MPTET Varg 1 Mathematics Mock Test - 3 Free Online Test 2026

Let's evaluate each statement based on the provided information:

1. India's nuclear energy strategy involves the development of Pressurized Heavy Water Reactors (PHWRs) as a central component. This statement is correct. According to India's nuclear strategy, PHWRs are indeed a central component. They form the backbone of India's domestic nuclear capabilities, which are crucial for scaling up the nuclear capacity as part of the broader energy goals.

2. The Three-Stage Nuclear Programme in India is hindered by the slow progress of Thorium-based Molten Salt Reactors (MSRs). This statement is correct. The Three-Stage Nuclear Programme, which was envisioned to utilize India's abundant thorium resources, faces challenges due to the slow progress in developing Thorium-based MSRs. This is a known bottleneck in advancing to the later stages of the program.

3. India's target for achieving net zero carbon emissions by 2070 necessitates a reliance on imported uranium to meet its nuclear energy demands. This statement is incorrect. While India does require uranium for its nuclear reactors, the current focus is on reducing dependence on imported uranium by developing fuel recycling technologies and enhancing domestic capabilities. The strategy aims to achieve energy security through self-reliance rather than increased importation.

Thus, the correct statements are 1 and 2, making Option B the right answer.

The correct answer is Sikandar Lodī.
Explanation:
Sikandar Lodi (1489-1517 AD)​:

• Modern Agra was founded by Sikandar Lodhi in the 16th century.
• He shifted the capital from Delhi to Agra in 1506.
• Babur, the founder of the Mughal dynasty, also stayed for some time in Agra and introduced the concept of square Persian-styled gardens here.
• Emperor Akbar built the Agra fort and Fatehpur Sikri near Agra.​

Other Related Points
Babur

• He is the founder of the Mughal Empire.
• He fought the first Battle of Panipat in 1526 with Ibrahim Lodhi and laid the foundation of the Mughal Empire.
• His autobiography Tuzuk-i-Babri was written in the Turkish Language.
• In 1527 his general Mir Baki built Babri Maszid.
• He died in 1530.

Akbar

• He was born in 1542 at Amarkot.
• He fought the second battle of Panipat in 1556 with Hemu and he defeated Hemu.
• He abolished the religious tax Jizya.
• He founded a new religion "Din-i- Allahi".
• His biography was written by Abul Fazal " Ain-i-Akbari".

Raja Bhoj

• He was a famous scholar and poet of Sanskrit.
• He ruled 'Dhar' from 1000 AD to 1055 AD.
• He was a descendant of "Chakravarti Emperor Vikramaditya".
• He was born in 980 AD in Ujjaini, the city of Maharaja Vikramaditya.
• He made 'Dhar' his new capital in place of 'Ujjain'.​

Bloom and his associates classified educational behavior from simple to complex, based on the level of learning, and developed a taxonomy of learning outcomes.
Explanation:
Objectives can be written for any type of learning. The three domains as propounded by Benjamin Bloom are Cognitive means thought or knowledge, Affective means feelings or choices, Psychomotor means physical skills.

Thus, it is concluded that the three domains of Bloom's taxonomy are Cognitive, Affective, and Psychomotor.

Educational planning generally is understood to be the identification, development, and implementation of strategies designed to attain, efficiently and effectively, the educational needs and goals of students and society.

Explanation

• Educational planning relates to choosing the best course of action for achieving educational objectives.
• It may be defined as a systematic design of action for the realization of educational aims and objectives for individual and social development through maximum utilization of available resources.
• It is essential for the best utilization of available resources.
• Through proper planning in education, education can be the best means by which society will preserve and develop its future value system, way of life of an individual, knowledge, skills, applications, and culture of the country.
• Through proper educational planning, the means and ends of society can properly interact through the educational system.
• It is highly essential for preparing a blueprint or plan of action for every program of an educational institution or organization.
• Planning in education is necessary for making one’s educational journey goal-oriented and purposeful.
• It is essential to maintain, sustain and enhance the thinking process of an individual, institution, or organization.

Thus, it is concluded that educational planning relates to choosing the best course of action for achieving educational objectives.

