HPSC PGT Mathematics Mock Test - 4 - HPSC TGT/PGT MCQ

The correct answer is Komal Gurjar.

Key Points

Anita Kundu

• She is a sub-inspector of police and passionate mountaineer, is the first woman to scale Mount Everest from both Nepal and China sides.
• She hails from village Faridpur, Hisar (Haryana).
• She was born in Hisar, Haryana, Kundu joined the Haryana Police in 2008.
• From an early age, she had been drawn to challenging sports - as a pre-teen, she had trained in boxing. Hence, Option 1 is correct.

Mamta Sodha

• She is an Indian sportsperson, known for her successful 2010 attempt to scale Mount Everest.
• She was honoured by the Government of India, in 2014, by bestowing on her the Padma Shri, the fourth highest civilian award, for her services to the field of mountaineering sport.

Komal Gurjar

• Komal Gurjar is a junior women wrestler.

Santosh Yadav

• She is an Indian mountaineer.
• She is the first woman in the world to climb Mount Everest twice and the first woman to successfully climb Mount Everest from Kangshung Face.
• She climbed the peak first in May 1992 and then again in May 1993 with an Indo-Nepalese Team

The correct answer is Punch marked coins.

• Punch marked coins have been discovered form the mounds of Hisar and Agroha.

Key Points

• A collection of coins was found from the site, including 4 Indo Greek, one punch-marked and 51 other coins from Agrodaka.
  • The discovery of coins of Agraya Janapada (Republic) during excavation and its ancient name in literature as Agrodka is sufficient to prove it as the headquarters of a republic.
  • The site was excavated in 1978-84 by Mr. PK Sharan and Mr. JS Khatri of the Department of Archeology and Museums, Government of Haryana.
  • Apart from the residential and community houses, made of baked bricks, the remains of a Buddhist stupa and a Hindu temple existing side by side indicated co-existence and respect of communal harmony.

Additional Information

• The city of Hisar was founded by a Muslim ruler Firoz Shah Tughlaq in 1354 AD.
  • 'Hisar' is an Arabic word which means 'fort'.
  • The city which we know today as 'Hisar' was originally called 'Hisar Firoza (Hisar-e-Firoz)' or in other words 'Fort of Firoz'.
  • But as the days passed, the word 'turquoise' was dropped from its original name.

The correct answer is Nakula - Digvijyams.

• Rohtak derives its name from its headquarters town, Rohtak, which is called the Reformation of Rohtashgarh, a name still applied to the ruined sites of the two old sites.
  • Traditionally, it is named after King Rohtash, in whose days the city is said to have been built.
  • It is also claimed that the city derives its name from the Roherra (Tacoma undulate) tree which is called Rohitaka in Sanskrit.
  • It is said that before the city came into existence, it was the site of a forest of Rohitaka trees and hence the name Rohtak.
  • Another version links Rohtak with Rohitaka, which is mentioned in the Mahabharata in connection with the campaign of the Pandava warrior Nakula. 

Additional Information

• Divyavadana (meaning Divine Tales) is a text of Buddhist tales.
  • The origin of many of these stories is the original Mool Sarvastivada Vinaya Granth.
• Majjima Nikaya is a Buddhist text, the second of the five Nikayas or collections in the Sutta Pitaka, one of the "three baskets" that compose the Pali Tipitaka of Theravada Buddhism. Composed between the 3rd century BCE and the 2nd century CE.

The correct answer is the Fatehabad district.

About Fatehabad District:

• The Fatehabad city recognised as the Fatehabad district on 15th July 1997. It was separated from the Hisar district of Haryana state and recognized as the new separate district.
• It is one of the smallest districts in the Haryana State and covers 5.69 % areas of the state. Fatehabad district is surrounded by Sirsa district in the West direction, Punjab state in the North, Jind district in the East, Hisar district, and Rajasthan state in the South. 
• The Geographical area of the Fatehabad district is 2520 km, which is 5.4% of the total state share. Fatehabad is connected by metalled roads with important cities of the Haryana and Delhi state. 
• According to Puranas, the areas of Fatehabad district remained a part of the Nanda Empire. It is situated on the banks of Saraswati and the Drishadvati river.
• The discovery of Ashokan pillars at Fatehabad and Hisar shows that the area of the district remained a part of the Mauryan Empire.

