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

School is a place designed to provide an adequate learning environment. In school, the opportunity a child gets to learn affects his cognitive development. The more opportunities a child gets, the better his cognition because he/she will be able to add to his/her mental capacities by learning through these opportunities. In school, during teaching-learning, many new concepts form, many new ideas develop, and many other mental processes take place.

Inclusive Education:

• Inclusive education is when all students, regardless of any challenges they may have, are placed in age-appropriate general education classes that are in their own neighbourhood schools to receive high-quality instruction, interventions and supports that enable them to meet success in the core curriculum.
• Inclusive education is a system in which children with and without disabilities learn and play together.

Objectives of Inclusive Education:

• To develop an individualized education program to impart personalized and need-based education systems and experiences to all students, abled, or disabled.
• To give appropriate vocational services to all students with disabilities and guide them towards their careers.
• To provide transition services and implement a post-secondary school plan for each student with a disability so that they can get help and adjust at higher levels of education.
• To give a coordinated and comprehensive instructional program for kindergarten through high school.

Conclusion:
Inclusive education is a process in way of thinking about how to be creative to make our schools a place where all children can participate in learning. Though there the children with disability can play, work and enjoy themselves with regular students, but its main motive is to help children with disability in learning. Hence, option (4) is correct.

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.

Childhood refers to the period of the human lifespan, ranging from birth to puberty. It consists of many developmental stages and 'adolescence' is one of them.

• 'Adolescence' comes from the Latin word 'Adolescere' which means 'to grow to become mature'. It a stage which lies between the age of '12 to 19 years'.
• Emotion is a mental state associated with fear, anger, love, etc. Emotional states of an individual are the combination of physiological arousal, psychological appraisal, and behavioral changes.
• Emotionally disturbed children can be defined in terms of certain observable characteristics such as withdrawn behavior, failure to achieve a level reasonably commensurate with ability or tendency towards fighting, and other aggressive behavior.

​Key Points

Emotional disturbances of adolescents:

• They show the same emotions as they used to show in their childhood  i.e., fear, disgust, but the emotions of adolescence are different in their patterns.
• The emotions at this stage are more uncontrolled, intensive, and irrational.  As they gradually grow, their emotional behavior improves and become mature.
• Children who suffer from emotional disturbances generally do nail-biting, thumb sucking, depression, pulling or stretching of hairs, fear of specific objects or things, etc.
• A child wearing a well-fitted dress is not a symptom of emotional disturbance but wearing proper clothes that fit them well makes them confident.

Hence, it could be concluded that wearing a well-fitted dress is not found in the symptoms of emotional disturbances of adolescents.

Calculation:

The set X is given by

{2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50, 54, 58, 62, 66, 70, 74, 78, 82, 86, 90, 94, 98}.

We want to find the maximum size of a subset S of X such that no two

elements sum to 100.

The pairs in X that sum to 100 are

(2, 98), (6, 94), (10, 90), (14, 86), (18, 82), (22, 78), (26, 74), (30, 70), (34,

66), (38, 62), (42, 58), (46, 54).

Therefore, 

S = {2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50}

∴ The maximum possible number of elements in S be 13.

He can invite one or more friends by inviting 1 friend, or 2 friends or 3 friends, or all the 6 friends.
1 friend can be selected out of 6 in 6C1 = 6 ways
2 friends can be selected out of 6 in 6C2 = 15 ways
3 friends can be selected out of 6 in 6C3 = 20 ways
4 friends can be selected out of 6 in 6C4 = 15 ways
5 friends can be selected out of 6 in 6C5 = 6 ways
6 friends can be selected out of 6 in 6C6 = 1 ways
Therefore the required number of ways (combinations) = 6 + 15 + 20 + 15 + 6 + 1 = 63

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.

Let A and B be two sets. Then a relation R from set A to set B is a subset of A × B. Thus, R is a relation from A to B ⇔ R ⊆ A × B.

a = cos α + i sin α, z = iy, y ∈ R
-(cos α + i sin α) y² + iy + 1 = 0
⇒ y² cos α = 1, sin α = y
y sin α = 1, sin cey ≠ 0
∴ sin² α =  cos α, cos² α+ cos α - 1 = 0

Required area :

Geometric mean of two numbers a and b is √ab
As two numbers are 14 and 56
Geometric mean is √14×56
= ± √2×7×2×2×2×7
= ±(2×2×7)
= ±28

Slope of the normal is given to be -1. We know that, foot of the normal is (am², -2am). Here a = 2, m = -1. Hence the required point is (2, 4). 

tan⁻¹ x = θ , so that  x=tanθ . We need to determine  cosθ .
sec²θ = 1 + tan²θ = 1 + x² 
∴s ecθ = ±√(1+x²) 
Then, cos(tan⁻¹x) = cosθ=1/secθ = ±1/√(1+x²)

For n=1, we have 7² +16 -1 = 64 which is divisible by 64
 
For n=k, we assume that 7^2k + 16k-1 is divisible by 64
 
We must check for n=k+1, is the statement true
7^2(k+1) +16(k+1) -1 = 49 * 7^2k + 16k +15
 
we adjust this expression by adding and subtracting terms, that is
we add 49*16k, -49 but we have to subtract 49*16k and -49
 
we have the following
49*7^2k +16k +15 = 49*7^2k + 49*16k -49 -49*16k -49 +16k +15 =
49(7^2k +16k -1) - 49(16k-1) +16k +15 =
49(7^k +16k -1) - 48*16k +64
all these terms are divisible by 64

Number of tickets selected from first station =20
from second =19
.... for last station =0
We have to select 2 consecutive stations
so total number of possible tickets = P(20,2)

Option d is correct, because it is the property of definite integral
 ∫02a f(x) dx = ∫0a f(x) dx + ∫0a f(2a – x) dx

Let
If α, β,and γ are the three roots then we must have
ax³ + bx² + cx + d = a(x−α)(x−β)(x−γ)
Comparing coefficients of various power of x on both sides leads to
α + β + γ = −b/a −−−−(1)
αβ + βγ + γα = c/a −−−−(2)
αβγ= −d/a−−−(3)
On substituting β = αr and γ = αr2 we get
α(1+r+r²) = −b/a−−−(4)
α²(r+r² + r³) = c/a−−(5)
α³/r³ = -d/a−−−−−(6)
Now, dividing (5) by (4) to αr=− c/b and substituting this in(6)weget
(−c/b)³ = -d/a
⇒ c³/b³ = -d/a
∴ c³a = b³d

We have, 4x + 8y + z – 8 = 0 and y + z – 4 = 0. Let be the angle between the planes, then 

 has a local maximum value = - 2 at x = - 1 and a local minimum value = 2 at x = 1.

As x R y if x + 2y = 8, therefore, domain of the relation R is given by x = 8 – 2y ∈ N.
When y = 1, 
⇒ x = 6 ,when y = 2, 
⇒ x = 4, when y = 3, 
⇒ x = 2.
therefore domain is {2, 4, 6}.

2 + 4 + 8 + …..1024

tan3θ = (3tanθ - tan³θ)/(1 - 3tan²θ)
tan θ = 1/2
= [3(1/2) - (1/2)³]/[1 - 3(1/2)²]
= [3/2 - 1/8]/[1 - 3/4]
= [11/8]/[1/4]
= 11/2