HTET TGT Maths Mock Test - 4 - HTET MCQ

For Vygotsky, although biological factors constitute the necessary pre-requisite for elementary processes to emerge, socio-cultural factors are indispensable for elementary natural processes to develop. He argues for the uniqueness of the social milieu and regards socio-cultural settings as the primary and determining factor in the development of higher forms of human mental activity such as voluntary attention, intentional memory, logical thought, planning, and problem solving.

Arnold Gesell gives the Maturational Theory. He explained that mature is a genetic factor, and it plays an important role in children's development; the influence of the environment is not more important than genes. Children take birth by genes; their habits are like their parents, their sitting posture, how they eat, how they walk, etc., all depends on their parents. He believes that development is a biological process that occurs automatically in sequential stages, which means genes play an important role in developing a child.
Hence, the correct answer is "Maturational Theory".

When the children do new things that they did differently from what they did before, creativity, creativity is a skill that helps children to enhance their skills of thoughts, ideas, feeling, etc. It's just that they make new things on their own. Creativity is something kids like when they play or make connections for different reasons.
Hence, the correct answer is "Play".

Learning is a process where we live, talk with others, learn new things which means we are interacting with others, and learning different skills like sitting, walking, eating, etc. We gain knowledge of what we are not aware of and also learn various skills that help us in the future.
Hence, the correct answer is "All of the above".

Student-Centered Learning, also known as learner-centered education, broadly encompasses methods of teaching that require learners to actively construct their own knowledge and puts the responsibility for learning on them. 

A diagnostic approach would be the most effective part of addressing students who run away from class.

• A diagnostic approach involves identifying and understanding the root causes and factors contributing to the student's behavior.
• It may involve gathering information, assessing the student's needs, and identifying any underlying challenges or issues they may be facing.
• This information can be helpful in developing appropriate strategies and interventions to support the student.

मयंक शब्द 'सूर्य' का पर्यायवाची नहीं है। मयंक चन्द्रमा का पर्यायवाची होता है।

Key Points

• चन्द्रमा : शशि, हिमकर, राकेश, रजनीश, हिमांशु, चाँद, मयंक, विधु, सुधाकर, कलानिधि, निशापति, शशांक
• सूर्य : अंशुमान, सूरज, रवि, दिनकर, दिवाकर, प्रभाकर, भास्कर।
• सविता : भानु, दिनेश, मार्तण्ड, अंशुमाली, सूर्य, रवि, सूरज, दिनकर, प्रभाकर, दिवाकर, भास्कर
• दिनकर : प्रभाकर, दिवाकर, भास्कर, सविता, भानु, दिनेश, मार्तण्ड, अंशुमाली, सूर्य
• भास्कर : सूरज , फतिंगा , पतंगा पतंग , सूर्य

The correct preposition to use in this sentence is between.

• Option A: 'in' is incorrect because it implies that the book is inside the mug or the pen, which doesn't make sense in this context.
• Option B: 'between' is the correct preposition because it indicates that the book is located in the middle of the mug and the pen.
• Option C: 'on' is incorrect because it suggests that the book is placed on top of the mug or the pen, which is not the intended meaning.
• Option D: 'above' is incorrect because it implies that the book is in a higher position than the mug and the pen, which is not the intended meaning.

Therefore, the correct sentence is: The book is between the mug and the pen.

As we know that ,
Delectable is delightful or pleasant, distasteful is something unpleasant, agonizing is something that gives pain and anxiety, appetizing means to rouse or increase appetite, laborious means something that needs or shows much effort.
Thus , Distasteful is antonym ( opposite ) word of Delectable . 

The idiom head over heels is used to imply that she was very much excited about her relationship with her boyfriend. She was overwhelmed in joy because of that relationship but it ended in despair only. Among the given options, we can choose option D since it provides us with the correct meaning of the given idiom as underlined.

This makes option D the correct choice among the given options.

Let us first learn the meanings of the given words:

Tag (verb): give a specified name or description to.

Label (verb): assign to a category, especially inaccurately or restrictively.

As the sentence is in passive voice, only a verb in third form will fit the blank. Thus, ‘labelled’ is an apt fit while ‘tag’ and ‘design’ are incorrect.

Hence, the correct answer is option B.

Venue: This term typically refers to a specific location where an event or activity takes place. It often carries the connotation of being chosen or designated for a particular purpose, such as a concert venue, a wedding venue, or a courtroom.

Place: "Place" is a more general term that can refer to any location or area, whether specific or undefined.

The correct answer is The Ramayana.

About Maharaja Tej Singh Palace:

• The Palace of Maharaja Tej Singh which is situated in the Mirpur Gurgaon city of Haryana state. The Palace is adorned with Paintings done in Rajput style. The patterns on the walls of the Palace have depicted scenes from the Epic Ramayana.
• Gurgaon, now officially known as Gurugram, is a district located in Haryana state. It is located near the Delhi-Haryana border, about 30 km southwest of the capital New Delhi and 268 km south of Chandigarh.
• The region of Gurgram originally fell under the Kuru dynasty. This place was the part of the Mahabharata, It was described as the village of Guru Dronacharya, the Guru of the Pandavas and Kauravas.

Map view of Gururam district (Haryana state):

The correct answer is Jainism.

• The compositions of Poet Pushpadant are related to Jainism.

