HPSC PGT Mathematics Mock Test - 8 - HPSC TGT/PGT MCQ

The correct answer is 98.4° F,

Key Points

• The normal temperature of the human body is 37 °C (98.4° F). 
• A temperature above 100.4 °F(38°C) most often means you have a fever.
• Normal body temperature is also called the normal or core body temperature, which is based on Homeostasis.
• ​Homeostasis is a phenomenon in which the body regulates its functions to keep the internal conditions as stable as possible.
• The normal human body temperature border is typically stated as 36.5–37 °C (97.7–98.6 °F).
• Human body temperature depends on gender, age, time of day, exertion level, health status
  • Also in what part of the body the measurement is taken at.
• Body temperature is kept in the normal range by thermoregulation, in which adaptation of temperature is triggered by the central nervous system.

Thus, the Body temperature of a normal human being is 98.4° F.

Concept:

The nucleus of the Cell

• The cell is the basic building block of all living beings. 
• The nucleus is the special structure located generally at the center of the cell. 
• The nuclear membrane separates the cytoplasm from the nucleus. 
• The nucleus contains a central part called nucleolus and chromosomes. 
• Chromosomes and nucleolus coexist in the nucleus. Chromosomes are not present in the nucleolus.

Chromosomes and Genes

• Chromosomes carry genes and help in the inheritance or transfer of characters from the parents to their children. 
• The chromosomes can be seen only when the cell divides.
• The credit for the discovery of chromosomes goes to Strasburger as he first described the chromosome structure seen in the nucleus during cell division.
• Usually, chromosomes are rod-shaped, elongated, or dot-like in shape with sizes varying from 0.5 to 32 micrometer.
• In humans, 23pairs of chromosomes are found i.e. 46 in numbers.

Conclusion:

• Chromosomes are present in the nucleus of the cell. So, all the given options are wrong. 
• The correct option is None of these.

Additional Information

• The cell membrane is also known as the plasma membrane. 
  • It is a porous structure that allows the exit and entry of certain materials according to its requirement. 
  • This is the major function of the cell membrane. 
• The cytoplasm is between the cell and the nucleus.
  • Cell organelles like mitochondria, Golgi bodies, ribosomes, etc. are present in the cytoplasm. 
  • Mitochondria is called the powerhouse of cells and is responsible for cellular respiration. 
  • vacuole is open space between cell which is used for storing food. 
• Plastids are double-membrane organelles found in the plant cell. They contain pigments that help the plant in photosynthesis. 

The correct answer is Rakhigarhi.

Key Points

• Rakhigarhi site of Indus valley civilisation situated at Rakhigarhi village in Hisar district.
• The site is located in the Sarasvati river plain, some 27 km from the seasonal Ghaggar river.
• The Global Heritage Fund declared Rakhigarhi one of the 10 most endangered heritage sites in Asia.
•  A team of Indian and South Korean researchers carried out excavations in Rakhigarhi.
• The team unearthed a fire altar, parts of a city wall, drainage structures as well as a hoard of semi-precious beads.

Additional InformationImportant Sites of Harappan Civilization:

The correct answer is Shairishkam.

Key Points

• Sirsa:
  • It is said to be one of the oldest places in North India and its ancient name was Saarishka, which is mentioned in the Mahabharata, Panini's Ashtadhyayi and Divyavadana.
  • In the Mahabharata, Sharishaka is described as being taken by Nakula in his conquest of the western quarter.
  • During the medieval period, the town was known as Sarsuti.
  • In the ancient period, Sirsa was also known as Sirsapattan.

The correct answer is Baba Farid.

Key Points

• Faridabad was founded in the year 1607 AD by Sheikh Farid also addressed as Baba Farid.
• He was a treasurer of the Mughal Emperor Jehangir.
• The city was also famous for a Fort, a Tank, and a Mosque which over a period of time has got into ruins.
• Baba Farid built the first travelers lodge commonly referred to as Sarai which used to be the last stop for people traveling to New Delhi.
• After the rebellion of 1857, their estate was confiscated by the British Raj & The territory of Ballabhgarh was added to the Delhi district.
• It was constituted as a municipality in 1867.
• Faridabad became the 12th district of Haryana state on 15th August 1979.

Additional Information

• ​Faridabad is also a major industrial hub of Haryana.
• Faridabad is famous for henna production from the agricultural sector.
• Tractors, motorcycles, switch gears, refrigerators, shoes, tyres and garments constitute its primary industrial products.
• Famous Places
  • Raja Nahar Singh Palace 
  • Badkhal Lake 
  • Suraj Kund

The correct answer is Prithviraj Chauhan. 

Key Points

• The fort of Asigarh in Hansi town of Hisar district was built by Prithviraj Chauhan.
• This fort is also known as Prithvi Raj Chauhan Ka Qila.
• It is located on the eastern bank of Amti lake.

