UP TGT Mathematics Mock Test - 5 - UPTET MCQ

• The Kingdom of Anga was indeed a powerful entity before being conquered by Bimbisara, making statement 1 correct.
• Avanti, with its capital at Ujjain, was a significant competitor, partly due to its access to iron deposits, which made it a rival in military capabilities, so statement 2 is correct.
• Magadha's economic prosperity was significantly boosted by its agricultural productivity due to the fertile alluvial soil of the middle Gangetic plain, contrary to statement 3, which underestimates agriculture's role.

• Division of Indian society into seven castes: The source that informed us about the division of Indian society into seven castes is the Indica.

• The Indica is a work by the Greek historian Megasthenes, who was an ambassador of Seleucus Nicator in the court of Chandragupta Maurya.


• In his work, Megasthenes describes the social structure of India, including the division of society into seven castes.


• He mentions that the Indian society was divided into seven principal castes, each with its own occupation and duties.


• This division of society into seven castes is an important insight into the social structure of ancient India.

Tenali Ramakrishna (born Garlapati Ramakrishnayya; also known as Tenali Rama or Tenali Raman) was an Indian poet, scholar, thinker and a special advisor in the court of Sri Krishnadevaraya. He was a Telugu poet who hailed from what is now the Andhra Pradesh region, generally known for the folk tales which focus on his wit. He was admired by everyone for his sense of humor and wit.

The correct answer is option (A). During Aurangzeb's reign, Muntakhab-ul-Lubab was written by Khafi Khan, a historical writer. The book covers the history of Mughal period till the reign of Mughal emperor Aurangzeb.

Collateral refers to an asset owned by the borrower, such as land, building, vehicle, livestock, or deposits with banks, that is offered as security to the lender until the loan is repaid. In the event of loan default, the lender has the right to sell the collateral to recover the outstanding amount. Collateral provides a level of assurance to the lender and reduces the risk associated with lending, making it a common practice in credit agreements.

• Pollution is any undesirable change in physical, chemical, or biological characteristics of air, land, water, or soil.
• Agents that bring about such an undesirable change are called pollutants.
• Pollution can be broadly classified into 4 types - Air pollution, Water pollution, Noise pollution & Soil pollution.

Important Points

• Water pollution occurs when there is an addition of undesirable substances into water bodies.
• The sources of water pollution can be divided into 2 major groups:
  • Point source - It refers to the single identifiable sources like discharge pipes from a factory or sewage plant.
  • Non-point source - It refers to sources that do not originate from a single point. Example - agricultural run-offs.
• Factories and sewage treatment plants have proper drainage pipes leading to a water body for disposing their wastes and thus they are point sources of water pollution.
• Urban and Suburban lands cause water pollution by the field run offs and thus do not originate from a single point. Hence, they are non-point sources of water pollution.

Israel Aerospace Industries (IAI) is the aerospace and defence company that set up AeroSpace Services India (ASI) as its Indian subsidiary. This move signifies IAI's expansion into the Indian market, with ASI being authorized as the sole OEM's Technical Representative for the medium-range surface-to-air Missile (MRSAM) system in India.

Given: 

Mean of the observations 25, 29, 25, 32, 24 and x is 27.

Concept used: 

Mean = sum of all observations / total number of observations

Median:

The median formula of a given set of numbers, say having 'n' odd number of observations, can be expressed as:
 term

The median formula of a given set of numbers say having 'n' even number of observations, can be expressed as:
term

Calculation: 

Mean of the observations =  25 + 29 + 25 + 32 + 24 + x / 6 = 27

x = 27 × 6 - 135

⇒ 162 - 135 

⇒ 27

Arranging the terms in ascending order,

24, 25, 25, 27,29, 32 

Here, number of terms are even, 

∴ Median = (25 + 27) / 2 

⇒ 26

 ∴ Option 3 is correct.

