HPSC PGT Mathematics Mock Test - 1 - HPSC TGT/PGT MCQ

The correct answer is Mela Baba Ramakshah.

Key Points

• Baba Ramakshah Fair is organized at Khubru situated in district Sonepat during the months of February-March.
  • The fair is organized on Pooranmasi in the month of Phalguna (February-March).
• Baba Ramakshah was a saint belonging to Sayeed Community.
  • He used to treat diseases of people with herbs.
  • He was worshipped by many folks.
  • After his death, his grave was made by Hindu and Muslim devotees.

Additional Information

•  Khubru is a Village in Ganaur Tehsil in Sonipat District of Haryana State.
  • It belongs to Rohtak Division.
  • As per census 2011, Khubru has a total population of 3,578 peoples, out of which male population is 1,911 while female population is 1,667.

The logic follows here is :

Sum of place values of alphabets of the given word = Given number.

MONK → M + O + N + K = 13 + 15 + 14 + 11 = 53.

Similarly,

TUTOR → T + U + T + O + R = 20 + 21 + 20 + 15 + 18 = 94.

Hence, the correct answer is "94".

A man walks 30 meters towards South, then turning to his right, he walks 30 meters, then turning to his left, he walks 20 meters, again turning to his left, he walks 30 meters. The figure are given below -

50 meters far is he from his starting point.

Hence, "50 meters" is the correct answer.

The meaning of synonym is Similar.

Similarly;

The meaning of Opposite is Adverse.

Hence, "Adverse" is the correct answer.

Additional Information

Hence. the person will be allowed to appear in the interview.

National Education Policy 2020 is the first education policy of the 21st century and aims to address the many growing developmental imperatives of our country. This policy replaced the 34-year-old National Policy on Education (NPE), in 1986.

Key Points

•  National Education Policy 2020 lays emphasis on conceptual understanding and improving educational outcomes so that children can implement the acquired knowledge and skills in real-life.
• It suggests that assessment is important to find out a child's progress over a certain period of time. 
• Emphasizes on transforming assessment for optimizing learning and development of all students.
• NEP aims to provide a holistic report on student’s progress and learning.
• It should be able to depict the actual level of knowledge and performance of each student.

Thus, it is concluded that National Education Policy 2020 suggests that assessment is important to find out a child's progress over a certain period of time.

Diagnostic evaluation has a placement function and so takes place before instruction. It is used to determine underlying causes of learning difficulties. Diagnostic evaluation may include administration of standardised achievement tests, standardised diagnostic tests, teacher-made tests, observation and checklists. Scoring and interpretation can be normative (based on norms) or criterion-referenced. Results are usually shown as an individual profile of sub-skills.

A first order linear differential equation. Is a differential equation of the form  + Py = Q or  + Px = Q

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.

Any point on the given parabola is (t², 2t). The equation of the tangent at (1,2) is x-y +1 = 0. The image (h,k) of the point (t²,2t) in x-y + 1 = 0 is given by 

∴ h = t² - t² + 2t - 1 = 2t - 1
Eliminating t from h = 2t – 1 and k = t²+1
we get, (h+1)² = 4(k-1)
The required equation of reflection is (x+1)² = 4(y-1)

 lim (x²-9)/x³+9x-6x²
Lim x=0
lim[(0)² - 9]/ (0)³ + 9(0) - 6(0)²
-9/0 (which is not defined)

sin(n+1)x cos(n+2)x - cos(n+1)x sin(n+2)x
⇒ sin[(n+1)x - (n+2)x] 
As we know that sin(A-B) = sinA cosB - cosA sinB
⇒ sin(n+1-n-2)
sin(-x) 
= -sinx

The determinant of a matrix A and its transpose always same.

Let the two numbers be α and β. Given that 

∴ Required equation is x² – 18x + 16 = 0 

(i.e. angle subtended at the centre of a circle is double the angle subtended in the alternate segment).
Hence (c) is the correct answer.

cos 9x + cos (-10) x = cos 9x + cos10x

f (0) = cos0 + cos0 = 2,f (π/4) = cos9 π/4 + cos5π/2 = cos π/ 4 =
f (π/2) = cos9π/2 + cos5π = 0 -1 = -l,f (π) = cos9π + cos10π = - l +1 = 0.

Minimum value of |z₁ - z₂| = 12 - 10 = 2

is a vector joining two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂). If  Direction cosines of  are given by : 

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

log (x - 3) (5 - x) exists ⇒ (x - 3) (5 - x) > 0
⇒ (x -3)(x -5) < 0 ⇒ 3< x < 5
∴ Domain = (3, 5).

The period of y=tanx is π.
Since the function y = tan x is periodic of period π hence each branch is simply a repetition of the branch from - 
(-π/2 , π/2).

(1/1² + 1/2² + 1/3² + ........∞) = π²/6
⇒ (1/1² + 1/3² + 1/5² + ........∞) + (1/2² + 1/4² + 1/6² + ........∞) = π²/6
=> (1/1² + 1/3² + 1/5² + ........∞) + 1/2²(1/1² + 1/2² + 1/3² + ........∞) = π²/6
=> (1/1² + 1/3² + 1/5² + ........∞) + 1/2²(π²/6) = π2/6
=> (1/1² + 1/3² + 1/5² + ........∞) + π²/24 = π²/6
= π²/8

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 :

Required area :

(1 + x)ⁿ > 1 + nx
For n = 1, (1 + x)ⁿ = (1 + x)¹ = 1 + x
1 + nx = 1 + x
(1 + x)ⁿ = 1 + nx        
For n = 2, (1 + x)ⁿ = (1 + x)² = 1 + x² + 2x
1 + nx = 1 + 2x
(1 + x)ⁿ ≥ 1 + nx (Equality when x = 0)
So, (1 + x) ⁿ > 1 + nx for n ≥ 2