TG EAMCET 2025 Mathematics Chapter-Wise Questions with Solutions

TG EAMCET 2025 Mathematics Chapter-Wise Questions with Solutions: Mathematics is one of the most important sections in TG EAMCET exam as it carries the maximum weightage in terms of marks i.e. 80 marks. The Mathematics section in TG EAMCET includes 80 multiple-choice questions out of a total of 160 questions. These questions cover topics from TG EAMCET Mathematics syllabus 2025 , like Matrices, Complex Numbers, Trigonometry, Vectors, and Geometry. For each question, candidates will have four answer choices, of which only one will be correct. To prepare well for the upcoming examination, it is advisable to practice TG EAMCET 2025 Mathematics chapter-wise questions with solutions. This will allow students to focus on one topic at a time, understand the types of questions they might face, and see where they need to improve. By practicing TG EAMCET 2025 Mathematics questions and reviewing the step-by-step solutions, they can get a better grasp of the chapters, improve their problem-solving skills, and feel more confident. In the section below candidates can view a few TG EAMCET 2025 Mathematics chapter-wise questions with solutions. They can also download these questions and answers in PDF format from this page to keep practising later.

Also Check - TG EAMCET 2025 Syllabus: Subject-Wise Syllabus Topics

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TG EAMCET 2025 Mathematics Chapter Wise Question with Solutions

As students already know that MCQ- type questions will be asked in the exam, they can start practising the TG EAMCET 2025 Mathematics questions with solutions to solidify their understanding of the fundamental concepts. This will also help them manage their time better during the exam, as they will be familiar with how to quickly approach each question. Overall, knowing what to expect reduces stress and improves their chances of doing well. Below are some sample TG EAMCET 2025 Mathematics chapter-wise questions with solutions to help students practice and prepare.

Algebra

Q 1. If A={1,2,3,4,5,6} , then the number of subsets of A which contain at least two elements is:

a) 64

b) 63

c) 57

d) 58

Solution:
Total number of subsets of A is 2n(A)=26=64

Number of subsets of A which contain at least two elements is

64−(6C0​+6C1​)

=64−(1+6)

=57

Q 2. As per the principle of mathematical induction, for all n∈N, the expression 3²ⁿ⁺² - 8n - 9 is divisible by:

a) 9

b) 8

c) 7

d) 5

Solution:
We can write,

P (n) : 3 ^{2n + 2} - 8n - 9 is divisible by 8

We note that

P (1) : 3 ^{2.1 + 2} - 8n - 9 = 34 - 8 - 9 = 81 - 17 = 64, which is divisible by 8.

Thus P (n) is true for n = 1

Let P (k) be true for some natural number k.

i.e., P (k) : 3 ^{2k + 2} - 8k - 9 is divisible by 8

We can write

3 ^{2k + 2} - 8k - 9 = 8a ....(1)

where a ∈ N

Now, we will prove that P (k + 1) is true whenever P (k) is true.

Now,

3 ^{2(k+ 1) + 2} - 8(k + 1) - 9

= 3 ^2k + 4 - 8k - 8 - 9

= 3 ² .3 ^2k+ 2 - 8k - 17

= 3 ² (3 ^2k+ 2 - 8k - 9 + 8k + 9) - 8k - 17 (added and subtracted 8k and 9)

= 3 ² (3 ^2k+ 2 - 8k - 9) + 3² (8k + 9) - 8k - 17

= 3 ² .8a + 72k + 81- 8k - 17 .... [from (1)]

= 9.8a + 64k + 64

= 8(9a + 8k + 8)

From the last line, we see that

8(9a + 8k + 8) is divisible by 8.

Q 3. If (24, 92) = 24 m + 92 n, then (m, n) is

a) (-4, 3)

b) (4, -3)

c) (4, -1)

d) (-1, 4)

Solution:
Since, 92=3−24+20

24=1⋅20+4

20=4⋅5+0

∴(24,92)=4

=24−1⋅20

=24−1⋅(92−3⋅24)

=24−92+3⋅24

=4⋅24−92…(i)

But (24,92)=24m+92n…(ii)

∴ From Eqs. (i) and (ii) , we get

m=4 and n=−1

∴(m,n)=(4,−1)

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Trigonometric Functions

Q 1. Find the values of the trigonometric functions in sin 765°.

a) 1/√2

b) 1/√6

c) 1/√4

d) 1/√3

Solution:
We know that the value of sun x repeats after an interval of 2π or 360°.