Formula Used:

sec2θ - tan2θ = 1

If tanx = tanθ then x = nπ + θ 

Calculations:

secθ + tanθ = √3        ----(1)

According to the formula used

sec²θ - tan²θ = 1 

⇒ (secθ + tanθ) (secθ - tanθ) = 1     ----(2)

From (1) and (2), we get 

⇒ secθ  - tanθ = 1/√3        ----(3)

So, after solving, 

secθ = 2/√3 and tan θ = 1/√3

Now, we will find the values of θ in the given domain which will satisfy both of the above.

⇒  secθ = 2/√3 ⇒ cos θ = √3/2  ⇒ θ = π/6,  2π +π/6 (Can not be included) 

⇒  tanθ = 1/√3  ⇒ θ = π/6,  π +π/6 = 7π/6  (Can be included) 

So, the common value of θ will be only π/6.

So, there is one solution.

Hence, The correct answer is option 2).

Concept:

Non-homogeneous equation of type AX = B has infinite solutions;

if ρ(A | B) = ρ(A) < Number of unknowns

Calculation:

Given:

2x + 3y = 1

4x + 6y = λ

The augmented matrix is given by:

Applying R2 → R2 – 2R₁

For the system to have infinite solutions, the last row must be a fully zero row.

So if λ = 2 then the system of equations has infinitely many solutions.

Key Points:

Remember the system of equations

AX = B have

1) Unique solution, if ρ(A : B) = ρ(A) = Number of unknowns.

2) Infinite many solutions, if ρ(A : B) = ρ(A) < Number of solutions

3) No solution, if ρ(A : B) ≠ ρ(A).

Concept:

Let G be a non-empty set and 'o' be the binary operation, then an algebraic structure (G, o) is said to be a abelian group if the binary operation 'o' satisfies the following axioms:

(1) Closure Axiom: G is closed under the operation o, 

a o b ∈ G, for every a, b ∈ G.

(2) Associative Axiom: The binary operation o is associative.

(a o b)o c = a o(b o c), for every a, b, c ∈ G.

(3) Identity Axiom: For every element a ∈ G, there exists an element e ∈ G such that a o e = e o a = a, for every a ∈ G.

(4) Inverse Axiom: Each element of G possess inverse. 

For every a ∈ G, there exists an element b in G such that 

a o b = b o a = e

Thus the element 'b' is called the inverse of a and written as b = a^-1

(5) Commutative Axiom: G is commutative under the operation 'o'

a o b = b o a, for every a, b ∈ G,

Calculation:

We have, G = {e, a, b, c}

We have the following observations: 

(1) Closure Axiom: We see that all the entries in the composition table are elements of the set G. 

Thus closure axiom is satisfied.

(2) Associative Axiom: Associative law always hold in modulo system.

(3) Identity Axiom: The order of elements in first row and first column is the same as row and column makings.

Hence, e is identity element.

(4) Inverse Axiom: From table we have the inverses of e, a, b, c are e, a, b, c.

(5) Commutative Axiom: From the above table we have that each row is identical to corresponding columns.

Hence, G is abelian.

∴ G is an abelian group.

Given:

sin2 θ = 2 sin θ - 1

0° ≤ θ ≤ 90°

Concept used:

(a - b)² = a² - 2ab + b²

Calculation:

sin2 θ = 2 sin θ - 1

⇒ sin2 θ - 2 sin θ + 1 = 0

⇒ (Sin θ - 1)² = 0

⇒ (Sin θ - 1) = 0

⇒ Sin θ = 1

⇒ θ = 90°

Now,

Cos 90° = 0

Cosec 90° = 1

Now,

⇒ (1 + 1) / (1 - 0)

⇒ 2

∴ The required value of  is 2.

Given:

Length of a chord is 30 cm

Chord is 8 cm away from the centre of the circle

Calculation:

Let AB be the chord and O be the center of the circle.

Also, OP is perpendicular to AB

In Δ OPB,

OB² = PB² + OP²

OB² = 15² + 8²

OB² = 225 + 64

OB² = 289

OB = √289 

OB = 17

Diameter = 2 × OB = 2 × 17 = 34

∴ The diameter of the circle is 34 cm.

Concept used:

ⁿP_r = n!(n – r)!

