# CBSE Class 11 Applied Maths Algebra 2021

CBSE board has introduced a new topic named Applied Mathematics to help create and progress your mathematical skills and techniques. Algebra is one of the chapters among the Class 11 Applied Maths syllabus.

The algebra unit holds a weightage of 10 marks out of 80 marks in the theory exam. This post takes you through detailed information about CBSE Class 11 Applied Maths Algebra.

## CBSE Class 11 Applied Maths Algebra Important Topics

Applied mathematics course prepares you to choose algebraic methods as a means of representation and as a problem-solving tool. Algebra mainly focuses on topics like sets, relations, Venn diagrams, relation between arithmetic and geometric progression, etc.

Go through the table below to know the detailed** CBSE Class 11 Applied Maths Syllabus **for algebra**.**

**Name of the topics** |

1. Sets |

2. Types of sets |

3. Venn diagram |

4. De Morgan’s Law |

5. Problem solving using Venn Diagram |

6. Relations and types of relations |

7. Introduction of sequences, series |

8. Arithmetic and Geometric progression |

9. Relationship between AM and GM |

10. Basic concepts of Permutations and Combinations |

11. Permutations, Circular Permutations, Permutations with restrictions |

12. Combinations with standard results |

Algebra is basically the study of unknown quantities. Some of the topics of algebraic expressions and formulas are:

**Algebraic identities**

Various equality equations are consisting of different variables in algebraic identities.

**a) Linear Equations in One Variable:** A linear equation in one variable has the maximum of one variable present in order 1. It is depicted in the form of ax + b = 0, where x is represented as the variable.**b) Linear Equations in Two Variables:** A linear equation in two variables consists of the utmost two variables present in order 2. The equation is depicted in the form: ax2 + bx + c = 0. The two variables are quite important because your coursebook has a lot of questions based on it. So, you need to stay focused on important algebra formulas to find the solution.

**Some basic identities to note are:**

- The combination of literal numbers obeys every basic rule of addition, subtraction, multiplication and division.
- x × y = xy; such as 5 × a = 5a = a × 5.
- a × a × a × … 9 more times = a12
- If a number is x8, then x is the base, and 8 is the exponent.
- A constant is a symbol with a fixed numerical value.

**Law of exponent**

The degrees and powers in any mathematical expressions are known as exponent. Some of the laws of exponent are:

- a0 = 1
- a-m = 1/am
- (am)n = amn
- am / an = am-n
- am x bm = (ab)m
- am / bm = (a/b)m
- (a/b)-m =(b/a)m
- (1)
*n* = 1 for infinite values of *n*

**Quadratic equations**

The linear equations in two variables are known as quadratic equations.

The roots of the equation ax2 + bx + c = 0 (where a ≠ 0) can be given as:

−*b*±*b*2−4*ac*√2*a*

Below given are some important points about the equation as a part of important algebra formulas:

- Δ = b2 − 4ac is also known as a discriminant.
- For roots;

Delta; > 0 happens when the roots are real and distinct

For real and coincident roots, Δ = 0

Delta; < 0 happens in the case when the roots are non-real

- If α and β are the two roots of the equation ax2 + bx + c then,

α + β = (-b / a) and α × β = (c / a). - If the roots of a quadratic equation are α and β, the equation will be

(x − α)(x − β) = 0.

The general algebra formulas can be given as:

**n is a natural number:** an – bn = (a – b)(an-1 + an-2b+…+ bn-2a + bn-1)**If n is even:** (n = 2k), an + bn = (a – b)(an-1 + an-2b +…+ bn-2a + bn-1)**n is odd:** (n = 2k + 1), an + bn = (a + b)(an-1 – an-2b +an-3b2…- bn-2a + bn-1)**General square Formula:** (a + b + c + …)2 = a2 + b2 + c2 + … + 2(ab + ac + bc + ….)