About Humayun Mosque:

• There is a Small Mosque built by him known as the Humayun mosque in the city of Fatehabad. The reason behind the Mosque was built by the 2nd Mughal Emperor Humayun, who in his fight after his defeat at the hands of Sher Shah Suri happened to pass through the city of Fatehabad.

Image of the Humayun Mosque:

Given

Ratio of principal and amount is 4 : 5 at a certain time. If it becomes 5 : 7 after 3 years

Concept used

Rate = (interest × 100)/(Principal × time)

Calculation

Principal : Amount = 4 : 5

Principal be equal 

Multiply the first ratio with 25 and second with 20

⇒ (4 : 5) × 25 

⇒ (5 : 7) × 20

Principal : Amount = 100 : 125

Principal : Amount = 100 : 140

Difference in amount = 140 - 125

Difference in Amount = 15

3 year = 15

1 year = 5

Rate = (interest × 100) / Principal × time

Rate = 

Rate = 5%

∴ The required simple rate of interest is 5%.

Trapezoid, Rhombus, and Parallelogram are the 2 dimensional figure.

While;

Prism is a 3-dimensional figure.

Hence, "Prism" is the correct answer.

Important Points

f (x) = x² -4x +5 = (x -2)² +1
x ∈ [2,∞) ⇒ x > 2 ⇒ x - 2 > 0 ⇒ (x - 2)² +1 > 1 ⇒ B = [1,∞). 

In Cartesian coordinate system Shortest distance between the lines

ρ = (b(xy) * b(yx))
But sign of ρρ is same as sign of b(xy), b(yx)
Therefore, ρ = 2/7

Comparing the equation with the standard form ax²+2hxy+by²+2gx+2fy+c=0
a=2,h=−3/2,b=5,g=3,f=−3/2,c=5
Δ=abc+2fgh−af²−bg²−ch²
=(2)(5)(5)+2(−3/2)(3)(−3/2)−(2)(−3/2)²−(5)(3)²−(5)(−3/2)²
=50+27/2−9/2−45−225/4
=−169/4 is not equal to 0
Descriminant =h²−ab
=(−3/2)²−(2)(5)
= 9/4−10
= −31/4<0
So, the curve represents either a circle or an ellipse
a is not equal to b and  
Δ/a+b = −(169/4)/2+5
=−169/28<0
So, the curve represents a ellipse.

Because, the element b in the domain A has no image in the co-domain B.

A³ = I ⇒  Pre - multiplying both sides by A⁻¹,A⁻¹, A³ = A⁻¹ I ⇒ A² = A⁻¹

Centre of the ellipse is the intersection point of 
x+y−2=0.........(1) 
x−y=0............(2)
Substituting x from equation 2 in equation 1 two equations, we get,
2y=2 which results in y=1 
Replacing, we get x=1
⇒(1,1) is the centre

 Let Tr and Tr+1 denote the rthand(r+1)th terms in
the expansion of (1+x)¹⁹. 
 Tr = ¹⁹C_r-1 x^r-1 and T_r+1 = ¹⁹C_r x^r .
∴T_r+1/T_r = ¹⁹C_r xr/(19C_r-1 x^r-1)
⇒T_r+1/T_r = ¹⁹C_r  ¹⁹C_r-1 x
⇒T_r+1/T_r = 19!/(19−r)!r! × x[(19−r+1)!(r−1)]/10!
⇒T_r+1/Tr = x(20−r)/r
⇒T_r+1/T_r = (20−r)/r × 1/2   [∵x not equal to 1/2]
Now
T_r+1/T_r > 1
⇒ (20−r)/r × 1/2 > 1
⇒ 20 > 3r
r > 20/3
∴ (6+1)^th i.e. 7th term is the greatest term.

