HPSC PGT Mathematics Mock Test - 1 - HPSC TGT/PGT MCQ

The correct answer is sheep and goat.

Key Points

• The earliest domesticated animals were sheep and goats.
• Domestication is the process by which people grow plants and look after animals.
• All the animals which we use today as food are the result of domestication.
• It started about 12,000 years ago.

 Thus, we can say that the earliest domesticated animals include sheep and goats.

Additional Information

•  The earliest domesticated plants were wheat and barley.

Key PointsBiogas:

• Biogas is an emerging renewable, non-conventional energy source that is obtained through the degradation of organic matter by bacteria under anaerobic conditions.
• The major constituent of biogas is Methane.
• Biogas is usually made up of around 50-70% methane (CH4) and 25-45% carbon dioxide (CO2), with other gases such as hydrogen (H2), hydrogen sulphide (H2S), watervapour (H2O), Nitrogen (N2), oxygen (O2), ammonia (NH3) making up the rest.
• It is released when Cow, Buffalo, and Pig manure is processed anaerobically i.e. in the absence of Oxygen.
• Biogas can be used for Space Heating, Generation of Electricity, Fuel for Cooking, etc.

Important Points

Advantages of Biogas - 

• ​Biogas burns without smoke; hence no harmful gas such as CO₂, CO, NO₂, and SO₂ are evolved.
• The slurry produced after the production of biogas is used as manure in fields.
• The method of disposal is safe and efficient and hence no space is wasted in the form of landfills.
• Biogas plants require very little installation costs and become self-sufficient in a span of 3-4 months.
• Work opportunity for thousands of people is created, especially in rural areas.

Thus, Biogas is a non-conventional type of natural resource.

Additional Information

There are basically two sources of energy:

The logic follows here is :

Sum of place values of alphabets of the given word = Given number.

MONK → M + O + N + K = 13 + 15 + 14 + 11 = 53.

Similarly,

TUTOR → T + U + T + O + R = 20 + 21 + 20 + 15 + 18 = 94.

Hence, the correct answer is "94".

Hence. the person will be allowed to appear in the interview.

National Education Policy 2020 is the first education policy of the 21st century and aims to address the many growing developmental imperatives of our country. This policy replaced the 34-year-old National Policy on Education (NPE), in 1986.

Key Points

•  National Education Policy 2020 lays emphasis on conceptual understanding and improving educational outcomes so that children can implement the acquired knowledge and skills in real-life.
• It suggests that assessment is important to find out a child's progress over a certain period of time. 
• Emphasizes on transforming assessment for optimizing learning and development of all students.
• NEP aims to provide a holistic report on student’s progress and learning.
• It should be able to depict the actual level of knowledge and performance of each student.

Thus, it is concluded that National Education Policy 2020 suggests that assessment is important to find out a child's progress over a certain period of time.

As the curve goes from c to d and the equation is x = f(y)
So the shaded area is ∫(c to d)f(y)dy

Any point on the given parabola is (t², 2t). The equation of the tangent at (1,2) is x-y +1 = 0. The image (h,k) of the point (t²,2t) in x-y + 1 = 0 is given by 

∴ h = t² - t² + 2t - 1 = 2t - 1
Eliminating t from h = 2t – 1 and k = t²+1
we get, (h+1)² = 4(k-1)
The required equation of reflection is (x+1)² = 4(y-1)

Apply , C₁ → C₁ - C₂, C₂ → C₂ - C₃,

Because here row 1 and 2 are identical

(i) |a – a| = 0 < 1 is always true
(ii) a R b ⇒ |a – b| < 1 ⇒ |-(a – b)| < ⇒ |b – a| < 1 ⇒ b R a.
(iii) 2R 1 and
But, 2 is not related to 1/2. So, R is not transitive. 

C₁C₂ < r₁+ r₂

(i.e. angle subtended at the centre of a circle is double the angle subtended in the alternate segment).
Hence (c) is the correct answer.

cos 9x + cos (-10) x = cos 9x + cos10x

f (0) = cos0 + cos0 = 2,f (π/4) = cos9 π/4 + cos5π/2 = cos π/ 4 =
f (π/2) = cos9π/2 + cos5π = 0 -1 = -l,f (π) = cos9π + cos10π = - l +1 = 0.

