TS SSC Maths Answer Key 2026 LIVE: Paper Difficulty Level; Detailed Solutions by Subject Expert

10 30 AM IST - 28 Mar'26

Midway Through Exam

The first hour of the exam has passed. Students are concentrating on solving questions. Invigilators report a smooth flow inside all centres.

09 30 AM IST - 28 Mar'26

TG SSC Maths Exam 2026 Exam Begins!

The exam has officially started across Telangana. Students are seated and beginning to attempt the paper. 

09 00 AM IST - 28 Mar'26

Students Gathered Near Exam Centres

Students have started arriving at the exam centres. Long queues and gatherings can be seen outside schools. Security personnel and teachers are managing the entry process efficiently.

08 00 AM IST - 28 Mar'26

Students Leaving Homes

Students are leaving their homes in large numbers. Parents and guardians are accompanying younger students. The city roads are gradually filling up with commuters heading to exam locations.

07 00 AM IST - 28 Mar'26

Preferred Reaching Time

Students are advised to reach their exam centres at least 40 minutes before the exam begins. The exam starts at 9:30 AM sharp, so make sure to arrive by 8:30 or 8:45 AM. 

06 00 AM IST - 28 Mar'26

Morning Traffic Alert

Traffic is picking up near major exam centres as buses, autos, and private vehicles move towards schools. Authorities are ensuring proper flow to avoid delays for students arriving on time.

05 00 AM IST - 28 Mar'26

Good Morning, Students!

The much-awaited day has arrived! Students across Telangana are prepping up for the TG SSC Maths Exam 2026. 

04 00 AM IST - 28 Mar'26

Find the HCF of 24 and 33 by using division algorithm

33 ÷ 24 = 1 remainder 9
24 ÷ 9 = 2 remainder 6
9 ÷ 6 = 1 remainder 3
6 ÷ 3 = 2 remainder 0

HCF of 24 and 33 = 3

03 00 AM IST - 28 Mar'26

Express 360 as a product of prime factors

360 ÷ 2 = 180
180 ÷ 2 = 90
90 ÷ 2 = 45
45 ÷ 3 = 15
15 ÷ 3 = 5

5 is prime.

Prime factorisation of 360:
360 = 2³ × 3² × 5

02 00 AM IST - 28 Mar'26

Essentials Checklist for Tomorrow’s Exam

Make sure you have everything ready for the exam:

• Admit card or hall ticket
• Stationery (pens, pencils, eraser, ruler)
• Water bottle
• Watch or small clock
• Face mask and hand sanitizer (if needed)
• Comfortable clothes and ID proof
• Snacks for break time

01 00 AM IST - 28 Mar'26

What not to do right now?

At this hour, avoid last-minute cramming or stressing about topics. Do not discuss anything with your friends as well. Stay calm, relax, and get ready for tomorrow.

12 00 AM IST - 28 Mar'26

TG SSC Maths Model Question Paper

You can now refer to the TG SSC Maths Model Question Paper here for last-minute practice. It helps in understanding the exam pattern, important question types, and marking scheme ahead of the exam.

11 40 PM IST - 27 Mar'26

Probability Practice Question - 3

From a well shuffled deck of cards if a card is selected randomly, then find the probability of getting

(i) A red coloured king
(ii) A black coloured face card
(iii) A diamond card with number 11 on it
(iv) Queen of clubs

11 20 PM IST - 27 Mar'26

Probability Practice Question - 2

In a bag, there are 5 Red balls, 2 Black balls and 3 White balls. If one ball is selected randomly from the bag, then find the probability of:
(i) getting a Red ball
(ii) getting not a Red ball

11 00 PM IST - 27 Mar'26

Probability Practice Question - 1

Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is:

(i) 8?

(ii) 13?

(iii) less than or equal to 12?

(iv) 5 will not come up either time?

(v) 5 will come up at least once?

