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

Nigeria leads the global fight against meningitis, introducing the groundbreaking Men5CV vaccine, endorsed by WHO, targeting five strains in one shot. This surpasses previous vaccines, vital as bacterial meningitis, prevalent in Africa, poses serious health risks. Africa’s “meningitis belt,” including Nigeria, faces escalating cases, with a 50% rise reported last year by WHO. Nigeria’s pioneering move sets a precedent for broader adoption worldwide, marking a significant stride in public health.

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.

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.

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.

3sin^-1(x) = sin^-1(3x - 4x^3) when -1/2<=x<=1/2
Definitely 0.4 comes in this range of x and so

3sin^-1(0.4) = sin^-1[3*0.4 - 4*0.4^3]
3sin^-1(0.4) = sin^-1[1.2 - 4*0.064]
3sin^-1(0.4) = sin^-1[1.2 - 0.256]
3sin^-1(0.4) = sin^-1[0.944]

Finally , sin(3sin^-1(0.4)) = sin{sin^-1(0.944)} = 0.944

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

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

a + b = a₁ +a_2n = a₂ + a_2n-1 = and ab = g₁g_2n = g₂g_2n-1 .... and

Solution we know that if f(t) is an odd function the  dt is given integral since   is odd, the given integral is even

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