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Bihar PGT Mathematics Mock Test - 6 - Bihar PGT/TGT/PRT MCQ


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30 Questions MCQ Test Bihar PGT Exam Mock Test Series 2024 - Bihar PGT Mathematics Mock Test - 6

Bihar PGT Mathematics Mock Test - 6 for Bihar PGT/TGT/PRT 2024 is part of Bihar PGT Exam Mock Test Series 2024 preparation. The Bihar PGT Mathematics Mock Test - 6 questions and answers have been prepared according to the Bihar PGT/TGT/PRT exam syllabus.The Bihar PGT Mathematics Mock Test - 6 MCQs are made for Bihar PGT/TGT/PRT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Bihar PGT Mathematics Mock Test - 6 below.
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Bihar PGT Mathematics Mock Test - 6 - Question 1

The question below consists of a set of labeled sentences. Out of the four options given, select the most logical order of the sentences to form a coherent paragraph.
P: The climate has been changing drastically over the past years.
Q: If we don't correct the situation now, maybe we will never get a chance again.
R: Human activities itself are the main reason behind this.
S: The redressal of this problem is the biggest challenge of our generation.

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 1

The passage says that there has been an immense change in the climate of the planet over the years. We humans are ourselves responsible for generating this problem. Finding the solution to this problem is the biggest challenge of our generation. Lastly, the passage concludes that if we don't correct this situation now, it will be too late to control the problem.
This sequence of events is shown only by option 1. Hence, it is the correct answer.

Bihar PGT Mathematics Mock Test - 6 - Question 2

Improve the bracketed part of the sentence with the parts given below.

Q. I requested the teacher (to go out) early.

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 2

The student requested the teacher to allow him to leave the class early. Thus, Option 1 depicts the correct answer as no other alternative makes correct grammatical sense.

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Bihar PGT Mathematics Mock Test - 6 - Question 3

In the following question, a sentence has been given in Active/Passive Voice. Out of the four alternatives suggested, select the one which best expresses the same sentence in Passive/Active Voice.

I have seen him.

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 3

Rule- The object in the active voice becomes the subject in the passive voice and the subject in the active voice becomes the object in the passive voice.This is valid in both ways (Active to passive and vice-versa). Further, the verb is changed to the past participle form

Bihar PGT Mathematics Mock Test - 6 - Question 4

कालिदास का खण्डकाव्य कौन सा है?

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 4

कालिदास संस्कृत भाषा के महान कवि ओर नाटककार थे। उन्होंने भारत की पौराणिक कथाओं को आधार बनाकर रचनाएं की, जिसमें भारतीय जीवन और दर्शन के विविध रूप और मूल तत्त्व निरूपित हैं। कालिदास अपनी इन्हीं विशेषताओं के कारण राष्ट्र की समग्र राष्ट्रीय चेतना को स्वर देने वाले कवि माने जाते हैं संसृत साहित्य में ही नहीं अपितु समग्र साहित्यिक संसार में उन्हें कविकुलश्रेष्ठ तथा कविशिरोमणि माना जाता है।

प्रस्तुत विकल्पों में से 'मेघदूतम्' यह कालिदास का खण्डकाव्य है।

Bihar PGT Mathematics Mock Test - 6 - Question 5

बौद्ध धर्म से जुड़े ग्रंथो की भाषा कौनसी थी?

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 5

"बौद्ध धर्म" से जुड़े ग्रंथो की भाषा पालि भाषा है। बौद्ध धर्म से जुड़े सभी ग्रंथ पालि भाषा में लिखे गए हैं। बौद्ध धर्म का प्रमुख ग्रंथ त्रिपिटक है।

Bihar PGT Mathematics Mock Test - 6 - Question 6

'अभिनव जयदेव' के नाम से विभूषित कवि है:-

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 6

"विद्यापति" को "अभिनव जयदेव" की उपाधि दी गई है। कीर्तिलता, कीर्तिपताका से राजा शिवसिंह ने बहुत प्रसत्र होकर विद्यापति को "अभिनव जयदेव" की उपाधि और "विसपी" नामक ग्राम उपहार में दिया।

Bihar PGT Mathematics Mock Test - 6 - Question 7

Select the Venn diagram that best represents the relationship between the following classes.

