UP LT Grade Teacher Mathematics Mock Test - 3 Free Online Test 2026

The Gandhi-Irwin Pact, also known as the Delhi Pact, was signed on 5th March 1931 between Mahatma Gandhi and Lord Irwin, who was the Viceroy of India at that time. The pact was a compromise reached during the Civil Disobedience Movement, allowing for the release of political prisoners and permitting Indians to participate in the Second Round Table Conference in London to discuss constitutional reforms.

Charter Act 1833 or the Saint Helena Act 1833  was passed by the British Parliament to renew the charter of East India Company which was last renewed in 1813. Via this act, the charter was renewed for 20 years but the East India Company was deprived of its commercial privileges which it enjoyed so far.

It is compulsory for every citizen of the country to register in a National Register of Indian Citizens (NRIC) as per Section 14A of the Citizenship Act 1955 as amended in 2004. It is a Register of usual residents of the country.The objective of NPR is to create a comprehensive identity database of every usual resident in the country.

Correct option is A. It divides India into northern and southern parts
The Tropic of Cancer passes through the central part of India, approximately dividing the country into two halves, namely the northern and southern halves. This line of latitude is significant for India's geography and climate, as it marks the northern boundary of the tropical region and influences various climatic phenomena such as the monsoon. However, it does not act as the standard Time Meridian of India, nor is it the southernmost point of the Indian mainland, nor does it serve as the northern boundary of India.

Every type of particle has a corresponding antiparticle, for example; the positron is the antiparticle of the electron. the antiproton is the antiparticle of the proton. the antineutron is the antiparticle of the neutron. the antineutrino is the antiparticle of the neutrino.

Concept

• 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.

Explanation

• In order to control environmental pollution, the Government of India has passed the Environment (Protection) Act, 1986 to protect and improve the quality of our environment (air, water, and soil).
• In India, the Air (Prevention and Control of Pollution) Act came into force in 1981 but was amended in 1987 to include noise as an air pollutant.
• Realizing the importance of maintaining the cleanliness of the water bodies, the Government of India has passed the Water (Prevention and Control of Pollution) Act, 1974 to safeguard our water resources.

The Textiles and Apparel sector is likely to be most affected by the U.S. tariffs. This sector operates on thin margins, making it particularly vulnerable to price increases resulting from tariffs. The impact could lead to higher costs for consumers and reduced competitiveness in the U.S. market.

CONCEPT:

• If the two vectors are assumed as a̅  and b̅  then the dot created is articulated as a̅. b̅.
• The angle between the two vectors is θ.

• We know that the dot product of two product is given as:

• Thus, the angle between two vectors formula is given by:

Where θ is the angle between a̅ and b̅.

CALCULATION:

Given

a̅ = 2i + 4j - k and b̅ = 4i -j + k.

The dot product is defined as:

a̅ ⋅ b̅ = (2i + 4j - k) ⋅ (4i -j + k) = (2) (4) + (4)(-1) + (-1)(1) = 8 - 4 -1 = 3

The magnitude of vectors is given by:

Given:

Equation of right circular cone whose vertex is origin, axis Z - axis and semi - vertical Angle α is

Calculations:

Let us assume the right circular cone with base radius R and height H, centered at the origin and its axis is along the axis, and axes pass through the base.

In this scenario, we can express it as a series of circles or disks placed one on top of the other, uniformly decreasing in the radius from bottom to top.

So the circle's radius at a height h from the top will be , where $\text{Unexpected text node: 'θ'}$ is the semi-vertical angle.

The equation of such a circle will be x2 + y2 = r2

The equation of such a circle will be x2 + y2 = z2 tan2(θ)

: a particular case for equation of cone which is x² + y² = z² tan² α

Concept:

Differentiation formula

Quotient rule:

Calculation:

f(x) = 

⇒ f^'(x) = 

⇒ f′(x) = 0 at x = e

⇒ f′(x) > 0 when x < e

⇒ f′(x) < 0 when x > e

So, f(x) is maxima at x = e

⇒  f(e) = 

∴ The maximum value of  is 

Concept:

Integral of the product of two functions

Calculation:

Given, I = ∫ex{1 + ln x + x ln x}dx

⇒ I = ∫ex dx + ∫ e^x ln x dx + ∫ e^x x ln x dx

⇒ I = ex + ∫ ex ln x dx + ∫ ex x ln x dx

Computing the last integral ∫ ex x ln x dx by taking f(x) = x ln x and g(x) = e^x

⇒ I = ex + ∫ ex ln x dx + [x ln x ∫ ex dx - ∫(x ln x)' ∫ ex dx dx]

⇒ I = ex + ∫ ex ln x dx + x ln x ex - ∫(1 + ln x)ex dx

⇒ I = ex + ∫ ex ln x dx + x ln x ex - ∫ex dx - ∫ex ln x dx

⇒ I = ex + x ln x ex - ex 

⇒ I = x ln x ex + c

∴ The correct option is (1).

Concept:

Mean value theorem for integrals:

Let f be continuous on [a, b]. Then there is a point x₁ in (a, b) such that,

f' (x1) = [f (b) - f (a)]/(b - a) 

Calculation:

Given: f (x) = $\frac{1}{x}$

Given:

If θ is the angle between any two vectors  and 

Formula Used:

|a→×b→|=|a→||b→|sinθ

Calculation:

Concept:

If aî + bĵ + ck̂ and pî + qĵ + rk̂ are 2 different sides of the rhombus.

Suppose  and B = 

Then anyone diagonal of the rhombus is given by 

The other diagonal is given by 

The magnitude of the vector  

Calculation:

Concept:

If ω is the cube root of unity, i.e. ω³ = 1.

then 1 + ω + ω² = 0.