A square matrix A for which Aⁿ= 0 , where n is a positive integer, is called a Nilpotent matrix.

All the elements in set-A are presented in set-B. So "A" is a subset of "B".

Concept:

The null set is a subset of every set. (ϕ ⊆ A)

Every set is a subset of itself. (A ⊆ A)

The number of subsets of a set with n elements is 2ⁿ.

The number of proper subsets of a given set is 2n - 1

Calculation:

Statement I: If A = {x: x is an even natural number} and B = {y: y is a natural number}, A subset B.

A = {2, 4, 6, 8, 10, 12, ...} and A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, ...}.

It is clear that all the elements of set A are included in set B.

So, set A is the subset of set B.

Statement I is correct.

Statement II:  Number of subsets for the given set A = {5, 6, 7, 8) is 15.

Given: A = {5, 6, 7, 8}

The number of elements in the set is 4

We know that,

The formula to calculate the number of subsets of a given set is 2ⁿ

 = 2⁴ = 16

Number of subsets is 16

Statement II is incorrect.

Statement III: Number of proper subsets for the given set A = {5, 6, 7, 8) is 15.

The formula to calculate the number of proper subsets of a given set is 2ⁿ - 1

 = 2⁴ - 1

 = 16 - 1 = 15

The number of proper subsets is 15.

Statement III is correct.

∴ Statements I and III are correct.

When a = b = 1, c = 2, it gives minimum value (since a,b,c not all equal)

A = {2, 4}
B = {5, 6} 
A x B = {(2, 5), (2, 6), (4, 5), (4, 6)}
(A x B) U Φ = A x B = {(2, 5), (2, 6), (4, 5), (4, 6)}

Solution we know that if f(t) is an odd function the  dt is given integral since   is odd, the given integral is even

Let the required circle is x² + y² - 6x - 4y + k = 0 and it is passing through (-2, 14)

(1/1² + 1/2² + 1/3² + ........∞) = π²/6
⇒ (1/1² + 1/3² + 1/5² + ........∞) + (1/2² + 1/4² + 1/6² + ........∞) = π²/6
=> (1/1² + 1/3² + 1/5² + ........∞) + 1/2²(1/1² + 1/2² + 1/3² + ........∞) = π²/6
=> (1/1² + 1/3² + 1/5² + ........∞) + 1/2²(π²/6) = π2/6
=> (1/1² + 1/3² + 1/5² + ........∞) + π²/24 = π²/6
= π²/8

As the value of both determinants are equal,
∴ 1 = x²
x = ±1

Suppose that the Reqd. Limit L = lim x→0 (sinx − x)/x³.
Substiture x = 3y , so that, as x→0 , y→0.
∴ L = lim y→0 sin3y − 3y/(3y)³,
= lim y→0 (3siny − 4sin3y)− 3y)/27y³,
= lim y→0 {3(siny − y)/27y³) − (4sin³y/27y³)},
⇒ L = lim y→0 1/9 * (siny − y)/y³) − 4/27* (siny/y)³...(∗)
.Note that, here,
= lim y→0(siny − y)/y³)
= lim x→0 (sinx−x)/x³) = L.
Therefore, (∗) ⇒ L = 1/9*L − 4/27
or,  8/9L = −4/27
Hence, L = −4/27*9/8
=−1/6

A 3-digit no.: _ _ _
Possible digits: 0, 1 ,2, 3, 4, 5, 6, 7, 8, 9
In the first place there are 9 possible digits (0 cannot be the 1st digit)
In the second and third place there are 10 possible digits.
Total no. of 3-digit numbers = 9 *10 *10 = 900 
For 3-digit no. without digit 7,
Possible digits: 0, 1 ,2, 3, 4, 5, 6, 8, 9
In the first place there are 8 possible digits (0 cannot be the 1st digit)
In the second and third place there are 9 possible digits.
Total no. of 3-digit numbers = 8 *9 *9 = 648 
No. of 3-digit numbers with at least one digit 7 = 900 – 648 = 252

f(x + y)=f(x) f(y) wrt x we get:
f′(x+y) (1 + y′) = f(x) f′(y) (y′) + f′(x) f(y)
Now on substituting x = 0 and y = 3, we get :
f′(3) (1 + 0)=f(0) f′(3) 0+ f(3) f′(0)
f′(3) = 3 × 11 = 33