Telugu, Urdu, Poetry

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 7

The logic followed here is:

Poetry can be written in Telugu or Urdu.

Telugu and Urdu both are different languages.

Venn diagram is given below:-

Hence, "option 4" is the correct answer.

Bihar PGT Mathematics Mock Test - 6 - Question 8

The splitting of white light into its component colours is called:

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 8

The correct answer is Dispersion.

Key Points

  • When a beam of white light is made to fall on one refracting face of the prism, it splits into seven colours i.e. violet, indigo, blue, green, yellow, orange and red (VIBGYOR) from the base.
  • The phenomenon of the splitting of white light into its constituent colours is called dispersion of light.
  • The pattern of colour components of light is called the spectrum of light.
  • The red light bends the least, while the violet light bends the most
  • The rainbow is an example of the dispersion of sunlight by the water drops in the atmosphere. 

Bihar PGT Mathematics Mock Test - 6 - Question 9
In which part of our body RBC formed?
Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 9

The correct answer is Bone-marrow.

Key Points

Red blood cell:

  • Red blood cells (RBCs) are the most common type of blood cell and the vertebrate's principal means of delivering oxygen to the body tissues via blood flow through the circulatory system.
  • RBC is formed in Bone Marrow.
  • RBCs take up oxygen in the lungs, or in fish the gills, and release it into tissues while squeezing through the body's capillaries.
  • The cytoplasm of erythrocytes is rich in hemoglobin, an iron-containing biomolecule that can bind oxygen and is responsible for the red color of the cells and the blood.
  • Each human red blood cell contains approximately 270 million of these hemoglobin molecules.
  • The cell membrane is composed of proteins and lipids, and this structure provides properties essential for physiological cell function such as deformability and stability while traversing the circulatory system and specifically the capillary network.

Additional Information

  • Red blood cells are thus much more common than the other blood particles: there are about 4,000–11,000 white blood cells and about 150,000 - 400,000 platelets per microliter.
  • Human red blood cells take on average 60 seconds to complete one cycle of circulation.
  • The blood's red color is due to the spectral properties of the hemic iron ions in hemoglobin.
  • Each hemoglobin molecule carries four heme groups; hemoglobin constitutes about a third of the total cell volume.
  • The normal pH range of blood is between 7.35 to 7.45. This means that blood is naturally slightly alkaline or basic.
Bihar PGT Mathematics Mock Test - 6 - Question 10

Find the wrong term in the given series?

50, 25, 15, 6.25, 3.125

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 10

The pattern followed here is;

15 is the wrong term that doesn't follow the pattern here.

There should be 12.5 in place of 15.

Hence, "15" is the correct answer.

Bihar PGT Mathematics Mock Test - 6 - Question 11

Which of the following is NOT an example of a biomass energy source?

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 11

The correct answer is Atomic Energy.

Key Points

  • A biomass energy source is one that is derived from organic matter.
  • The source of atomic energy is radioactive metals which are of non-organic nature and thus cannot be classified as a biomass energy source.
Bihar PGT Mathematics Mock Test - 6 - Question 12

Some equations are based on the basis of a certain system. Using the same solve the unsolved equation.

If 10 - 3 = 12, 12 - 4 = 13, 14 - 5 = 14, what is 16 - 6 = ?

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 12

The logic followed is:

Similarly,

Hence, the correct answer is "15".

Bihar PGT Mathematics Mock Test - 6 - Question 13

The Supreme Court of India came into being on ___________.

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 13

The correct answer is 28th January 1950.

Key Points

  • The Supreme Court in India was established under the Regulating Act, of 1773.
  • The Regulating Act of 1773 established a Supreme Court at Fort William, Calcutta.
  • Harilal Jekisundas Kania was the first Chief Justice of India.
  • On the 28th of January, 1950, two days after India became the Sovereign Democratic Republic, the Supreme Court came into being.
  • The inauguration took place in the Chamber of Princes in the Parliament building which also housed India's Parliament, consisting of the Council of States and the House of the People.
  • In the Chamber of Princes, the Federal Court of India had sat for 12 years between 1937 and 1950.
  • This was to be the home of the Supreme Court for years that were to follow until the Supreme Court acquired its own present premises.
  • It replaced both the Federal Court of India and the Judicial Committee of the Privy Council.
  • The First proceedings took place on 28 January 1950 at 9:45 am.
Bihar PGT Mathematics Mock Test - 6 - Question 14

The disease Blast is related to which among the following crops?

