AWES TGT Maths Mock Test - 1 - AWES TGT/PGT MCQ

The Asian Wrestling Championship 2024 is being held in Bishkek, Kyrgyzstan. In this championship, India's Udit won the silver medal in men's freestyle 57 kg. While Abhimanyu gave India the second medal in 70 kg.  

King Charles III is the monarch depicted on the new UK banknotes, which will co-circulate with those featuring Queen Elizabeth II's portrait starting from June 5, 2024.

Pradip Kumar Das, Chairman and Managing Director, Indian Renewable Energy Development Agency Limited (IREDA) participated in the 26th edition of the World Energy Congress held in Rotterdam, Netherlands. He also said that India ranks fourth globally in renewable energy installed capacity.      

IIT Bombay is ranked 45th globally in engineering according to the QS World University Rankings 2024, highlighting its excellence in the field and its leading position in South Asia.

Tarun Bajaj, the former Revenue Secretary and Secretary of Economic Affairs to the Government of India, has been appointed as the chairman of the US-India Tax Forum by USISPF, showcasing his expertise in economic matters and international relations.

Cooperation teaches the values of committing oneself to the common good, seeing success as depending on the efforts of all participants, feeling happy when others succeed, seeing others as resources to help one succeed, viewing worth as unconditional, and viewing diverse others in positive ways.
Cooperative learning is a teaching way for students to learn essential interpersonal life-skills and develop the ability to work collaboratively, a skill now greatly in demand in the workplace. Thus, it provides faster and efficient learning. In cooperative learning, the diverse students in the school are to be transformed into a symphony. Students need positive self-esteem, the psychological health to face conflict and challenge, and the social competencies required to work effectively with diverse peers. Personal and superordinate identities are developed through group processes.

Key Points 

• It is well knowledge that young infants acquire significant concepts more quickly in their mother tongue or home language.
• The mother tongue or the language used by the neighborhood is typically the same as the home language.
• However, there may occasionally be a home language used by other family members in multilingual households that is distinct from the mother tongue or the local vernacular.
• National Education Policy 2020 proposes that wherever feasible, the home language, mother tongue, local language, or regional language will be used as the medium of teaching until at least Grade 5, but preferably until Grade 8 and beyond.

Hence, At the primary level, National Education Policy 2020 proposes mother tongue/Home language as the medium of instruction across the nation.

Learning disabilities in Mathematics can be assessed most appropriately by the diagnostic test. 
A diagnostic test is a test that is used to measure the level of understanding a student possess regarding any subject area. Teachers conduct diagnostic tests for reading and maths skills and use these results to provide remedial instructions or classes to students as per their level.
Hence, the correct answer is, 'Diagnostic Test.'

Key Points 

• The curricular and pedagogical structure and the curricular framework for school education will therefore be guided by a 5 + 3 + 3 + 4 design, consisting of the stages:
  • Foundational (3 years of preschool + Grades 1-2):- The Foundational Stage will consist of play/activity-based learning. It will also include a focus on good behavior, courtesy, ethics, personal and public hygiene/cleanliness, teamwork, cooperation, etc. 
  • Preparatory (Grades 3- 5):-  The Preparatory Stage will transition gradually from play-based learning to more formal but interactive classroom learning, with the introduction of some (light) textbooks, in order to lay a solid groundwork across subjects, including reading, writing, speaking, physical education, art, languages, science, and mathematics.
  • Middle (Grades 6-8):- The Middle Stage will see the introduction of subject teachers for learning/discussion of the more abstract concepts in each subject that students will be ready for at this stage, across the sciences, mathematics, arts, social sciences, and humanities.
  • High school (Grades 9-12 in two phases, i.e. 9 and 10 in the first and 11 and 12 in the second) stages respectively, with an option of exiting at Class 10 and re-entering in the next phase:- The High School (or Secondary) Stage will comprise of four years of multidisciplinary study, building on the subject-oriented pedagogical and curricular style of the Middle Stage, but with greater depth, greater critical thinking, greater attention to life aspirations, and greater flexibility and student choice.

Thus, it is concluded that Primary classes (Classes I to V) will consist of  Preparatory and  Foundational Stages according to NEP, 2020.

To have confidence that a test is valid, and thus the inferences we make based on the test scores are valid, three kinds of validity evidence are:

Construct- It is the extent to which the content of the test matches the instructional objectives.

Criterion- It is the extent to which scores on the test are in agreement with or predict an external criterion.

Content- It is the extent to which an assessment corresponds to other variables, as predicted by some rationale or theory.

