [√(1 + x) (3 + √(1 – x²))

= √(1 – x²) / √(1 + x)

= √(1 + x)(1 – x) / √(1 + x)

= √(1 – x)

∴ [3 / √(1+x) + √(1-x)] : [3 / √(1 – x²) + 1] = √(1 – x)

Example 4.

Reduce to lowest terms — if possible 3x / 4a²b – 7 / 6ab⁵ – 5x / 2ab².

Solution:

Given fraction is 3x / 4a²b – 7 / 6ab⁵ – 5x / 2ab²

Find the L.C.M of all terms denominators.

L.C.M of 4a²b, 6ab⁵, 2ab² is 12a²b⁶.

= [3x . 3b⁵]/12a²b⁶ – [7 . 2a]/12a²b⁶ – [5x . 6ab⁴]/12a²b⁶

= [9xb⁵ – 14a – 30xab⁴]/ 12a²b⁶

∴ 3x / 4a²b – 7 / 6ab⁵ – 5x / 2ab² = [9xb⁵ – 14a – 30xab⁴]/ 12a²b⁶

In this article, you will learn how to find Compound Interest when Interest is Compounded Quarterly. You may feel the process of calculating the Compound Interest using the growing principal a bit difficult if the time duration is long. Refer to Solved Examples on finding Quarterly Compounded Interest and learn how to solve related problems. We even provided the solutions for the sample problems on calculating the Compound Interest when Interest is Compounded Quarterly in the coming modules.

How to find Compound Interest when Interest is Compounded Quarterly?

If the Rate of Interest is Annual and Interest is Compounded Quarterly then the number of years is multiplied by 4 i.e. 4n and the annual interest rate is cut down by one-fourth. In such cases, Formula for Quarterly Compound Interest is given as under

Let us assume the Principal = P, Rate of Interest = r/4 %, and time = 4n, Amount = A, Compound Interest = CI then

A = P(1+(r/4)/100)⁴ⁿ

In the above formula rate of interest is divided by 4 whereas the time is multiplied by 4.

We know CI = A – P

= P(1+(r/4)/100)⁴ⁿ – P

= P{1+(r/4)/100)⁴ⁿ – 1}

If you are aware of any of the three then you can automatically find the other one.

Solved Problems on finding Compound Interest when Compounded Quarterly

1. Find the compound interest when $1,00, 000 is invested for 6 months at 5 % per annum, compounded quarterly?

Solution:

Principal Amount = $1,00, 000

Rate of Interest = 5% per annum

n = 6 months = 1/2 year

Since Interest Rate is Compounded Quarterly divide the interest rate by 4 i.e. r/4 and multiply the time by 4 i.e. 4n

Amount A = P(1+(r/4)/100)⁴ⁿ

Substitute the Inputs in the above formula to find the amount

A = 1,00,000(1+(5/4)/100)^4*1/2

= 1,00,000(1+5/400)²

= $ 1,02,515

CI = A – P

= $ 1,02,515 – $ 1,00,000

=$2515

2. Find the amount and the compound interest on Rs. 12,000 compounded quarterly for 9 months at the rate of 10% per annum?

Solution:

Principal Amount = Rs.12, 000

Rate of Interest = 10% per annum

n = 9 months = 9/12 = 3/4 year

Since Interest Rate is Compounded Quarterly divide the interest rate by 4 i.e. r/4 and multiply the time by 4 i.e. 4n

Amount A = P(1+(r/4)/100)⁴ⁿ

Substitute the Input Values in the above formula to find the amount

A= 12,000(1+(10/4)/100)^4*3/4

= 12,000(1+10/400)³

= 12,000(1+0.025)³

= 12,000(1.025)³

= Rs. 12922

CI = A – P

= 12922 – 12000

= Rs. 922

3. Calculate the compound interest (CI) on Rs. 4000 for 1 year at 10% per annum compounded quarterly?

Solution:

Principal Amount = Rs. 4,000

Rate of Interest = 10% per annum = 10/4 %

n = 1 year

Since Interest Rate is Compounded Quarterly divide the interest rate by 4 i.e. r/4 and multiply the time by 4 i.e. 4n

Amount A = P(1+(r/4)/100)⁴ⁿ

Substitute the Input Values in the above formula to find the amount

A = 4000(1+(10/4)/100)^4*1

= 4000(1+10/400)⁴

= 4000(1.1038)

= Rs. 4415.25

CI = A – P

= 4415.25 – 4000

= Rs. 415.25

Formula and Framing the Formula are the concepts used to explain the relationship between different variables. Let us discuss deeply every individual concept of Formula and also Framing the Formula. It is easy to find the relation between two variables by framing a formula with simple steps. Every concept is clearly explained with the solved examples in the below article. Therefore, completely read the entire article and follow every step to get complete knowledge on Formula and Framing the Formula.

