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Bihar STET Paper 1 Mathematics Mock Test - 8 - Bihar PGT/TGT/PRT MCQ


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30 Questions MCQ Test - Bihar STET Paper 1 Mathematics Mock Test - 8

Bihar STET Paper 1 Mathematics Mock Test - 8 for Bihar PGT/TGT/PRT 2024 is part of Bihar PGT/TGT/PRT preparation. The Bihar STET Paper 1 Mathematics Mock Test - 8 questions and answers have been prepared according to the Bihar PGT/TGT/PRT exam syllabus.The Bihar STET Paper 1 Mathematics Mock Test - 8 MCQs are made for Bihar PGT/TGT/PRT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Bihar STET Paper 1 Mathematics Mock Test - 8 below.
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Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 1

In the given figure, AC is a diameter of the given circle and ∠BCD = 75o. Then, ∠EAF−∠ABC is equal to

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 1

Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 2

The 21st term of the AP whose first two terms are -3 and 4, is

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 2

a1=-3,
a2=4,
d=4-(-3)=7,
a21=?,
a21=a+(n-1)d
=-3+(21-1)7
= -3+(20)(7)
= -3+140
=137

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Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 3

x = 0 is the equation of

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 3
Vertical lines cannot be written in standard form because the slope is undefined. The equation of the line is given by the x-intercept of the line. In this case x=0 goes though the origin (x=0), which means this is the equation of the y-axis.
Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 4

A garden roller has a circumference of 4 m. The number of revolutions it makes in moving 40 metres are:

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 4

We have circumference which is equal to the boundary of the circle. So we have one revolution equal to circumference of the circle which is equal to 4
Distance covered by the roller=circumference of the roller*number of revolutions
⇒ number of revolutions = distance covered by the roller/circumference of the roller
= 40/4 = 10

Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 5

Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:

Assertion: If the sum of the zeroes of the quadratic polynomial x2-2kx+8 is 2 then value of k is 1.
Reason: Sum of zeroes of a quadratic polynomial ax2+bx+c is -b/a

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 5


Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 6

The solution of the trigonometric equation 

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 6


cos2θ = 3(cot2θ-cos2θ)
4cos2θ = 3cot2θ
4=3(1/sin2θ)
sin2θ = 3/4
sinθ = √3/2
θ = 60

Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 7

Read the following text and answer the following questions.

A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year, assuming that the production increases uniformly by a fixed number every year.

Q. Find the production in the first year?

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 7
Let a be the production in first year and d be the difference between the productions of two consecutive years.

a3 = a + 2d = 600 …(i)

a7 = a + 6d = 700 …(ii)

Subtracting (i) from (ii),

d = 25

Put value of d in (i), we get

a = 550

Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 8

In the given figure, the value of x which makes POQ a straight line is :

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 8
2x+ 30+ 4x = 180 6x = 150 x = 25
Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 9

If sin θ = 5/13 then cos θ =

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 9


 

Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 10

The boundaries of surfaces are

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 10

Boundaries of surfaces are known as curves whereas boundaries of solids are known as surfaces. Therefore, the boundaries of surfaces are curves.

Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 11

In isosceles ΔPQR, PQ = PR, M is the mid point of QR. LM ⊥ PQ, MN ⊥ PR. By which criterion of congruency is ΔQLM 0 ≅ ΔMNR.

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 11

∠LQ = ∠MNR,

∠Q = ∠R

QM = MR,

Hence, ΔQLM ≅ ΔMNR (by AAS)

Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 12

The probability that a leap year has 53 Sundays is​

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 12

There are 366 days in a leap year, i.e, 1 more than a normal year.

Now, 52 weeks make up 344 days (52 x 7 = 344)
That means that we already have 52 sundays for sure.

Then, we are left with 2 days. Now, these days can be any from a pair of- mon-tues,tues-wed,wed-thurs,thurs-fri,fri-sat,sat-sun,sun-mon. Here favourable cases are sat-sun and sun-mon i.e, 2 cases and total number of cases is 7.

So, Probability=number of favourable cases/Total number of cases.

Therefore, Probability= 2/7.

Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 13

If 8 is a root of the equation x2 – 10x + k = 0, then the value of k is:​

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 13

Let, p(x) = x²-10x+k
since, 8 is the root of p(x)
∴ p(8) = 0
8²-10(8)+k = 0
64-80+k = 0
-16+k = 0
⇒ k = 16 .

Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 14

What value/s can x take in the expression k(x – 10) (x + 10) = 0 where k is any real number.

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 14

k(x – 10) (x + 10) =0
⇒ either k=0
Or x-10=0
Or x+10=0
Since we don’t know the value of k 
So either x-10=0
x=10
Or x+10=0
x=-10
So values of x can be 10,-10

Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 15

Direction: India's national animal Tiger (which has also been under the radar of the government as its population declined in the country) has witnessed an increase in its population.

A survey was done by the Ministry of Environment, Forests and Climate change, according to which tiger population is on rise at the rate of 6 per cent every year from 2006 and 2018.

Pranav has found a report on status of tigers in India in last 100 years. He has some queries as follow :

Q. In which year, number of tigers were minimum?

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 15
In 2007, number of tigers were minimum.
Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 16

If two parallel lines are intersected by a transversal then the bisectors of the interior angles form a :

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 16
if two parallel lines are intersected by a transversal then prove that bisectors of the interior angles form a triangle.if two parallel lines are intersected by a transversal then prove that bisectors of the interior angles form a rectangle.

Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 17

Given an A.P. few of whose terms are x, y, 2, 4, 6, 8,………. What must be the values of x and y?​

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 17

Arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant,d
AP is x, y, 2, 4, 6, 8,…
So we have 4-2=2,6-4=2 so we have d=2
So y=2-2=0
And x=0-2=-2

Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 18

The way in which students bring their finding to others and apply their understanding to new and unfamiliar circumstances grows out of

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 18

The way in which students bring their finding to others and apply their understanding to new and unfamiliar circumstances, grows out of excitement for what they have accomplished. Student would be excited to reveal his finding.

Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 19
Which of the following can be considered as an activity based method to teach the concept of probability in middle school?
Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 19

By physically conducting experiments and recording outcomes, students actively participate in the learning process, making it more meaningful and memorable.

Key Points

  • Through this hands-on approach, students develop a concrete understanding of probability concepts like sample space, equally likely outcomes, and relative frequency.
  • They observe patterns and begin to grasp the concept of probability intuitively.
  • This activity also encourages critical thinking as students analyze the data they collect and draw conclusions about the likelihood of different outcomes.
  • Moreover, the activity promotes teamwork and communication skills as students discuss their findings and observations with their peers.
  • It creates an interactive and dynamic learning environment, fostering curiosity and a deeper appreciation for mathematics.

Hence, it can be concluded that Asking students to throw a dice or flip a coin and note down the number of times each possibility occurs is the best approach.

Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 20

Which of the following learner characteristics is highly related to the effectiveness of teaching?

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 20

Learners Characteristics:

  • The concept of learner's characteristics is used in the science of learning and cognition to designate a target group of learners and define those aspects of their persona, the academic, social, or cognitive self that may influence how and what they learn.
  • Learners' Characteristics are important for an instructional designer as they allow them to design and create tailored instruction for a target group. It is expected by taking into account of the characteristics of learners, more efficient, effective, and/or motivating instructional materials can be designed and developed.
  • Learners' Characteristics can be: Personal, Academic, Social and emotional, and Cognitive.

Important Points

One of the important characteristics that is highly related to the effectiveness of teaching is

1. Prior experience of the learner:

  • Prior experience of the learner is based on the learner's Academic characteristics. 
  • To facilitate learning, one of the fundamental principles instructors employ is understanding students' prior knowledge.
  • It is also important to assess prior knowledge and skills early since such information could be used to help foster student engagement and critical thinking in the course.

Key Points

Other characteristics that affect the learning experience of the learner are:

1. Educational status of the parents of the learner:

  • This is based on the learner's characteristics.
  •  Parents play an important role in their children's learning.
  • Aside from being actively involved in their children's educationparents also provide a home environment that can affect learning.

2. Peer groups of the learner: 

  • The term peer learning refers to situations where peers support each other in learning processes.
  • Peer learning occurs among peers from similar social groupings, who are not professional teachers, helping each other to learn and in doing so, learning themselves.

3.  Family size from which the learner comes:  

  • Large numbered families whether rich or poor are difficult to maintain, they are characterized by a high number of children this does not create convenience for learning.
  • They also create in the upbringing of children some identified problems such as feeding, poor clothing, insufficient funds, lack of proper attention for children, disciplinary problems, and malnutrition which impact negatively children's academic performance.​

Therefore, the prior experience of the learner is highly related to the effectiveness of teaching.

Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 21

The learner’s purpose of learning is the most significant factor in a learning process because:

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 21

Learning is defined as a process of behavior modification through experiences, exercise, and efforts. Learning occupies a very important place in our life. It provides a key to the structure of our personality and behavior. Experience, direct or indirect, plays a very important and dominating role in molding and shaping the behavior of the individual from the very beginning.

Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 22

What is the most effective way to discipline a mischievous student?

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 22

Students feel humiliated being criticised by a teacher in front of their peers. So, handling discipline issues without an audience allows for better, more genuine exchanges, since the student responses will not be witnessed by classmates.

Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 23

The primary duty of the teacher is to

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 23

Teachers play the most crucial role in moulding the behaviour of learners. This shaping of behaviour is achieved through planning and dedication.

The Primary duties of a teacher include:

  • to achieve all-round development of students by guiding the students in their learning process.
  • to provide motivation for learning.
  • to stimulate learners in order to realize their optimum potential.
  • to plan the teaching and learning process to suit the needs of the students.
  • to promote learning by providing a conducive environment.
  • to encourage learners to become active learners.

Although imparting knowledge and values into the students are important duties of a teacher. However all round development of students is the most important and vital duty of a teacher.

Thus, the primary duty of the teacher is to help in all round development of the students.

Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 24

Mrs. Sharma makes her assessment useful and interesting. She does this by

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 24

Mrs. Sharma makes her assessment useful and interesting by using various ways to collect the information. Each way provides a different piece of information or same information with some additional inputs. For example, a written question paper gives us the idea of the person's knowledge, an oral assessment gives us an idea of the person's verbal ability and articulation, and activities may prove useful in testing child's ability to apply the knowledge practically. Activities also make assessment interesting for children.

Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 25

Learning disabilities can interfere with learning basic skills such as reading, writing and/or math. These can arise due to all listed reasons except:

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 25

Learning disabilities are neurologically-based processing problems. These processing problems can interfere with learning basic skills such as reading, writing and/or math. They can also interfere with higher level skills such as organisation, time planning, abstract reasoning, long or short term memory and attention. They are not caused due to the method of teaching used by the teacher. They may be present due to improper brain development, or infection in brain, etc. Constant fights amongst parents also lead to mental strain and anxiety in students, which may lead to learning disabilities.

Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 26

Sita is a visual learner, i.e. she learns best by seeing. She may get the least help from

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 26

Kinesthetic/tactile persons learn best through a hands-on approach, actively exploring the physical world around them. Thus, using the hands-on approach may not help Sita much.

Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 27
 Delhi became the capital of a kingdom for the first time under ______. 
Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 27

The Correct Answer is Option 1 i.e Tomara Rajputs.

  • Tomara Rajputs (8th Century -12th Century):
    • Anangapala was considered to be the founder of this dynasty.
    • Delhi became the capital of a kingdom for the first time under the Tomara Rajputs.
    • They were defeated by Chauhans in the middle of the twelfth century.
    • The name of Anangpala is inscribed in the Iron Pillar of Delhi. 
  • Chauhans (1165-1192):
    • Prithviraj Chauhan was the most important ruler of this dynasty.
    • In the first battle of Tarain 1191 AD, he defeated Muhammad Ghori.
    • In the second battle of Tarain 1192 AD, he was defeated by Muhammad Ghori.
  •  Aurangzeb (1658-1707):
    • He was the sixth Mughal Emperor.
    • He executed the ninth Sikh guru Guru Tegh Bahadur in 1675 AD.
    • He re-imposed Jaziya in 1679 AD.
Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 28

In which country did the fertility rate hit a new low of 0.72 in 2023?

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 28

South Korea experienced a further decline in its fertility rate, reaching a record low of 0.72 in 2023. This exacerbates concerns about population decline and highlights socio-economic factors influencing birth rates, such as career advancement concerns and financial burdens.

Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 29

The concept of 'seed germination' can be taught best by

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 29

The activity will help the students to gain first-hand experience. This helps in better retention of the concepts.

Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 30

The complement of 30°20′ is:

Detailed Solution for Bihar STET Paper 1 Mathematics Mock Test - 8 - Question 30

Complement of 30°20′ = 90° – ( 30°20′ ) = 90° – ( 30° + 20′ )
= (89° – 30°) + (1° – 20′)
= 59° + 60′ – 20′ [ ∴ 1° = 60°′]
= 59° + 40′ = 59°40′.

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