**List of important formulas**

- (a + b)2 = a2 + 2ab + b2
- (a – b)2 = a2 – 2ab + b2
- (a + b) (a – b) = a2 -b2
- (x + a) (x + b) = x2 + (a + b) x + ab
- (x + a) (x – b) = x2 + (a – b) x – ab
- (x – a) (x + b) = x2 + (b – a) x – ab
- (x – a) (x – b) = x2 – (a + b) x + ab
- (a + b)3 = a3 + b3 + 3ab (a + b)
- (a – b)3 = a3 – b3 – 3ab (a – b)
- (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
- (a – b)4 = a4 – 4a3b + 6a2b2 – 4ab3 + b4
- (x + y + z)2 = x2 + y2 + z2 + 2xy +2yz + 2xz
- (x + y – z)2 = x2 + y2 + z2 + 2xy – 2yz – 2xz
- (x – y + z)2 = x2 + y2 + z2 – 2xy – 2yz + 2xz
- (x – y – z)2 = x2 + y2 + z2 – 2xy + 2yz – 2xz
- x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz -xz)
- x2 + y2 = 12
- [(x + y)2 + (x – y)2]
- (x + a) (x + b) (x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc
- x3 + y3 = (x + y) (x2 – xy + y2)
- x3 – y3 = (x – y) (x2 + xy + y2)
- x2 + y2 + z2 – xy – yz – z [(x – y)2 + (y – z)2 + (z – x)2]

### CBSE Class 11 Applied Maths Algebra Important Questions

To ease out your preparation, we have provided topic-wise important questions for class 11 Applied Maths Algebra in the post below.

#### Sets

**Question 1:**

Write the following sets in the roaster form.

(i) A = {x | x is a positive integer less than 10 and 2x – 1 is an odd number}

(ii) C = {x : x2 + 7x – 8 = 0, x ∈ R}

**Question 2:**

Write the following sets in roster form:

(i) A = {x : x is an integer and –3 ≤ x < 7}

(ii) B = {x : x is a natural number less than 6}

Sets Important Questions PDF

#### Types of sets

**Question 1:**

Let U = {x : x ∈ N, x ≤ 9}; A = {x : x is an even number, 0 < x < 10}; B = {2, 3, 5, 7}. Write the set (A U B)’.

**Question 2:**

Let U = {1, 2, 3, 4, 5, 6, 7}, A = {2, 4, 6}, B = {3, 5} and C = {1, 2, 4, 7}, find

(i) A′ ∪ (B ∩ C′)

(ii) (B – A) ∪ (A – C)

#### Venn diagram

**Question 1:**

In a class of 50 students, 10 take Guitar lessons and 20 take singing classes, and 4 take both. Find the number of students who don’t take either Guitar or singing lessons.

#### De Morgan's laws

**Question 1:**

Prove De Morgan’s Laws by Venn Diagram

(i) (A∪B)’= A’∩ B’

#### Relations and types of relations

**Question 1:**

Write the range of a Signup function.

**Question 2:**

The Cartesian product A × A has 9 elements, among which are found (–1, 0) and (0,1). Find the set A and the remaining aspects of A × A.

Relations Important Questions PDF

#### Introduction of Sequences, Series

**Question 1:**

The sums of n terms of two arithmetic progressions are in the ratio 5n+4: 9n+6. Find the ratio of their 18th terms.

**Question 2:**

Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

#### Arithmetic and Geometric progression

**Question 1:**

Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

**Question 2:**

What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?

- 897
- 1,64,850
- 1,64,749
- 1,49,700

Arithmetic & Geometric Progression Important Questions PDF

#### Relationship between AM and GM

**Question 1:**

Find the AM, GM, and HM between 12 and 30

#### Basic concepts of Permutations and Combinations

**Question 1:**

Find the 3-digit numbers that can be formed from the given digits: 1, 2, 3, 4 and 5, assuming that

- a) digits can be repeated.
- b) digits are not allowed to be repeated.

**Question 2:**

A coin is tossed 6 times, and the outcomes are noted. How many possible results can be there?

Permutations & Combinations Important Questions PDF

#### Permutations, Circular Permutations, Permutations with restrictions

**Question 1:**

From a team of 6 students, in how many ways can we choose a captain and vice-captain assuming one person cannot hold more than one position?

**Question 2:**

Find the number of ways in which 10 beads can be arranged to form a necklace.

**Question 3:**

Find the number of ways in which four girls and three boys can arrange themselves in a row so that none of the boys is together? How is this arrangement different from that in a circular way?

#### Combinations with standard results

**Question 1:**

How many words can be formed each of 2 vowels and 3 consonants from the letters of the given word – DAUGHTER?

**Question 2:**

Find the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.

### CBSE Class 11 Applied Maths Algebra Sample Questions with Solutions

To help you get an idea about the type of questions that can be asked in the exam, we have provided some sample questions for your reference here.

**Question 1**

How many words can be formed each of 2 vowels and 3 consonants from the given word's letters – DAUGHTER?

**Solution:**

No. of Vowels in the word – DAUGHTER is 3.

No. of Consonants in the word Daughter is 5.