List of Topics for Formula and Framing the Formula

• Change the Subject of a Formula
• Changing the Subject in an Equation or Formula
• Practice Test on Framing the Formula

Formula

A formula is a relation between different variables. The formula can be written as an equation with the help of variables and also mathematical symbols. When you look at the equation, it is clearly stating how a variable is related to another variable.

Example:

1. Let us consider a square which is of side a and the perimeter of it is p, then the formula will be p = 4a.
Here the formula shows the exact relation between the perimeter of a square and also the side of a square. It is easy to find the unknown quantity using the known quantity when the values of all the quantities are known.
2. If the perimeter of a rectangle p is twice the sum of its breadth b and length l, then the formula will represent as p = 2(l + b)
3. The volume of a cube is V and its side is a. Then, the formula is V = a^3.
4. We can also write a formula to express the relation between force, mass, and acceleration. Force of an object F is the product of mass “m” and acceleration “a” of that object. The formula is F = ma.
5. If the sum of two unknown variables is 15, then the formula is a + b = 15, where a and b are unknown variables.

Framing a Formula

Framing a formula is arranging the formula of the given mathematical statements with the help of symbols and literals.

1. Firstly, select variables need to form an equation. Also, decide the symbols that need to use for the equation. Some of the symbols and letters are already in the use to represent certain quantities. For example, p is used to represent the principal.
2. Finally, understand the conditions to write an equation and frame the formula.

Subject of a formula

When a quantity is expressing in terms of other quantities, then that particular quantity expressed is defined as the Subject of a formula. Generally, the Subject of the formula is written on the left-hand side and other constants are written on the right-hand side of the equality sign in a formula.

Example:
If z = x + y, then z is expressed in terms of the sum of the x and y. Here, z is the subject of this formula.
x = z – y. Here x is the subject of this formula.

Substitution in a formula

If the variable of an algebraic expression is assigned with certain values, then the given expression gets a particular value. This process is known as substitution.
1. Note down the unknown quantity as the subject of the formula.
2. Substitute different values of the known quantity in the formula to find the value of the subject.

Examples of Framing of a Formula:

1. The total amount A is equal to the sum of the Interest (I) and Principal (P).

Solution:
Formula: A =  I + P

2. One-third of a number subtracted from 4 gives 2.

Solution:
Formula: 4 – 1/3 x = 2

3. The sum of the three angles (∠a, ∠b, ∠c) of a triangle is equal to two right angles.

Solution:
Formula: ∠a + ∠b + ∠c = 180°

4. The area of the rectangle (A) is equal to the product of the breadth (B) and length (L) of the rectangle.

Solution:
Formula: A = B × L

Solved Examples on Formula and Framing the Formula

1. Express the following as an equation. Arun’s father’s age is 3 years more than 4 times Arun’s age. Father’s age is 39 years.

Solution:
Given that Arun’s father’s age is 3 years more than 4 times Arun’s age. Father’s age is 39 years.
Let Arun’s age is s years.
Four times his age = 4s.
Father’s age 3 + 4s
Given father’s age = 39 years
3 + 4s = 39

The final equation is 3 + 4s = 39

2. Write the formula for the following statement. One-fourth of the weight of an apple (A) is equal to one-fifth of the difference between Orange (O) and 2.

Solution:
Given that One-fourth of the weight of an apple (A) = 1/4
Difference between Orange (O) and 2 = O – 2
One-fifth of the difference between Orange (O) and 2. = 1/5(O – 2)
One-fourth of the weight of an apple (A) is equal to one-fifth of the difference between Orange (O) and 2.
1/4 = 1/5(O – 2)

3. Change the following statement using expression into a statement in ordinary language.
(a) Cost of a Desk is Rs. X and Cost of a Box is Rs. 4X
(b) Sam’s age is p years. His brother’s age is (5p + 2) years.