10 40 PM IST - 27 Mar'26

Probability Solved Question - 3

Two dice are thrown at the same time. Find the probability that:

(i) the sum of the numbers is 9

(ii) the sum is a multiple of 3

(iii) both numbers are even

(iv) at least one number is 6

(v) the sum is less than 4

(i) Sum = 9

Possible outcomes:
(3,6), (4,5), (5,4), (6,3) → 4 outcomes

Probability = 4/36 = 1/9

(ii) Sum is a multiple of 3

Multiples of 3: 3, 6, 9, 12

Sum = 3 → (1,2), (2,1) → 2
Sum = 6 → (1,5), (2,4), (3,3), (4,2), (5,1) → 5
Sum = 9 → 4 outcomes
Sum = 12 → (6,6) → 1

Total = 2 + 5 + 4 + 1 = 12

Probability = 12/36 = 1/3

(iii) Both numbers are even

Even numbers: 2, 4, 6 → 3 options each

Total outcomes = 3 × 3 = 9

Probability = 9/36 = 1/4

(iv) At least one number is 6

Use complement:
No 6 → 5 × 5 = 25 outcomes

So, at least one 6 = 36 − 25 = 11

Probability = 11/36

(v) Sum is less than 4

Possible sums: 2, 3

Sum = 2 → (1,1) → 1
Sum = 3 → (1,2), (2,1) → 2

Total = 3

Probability = 3/36 = 1/12

10 20 PM IST - 27 Mar'26

Probability Solved Question - 2

A bag contains 6 red balls, 4 blue balls, and 5 green balls. One ball is drawn at random. Find the probability that the ball drawn is:

(i) blue

(ii) not green

Total balls = 6 + 4 + 5 = 15

(i) Probability of blue ball = 4 / 15

(ii) Not green = red + blue = 6 + 4 = 10
Probability = 10 / 15 = 2/3

10 00 PM IST - 27 Mar'26

Probability Solved Question - 1

A card is drawn at random from a well-shuffled deck of 52 playing cards. Find the probability that the card drawn is:

(i) a king

(ii) not a king

Total cards = 52
Number of kings = 4

(i) Probability of getting a king = 4 / 52 = 1/13

(ii) Probability of not getting a king = 1 − 1/13 = 12/13

09 40 PM IST - 27 Mar'26

Statistics Practice Question - 3

The following data shows the marks obtained by students in a test:

Find the mean marks using the assumed mean method.

09 20 PM IST - 27 Mar'26

Statistics Practice Question - 2

Find the Arithmetic mean from the following data.

Class Interval: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70

Frequency: 11, 14, 15, 20, 15, 18, 12

09 00 PM IST - 27 Mar'26

Statistics Practice Question - 1

Find the mode of the following data :

Class Interval: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70, 70-80
Frequency: 7, 14, 13, 12, 20, 11, 15, 8

08 40 PM IST - 27 Mar'26

Statistics Solved Question - 3

The following table shows the distribution of weights of students:

Find the median weight of the students.

Total N = 40

N/2 = 20

Median class = 60–70

l = 60, cf = 14, f = 14, h = 10

Median = 60 + [(20 − 14)/14] × 10
= 60 + (6/14) × 10
= 60 + 4.29
= 64.29

Final Answer: Median ≈ 64.29 kg

08 20 PM IST - 27 Mar'26

Statistics Solved Question - 2

The following data shows the number of absentees in a class during 6 days:
2, 5, 3, 4, 6, 2

Find the mean number of absentees.

Sum = 2 + 5 + 3 + 4 + 6 + 2 = 22
Number of observations = 6

Mean = 22 / 6 = 3.67

Final Answer: Mean = 3.67

08 00 PM IST - 27 Mar'26

Statistics Solved Question - 1

Write the formula to find the median of a grouped data set and explain the terms in it.