No of ways to select a vowel = 3c2 = 3!/2!(3 – 2)! = 3

No. of ways to select a consonant = 5c3 = 5!/3!(5 – 3)! = 10

Now you know that the number of combinations of 3 consonants and 2 vowels = 10x 3 = 30

Total number of words = 30 x 5! = 3600 ways.

**Question 2**

It is needed to seat 5 boys and 4 girls in a row to get the even places. How many are such arrangements possible?

**Solution:**

5 boys and 4 girls are to be seated in a row to get the even places.

The 5 boys can be seated in 5! Ways.

For each of the arrangement, the 4 girls can be seated only at the places which are cross marked to make girls occupy the even places).

B x B x B x B x B

So, the girls can be seated in 4! Ways.

Hence, the possible number of arrangements = 4! × 5! = 24 × 120 = 2880

**Question 3**

Find the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.

**Solution:**

Take a deck of 52 cards,

To get exactly one king, 5-card combinations have to be made. It should be made in such a way that in each selection of 5 cards, or a deck of 52 cards, there will be 4 kings.

To select 1 king out of 4 kings = 4c1

To select 4 cards out of the remaining 48 cards = 48c4

To get the needed number of 5 card combination = 4c1 x 48c4

= 4x2x 47x 46×45

= 778320 ways.

**Question 4**

Find the number of 6 digit numbers that can be formed by using the digits 0, 1, 3, 5, 7, and 9. These digits shall be divisible by 10, and no digit shall be repeated?

**Solution:**

The number which has a 0 in its unit place is divisible by 10.

If we put 0 in the unit place, _ _ _ _ 0, there will be as many ways to fill 5 vacant places. (1, 3, 5, 7, 9)

The five vacant places can be filled in 5! ways = 120.

**Question 5**

Evaluate 10! – 6!

**Solution:**

10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x1 = 3628800

6! = 6 X 5 x 4 x 3 x 2 x 1 = 720

10! – 6! = 3628800 – 720 = 3628080

**Question 6**

Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.

**Solution:**

Let a and d be the first term and the common difference of the A.P. respectively. It is known

that the kth term of an A.P. is given by

ak = a +(k -1)d

Therefore, am+n = a +(m+n -1)d

am-n = a +(m-n -1)d

am = a +(m-1)d

Hence, the sum of (m + n)th and (m – n)th terms of an A.P is written as:

am+n+ am-n = a +(m+n -1)d + a +(m-n -1)d

= 2a +(m + n -1+ m – n -1)d

=2a+(2m-2)d

=2a + 2(m-1)d

= 2 [a + (m-1)d]

= 2 am [since am = a +(m-1)d]

Therefore, the sum of (m + n)thand (m – n)th terms of an A.P. is equal to twice the mth term.

**Question 7**

Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

**Solution:**

The integers from 1 to 100, which are divisible by 2, are 2, 4, 6 ….. 100.

This forms an A.P. with both the first term and common difference equal to 2.

⇒ 100=2+(n-1)2

⇒ n= 50

Therefore, the sum of integers from 1 to 100 that are divisible by 2 is given as:

2+4+6+…+100 = (50/2)[2(2)+(50-1)(2)]

= (50/2)(4+98)

= 25(102)

= 2550

The integers from 1 to 100, which are divisible by 5, 10…. 100

This forms an A.P. with both the first term and common difference equal to 5.

Therefore, 100= 5+(n-1)5

⇒5n = 100

⇒ n= 100/5

⇒ n= 20

Therefore, the sum of integers from 1 to 100 that are divisible by 2 is given as:

5+10+15+…+100= (20/2)[2(5)+(20-1)(5)]

= (20/2)(10+95)

= 10(105)

= 1050

Hence, the integers from 1 to 100, which are divisible by both 2 and 5 are 10, 20, ….. 100.

This also forms an A.P. with both the first term and common difference equal to 10.

Therefore, 100= 10+(n-1)10

⇒10n = 100

⇒ n= 100/10

⇒ n= 10

10+20+…+100= (10/2)[2(10)+(10-1)(10)]

= (10/2)(20+90)

= 5(110)

= 550

Therefore, the required sum is:

= 2550+ 1050 – 550

= 3050

Hence, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.

**Question 8**

Let U = {x : x ∈ N, x ≤ 9}; A = {x : x is an even number, 0 < x < 10}; B = {2, 3, 5, 7}. Write the set (A U B)’.