E.L. Thorndike is one of the well-known American learning theorists who propounded a theory called connectionism. Thorndike posited that learning was a process of "stamping in" or forming connections between a stimulus and a response.

Explanation Connectionism is a theory of learning which comes under the Behaviorist School of Thought on learning.

• Connectionism was propounded by Edward Lee Thorndike.
• According to him, learning is the result of the neural connection between stimulus (S) and response(R) through trial and error.
• For him, trial and error mean selecting and connecting.
• His notion of learning was arrived at after his famous experiment conducted on animals.
• He used to put an animal in a cage that was arranged in such a way that the animal made a certain kind of response to escape from the cage.
• Based on his theory, Thorndike formulated three important laws of learning. These are the Law of Readiness, the Law of Exercise, and the Law of Effect.

Thus, it is concluded that Thorndike’s theory is known as connectionism.

Ivan Petrovich Pavlov was a Russian physiologist. During the late nineteenth and early twentieth century, Pavlov and his associates were working in a laboratory in St. Petersburg on the digestive process of dogs.

Explanation

• Pavlov’s learning is based on the assumption that the behavior of the living organism is mechanical.
• According to this viewpoint, learning is the ‘formation of conditioned reflexes’ or “acquisition of involuntary anticipatory adjustment” or “habit formation”.
• This controlled learning is conditioned learning.
• In this the thing to be learned is termed as, ‘cue stimulus,’ and this is then followed by a satisfying stimulus.
• This combination is repeated again and again and ultimately the learning of the “cue stimulus” is established. This is conditioned learning.
• Conditioned learning is also accompanied by a generalization of the experience.

Thus, it is concluded that Pavlov’s learning is based on the assumption that the behavior of the living organism is mechanical.

Inclusive education is defined as a learning environment that encourages all students to reach their full potential in terms of personal, academic, and professional growth regardless of race, class, color, gender, disability, or language.

ExplanationPrinciples of inclusive education-

• Inclusive education allows students to form friendships with their peers and experience less social anxiety as a result of their backgrounds or disabilities.
• It provides equal educational benefits for all types of students.
• It emphasizes no discrimination between students.
• Individual differences among children are a source of variety and diversity, rather than a source of conflict.
• Children with special educational needs (SEN) having visual, hearing locomotor, or intellectual disabilities, a learner from disadvantaged and deprived communities learn more in a regular classroom environment with the help of inclusive education.

Hence, we conclude that according to contemporary policies of education, inclusive education is the most suggested model of schooling for children with disabilities.

Other Related Points

• Vocational education refers to the education that prepares people to work as technicians, tradespeople, or artisans in a skilled craft or trade.
• Separate education refers to the practice of instructing children with special education needs in separate buildings or schools.

Communication refers to sharing of ideas, facts, opinions, information, and understanding. Simply stated, communication is the dissemination of information and understanding from one person to another.

• Communication does not mean a good grip over vocabulary or grammar, it simply defines how well or effectively a person can pass the information.
• Communication skills are the abilities that we use while giving and receiving different kinds of information.

Explanation

• ​Communication may not be always effective. There are many factors that hamper communication. The factors that hamper and distort communication are called noise or barrier.
• In classroom communication mostly the teacher is the sender, the student is the receiver, subject knowledge is the message, and medium could be oral, written, etc.

Following are the teacher-centric barriers in communication:

• Through the use of non-verbal language, teachers draw the student’s attention to more understanding and motivate the students and even bring excitement to the bored students.
• Non-verbal communication includes facial expressions, eye contact, tone, the pitch of the voice, and overall condition of the body and hands movement.
• Gestures & tone: The gestures and tone of the teacher must be effective for effective communication.
• Lack of interest: When the teacher is not interested in the subject matter for which the message is based, communication is bound to be ineffective.
• Language: It Opens in new windowplays an important role in any inter-personal communication where both the sender and the receiver must be reasonably comfortable in the language used. If the language is not understood, it will be a barrier.
• Attitude: Teachers must have a positive attitude while sharing a message with students.

