HPSC PGT Mathematics Mock Test - 6 - HPSC TGT/PGT MCQ

The Correct Answer is Option 3 i.e India.

• India:
  • It shares a border with seven countries, namely China, Pakistan, Bhutan, Myanmar, Afghanistan, Nepal, and Bangladesh.
  • India shares the longest border with Bangladesh.
  • It is the largest democracy in the world, the representative with the most number of votes gets the power.
  • New Delhi is the capital of India.
  • Droupadi Murmu is the current President of India and Narendra Modi is the current PM of India.
• Sweden:
  • It is a country in Europe.
  • Sweden is a constitutional monarchy with a parliamentary democracy.
  • Stockholm is the capital of Sweden.
  • Incumbent Magdalena Andersson is the current PM of Sweden.
• Iceland:
  • It is an Island country in Europe.
  • Iceland is a constitutional republic with a multi-party system and is considered as the worlds' oldest parliamentary democracy.
  • Reykjavik is the capital of Iceland.
  • Guoni Thorlacius Johannesson is the current President of Iceland.
• Norway:
  • It is located in Northwestern Europe.
  • It is a constitutional monarchy.
  • Jonas Gahr Støre is the current Prime Minister of Norway.
  • Oslo is the capital city of Norway.

The correct answer is Inscriptions.

Key Points

• Writings on hard surfaces like stone or metal are called Inscriptions.
• Often kings got their orders inscribed so that people could see, read and obey them.
• Other inscriptions have the activities done by men and women recorded on them.
• Kings kept record of their victories in the form of inscriptions.

 Thus, we can say that writings on hard surfaces like stone and metals are called Inscriptions.

Additional Information

•  Manuscripts were written on palm leaves or birch.

The correct answer is Sonipat.

Key Points

• Anju Dahiya -
  • She is a teacher from the Sonipat district of Haryana.
  • She has been honored with the 'National Teacher Award 2022.​

Additional Information

• ​On August 25, 2022, the Ministry of Education, Government of India released the final list of National Teacher Awards 2022.
• The name of Haryana's teacher Anju Dahiya is also included in this.
• President Draupadi Murmu awarded the selected teachers at Vigyan Bhavan in New Delhi.
• This award was given on September 5, 2022, on the occasion of Teachers' Day.
• As an award, teachers will be given an amount of Rs 50 thousand and a silver medal.

The correct answer is Haryana.

Key Points

• The Haryana government has appointed international wrestler Babita Phogat and Kabaddi player Kavita Devi as deputy directors in its Sports and Youth Affairs Department.
• They were appointed under the Haryana Outstanding Sportspersons (Recruitment and Conditions of Service) rules, 2018.
• As per the orders, both have to join the department within a month.

Additional Information

• Haryana:
  • Lok Sabha seats - 10.
  • Rajya Sabha seats - 5.
  • National Parks - Kalesar National Park, Sultanpur National Park.
  • Dams - Kaushalya Dam (Yamuna River), Ottu barrage (Ghaggar River), Anagpur Dam (Hakra River).

All exams must be conducted for the backward children. Emphasis must be paid on providing proper education to them so that they can easily face all examinations.

Multiple means of discrimination is not a part of universal design for learning guidelines.

(2,4) written in set builder form is:
{x : 2 < x < 4}

Step-by-step Explanation:
Since we have given that,
Percentage of families own a cell phone = 25%
Percentage of families own a scooter = 15%
Percentage of families neither own = 65%
So, it becomes,
P(C ∪ S)^' = 1 - P(C ∪ S)
0.65 = 1 - P(C ∪ S)
1 - 0.65 = P(C ∪ S)
P(C ∪ S) = 0.35
So, it becomes, 
P(C ∪ S) = P(C) + P(S) - P(C ∩ S)
0.35 = 0.25 + 0.15 - x
0.35 = 0.40 - x
x = 0.05
Thus, 0.05 x Total number of families in town = 1500
Total number of families in the town =
1500 / 0.05 = 30,000
Hence, the total number of families in the town is 30,000.

For f to be defined term under square root has to be greater than zero.
⇒ cosx − 1 ≥ 0
⇒ cosx ≥ 1
Only possible if cosx = 1
⇒ x = 2nπ ∀ n∈Z

f(x) = 3.sin(√π²/16 − x²)
since the quantity within the square root can not be negative.
therefore
π²/16 − x² ≥ 0
x² ≤ π²/16
−π/4 ≤ x ≤ π/4
the minimum value of the function is 3.sin0=0
and the maximum value of the function is 3sinπ/4 = 3/√2
therefore range is [0, 3/√2]

n! < [(n+1)/2]ⁿ
For(n=1) 1! < [(1+1)/2]¹
= 1 < (2/2)¹
1 < 1 (False)
 
For(n=2) 2! < [(2+1)/2]²
= 2 < (3/2)²
1 < 6/4 
1 < 3/2
1 < 1.5 (True)
Therefore smallest number will be 2.