Replacing x by -x, 


Subtracting and putting

P(x) when divided by x−1, x-2 and x−3 leaves remainder 3,7 and 13 respectively.
From polynomial-remainder theorem, P(1) = 3 and P(2) = 7 P(3) = 13
If the polynomial is divided by (x−1)(x−2)(x-3) then remainder must be of the form ax+b (degree of remainder is less than that of divisor)
 
⇒P(x) = Q(x)(x−1)(x−2)(x-3)+(ax² + bx + c), where Q(x) is some polynomial. 
Substituting for x=1, x=2 and x=3:
P(1) = 3 = a+b+c
P(2) = 7 = 4a+2b+c
P(3) = 13 = 9a+3b+c
=> Subtract P(2) from P(1)
-3a - b = -4.............(1)
Subtract P(3) from P(2)
5a + b = 6..............(2)
From (1) and (2) Solving for a and b, we get 
a = 1 and b = 1, c = 1
⇒Remainder = x² + x + 1

If 2 tan^-1 (cos x) = tan ^-1(2 cosec x),

2tan^-1(cos x) = tan^-1 (2 cosec x)

= tan^-1(2 cosec x) 

= cot x cosec x = cosec x = x = π/4

(acos x + bsin x)² + (asin x – bcos x)² = a² + b²
⇒ c² + (asin x – bcos x)² = a² + b²
⇒ (asin x – bcos x)² = a² + b² – c²

If AB = BA = I , then A and B are inverse of each other. i.e. A is invers of B and B is inverse of A.

The nth term of an AP is given by T_n = S_n − S_n−1.
First compute S₄ = 2×4×4 + 5×4 = 32 + 20 = 52.
Then compute S₃ = 2×3×3 + 5×3 = 18 + 15 = 33.
So T₄ = S₄ − S₃ 
= 52 − 33
= 19.

f (x + 2y) = f (x) + f (2 y) + 4xy for x, y ∈ R putting x = y = 0, we get f (0) = 0

In a ΔABC, we know that
tanA + tanB + tanC = tanA tanB tanC
∴ tanB + tanC = tanA(tanBtanC – 1) 


⇒ tan² B - (K - 1) tan B + K = 0
For real values of tan B, Disc.
(K – 1)² – 4K > 0 
K² – 6K + 1 > 0 

A (1,1,1), B (2,3,5), C(-1,0,2) direction ratios of AB are < 1,2,4 >
Therefore, direction ratios of normal to plane ABC are < 2, -3,1 >
As a result, equation of the required plane is 2x – 3y +z = k then 

Hence, equation of the required plane is 

 By formula 
Sn = n/2(a+l)

When a = b = 1, c = 2, it gives minimum value (since a,b,c not all equal)

The equation y = 2x − x² i.e. y – 1 = - (x - 1)² represents a downward parabola with vertex at (1, 1) which meets x – axis where y = 0 .i .e . where x = 0 , 2. Also , the line y = - x meets this parabola where – x = 2x − x² i.e. where x = 0 , 3. 
Therefore , required area is :

As β ≠ 1 is afourth root of unity,
β4 = 1 ⇒ (1 - β) (1 + β + β² + β³) = 0


∴ z = 0

Gen equation of a circle is (x−h)²+(y−k)²=r² (3,0),(1,−6),(4,−1) circle passes through these points
∴ These three points should satisfy the equation
of circle.
(3−h)²+k²=r² ___ (I)
(1−h)² +(−6−k)² = r²(1−h)²+(6+k)²=r²  ___ (II)
From (I) & (II)
(3−h)²+k² =(1−h)²+(6+k)²
9+h²−6h+k² =1+h² 2h+36+k²+12k
9−6h=1−2h+36+12k
9−37=4h+12k
−28=4h+12k
h+3k=−7
h=−7−3k ___ (III)
(4−h)²+(−1−k)²=r² ___ (IV)
From (I) & (IV)
(3−h)²+k² = (4−h)² +(1+k)²
9+h² −6h+k²
 =16+h² −8h+1+k²+2k
9−6h=17−8h+2k
9−17=−2h+2k
−8=−2h+2k
−h+k=−4
k=−4+h ___ (V)
Put (V) in (III)
h=−7−3(−4+h)
h=−7+12−3h
h=5/4,k=−11/4,r=170/16
Eq of circle: (x−5/4)² +(y+11/4)² =170/16
x²+25/16−10x/4+y² +121/16+22y/4=170/16
Simplifying, we get 2x²+xy²−5x+11y−3=0

The lady can select one cotton saree out of 15 cotton varieties in 15 ways since
any of 15 varieties can be selected. Corresponding to each selection of a cotton saree, she can
choose a polyester saree in 13 ways. Hence the two sarees (one cotton and one polyester), by
multiplication principle of counting, can be selected in 15 x 13= 195 ways

n(A × B) = n(A) × n(B)

lim xsecx    x=0
lim x(1/cosx) = lim (x/cosx)
Put x = 0
= 0/cos0
= 0/1 = 0