Thus,it is concluded that Gestures, tone, language, attitude, and interest are the teacher-centric barriers in communication.

ICT refers to Information and Communication Technology which is defined as a diverse set of technological tools and resources used to communicate, and to create, disseminate, store, and manage information. It includes physical telecommunication systems, networks, and the services that utilize these.

Explanation

Advantages of ICT in education at the school level:

• Teaching-learning: It provides information in a comprehensive manner with different examples, therefore, expanding the learner’s information base. ICT provides variety in the presentation of content to which they can learn according to their learning pace. Further, it builds the interaction between the user and makes the learning more active.
• Collaborative Learning: Students work in teams together to explore the assigned project. The students are accountable to one another and guide the project in a meaningful direction.
• Diagnostic Testing and Remedial teaching: ICT can help the teachers as well as students in identifying the problem area. Tests can be made available on the website of the school and students can access them from home also.
• Evaluation activities: Online tests can be used by individual students to evaluate his/her learning. Students can instantaneously get feedback about the status of his/her understanding.
• Psychological analysis of learners: ICT can be used in psychological testing also as it makes it easy to digitalize all the psychological tests including the scoring process and evaluation.
• Development of reasoning and thinking abilities among students: ICT provides students with a variety of instructional materials and they can choose those that suit them the best. ICT can be used for developing reasoning and thinking abilities among students belonging to different age groups.
• Instructional material development: The instructional technique of the teachers who teach very well, their lectures can be recorded in CD-ROM and should be made available for everyone to access. It enhances the quality of instruction in the classrooms.

Thus, we can conclude that both it encourages interactions and develops collaborative culture and utilizes active learning and introduces feedback in a proper way are the benefit of using ICT in classroom teaching.

Concept Used:-

When objects or things are arranged in a definite order, then it is called permutation. In permutation, order of arrangement matters.

The total number of permutation in which n things can be arranged, taken r at a time, is given as,

Explanation:-

Given word is ‘TRICK’

There are total 5 letters in this word and all are different from each other. So, the number of 3-letter words [with or without meaning] which can be formed with a given word are,

⇒ ⁵P₃ = 5!/(5 - 3)!

⇒ 5P3 = (5 × 4 × 3 × 2!)/2!

⇒ 5P3 = 5 × 4 × 3

⇒ 5P3 = 60

Thus, the number of 3-letter words with or without meaning that can be formed using letters of the word ‘TRICK’ is 60.

Concept Used:

also, sin (π/2 - x) = cos x
and cos (π/2 - x) = sin x

Also,

Calculation:

Let
⇒
⇒
Adding (1) and (2)
⇒
⇒

Now, let

⇒

⇒

Put tan x/2 = t
⇒
⇒

⇒ I₁
⇒ I1
⇒ I1
⇒ I1
⇒ I1
⇒I1
⇒ I1
Using I₁, in I

Concept:
Domain of a Relation: Let R be a relation from set A to set B. Then, the set of all first components of the ordered pair belonging to relation R forms the domain of the relation R i.e Domain (R) = {a: (a, b) ∈ R}.
Range of a Relation: Let R be a relation from set A to set B. Then, the set of all second components of the ordered pair belonging to relation R forms the range of the relation R i.e Range (R) = {b: (a, b) ∈ R}.
Calculation:
Given: A and B are the domain and range respectively for the relation R such that R = {(5, 3), (6, 4), (7, 5)}
As we know that, Domain (R) = {a: (a, b) ∈ R}.
⇒ A = {5, 6, 7}
We also know that Range (R) = {b: (a, b) ∈ R}.
⇒ B = {3, 4, 5}
Hence, option 3 is the correct answer.