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 14

The correct answer is Rice.

  • Rice blast disease, caused by Magnaporthe oryzae (Ascomycota), occurs in about 80 countries on all continents where rice is grown, in both paddy fields and upland cultivation.
  • The extent of damage caused depends on environmental factors, but worldwide it is one of the most devastating cereal diseases, resulting in losses of 10–30% of the global yield of rice.
  • Other common diseases of the crop of rice are - Blight, Bacterial leaf streak, Brown spot, Sheath blight, False smut, etc.

Additional Information

  • The biological agents that causing diseases to plants are known as pathogens. 
  • Some of the viral diseases in plants are - bud blight (soyabeans), Curly tops (Beans, tomato, etc.), Mosaic leaf (Tomato, pea, potato,tobacco, etc.).
  • Some of the bacterial diseases in plants are - Blights (vegetable crops), Bacterial wilts (Tobacco, potatoes, etc), Cankers (Woody plants), Leaf Spots, Soft Wrots, etc.
  • Some of the fungal diseases in the plants are - Cankers, Ergot, Rusts, Root rots, Scabs, Smuts, etc.
Bihar PGT Mathematics Mock Test - 6 - Question 15

Which figure from the answer figures will replace the question mark (?) in the problem figures?

Problem Figures:

Answer Figures:

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 15

The logic followed here is:

The figure in image 2 is water image of the figure in image 1. The figure in image C of answer figure will be the water image of the figure in image 3.

Hence, image C of answer figure is the correct answer.

Bihar PGT Mathematics Mock Test - 6 - Question 16
The book written by Aryabhatta is called?
Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 16

The correct answer is Aryabhatiyam.

Key Points

  • Aryabhatta was a well-known mathematician and astronomer.
  • He wrote a book in Sanskrit called Arabhatiyam.
  • In the book, he stated that day and night was caused by the rotation of Earth on its axis.
  • He even developed scientific explanations of eclipses.
  • He also found a way to calculate the circumference of a circle.

 Thus, we can say that Aryabhatta wrote a book called Aryabhatiyam.

Additional Information

  • Other mathematicians and astronomers were Varahamihira, Brahmagupta and Bhaskaracharya.
Bihar PGT Mathematics Mock Test - 6 - Question 17

A cow is tethered to a corner of a field with a rope of length 7 m. If she grazes on the length of 210 m, what is the angle through which the rope moves?

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 17

We know that in a circle of radius r units, if an arc of length l units subtends an angle theta radian at the centre, then θ = l/ r.
Here, r = 7 m (length of rope will be equal to radius) and l = 210 m (length of arc will be the length which the cow grazed)
Thus, θ = 210o/ 7 
= 30o

Bihar PGT Mathematics Mock Test - 6 - Question 18

The length of intercept on y-axis, by a circle whose diameter is the line joining the points (–4,3) and (12,–1) is

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 18

Equation of circle is given by,

Bihar PGT Mathematics Mock Test - 6 - Question 19

so that f(x) is continuous at then

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 19

 By L-Hospital rule

Bihar PGT Mathematics Mock Test - 6 - Question 20

The values of x which satisfy the trigonometric equation   are :

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 20



Bihar PGT Mathematics Mock Test - 6 - Question 21

The equation of the hyperbola whose foci are (6, 5), (–4, 5) and eccentricity 5/4 is

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 21

Let the centre of hyperbola be (α,β)
As y=5 line has the foci, it also has the major axis.
∴ [(x−α)2]/a2 − [(y−β)2]/b2 = 1
Midpoint of foci = centre of hyperbola
∴ α=1,β=5
Given, e= 5/4
We know that foci is given by (α±ae,β)
∴ α+ae=6
⇒1+(5/4a)=6
⇒ a=4
Using b2 = a2(e2 − 1)
⇒ b2=16((25/16)−1)=9
∴ Equation of hyperbola ⇒ [(x−1)2]/16−[(y−5)2]/9=1

Bihar PGT Mathematics Mock Test - 6 - Question 22

2(bc cos A+ ca cos B + ab cos C) =

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 22

Bihar PGT Mathematics Mock Test - 6 - Question 23

Find the distance travelled by a car moving with acceleration given by a(t)=Sin(t), if it moves from t = 0 sec to t = π/2 sec, and velocity of the car at t=0sec is 10 km/hr.