The x-axis touches at A(1, 0) and x = y touches at B(1, 1). Hence the equation to the curve through these points is given by y(y – x) + k(x – 1)² = 0. For this to represent a parabola, 4k = 1. The equation is x² – 4xy + 4y² – 2x + 1 = 0. Vertex  focus

x = t² +t +1andy = t² -t +1
⇒ x + y - 2 = 2t² and x - y = 2t
⇒ 2(x + y -2) = (x - y)² ⇒ x² + y² -2xy -2x -2y + 4 = 0
Comparing with the equation
ax² + 2hxy + by² + 2gx + 2fy + c = 0, 
∴ abc + 2fgh - af² - bg² - ch² ≠ and h² = ab

Since g is continuous at a , therefore , if g (a) > 0 , then there is a nhd.of a, say (a-e , a+ e) in which g (x) is positive .This means that f ‘ (x)>0 in this nhd of a and hence f (x) is increasing at a.

3A= A+ 2A
⇒ tan 3A = tan (A + 2A)
⇒ tan 3A = (tan A + tan 2A) / (1 – tan A . tan 2A)
⇒ tan A + tan 2A = tan 3A – tan 3A x tan 2A . tan A
⇒ tan 3 A – tan 2A – tan A = tan 3A . tan 2A . tan A

Calculation:

Given: A is a matrix of order 3 × 5 and B is a matrix of order 5 × 3

Number of rows in A = 3

Number of column in A = 5

Number of rows in B = 5

Number of column in B = 3

The order of AB = number of row is A × number of columns in B
= 3 × 3
And, 
The order of BA = number of row is B × number of columns in A
= 5 × 5
Hence, option (3) is correct.

Other two vertices will make two right angled triangles with AB as the common hypotenuse. So they must lie on the circle with AB as the diameter and O as the centre. Radius of that circle will be 5 units.
There will be 5 such pairs in which both the coordinates are integers.
[(5, 0), (–5, 0), [(4, 3), (4, – 3)],
[(–3, 4), (3, –4)] [(–3, –4), (3, 4)] and [(0, 5), (0, –5)]

Let  E₁ : Head appears
E₂ : Tail appears
A : A reports that head appears

∴ Rwquired probability = 
By Bayes’ Theorem

Let (x₁ y₁) be point on 
chord to 

Area formed by the lines = 4ab

Let A be the set of students who have passed in Mathematics, and B be the set of students who have passed in Physics.
We are given that |A| = 55 and |B| = 67, where |A| and |B| represent the number of students in sets A and B, respectively.
We are also given that there are 100 students in total. We need to find the number of students who have passed in Physics only, which means we need to find |B - A|.
First, let's find the number of students who have passed in both Mathematics and Physics, which can be represented by |A ∩ B|.
Using the principle of inclusion-exclusion, we have:
|A ∪ B| = |A| + |B| - |A ∩ B|
Since there are 100 students in total, we can say that |A ∪ B| = 100.
Now, we can find |A ∩ B|:
100 = 55 + 67 - |A ∩ B|
100 = 122 - |A ∩ B|
|A ∩ B| = 22
Now, we can find the number of students who have passed in Physics only, which is |B - A|:
|B - A| = |B| - |A ∩ B|
|B - A| = 67 - 22
|B - A| = 45
so, the correct answer is 45.

n(A) = 4
B = {5, 6}
n(B) = 2
n(A x B) = n(A)*n(B) = 4*2 = 8

No. of ways to select 4 senior and 3 U-19 players = ⁶C₄ * ⁵C3 = 150
No. of ways to select 5 senior and 2 U-19 players = ⁶C₅ * ⁵C2 = 60
No. of ways to select 6 senior and 1 U-19 players = ⁶C₆ * ⁵C1 = 5 
Total no. of ways to select the team = 150 + 60 + 5 = 215