X coordinate of A can be given as, Using Section formulae

⇒ X coordinate of A = {4 × (7/2) + 1 × 6}/(4 + 1) = 4

Y coordinate of A can be given as, Using Section formulae

⇒ Y coordinate of A = {4 × (6) + 1 × 1}/(4 + 1) = 5

∴ Coordinates of point A are (4, 5)

Concept:
In case of a moderately skewed distribution, the difference between mean and mode is almost equal to three times the difference between the mean and median.
Mean – Mode = 3 (Mean – Median)
Calculations:
Given, Mean = 12.5
Median = 10
Mode = x
(Mean – Mode) = 3 (Mean – Median)
∴ (12.5 – x) = 3 (12.5 – 10)
Mode, x = 5

Concept:
If we position the vector of two points A and B, then the position vector of AB will be given by:
AB = OB - OA
Given:
The position vectors of points P, Q, R, and S are a, b, (2a + 3b), and (a – 2b).
Calculation:
OP = a, OQ = b, OR = (2a + 3b), OS = (a - 2b)
Now, SQ = (b - (a - 2b)
⇒ (b - a + 2b) = (3b - a)

Concept:

• A skew-symmetric (or anti-symmetric or anti-metric) matrix is a square matrix A = [a_ij] such that a_ij = -a_ji for every i, j.
• The transpose of a skew-symmetric matrix equals its negative: A^T = -A.
• The inverse of the transpose of a matrix is equal to the transpose of its inverse: (A^T)^-1 = (A^-1)^T.

Calculation:
Since A is an skew-symmetric matrix, we must have:
AT = -A
Because A is given to be invertible, on taking the inverse of both sides, we get:
(AT)^-1 = (-A)^-1
We also know that (AT)-1 = (A-1)T.
∴ (A-1)T = -(A-1)
Since the transpose of A^-1 is equal to its negative, A^-1 is a skew-symmetric matrix.Other Related Points

• The elements on the diagonal of a skew-symmetric matrix are zero. i.e. a_ij = 0, for i = j.
• For a n×n skew-symmetric matrix A: det(A^T) = det(-A) = (-1)ⁿ det (A).
• All odd dimension skew symmetric matrices are singular as their determinants are always zero.

Explanation:

We have, Δ =

Applying C₁ → C₁ - C₃

⇒ Δ =

Applying C₂ → C₁ + C₂

⇒ Δ =

⇒ Δ =

⇒ Δ = - [b(ac - b²) - c(c² - ab) + a(bc - a²)]

⇒ Δ = - abc + b³ + c³ - abc - abc +a³

∴ Δ = a3 + b3 + c³ - 3abc

Explanation:
Lami's theorem:
It states that if three forces acting at a point are in equilibrium, each force is proportional to the sine of the angle between the other two forces.
Consider three forces FA, FB, FC acting on a particle or rigid body making angles α, β and γ with each other.

Then from Lami's theorem,

The domain is {-3, -2, 5, 6} and range = {-8, -6, 3, 4, 7}.
Here we can see, for x = 5, there are two corresponding y values -6 and 3. Hence, this relation is not a function.

Also, there are four elements in the domain and five elements in range. This implies two values in the range is associated with one value in the domain.
Hence, the given domain and range will not form a function.

Given:

an and bn are bounded sequences

Concept used:

A sequence is bounded if all the terms of the sequence are ≤ N s.t there does not exist any term in the sequence which is > N.

Calculations:

an and bn are bounded sequences

⇒ a_n < M and b_n < M'

⇒ a_n + b_n < M + M'

∴ all the term of the sequence of a_n + b_n < M + M'

∴ a_n + b_n is also bounded.

Concept:
Trigonometric Identities:

• sin2 θ + cos2 θ = 1.
• sin 2θ = 2 sin θ cos θ.
• cos 2θ = cos2 θ - sin2 θ.

Calculation:
Given: sin θ + cos θ = 1
Squaring both sides, we get:
⇒ sin2 θ + cos2 θ + 2 sin θ cos θ = 1
⇒ 1 + sin 2θ = 1
⇒ sin 2θ = 0
∴ The value of sin 2θ is equal to 0.

Concept:
Fermat’s little theorem: It states that if p is a prime number, then for any integer a, the number a p – a is an integer multiple of p.

• Here p is a prime number a^p ≡ a (mod p).
• If 2^p ≡ 2 (mod p) i,e, 2p - 2 is divisible by p.

Concept:

Projection Formula:- the length of any side of a triangle is equal to the sum of the projections of the other two sides on it.