ω⁴ = ω³ω = ω [∵ ω³ =1]

Calculation:

Given:

= 0

C'₁ = C₁ + C₂

Expanding along first column:

Calculation:

ω15 + ω20+ ω25

= ω¹5 (1+ ω⁵ +ω¹⁰)

= ω¹5 × (1 + ω³.ω² +ω9.ω)

= (ω³)⁵ × (1 + ω² + ω)

= 1 × (1 + ω + ω2)

= 1× 0

= 0 

Concept:

Mean: Mean (average) is the sum of all quantities divided by no of quantities of the data set.

Mode: It is the most repeated value in the data set.

Median: Median is the middlemost value of the data set.

Skewness:

• The skewness is a measure of the asymmetry of the probability distribution assuming a unimodal
• distribution (assuming there is only one peak in the distribution of values). We can say that the
• skewness indicates how much our underlying distribution deviates from the normal distribution
• since the normal distribution has a skewness of 0. Generally, we have three types of skewness.

Symmetrical: When the skewness is close to 0 and the mean is almost the same as the median

Negative skew:

• When the left tail of the histogram of the distribution is longer and the majority of the observations are concentrated on the right tail.
• In this case, we can use also the term "left-skewed" or "left-tailed" and the median is greater than the mean.

Positive skew:

• When the right tail of the histogram of the distribution is longer and the majority of the observations are concentrated on the left tail.
• In this case, we can use also the term "right-skewed" or "right-tailed" and the median is less than the mean.

Explanation:

• We know that when the left tail of the histogram of the distribution is longer and the majority of the observations are concentrated on the right tail. In this case, we can use also the term "left-skewed" or "left-tailed". and the median is greater than the mean.
• So, the frequency curve corresponding to the data obtained is skewed to the left and the graph obtained is as below:

For the above left-skewed curve, Mode > Median > Mean.

∴ Mode > Median > Mean for left-skewed curve.

Concept:

The equation of parabola is y² = 4ax and focal distance is x + a.

Solution:

Given that, The focal distance of a point on the parabola y² = 8x is 10.

y² = 8x

⇒ 4a = 8 ⇒ a = 2

Now The focal distance is x + a = 10

⇒ x = 10 - 2 ⇒ x = 8

and y² = 8 × 8 = 8² ⇒ y = ± 8

So, its coordinates are (8,± 8)

∴ The correct option is (4)

Given:

f(x) = (x - 1)(x - 2)(x - 3)

Concept:

For maximum value of a differentiable function f(x) -

• f'(x) = 0
• f''(x) < 0 at x where f'(x) = 0

Solution:

f(x) = (x - 1)(x - 2)(x - 3)

⇒ f(x) = x3 - 6x2 + 11x - 6

⇒ f'(x) = 3x2 - 12x + 11

Putting f'(x) = 0

⇒ 3x2 - 12x + 11 = 0

⇒ x = 2 ± √(1/3)

⇒ x ≈ 1.423 , 2.577

Now, f''(x) = 6x - 12

f"(1.423) < 0

Hence, the maximum value occur at x = 2 - √(1/3)

⇒ f(2 ± √(1/3)) = (1 - √(1/3))(-√(1/3))(-1 - √(1/3))

= 

Given:

Concept:

• A parabola is a U-shaped plane curve where any point is at an equal distance from a fixed point (known as the focus) and from a fixed straight line which is known as the directrix.
• The simplest equation of a parabola is y2 = x when the directrix is parallel to the y-axis.
• The four equations of a parabola are given as: y2 = 4ax, y2 = – 4ax, x2 = 4ay, and x2 = – 4ay.
• 

⇒ (x - a)2 + y2 = (x + a)²

⇒ y2 = 4ax ( standard equation of parabola)

Calculation:

Let focus be S = (- 2, 0) and the directrix given by x + 3 = 0 

Let P(x, y) be any point on the parabola and PM be the line perpendicular line drawn from P to directrix as shown in the below image.

According to definition of parabola we have,

Distance between SP = |PM|

Using the distance formula we have

Squaring on both sides we get,

(x + 2)² + y² = (x + 3)²

⇒ x² + 4 + 4x + y² = x² + 9 + 6x

⇒ y2 - 2x - 5 = 0 or

⇒ y2​ = 2x + 5

∴ The equation of the parabola is y2​ = 2x + 5.

Hence, the correct answer is option 3).

Given:

A is a square matrix of order 3 and |A| = -2

Concept used:

 = 

Calculations:

∵ AA^-1 = I

⇒ |AA^-1| = 1 

⇒ |A| |A^-1| = 1

∴ |A-1 |  = 

Given that , |A| = -2

∴ option 2 is correct 

Given:

There are 5 women and 3 men.

Two particular men sit together.

Concept:

These types of questions are usually done by taking two particular men as a single unit, applying permutations within them, and then making permutations of all the units.

Formula:

Ways of arranging n distinct things taken all at a time = n!

Explanation:

Let the two particular men be M₁ and M₂

Number of ways of choosing and arranging the 2 men out of 3 = ³C₂ × 2! = 6

Now, taking these chosen 2 men as a single unit, the number of ways of arranging 1 man, 2 men as a single unit, and 5 women taking at a time = (3- 2 +1 + 5)! = 7! = 5040 

So, the number of ways can 5 women and 3 men are seated in a row so that two men are together = 3C2 × 2! × (3- 2 +1+ 5)! = 6 × 5040 = 30240

Cost of one = Rs 80

Cost of 5 = 80*5 = 400