Solution:
(a) Given that Cost of a Desk is Rs. X and Cost of a Box is Rs. 4X
The Cost of a Box is 4 times the Cost of a Desk.
(b) Sam’s age is p years. His brother’s age is (5p + 2) years.
Sam’s brother’s age is two years more than five times his age.

4. A rectangular box is of height h cm. Its length is 7 times its height and the breadth is 5 cm less than the length. Express the length, breadth, and height.

Solution:
Given that A rectangular box is of height h cm. Its length is 7 times its height and the breadth is 5 cm less than the length. Express the length, breadth, and height.
Height = h, length = l, breadth = b
The length of the rectangle is 7 times the height.
The Length of the rectangle = 7h
The breadth of the rectangle is 5 cm less than the length
The Breadth of rectangle = l – 5
As l = 7h, b = 7h – 5.

Height h
Length l = 7h
Breadth b = 7h – 5

One of the main operations performed for Algebraic Expression is Division. Deeply understand how division operations are performed on Algebraic Expressions and learn the easy way to solve Algebraic Expression Division Problems. Solving Division of Algebraic Expression Problems is the opposite process of Multiplication of Algebraic Expression.

Rule to Find Division of Algebraic Expression

If x is a variable and a, b are positive integers such that a > b then (x^a ÷ x^b) = x^(a − b).

Types of Algebraic Expression Division

There are different types of Division when it comes to Dividing Algebraic Expressions. They are as such

1. Division of a Monomial by a Monomial
2. Division of a Polynomial by a Monomial
3. Division of a Polynomial by a Polynomial

How to Find Division of a Monomial by a Monomial?

1. The coefficients of the quotient of two monomials are equal to the quotient of their numerical coefficients which are multiplied by the quotient of their coefficients.
2. The variable part of the quotient of two monomials is equal to the quotient of the variables in the given monomials.

Rule:

Quotient of two monomials = (quotient of their numerical coefficients) x (the quotient of their variables)

Solved Examples

1. Divide 6x²y³ by -4xy

Solution:

Given that Divide 6x²y³ by -4xy
Use quotient law to solve the given problem.
x^m ÷ xⁿ = x^{m – n}
x²/x = x^(2-1) = x
y³/y = y^(3-1) = y²
-(6/4) = -3/2
-3/2 x y²

The required expression is -3/2xy²

2. Divide 48x³yz² by -8xyz

Solution:

Given that Divide 48x³yz² by -8xyz
Use quotient law to solve the given problem.
x^m ÷ xⁿ = x^{m – n}
x³/x = x^(3-1) = x²
y/y = 1
z²/z = z^(2-1) = z
(48/-8) = -6
-6x²z

The required expression is -6x²z

3. Divide -24x³yz³ by -6xyz²

Solution:

Given that Divide -24x³yz³ by -6xyz²
Use quotient law to solve the given problem.
x^m ÷ xⁿ = x^{m – n}
x³/x = x^(3-1) = x²
y/y = y^(1-1) = y^0 = 1
z³/z² = z^(3-2) = z
(-24/-6) = 4
4x²z

The required expression is 4x²z

How to Find Division of a Polynomial by a Monomial?

Division of a Polynomial by a Monomial can be calculated by dividing each term of the polynomial by the monomial.

1. Note down the polynomial and the monomial.
2. Consider polynomial as Dividend and monomial as the divisor.
3. Separate the terms of the polynomial (dividend) in the descending order of their exponents.
4. Finally, divide every term of the polynomial (dividend) by monomial.
5. Simplify and write the required expression.

Solved Examples

1. Divide 8x⁵ + 20x⁴ – 16x² by 4x²

Solution:

Given that Divide 8x⁵ + 20x⁴ – 16x² by 4x²
Here the dividend is 8x⁵ + 20x⁴ – 16x² and the divisor is 4x²
(8x⁵ + 20x⁴ – 16x²) ÷ 4x²
Divide every term of the polynomial (dividend) by monomial.
(8x⁵ ÷ 4x² ) +( 20x⁴ ÷ 4x² ) – ( 16x² ÷ 4x²
(8/4)(x^5/x^2) + (20/4)(x^4/x^2) – (16/4)(x^2/x^2)
2(x^5-2) + 5(x^4-2) – 4(x^2-2)
2x^3 + 5x^2 – 4