In triangle ABC

1. a = b cos C + c cos B

2. b = c cos A + a cos C

3. c = a cos B + b cos A

Law of cosine:-

a² = b² + c² - 2bc cos A

b² = c² + a² - 2ac cos B

Calculation:

We have,

In triangle ABC, a, b and c are sides of triangle.

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

∴ The ratio of sin (A - B) to sin (A + B) is

The conditions to apply mean value theorem are
(i) f(x) is continuous
(ii) f(x) is differentiable
Given function, f(x) = |x|
f(x) is continuous

f(x) is not differentiable at x = 0.
Therefore, mean value theorem is not applicable.

Given:

Point P is the midpoint of segment AB.

Co-ordinates of P are (-5, -2) and A are (-2, 0)

Concept used:

If P is the midpoint of segment AB where Co-ordinates of A is (x₁, y₁)

and Co-ordinates of B is (x₂, y₂), then, Co-ordinates of P = {(x₁ + x₂)/2, (y₁ + y₂)/2}

Calculation:

Let the co-ordinates of B are (x, y)

According to the given formula:

Co-ordinates of P = {(x1 + x2)/2, (y1 + y2)/2}

⇒ (-2 + x)/2 = -5, (0 + y)/2 = -2

⇒ -2 + x = -10, 0 + y = -4

⇒ x = (-10 + 2), y = (-4 – 0)

⇒ x = -8, y = -4

∴ The co-ordinates of B is (-8, -4).

Explanation:
The given ordinary differential equation (ODE) is

The given equation can be written as
D = d/dx
(D² +D - 6) y = 0
The auxiliary equation is D² + D - 6 = 0 ⇒ (D - 2) × (D + 3) = 0
The roots of auxiliary equation are -3 and 2
The complementary function C.F = c₁e^-3x + c₂e^2x
Since the RHS = 0, there in no particular integral
The solution is y = c₁e^-3x + c₂e^2x

Concept:
The function f(x) is continuous at x = a if
f(a-) = f(a) = f(a+)
Calculation:

Given, is continuous at x = 0,

⇒ f(0-) = f(0) = f(0+)
⇒ k(2(0) - 0) = cos 0
⇒ 0 = 1
So, there does not exist any k for which f(x) is continuous at x = 0
∴ The correct answer is option (4).

Given:

Identify 3-digit numbers where all digits are prime.

Prime digits: 2, 3, 5, 7

Calculation:

Each digit of the 3-digit number can be any of the 4 prime numbers.

Hundreds place: 4 choices (2, 3, 5, 7)

Tens place: 4 choices (2, 3, 5, 7)

Units place: 4 choices (2, 3, 5, 7)

Total number of 3-digit numbers = 4 × 4 × 4

⇒ 4³ = 64

∴ The number of 3-digit numbers where all digits are prime is 64.

Concept Used:

Consider a pair of linear equations in two variables x and y.

Here a1, b1, c1, a2, b2, c2 are all real numbers.
If a1/a2 ≠ b1/b2, then the unique solution for the linear pair. (consistent)

Graph: two intersecting lines.

If a1/a2 = b1/b2 ≠ c1/c2, then no solution for the linear pair. (inconsistent)

Graph: two parallel lines.

If a1/a2 = b1/b2 = c1/c2, then infinite solution for the linear pair.(consistent)

Graph: two coincident lines.

Calculation:

For equations kx - 2y - 3 = 0 and 3x + y - 5 = 0

Here, a1 = k, b1 = -2, c1 = -3, a2 = 3, b2 = 1 and c2 = -5

a1/a2 = k/3 , b1/b2 = -2/1 = -2

For a unique solution,

a1/a2 ≠ b1/b2 ⇒ k/3 ≠ -2

⇒ k ≠ -6

Hence, k can be any real value except -6.

Given:
ax2 + 7x + b = 0
α + β = -b/a
Formula used:
α + β = -b/a
α β = c/a
Solution:
2/3 + (- 3) = -7/a
a = 3
α β = b/a

(2/3)(-3) = b/3

b = -6
Hence, the correct option is 1.