Calculation:

Number of ways of arranging the vowels in odd positions = ³P₂ = 6

Number of ways of arranging the remaining letters =3! = 6

Required number of ways = 6 × 6 = 36

∴ Required answer is 36

Given:-

∠OCB = 60°  

Concept Used:

Diagonals of a rhombus are ⊥ bisectors to each other

cosθ = Adjacent Side/Hypotenuse

cos60° = 1/2 

Calculation:

ABCD is a rhombus, 

Let sides of a rhombus be a units,

In the given figure, Δ COB, 

cos ∠OCB = cos60° = 1/2 

⇒ 1/2 = OC/CB

⇒ 1/2 = OC/a

⇒ OC = a/2

Hence, AC = 2 × OC = 2 × a/2 = a

Now in Δ AOB, ∠ AOB = 90° 

Using Pythagoras theorem, 

AB² = OA² + OB²

⇒ a2 = (a/2)2 + OB2

⇒ OB2 = a2 - (a/2)2

⇒ OB2 = 3a²/4

⇒ OB = √3a/2

Hence, BD = 2 × OB = 2 × √3a/2 = √3a

AC : BD = a:√3a

⇒ AC : BD = 1:√3

∴ The ratio of AC: BD is 1 : √3 

Concept used:

Ratio test:

L = 

L > 1 the series is Divergent neither convergent or divergent 

L < 1 the series is Convergent 

L = 1 test fails Neither Convergent nor Divergent 

Limit Comparision test:

if an and bn are two positive series such that 

where c > 0 and finite then, either Both series converges or diverges together

P - Series test: 

∑  is convergent for p > 1 and divergent for p ≤  1 

Calculations:

Given:

The HCF of (x8 – y8) and (x7 – y7 + x5y2 – x2y5) 

Concept:

1. HCF of two numbers:  Product of the smallest power of each common prime factor in the numbers.

2. (aⁿ - bⁿ) = (a - b) (a^n-1 + ba^n-2 +  ....+ b^n-1)

Where n is a natural number. Hence,

For every value of n, (a - b) will be a factor of (aⁿ - bⁿ)

3. (a + b)(a  - b) = a² - b²

Calculation:

Using the  above formula

(x8 – y8) = (x⁴ - y⁴)(x4 + y4)

⇒ (x8 – y8) = (x² - y²)(x² + y²) (x⁴ + y⁴) 

⇒ (x8 – y8) = (x + y)(x – y)(x² + y²)(x⁴ + y⁴)

Now, (x7 – y7 + x5y2 – x2y5) 

⇒ x⁷ + x⁵y² – y⁷ – x²y⁵

⇒ x⁵(x² + y²) – y⁵(y² + x²)

⇒ (x² + y²) (x5 – y5)

Using the general formula of (aⁿ - bⁿ)

⇒ (x² + y²)(x - y)(x⁴ + yx³ + y²x² + y³x + y⁴)

Now, using the concept of HCF

The required HCF = (x – y)(x² + y²)

Multiplying the factor

⇒ (x3 – y3 – x2y + xy2) 

∴The required HCF is (x3 - y3 - x2y + xy2) 

Given:

11Pr = 11P_{r + 1}

Formula used:

ⁿP_r = n!/(n - r)!

Calculation:

11Pr = 11P_{r + 1}

⇒ 11!/(11 - r)! = 11!/(10 - r)

⇒ 1/(11 - r)(10 - r)! = 1/(10 - r)!

⇒ 11 - r = 1

⇒ r = 10

∴ The value of r is 10.

Concept used:

Ratio test:

L = 

L > 1 the series is Divergent neither convergent or divergent 

L < 1 the series is Convergent 

L = 1 test fails Neither Convergent nor Divergent 
Calculations:

Given:

if an  is unbounded and bn is unbounded 

Calculations:

take a_n  and b_n both equal to n 

then a_n + b_n  = n + n = 2n which is unbounded

⇒ 1st option is incorrect

take a_n  = n and b_n  = -n 

then a_n + b_n  = n - n = 0 which is a bounded sequence  contains all terms equal to 0 

⇒ 2nd option is incorrect 

∴ option 3 is correct 

Given:

y = f(x) = |cos x - sin x|

Concept:

To find derivative of piecewise functions such as GIF, modulus, etc.; it is better to check the Left-Hand Derivative and Right-Hand Derivative at the point of concern.