The required expression is 2x³ + 5x² – 4

2. Divide 21x³y + 12x²y² – 9xy by 3xy

Solution:

Given that Divide 21x³y + 12x²y² – 9xy by 3xy
Here the dividend is 21x³y + 12x²y² – 9xy and the divisor is 3xy
21x³y + 12x²y² – 9xy ÷ 3xy
Divide every term of the polynomial (dividend) by monomial.
(21x³y ÷ 3xy) + (12x²y² ÷ 3xy) – (9xy ÷ 3xy)
(21/3)(x³/x)(y/y) + (12/3)(x²/x)(y²/y) – (9/3)(xy/xy)
7(x^3-1)(y^1-1) + 4 (x^2-1)(y^2-1) – 3
7x^2 + 4xy – 3

The required expression is 7x² + 4xy – 3

3. 10x + 20x² + 30x⁴ – 50x⁵ ÷ by 5x

Solution:

Given that Divide 10x + 20x² + 30x⁴ – 50x⁵ by 5x
Here the dividend is 10x + 20x² + 30x⁴ – 50x⁵  and the divisor is 5x
Arrange the dividend in the descending order of their exponents.
( – 50x⁵ + 30x⁴ + 20x² + 10x ) ÷ 5x
Divide every term of the polynomial (dividend) by monomial.
(-50/5)(x⁵/x) + (30/5)(x⁴/x) + (20/5)(x²/x) + (10/5)(x/x)
(-50/5)(x^5-1) + (30/5)(x^4-1) + (20/5)(x^2-1) + (10/5)(x^1-1)
-10x⁴ + 6x³ + 4x + 2

The required expression is -10x⁴ + 6x³ + 4x + 2

How to Find Division of a Polynomial by a Polynomial?

Get to know the Procedure on How to Divide Polynomial by a Polynomial in the below modules. They are along the lines

1. Note down the polynomial and the monomial.
2. Consider polynomial as Dividend and monomial as the divisor.
3. Separate the terms of the polynomial (dividend) in the descending order of their exponents.
4. Divide the first term of the polynomial (dividend) by the first term of the polynomial (divisor) and get the first term of the quotient.
5. Multiply every term of the divisor by the first term of the quotient and then subtract the result from the dividend.
6. In the next step, take the remainder if you have any and consider it as a new dividend then go with the above process.
7. Repeat the same process till you get the remainder as 0 or a polynomial of degree less than that of the divisor.

Solved Examples

1. Divide 18 – 12a² – 8a by (3 + 2a)

Solution:

Given that Divide 18 – 12a² – 8a by (3 + 2a)
Here the dividend is – 12a² – 8a + 18 and the divisor is 2a + 3
Multiply 6a with (2a + 3) then subtract the resultant expression from – 12a² – 8a + 18
-6a (2a + 3) = – 12a² – 18a
(– 12a² – 8a + 18) + (- 12a² – 18a) = 10a + 18
Now, multiply 5 with (2a + 3) then subtract the resultant expression from 10a + 18
5(2a + 3) = 10a + 15
10a + 18 – 10a – 15 = 3
Quotient = -6a + 5
Remainder = 3

Verification:
Dividend = divisor × quotient + remainder
– 12a² – 8a + 18 = (3 + 2a) × (-6a + 5) + 3
= -18a + 15 – 12a² + 10a + 3
= – 12a² – 8a + 18
– 12a² – 8a + 18 = – 12a² – 8a + 18

The required answer is -6a + 5

2. Divide 8x² + 12x + 4 by (4x + 4).

Solution:

Given that Divide 8x² + 12x + 4 by (4x + 4)
Here the dividend is 8x² + 12x + 4 and the divisor is 4x + 4.
Multiply 2x with (4x + 4) then subtract the resultant expression from 8x² + 12x + 4
2x(4x + 4) = 8x² + 8x
(8x² + 12x + 4) – (8x² + 8x) = 4x + 4
Multiply 1 with (4x + 4) then subtract the resultant expression from 4x + 4
1(4x + 4) = 4x + 4
(4x + 4) – (4x + 4) = 0
Quotient = 2x + 1
Remainder = 0

Verification:
Dividend = divisor × quotient + remainder
8x² + 12x + 4 = (4x + 4) × (2x + 1) + 0
= 8x² + 4x + 8x + 4 + 0
= 8x² + 12x + 4
8x² + 12x + 4 = 8x² + 12x + 4