Sin (765°) = sin (2×360°+45°)

= sin 45°

= 1/√2

Q 2. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

a) In one second, the wheel will rotate an angle of 8π radians

b) In one second, the wheel will turn an angle of 6π radians

c) In one second, the wheel will spin an angle of 10π radians

d) In one second, the wheel will turn an angle of 12 π radians

Solution:
Given that a wheel makes 360 revolutions in one minute

Then, the number of revolutions in one second = 360/60 =6.

In 1 complete revolution the wheel turns 360°= 2 π radian.

So, In 6 revolutions, the wheel will turn 6×2π radian = 12π radian.

Hence, in one second the wheel will turn an angle of 12π radian.

Q 3. Find the values of the trigonometric functions in Cot (-15π/4).

a) 2

b) 1

c) 5

d) 7

Solution:
A (-15π/4) = cot (2 X 2π -15π/4) = cot (16π -15π/4) cot = cot π/4 = 1.

Also Check - TG EAMCET 2025 Exam Date: Application Form, Hall Ticket, Answer Key, Result Date

Probability

Q 1. What is the probability of rolling a sum of 7 with two fair six-sided dice?

a) 1/6​

b) 1/12

c) 1/8

d) 1/36

Solution:
Let A = sum of numbers is 7 = { (1, 6)(2, 5)(3, 4) (4, 3) (5, 2) (6, 1)}

n(A) = 6

P (Sum of numbers is 7) = n(A) / n(S)

= 6/36 = 1/6

Therefore, the probability of rolling two dice and getting a sum of 7 is 1/6.

Q 2. In a deck of 52 playing cards, what is the probability of drawing a heart or a king?

a) 13/52

b) 17/52

c) 16/52

d) 14/52

Solution:
Let A be the event of taking a heart.

Then P(A) = 13/52

Let B be the event of taking a queen card

Then P(B) = 4/52

No.of queen heart = 1 => P(AnB) = 1/52.

So probability of taking heart or queen

=p(AuB)= P(A)+P(B)-P(AnB)= 13/52 + 4/52 - 1/52

=16/52 or 4/13.

Q 3. If you have 5 red balls and 3 blue balls in a bag, what is the probability of randomly drawing a red ball?

a) 3/8

b) 5/8

c) 1/2

d) 3/5

Solution:
Correct option is A. 58

Total no.of balls is 5+3=8

No .of red balls is 5

The probabilty is 5/8

Straight Lines

Q 1. Find the slope of the line passing through the points (2, 3) and (4, 7).

a) 2

b) 3

c) 4

d) 5

Solution:
Given points are= A(4, 7) B(2, 3)

The slope = change in y coordinates/change in x coordinates

The formula to find the slope of the line is m = (y ₂ - y ₁ )/ (x ₂ - x ₁ )

Substituting the given points

m = (3 - 7)/(2 - 4)

m = -4/-2

m = 2

Q 2. What is the equation of the line that has a slope of -3 and passes through the point (1, 2)?

a) y=−3x+5

b) y=−3x−1

c) y=−3x+7

d) y=−3x−4

Solution:
The slope of line is 3

The point it passes through is (1,2)

The equation of line is y−2=3(x−1)

y−2=3x−3

3x−y−1=0

​​​​​​y= -3x-1

Q 3. If a line is perpendicular to the line with the equation 2x−3y+6=0, what is the slope of the perpendicular line?

a) 3/2

b) −2/3

c) 2/3

d) −3/2

Solution:
y=2/3x−2

Using the slope-intercept form, the slope is 2/3.

m=2/3

The equation of a perpendicular line must have a slope that is the negative reciprocal of the original slope.

M ^{perpendicular} =−1/ _2/3

Simplify the result.

M ^{perpendicular} =−3 / 2

Also Check - TG EAMCET 2025: Application Form, Exam Dates, Eligibility, Pattern, Preparation

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TG EAMCET 2025 Mathematics Questions with Solutions PDF

Candidates can download the TG EAMCET 2025 Mathematics chapter-wise questions with solutions PDF by clicking on the link below and boost their exam preparation for the better.

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