Additional Information

• About Hisar:
  • It is the second-largest district of Haryana.
  • It is one of the 7 districts which was constituted with the establishment of Haryana.
  • It shares borders with the Fatehabad, Jind, Rohtak, and Bhiwani districts of Haryana.
  • It also shares a border with the state of Rajasthan.
  • It was established by Firozshah Tuglaq by the name "HISAR-I-FEROZA" in 1354 A.D.
  • According to the 2011 census,
    • The total population of Hisar is 1,743,931.
    • The Sex ratio of Hisar is 872.
    • The Literacy rate of Hisar is 72.89%
    • The Population density of Hisar is 438.

Order = 3 , degree not defined ,because the function y’ present in exponential form.

General solution of a given differential equation contains arbitrary constants depending on the order of the differential equation.

use 

The vector whose magnitude is always 1 or unity is called a Unit Vector.

The total number of ways is same as the number of arrangements of 4 fingers, taken 3 at a time.
So, required number of ways = ⁴P₃ 
= 4!/(4-3)!
= 4!/1!
= 4! => 24

If A is anon singular matrix of order , then 

Let √(5 – 12i) = x + iy
Squaring both sides, we get
5 – 12i = x² + 2ixy +(iy)² = x² – y² + 2xyi.
Comparing real and imaginary parts , we get
5 = x² – y² ———– (1) and xy = – 6 ———— (2)
Squaring (1), we get
25 = (x² – y²)2 = (x² + y²)² – 4x²y²
⇒ 25 = (x² + y²)² – 4(– 6)²
⇒ (x² + y²)² = 169
⇒ x² + y² = 13 ———- (3)
Adding (1) and (3) we get
2x² = 18
⇒ x = ± 3.
Subtracting (1) from (3) we get
2y² = 8
⇒ y = ± 2.
Hence, square root of √(5 – 12i) is (3 – 2i)
Similarly, √(5 + 12i) is (3 + 2i)
√(5 + 12i) + √(5 – 12i)
⇒ (3 + 2i) + (3 - 2i)
⇒ 6

As we know that the length of the perpendicular from point 
P(x₁,y₁,z₁) from the plane a₁x+b₁y+c₁z+d₁ = 0 is given by: 

sin(n+1)Asin(n+2)A + cos(n+1)Acos(n+2)A = cos (n+1)Acos(n+2)A + sin(n+1)Asin(n+2)A = cos{A(n+2-n-1)} = cos (A.1) = cos A

(1 + a)²ⁿ = C_o + C₁a + C₂a² + … + C_2n a²n
Differentiate both sides w.r.t. ‘a’.

R is reflexive and transitive but not symmetric 

Integers contain all the natural numbers. So it can be a universal set for natural numbers. In other options, there are only some of the elements of natural numbers.

Since, (3, 3), (6, 6), (9, 9), (12, 12) ∈ R ⇒ R is reflexive relation.
Now, (6, 12)∉R but (12, 6) ∉ R  ⇒ R is not a symmetric relation.
Also, (3, 6),(6, 12) ∈ R ⇒ (3, 12) ∈ R
⇒ R is transitive relation.

number of type 3k → 10
3k + 1 → 10
3k + 2 → 10
Either all the no. are of the same type or one No. from each type

No.of choices for each question = 3
 
There are six questions so total = (3)¹²
So no.of ways = (3)¹² - 1

Because both sin 200⁰ and cos 200⁰ lies in 3rd quadrant. In 3rd quadrant values of sin and cos are negative.

2b = a + c ….. (i)
b⁴ = a² c² or b² = ± ac .....(ii)
Here 
a+b+c = 3/2
⇒ 3b = 3/2 or b = 1/2
from (i)
From (i), a+c = 1

From (ii) 
Solving this we get

Let A1, A2, A3, ........ , A40 be 40 A.M's between two numbers 'a' and 'b'.
Then, 
a, A1, A2, A3, ........ , A40, b is an A.P. with common difference d  = (b - a)/(n + 1) = (b - a)/41
[ where n = 40]
now, A1, A2, A3, ........ , A40  = 40/2( A1 + A40)
A1, A2, A3, ........ , A40 = 40/2(a + b)
[ a, A1, A2, A3, ........ , A40, b is an Ap then ,a + b = A1 + A40]
sum of 40A.M = 120(given)
120= 20(a + b)
=> 6 = a + b ----------(1)
Again, consider B1, B2, ........ , B50  be 50 A.M.'s between two numbers a and b.
Then, a, B1, B2, ........ , B50, b will be in A.P. with common difference = ( b - a)/51
now , similarly,
B1, B2, ........ , B50 = 50/2(B1 + B2)
= 25(6) ----------------from(1)
= 150

3

Expand using the formula 

cot^-1a - cot^-1b + cot^-1 b - cot^-1 c - cot^-1 a = 0