The required answer is 2x + 1

3. Divide 3x² + 18x + 24 by (3x + 12)

Solution:

Given that Divide 3x² + 18x + 24 by (3x + 12)
Here the dividend is 3x² + 18x + 24 and the divisor is (3x + 12)
Multiply x with (3x + 12) then subtract the resultant expression from 3x² + 18x + 24
x(3x + 12) = 3x² + 12x
(3x² + 18x + 24) – (3x² + 12x) = 6x + 24
Multiply 2 with (3x + 12) then subtract the resultant expression from 6x + 24
2(3x + 12) = 6x + 24
(6x + 24) – (6x + 24) = 0
Quotient = x + 2
Remainder = 0

Verification:
Dividend = divisor × quotient + remainder
3x² + 18x + 24 = (3x + 12) × (x + 2) + 0
= 3x² + 6x + 12x + 24 + 0
= 3x² + 18x + 24
3x² + 18x + 24 = 3x² + 18x + 24

The required answer is x + 2

4. Divide 18x – 12x² + 2x³ – 4 by (2x – 4).

Solution:

Given that Divide 18x – 12x² + 2x³ – 4 by (2x – 4)
Here the dividend is 2x³ – 12x² + 18x  – 4 and the divisor is (2x – 4)
Multiply x² with (2x – 4) then subtract the resultant expression from 2x³ – 12x² + 18x  – 4
x²  (2x – 4) = 2x³ – 4x²
(2x³ – 12x² + 18x  – 4) – (2x³ – 4x²) = – 8x² + 18x  – 4
Multiply -4x with (2x – 4) then subtract the resultant expression from – 8x² + 18x  – 4
-4x (2x – 4) = -8x² + 16x
(- 8x² + 18x  – 4) – (-8x² + 16x) = 2x – 4
Multiply 1 with (2x – 4) then subtract the resultant expression from 2x – 4
1 (2x – 4) = 2x – 4 – 2x + 4 = 0
Quotient = x² – 4x + 1
Remainder = 0

Verification:
Dividend = divisor × quotient + remainder
18x – 12x² + 2x³ – 4 = (2x – 4) × (x² – 4x + 1) + 0
= 2x³ – 8x² + 2x – 4x² + 16x – 4 + 0
= 2x³ – 12x² + 18x – 4
2x³ – 12x² + 18x – 4 = 2x³ – 12x² + 18x – 4

The required answer is x² – 4x + 1

5. Divide (87x – 18x² – 84) by (9x – 12).

Solution:

Given that Divide (87x – 18x² – 84) by (9x – 12)
Here the dividend is – 18x² + 87x  – 84 and the divisor is (9x – 12)
Multiply -2x with (9x – 12) then subtract the resultant expression from – 18x² + 87x  – 84
-2x (9x – 12) = – 18x² + 24x
(- 18x² + 87x  – 84) – (- 18x² + 24x) = 63x  – 84
Multiply 7 with (9x – 12) then subtract the resultant expression from 63x  – 84
7 (9x – 12) = 63x – 84
(63x  – 84) – (63x  – 84) = 0
Quotient = -2x + 7
Remainder = 0

Verification:
Dividend = divisor × quotient + remainder
(87x – 18x² – 84) = (9x – 12) × (-2x + 7) + 0
= – 18x² + 63x + 24x – 84
= 87x – 18x² – 84
87x – 18x² – 84 = 87x – 18x² – 84

The required answer is -2x + 7

6. Divide (20x³- 16x² + 12x + 72) by (12 – 8x + 4x²)

Solution:

Given that Divide (20x³- 16x² + 12x + 72) by (12 – 8x + 4x²)
Here the dividend is (20x³- 16x² + 12x + 72) and the divisor is (12 – 8x + 4x²)
Multiply 5x with (4x² -8x + 12) then subtract the resultant expression from 20x³- 16x² + 12x + 72
5x (4x² -8x + 12) = 20x³- 40x² + 60x
(20x³- 16x² + 12x + 72) – (20x³- 40x² + 60x) = 24x² – 48x + 72
Multiply 6 with (4x² -8x + 12) then subtract the resultant expression from 24x² – 48x + 72
6 (4x² -8x + 12) = 24x² – 48x + 72
(24x² – 48x – 72) – (24x² – 48x + 72) = 0
Quotient = 5x + 6
Remainder = 0