4^2x = 4^-6

2x = -6

So,  x = -3

He can invite any one  friend in ⁶C₁ ways = 6 ways:
He can invite any two friends in ⁶C₂ ways = 15 ways
He can invite any three friends in ⁶C₃ ways = 20 ways
He can invite any 4 friends in ⁶C₄ways = 15 ways
He can invite any 5 friends in ⁶C₅ ways = 6 ways
He can invite  all the 6 friends in ⁶C₆ ways= 1 way.
Since any one of these could happen total possibilities are, 6+15+20+15+6+1 = 63.

Correct Answer :- c

Explanation : As given condition is , 5 must be followed by 7. So only possible way is 57X where X denotes 0,2,4,6 and 8.

So,total no. of ways=1×1×5=5

Hence, total ways in which we can make a 3-digit even no. without violating given condition are:

360+5=365

⇒ αS = α + 2α² + 3α³ + ...+(n -1)α^n-1 +nαⁿ

On subtracting, we get

Since, α is nth root of unity, therfore
αⁿ = 1 

(a + ib)² = 5 + 12i
use (a + b)² = a² + b² + 2*a*b
=> a² + b²*i² + 2*a*b*i = 5 + 12i
i² = -1
⇒ a² - b² + 2*a*b*i = 5 + 12i
equate the real and complex coefficients
⇒ a² - b² = 5 and ab = 6
a = 6/b
substitute in a² - b² = 5
⇒ 36/b² - b² = 5
⇒ 36 - b⁴ = 5b²
⇒ b⁴ + 5b² - 36 = 0
⇒ b⁴ + 9b² - 4b² - 36 = 0
⇒ b²(b² + 9) - 4(b² + 9) = 0
⇒ (b² - 4)(b² + 9) = 0
⇒ b² = 4 and b² = -9
b is a real number, so we eliminate b² = -9
b² = 4
⇒ b = 2 and b = -2
a = 3 and a = -3
The required number can be 3 + 2i and -3 -2i

-5x + 2 < 7x - 4
2 + 4 < 7x + 5x
6 < 12x
x > 1/2

/* We know that,
If a,b,c are in A.P then
a+c = 2b */

/* Splitting the middle term, we get

Here , n=2 is not possible because number of terms are more than three (According to the problem )
Therefore, n = 7

for this term to be independent of x. we must have


So, the required coefficient is

¹⁰C₄(-1)⁴ = 210

Substitute ‘n’ and verify the options.

The series is 24,20,16...in AP
so a = 1st term=24 & common difference = 20-24 = -4
sum = 72 is of say n terms
then 72 = n/2[2*24+(n-1)*-4] = n/2[48-4n+4] = n/2[52-4n]
or 144 = 52n-4n²

or 4n²-52n+144 = 0
or n²-13n+36 = 0
or n²-9n-4n+36 = 0
or n(n-9)-4(n-9) = 0
or (n-9)(n-4) = 0
or n = 9 or 4

The sequence is the Fibonacci series
1+1 = 0
1+2 = 3
2+3 = 5
3+5 = 8
5+8 = 13
8+13 = 21
13+21 = 34
21+34 = 55
34+55 = 89
The 11th term will be 89.

aα² + bα + c = 0 and bα² + cα + a = 0 ⇒ α = 1 ∴ a + b + c = 0

Roots of (x - α) (x - β) = λ are γ and δ, 
i.e., (x - α) (x - β) - λ = (x - γ) (x - δ)
So, (x - γ) (x - β) + λ = (x - α) (x - β)

|A|=3((4)-(4))-2((-4)-(2))-1((-8)-(-4))
|A|= -2(-6)-(-6)
|A|= -2(0)
=0

Operate,

Apply R₃→R₃- R₁, R₂→ R₂ -R₁,

⇒ -6(5x² - 20) +15(2x-4) = 0

⇒ (x- 2)(x+1) = 0⇒x=2, -1

⇒-6(5x² - 20) + 15(2x-4) = 0

⇒(x-2)(x+1) = 0 ⇒ x=2, -1