Additional Information

• Jainism is a very ancient religion. According to some traditions, it is as old as the Vedic religion.
  • There is a succession of great teachers or Tirthankaras in the Jain tradition.
  • There were 24 Tirthankaras, the last of whom was Vardhaman Mahavira.
  • The first Tirthankara is believed to be Rishabhanatha or Rishabhdev.
  • The 23rd Tirthankara was Parshvanath who was born in Varanasi.
  • All Tirthankaras were Kshatriyas by birth.

Important Points

• Jainism – Jain Councils:
  • First council
    • Held at Pataliputra in the 3rd century BC.
    • Presided by Sthulabhadra.
  • Second Council
    • Held at Vallabhi in Gujarat in 512 AD.
    • Presided by Devardhi Kshemasarmana.

The correct answer is 15th July, 1997.

Key Points

• Jhajjar was separated from Rohtak district on 15th July 1997 (officially).
• Jhajjar district was created out of Rohtak district by carving out Jhajjar and Bahadurgarh tahsils on 25th June 1997.
• During the period of 1991-2001, the following jurisdictional changes were observed.:
  • Bahadurgarh tahsil contained 62 villages and Bahadurgarh Town.
  • Jhajjar tahsil contained 199 villages and two towns, namely, Jhajjar and Beri, and these were part of Rohtak district.
  • Beri tahsil was newly created on 27th July 1995 by transferring 39 villages and Beri Town from Jhajjar tahsil.
  • Bahadurgarh tahsil gained 9 villages from Jhajjar tahsil and lost 2 villages to Rohtak tahsil (Rohtak district).

Runs → A : B = (16 – 1) : 16 = 15 : 16

Hence, Option A is correct.

The correct answer is Bihar.

Key Points

• State with the highest population density: Bihar
• Union Territory with Highest Population Density: Delhi
• State with the lowest population density: Arunachal Pradesh
• Union Territory with Lowest Population Density: Andaman and Nicobar Islands
• Highest Density District: Mumbai City
• Lowest density district: Lahaul and Spiti

Important Points

• The ratio of people to units of the land area is generally referred to as population density.
• It is calculated by dividing the total population of a country by the total area of the country.
• India's Population Density in 2011 increased to 382 per square kilometer.

Hence, Option A is correct.

The altitudes of a triangle are the line segments drawn from each side of the triangle to the angle opposite that side, such that the line segment is perpendicular to the side it is drawn from. In an isosceles triangle, only the altitudes of the legs of equal length in the triangle are congruent. 

Given Equations

ax + by = 3    {1}

3a - 2b = 1     {2}

x = 2, y = 1 ( given) 

Equation {1} becomes,

2a + b = 3      {3}

Substract equation {3} from {2}, we get,

2a + b - 3a + 2b = 3 - 1

-a + 3b  = 2

3b = 2 + a

b = (2+a) / 3 

Put the value of b in equation {3}, we get

2a + (2 +a ) / 3 = 3

(6a +a +2) / 3 = 3

7a + 2 = 9

a = 1

Now put value of a in {3}

2 + b = 3

b = 1

hence we get value of a and b i.e. 1 and 1 

Let AB be the tree and AP be the shadow.
Let AB = x meters. Then AP = x√3 meters
Also ∠APB = θ
In right angled triangle ABP 

Therfore the angle of elevation of the Sun is 30°.

ANSWER :- b

Solution :- Area of rec. park=(120×90)m²=10800m²

Area of rec. park excluding lawn=2950m²

Area of circular lawn=(10800 - 2950)m²

=7850m²

Let the radius be = r

7850/3.14 = 3.14r²/3.14

→2500 = r²

→√2500 = √r²

→50 = r

→r = 50

Radius = 50m

we know sin(90 - a) = cos(a) 

cos(90 - a) = sin(a)

sin(a) = 1/cosec(a)

sec(a) = 1/cos(a)

cos40 = cos(90-50) = sin50

cosec40 = cosec(90-50) = sec50

so our expression becomes

sin50/sin50 + sec50/sec50 - 4cos50 / sin40

= 1 + 1 - 4(1)   since cos50 = sin40

= -2

a₂₁ -a₇ = {a + (21 - 1)d}

- {a + (7 - 1)d} = 84

a + 20d- a - 6d = 84

14d = 84

d = 18/14 = 6

d = 6

So, A is incorrect but R is correct

Number of possible outcomes = {1} = 1
Number of Total outcomes = 6
∴ Required Probability = 1/6

∵ Mode = 3 median - 2 mean
⇒ 12k = 3 median - 2 x 15k
⇒ 42k = 3 median
⇒ Median = 14k.

The probability of a sure event is always 1.

Since, HCF of 56 and 72, by Euclid’s divsion lemma,
72 = 56 × 1 + 16 ……….(i)
56 = 16 ×3 + 8 ……….(ii)
16 = 8× 2 + 0 ……….(iii) 
∴ HCF of 56 and 72 is 8. 
∴ 8 = 56 – 16× 3
8 = 56 – (72 – 56 ×1) ×3
[From eq. (i) : 16 = 72 – 56× 1]
8 = 56 – 3 ×72 + 56× 3
8 = 56 × 4 + (–3) × 72 
∴ x = 4,y = −3

Let the breath of the river be h meters.

= 100√2 meters Therefore, the breadth of river is