Verification:
Dividend = divisor × quotient + remainder
(20x³- 16x² + 12x + 72) = (4x² -8x + 12) × (5x + 6) + 0
= 20x³ + 24x² – 40x² – 48x + 60x + 72
= (20x³- 16x² + 12x + 72)
(20x³- 16x² + 12x + 72) = (20x³- 16x² + 12x + 72)

The required answer is 5x + 6

7. Using division, show that (2x – 2) is a factor of (2x³ – 2).

Solution:

Given that Using division, show that (2x – 2) is a factor of (2x³ – 2)
Divide (2x³ – 2) by (2x – 2)
Here the dividend is (2x³ – 2) and the divisor is (2x – 2)
Multiply x with (2x – 2) then subtract the resultant expression from (2x³ – 2)
x (2x – 2) = 2x³- 2x
(2x³ – 2) – (2x³- 2x) = 2x – 2
Multiply 1 with (2x – 2) then subtract the resultant expression from (2x – 2)
(2x – 2) – (2x – 2) = 0
Quotient = x + 1
Remainder = 0

(2x – 2) divides (2x³ – 2). Therfore, (2x – 2) is the factor of (2x³ – 2)

8. Find the quotient and remainder when (14 + 30x – 26x² + 10x³) is divided by (8 – 6x + 2x²)?

Solution:

Given that Divide (14 + 30x – 26x² + 10x³) by (8 – 6x + 2x²)
Here the dividend is (14 + 30x – 26x² + 10x³) and the divisor is (8 – 6x + 2x²)
Rearrange the given expressions.
divide (10x³ – 26x² + 30x + 14) by (2x² – 6x + 8)
Multiply 5x with (2x² – 6x + 8) then subtract the resultant expression from (10x³ – 26x² + 30x + 14)
5x (2x² – 6x + 8) = 10x³ – 30x² + 40x
(10x³ – 26x² + 30x + 14) – (10x³ – 30x² + 40x) = 4x² – 10x + 14
Multiply 2 with (2x² – 6x + 8) then subtract the resultant expression from 4x² – 10x + 14
2 (2x² – 6x + 8) = 4x² – 12x + 16
(4x² – 10x + 14) – (4x² – 12x + 16) = 2x – 2
Quotient = 5x + 2
Remainder = 2x – 2

The Quotient is 5x + 2 and the Remainder is 2x – 2

9. Divide (30x⁴ + 51x³ – 186x² + 90x – 9) by (6x² + 21x – 3)

Solution:

Given that Divide (30x⁴ + 51x³ – 186x² + 90x – 9) by (6x² + 21x – 3)
Here the dividend is (30x⁴ + 51x³ – 186x² + 90x – 9) and the divisor is (6x² + 21x – 3)
Multiply 5x² with (6x² + 21x – 3) then subtract the resultant expression from (30x⁴ + 51x³ – 186x² + 90x – 9)
5x² (6x² + 21x – 3) = 30x⁴ + 105x³ – 15x²
(30x⁴ + 51x³ – 186x² + 90x – 9) – (30x⁴ + 105x³ – 15x²) = -54x³ – 171x² + 90x – 9
Multiply 9x with (6x² + 21x – 3) then subtract the resultant expression from -54x³ – 171x² + 90x – 9
-9x (6x² + 21x – 3) = -54x³ – 189x² + 27x
(-54x³ – 171x² + 90x – 9) – (-54x³ – 189x² + 27x) = 18x² + 63x – 9
Multiply 3 with (6x² + 21x – 3) then subtract the resultant expression from 18x² + 63x – 9
3 (6x² + 21x – 3) = 18x² + 63x – 9
(18x² + 63x – 9) – (18x² + 63x – 9) = 0
Quotient = 5x² – 9x + 3
Remainder = 0

Verification:
Dividend = divisor × quotient + remainder
(30x⁴ + 51x³ – 186x² + 90x – 9) = (6x² + 21x – 3) × (5x² – 9x + 3) + 0
= 30x⁴ -54x³ + 18x² -105x³ – 189x² +63x – 15x² +27x – 9
= (30x⁴ + 51x³ – 186x² + 90x – 9)
(30x⁴ + 51x³ – 186x² + 90x – 9) = (30x⁴ + 51x³ – 186x² + 90x – 9)

The required answer is 5x² – 9x + 3