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 23

Acceleration is the derivative of velocity, so we integrate a(t) to get v(t):

v(t)=∫a(t)dt=∫sin(t)dt=−cos(t)+C

Now, we are given that the velocity at t=0 is 10 km/hr. We can use this information to find the constant C:

v(0)=−cos(0)+C=−1+C=10

Solving for C, we get C=11.

Now, we have the velocity function:

v(t)=−cos(t)+11

Finally, we integrate v(t) to get the displacement function s(t):

s(t)=∫v(t)dt=∫(−cos(t)+11)dt

s(t)=−sin(t)+11t+D

Now, we need to find the constant D. We are given that the car moves from t=0 to t=π/2, and we know that s(0)=0 (starting position). Plugging in these values, we can solve for D:

s(0)=−sin(0)+11(0)+D=0

D=0

So, the displacement function is:

s(t)=−sin(t)+11t

Now, to find the distance traveled, we evaluate s(t) over the given time interval:

Distance=s(π/2​)−s(0)

Distance=(−sin(π/2​)+11(π/2​))−(−sin(0)+11(0))

Distance=−1+11π​/2 = 16.27887 kilometers

Therefore, the distance traveled by the car from t=0 to t=π/2​ is 16.27887 kilometers​.

 

 

Bihar PGT Mathematics Mock Test - 6 - Question 24

If P is of order 2 × 3 and Q is of order 3 × 2, then PQ is of order

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 24

Here, matrix P is of order 2 × 3 and matrix Q is of order 2 × 2 , then , the product PQ is defined only when : no. of columns in P = no. of rows in Q. And the order of resulting matrix is given by : rows in P x columns in Q.

Bihar PGT Mathematics Mock Test - 6 - Question 25

The centre of the circle passing through origin and making intercepts 8 and –4 on x and y axes respectively is

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 25

Verify g2 - f2 = 12

Bihar PGT Mathematics Mock Test - 6 - Question 26

The minor Mij of an element aij of a determinant is defined as the value of the determinant obtained after deleting the​

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 26

A minor, Mij, of the element aij is the determinant of the matrix obtained by deleting the ith row and jth column.

Bihar PGT Mathematics Mock Test - 6 - Question 27

Domain of f(x) = sin−1x−sec−1x is

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 27

Since sin−1x is defined for |x|⩽1, and sec−1x is defined for |x|⩾1,therefore,f(x) is defined only when|x|=1.so, Df = {−1,1}.

Bihar PGT Mathematics Mock Test - 6 - Question 28

  , -1 ≤ x < 0 is continuous in the interval [–1, 1], then `p' is equal to:

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 28

lim f(x) = -½
Apply L hospital 
lim(x → 0) p/2(1+px)½ + p/2(1-px)½ = -½
⇒ 2p = -1 
p = -1/2

Bihar PGT Mathematics Mock Test - 6 - Question 29

If P(x) = ax2 + bx + c and Q(x) = -ax2 + dx + c, where ac ≠ 0, then P(x)Q(x) = 0 has 

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 29

Let D1 = b2 – 4ac and  D2 = d2  + 4ac
As ac ≠ 0, either ac < 0 or ac > 0
If ac < 0, then D1 > 0
In this case P(x) = 0 has distinct two real roots 
If ac > 0, the D2 > 0
In this case Q(x) = 0 has two distinct real roots.
Thus, in either case P(x)Q(x) = 0 has at least two distinct real roots. 

Bihar PGT Mathematics Mock Test - 6 - Question 30

A wheel makes 360 revolutions in 1 minute. Through how many radians does it turn in 3 seconds?

Detailed Solution for Bihar PGT Mathematics Mock Test - 6 - Question 30

Number of revolutions made by the wheel in 1 minute = 360
∴ Number of revolutions made by the wheel in 1 second = 360/(60) = 6
In 3 seconds = (6*3) = 18
In one complete revolution, the wheel turns an angle of 2π radian.
Hence, in 6 complete revolutions, it will turn an angle of 18 × 2π radian, i.e.,
36 π radian
Thus, in one second, the wheel turns an angle of 36π radian.

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