**Solution:**

Let U = {x : x ∈ N, x ≤ 9}; A = {x : x is an even number, 0 < x < 10}; B = {2, 3, 5, 7}

U = { 1, 2, 3, 4, 5, 6, 7, 8, 9}

A = {2, 4, 6, 8}

A U B = {2, 3, 4, 5, 6, 7, 8}

(A U B)’ = {1, 9}

**Question 9**

In a survey of 600 students in a school, 150 students were drinking Tea and 225 drinking Coffee, and 100 were drinking both Tea and Coffee. Find how many students were drinking neither Tea nor Coffee.

**Solution:**

Given,

Total number of students = 600

Number of students who were drinking Tea = n(T) = 150

Number of students who were drinking Coffee = n(C) = 225

Number of students who were drinking both Tea and Coffee = n(T ∩ C) = 100

n(T U C) = n(T) + n(C) – n(T ∩ C)

= 150 + 225 -100

= 375 – 100

= 275

Hence, the number of students who are drinking neither Tea nor Coffee = 600 – 275 = 325

### NCERT Solution for Class 11 Applied Math Algebra

NCERT Solutions for CBSE Class 11 Applied mathematics Algebra is an essential resource for preparing for the exam. You can verify your answers and understand how each problem is solved.

Practicing the exercise problems regularly will help in scoring more marks in the exam. The solution PDF is prepared by the subject experts to help you understand the explanation and the way it is solved.

NCERT Applied Maths Algebra solutions provide a straightforward way of illustrations and explanations. The format of these textbooks is straightforward and easy. The content of the book and its problem-solving procedures are straightforward. The more straightforward solutions do not compromise on the quality of the content.

### Complex Number Class 11 Notes with Examples

Complex Numbers Class 11 is defined when a number can be represented in form p + iq. Here, p and q are real numbers and *i*=−1−−−√. For a complex number z = p + iq, p is known as the real part, represented by Re z, and q is known as the imaginary part; it is represented by Im z of complex number z.

The topics and the subtopics taught in Complex numbers **CBSE class 11 applied maths algebra** are:

- Introduction
- Complex Numbers
- Algebra of Complex Numbers - Addition of two complex Numbers; The difference between two complex Numbers; Multiplication of two complex Numbers; Division of two complex Numbers; Power of i; The square root of a negative real number; Identities
- The Modulus and the Conjugate of Complex Numbers
- Argand Plane and Polar Representation - Polar Representation of Complex Numbers

The notes for class 11 give you detailed knowledge describing the concepts involved in complex numbers. Some of the examples are:

**Example 1: ** If 4x + i(3x – y) = 3 + i (– 6), where x and y are real numbers, then find the values of x and y.

Solution: Given,

4x + i (3x – y) = 3 + i (–6) ….(1)

By equating the real and the imaginary parts of equation (1),

4x = 3, 3x – y = –6,

Now, 4x = 3

⇒ x = 3/4

And 3x – y = -6

⇒ y = 3x + 6

Substituting the value of x,

⇒ y = 3(3/4) + 6

⇒ y = 33/4

Therefore, x = 3/4 and y = 33/4.

**Example 2**: Express (-√3 + √-2)(2√3 – i) in the form of a + ib.

Solution: We know that i2 = -1

(-√3 + √-2)(2√3 – i) = (-√3 + i√2)(2√3 – i)

= (-√3)(2√3) + (i√3) + i(√2)(2√3) – i2√2

= -6 + i√3(1 + 2√2) + √2

= (-6 + √2) + i√3(1 + 2√2)

This is of the form a + ib, where a = -6 + √2 and b = √3(1 + √2).

**Example 3: ** Find the multiplicative inverse of 2 – 3i.

Solution: Let z = 2 – 3i

*z*¯ = 2 + 3i

|z|2 = (2)2 + (-3)2 = + = 13

We know that the multiplicative inverse of z is given by the formula:

*z*−1=*z*¯|*z*|2

= (2 + 3i)/13

= (2/13) + i(3/13)

Alternatively,

Multiplicative inverse of z is:

z-1 = 1/(2 – 3i)

By rationalizing the denominator we get,

= (2 + 3i)/(4 + 9)

= (2 + 3i)/ 13

= (2/13) + i(3/13)

**Example 4: **Represent the complex number z = 1 + i√3 in the polar form.

Solution: Given, z = 1 + i√3

Let 1 = r cos θ, √3 = r sin θ

By squaring and adding, we get

r2(cos2θ + sin2θ) = 4

r2 = 4

r = 2 (as r > 0)

Therefore, cos θ = 1/2 and sin θ = √3/2

This is possible when θ = π/3.

Thus, the required polar form is z = 2[cos π/3 + i sin π/3].

Hence, the complex number z = 1 + i√3 is represented as shown in the below figure.