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

The correct answer is Ethyl mercaptan.

Key Points

• The stench due to leakage from the LPG cylinder because of Ethyl mercaptan.
  • LPG gas is basically propane and butane.
  • It is odourless in its natural state.
  • The smell that you notice when there is a leak is actually the stench of an entirely different agent, called ethyl mercaptan​.
    • This substance is added to the gas when it leaves the main storage terminals.
• Hence, option 4 is correct.

• Methane :
  • Methane is a chemical compound with the chemical formula CH₄.
  • Methane is a one-carbon compound in which the carbon is attached by single bonds to four hydrogen atoms
  • It is a group-14 hydride and the simplest alkane and is the main constituent of natural gas.
  • It is a colourless, odourless, non-toxic but flammable gas.
  • It has a role as a fossil fuel, a member of greenhouse gas, and a bacterial metabolite.
• Butane :
  • Butane or n-butane is an alkane with the formula C₄H₁₀.
  • Butane is a gas at room temperature and atmospheric pressure.
  • Butane is a highly flammable, colourless, easily liquefied gas that quickly vaporizes at room temperature.
  • Butane is a hydrocarbon and a highly flammable, colourless, odourless, easily liquefied gas.
  • It is typically used as fuel for cigarette lighters and portable stoves, a propellant in aerosols, a heating fuel, a refrigerant.
• Propane :
  • Propane is a three-carbon alkane with the molecular formula C₃H₈.
  • It is a gas at standard temperature and pressure, but compressible to a transportable liquid. 
  • Propane is a flammable hydrocarbon gas that is liquefied through pressurization and commonly used for fuel in heating, cooling, hot water, and vehicles.
  • Propane can also be used for refrigerants, aerosol propellants, and petrochemical feedstock.​ 

The correct answer is Femur.

Key Points

• The largest bone in the human body is the femur, the thighbone.
• It is the largest, longest, and strongest bone in the human body.
• The shortest bone in the human body is the stapes or stirrup bone in the middle ear.
• The femur runs from the keen to the hip.
• The Femur or thigh bone is the largest and heaviest bone in the human body.
• At the proximal end of the Femur, there is a rounded prominence which is called the Head of the Femur.
• The femur runs from the keen to the hip.
• It carries all the upper body weight.
• The other parts of the leg are the tibia, femur, fibula, and keep cap also known as patella.

Additional Information

• The human skeleton is made up of 206 bones, including bones of the
  • Skull:  including the jaw bone (mandible, maxilla)
  • Spine: cervical, thoracic and lumbar vertebrae, sacrum and tailbone (coccyx)
  • Chest: ribs and breastbone (sternum)
  • Arms: the shoulder blade (scapula), collar bone (clavicle), radius, humerus and ulna
  • Hands: wrist bones (carpals), metacarpals and phalanges
  • Pelvis: hip bones
  • Legs: the thigh bone (femur), the kneecap (patella), shin bone (tibia) and fibula
  • Feet: tarsals, metatarsals, and phalanges.

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.

Given:

Efficiency of Amit = 2 × Efficiency of Ramit

Concept used:

Total work = Efficiency × Time period

Calculation:

The efficiency of Amit = 2

Efficiency of Ramit = 1

Total work = (2 + 1) × 28 = 3 × 28 = 84

Number of days Ramit takes to complete the whole work = 84/1 = 84

∴ The number of days in Ramit will complete the work is 84 days.

The family chart is as follows:

i) A is father of C and D is son of B.

ii) E is brother of A.

iii) If C is sister of D

Drawing the family tree as per the given information:

Thus we can see B is the wife of A.

E is the brother of A.

Thus we can conclude B is the sister-in-law of E.

Hence, the correct answer is "Sister-in-law".

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.

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {2, 4, 6}
B = {3, 4, 6, 8, 10}
A ∩ B = {4, 6}
(A ∩ B)’ = U – (A ∩ B)
= {1, 2, 3, 5, 7, 8, 9, 10} 

Let S = {Persons who drive scooter}
C = {Persons who drive car}
Given, n(S ∪ C) = 80, n(S) = 35, n(C) = 60
Therefore, n(S ∪ C) = n(S) + n(C) - n(S ∩ C)
80 = 35 + 60 - n(S ∩ C)
80 = 95 - n(S ∩ C)
Therefore, n(S ∩ C) = 95 – 80 = 15
Therefore, 15 persons drive both
scooter and car.
Therefore, the number of persons who
drive a scooter only = n(S) - n(S ∩ C)
= 35 – 15
= 20
Also, the number of persons who drive car only = n(C) - n(S ∩ C)
= 60 - 15
= 45

If R is a relation defined by xRy : ifx ⩽ y, then R is reflexive and transitive But, it is not symmetric. Hence, R is not an equivalence relation.

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

For the graph y=x²
The least value of x² is zero and square of any number will also be zero.

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.

 Number 1 is odd. As any number is divisible by 1: 1
Do prime factorisation of 3600: 2*2*2*2*3*3*5*5
Select all odd numbers from above: 3,3,5,5
Try every possible products of these:
Single number: 3, 5
Two numbers: 3*3, 3*5, 5*5: 9,15,25
Three numbers: 3*3*5, 3*5*5: 45, 75
All four: 3*3*5*5: 225
The odd divisors are: 1,3,5,9,15,25,45,75,225
There are 9 odd divisors of 3600.

z =| z | -3+ 2i ⇒ | z |²= (| z | -3)² + 4 = | z |² -6 | z | +9 + 4Þ | z | 13/6
=
Thus,  z = - 5/6 + 2i

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

The ratio of the coefficients of x^p and x^q in the expansion of (1+x)^p+q is

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


So, the required coefficient is

¹⁰C₄(-1)⁴ = 210

d = 100
S_n = 21700
n = 14
S_n = n/2[2a + (n-1)d]
21700 = 14/2[2a + (14-1)100]
= 21700 = 7[2a + 1300]
= 3100 = 2a + 1300
⇒ 3100 - 1300 = 2a
⇒ 1800 = 2a
⇒ a = 900

1
2 3
4 5 6
7 8 9 10
..........
Series for starting number:
1, 2, 4, 7, ...
Let  Tn = an² + bn + c
a+b+c=1                         ......(1)
4a+2b+c=2                   ......(2)
9a+3b+c=4                   .......(3)
From eq (1), (2), (3), we get:
a= 1/2, b= −1/2, c=1
∴ T_n = n²/2 − n/2 + 1
∴ T₅₀ = (50)²/2 − 50/2 + 1 = 1226
Now, elements in 50th group:
1226, 1227, 1228,.....(50 terms)
∴ Sn = 50/2[2×1226+(50−1)×1]
= 62525

L.H.S. is positive for all x ∈ R

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)

The given system of equations has unique solution , if 


⇒1(18−2)−1(9−α) ⇒13−5α ≠ 0 ⇒ α ≠ 13/5 + 1(6−6α) ≠ 0
Therefore , unique solution exists for all integral values of α.

where n is order of matrix. Here n = 3.

The equation of a circle with center (h,k) and radius r is given by (x−h)²+(y−k)²=r² . 
For a circle centered at the origin, this becomes the more familiar equation x²+y²=r²