Formula:

LHD (at x=a)=limh→0f(a−h)−f(a)−h

Also, |x| = x if x ≥ 0 and -x if x < 0 

Calculation:

f(π/4) = 0

Here, for x → (π/4)⁺ , sin x > cos x so the mod will open as negative because the inner function will be negative.

So, f(x) = - (cos x - sin x) 

Here, for x → (π/4)^- , sin x < cos x so the mod will open as positive because the inner function will be positive.

So, f(x) =  (cos x - sin x) 

⇒ LHD = f'(x → (π/4)^-) = ​

⇒ 2b = 4

⇒ 

⇒ b² = 8

Shortcut Trick
Constructing this figure,

In this figure, 

∠AOB = 180° - 110° 

⇒ ∠AOB = 70°

In △OAB,

As, OA = OB (radius)

and ∠OAB = ∠OBA ( In a triangle, angles opposite to equal sides are equal)

Now,

∠OAB + ∠OBA + ∠AOB = 180° 

⇒ 2∠OAB = 110° 

∴ ∠OAB is 55°.

Alternate Method

Given: 

MA and MB are two tangents from a point M outside the circle with centre O.

And ∠AMB = 110°

Concept used:

Tangent to a circle is always perpendicular to its radius at the point of tangency.

Sum of all angles of a quadrilateral is 360° 

Sum of all angles of triangle is 180° 

In a triangle, angles opposite to equal sides are equal.

Calculation:

 As, MA and Mb are tangents,

Tangent to a circle is always perpendicular to its radius at the point of tangency.

⇒ ∠OAM = 90° 

⇒ ∠OBM = 90° 

Now In quadrilateral OAMB,

⇒ ∠OAM + ∠AMB + ∠OBM + ∠AOB = 360° 

⇒ 90° + 110° + 90° + ∠AOB =  360° 

⇒ ∠AOB = 360° - 290° 

⇒ ∠AOB = 70°

In △OAB,

As, OA = OB (radius)

and ∠OAB = ∠OBA ( In a triangle, angles opposite to equal sides are equal)

Now,

∠OAB + ∠OBA + ∠AOB = 180° 

⇒ ∠OAB + ∠OAB + 70° = 180° 

⇒ 2∠OAB = 110° 

⇒ ∠OAB = 55° 

∴ ∠OAB is 55°.

GIVEN:

x = {n; 2n² + 7n - 15 < 0}

CONCEPT:

Concept of sets and relation.

FORMULA USED:

No formula

CALCULATION:

x = {n; 2n² + 7n - 15 < 0}

⇒ 2n² + 7n - 15 < 0

⇒ 2n² + 10n - 3n - 15 < 0

⇒ 2n(n + 5) - 3(n + 5) < 0

⇒ (n + 5)(2n - 3) < 0

⇒ n = (-5, 3/2)

Calculation:

Concept:

Conditions of collinear vector:

1. Three points with position vectors  are collinear if and only if the vectors  and  are parallel.

⇒ 

2. If the points (x 1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be collinear then 

Concept:

(a + b)² = a² + b² + 2ab

(a - b)² = a² + b² - 2ab

Power of iota (i)

Depending upon the power of “i”, it can take the following values;

i4k+1 = i, i4k+2 = -1, i4k+3 =  -i, i4k = 1

Given:

(1 - ) + (1 - ) + (1 - ) + up to n term 

Formula used:

Sum of n terms = n(n + 1)/2 

Calculation:

(1 - ) + (1 - ) + (1 - ) + up to n term

⇒ 1 × n - 1/n(1 + 2 + 3 + ......n)

⇒ n - 1/n × n × (n + 1)/2 

⇒ n - (n + 1)/2 

⇒ (2n - n - 1)/2 

⇒ (n - 1)/2 

∴ (1 - ) + (1 - ) + (1 - ) + up to n term will result as: (n - 1)/2