Median = l + [(N/2 − cf) / f] × h

Where:
l = lower boundary of median class
N = total frequency
cf = cumulative frequency before median class
f = frequency of median class
h = class width

07 45 PM IST - 27 Mar'26

Part B Solved Questions - 4

Given below are a few more MCQ-type questions for you to practice before the exam:

The probability of getting a number greater than 4 when a die is thrown is:

A. 1/6

B. 1/3

C. 1/2

D. 2/3

Answer: B. ⅓

If the radius of a circle is doubled, then its area becomes:

A. Double

B. Four times

C. Half

D. Same

Answer: B. Four times

The number of zeroes of the polynomial represented by a straight line parallel to the x-axis and not touching it is:

A. 0

B. 1

C. 2

D. Infinite

Answer: A. 0

07 30 PM IST - 27 Mar'26

Part B Solved Questions - 3

Check below some more solved questions for your prep:

If α and β are the roots of x² - 7x + 10 = 0, then αβ =

A. 7

B. 10

C. -7

D. -10

Answer: B. 10

The value of sin²30° + cos²30° is:

A. 0

B. 1

C. ½

D. 2

Answer: B. 1

The coordinates of the midpoint of (2, 4) and (6, 8) are:

A. (4, 6)

B. (3, 5)

C. (5, 6)

D. (4, 5)

Answer: A. (4, 6)

07 15 PM IST - 27 Mar'26

Part B Solved Questions - 2

Some more 1-mark solved objective-type questions are given below:

07 00 PM IST - 27 Mar'26

Part B Solved Questions - 1

Some solved objective-type questions are given below, which usually show up in Part B of the maths exam:

06 45 PM IST - 27 Mar'26

Mensuration Practice Question - 4

A conical tent is 7 m high and has a base radius of 24 m. Find the length of the canvas required to make the tent (excluding the base).

06 30 PM IST - 27 Mar'26

Mensuration Practice Question - 3

A toy is in the form of a cone mounted on a hemisphere of the same diameter. The diameter of the base and the height of the cone are 6 cm and 4 cm respectively. Determine the surface area of the toy. 

06 15 PM IST - 27 Mar'26

Mensuration Practice Question - 2

Due to heavy floods in the state thousands were rendered homeless. The State Government decided to provide canvas tents for temporary shelter. The lower part of each tent is cylindrical of base radius 2.8 metres and height 3.5 metres with conical upper part. The radius of the base of the cone is same as that of the cylinder. The height of the conical portion is 2.1 metres. If the canvas used to make the tent costs Rs 100 per square metre, find the total cost of canvas used to make the tent.

06 00 PM IST - 27 Mar'26

Mensuration Practice Question - 1

A metallic vessel is in the shape of a cylinder surmounted over a hemisphere. The radii of cylinder and hemisphere are same and the height of the cylindrical part is 10 cm. If the outer surface area of the vessel is 748 cm², then find their radii.

05 45 PM IST - 27 Mar'26

Mensuration Solved Question - 4

A cylindrical vessel of radius 7 cm contains water up to a height of 10 cm. If a solid sphere of radius 3.5 cm is immersed completely, find the rise in the water level. 

Volume of sphere = (4/3)πr³
= (4/3)π(3.5)³

Rise in water level = h

Volume displaced = volume of cylinder rise

⇒ (4/3)π(3.5)³ = π(7)² × h

Cancel π:
⇒ (4/3)(3.5)³ = 49h

⇒ (4/3)(42.875) = 49h
⇒ 171.5/3 = 49h
⇒ 57.17 ≈ 49h

⇒ h = 57.17 / 49 ≈ 1.17 cm

Final Answer: Rise in water level ≈ 1.17 cm

05 30 PM IST - 27 Mar'26

Mensuration Solved Question - 3

A solid sphere is melted and recast into smaller spheres of equal radius. If the radius of each small sphere is half of the original, how many small spheres are formed?

Volume of sphere = (4/3)πr³

Let original radius = r
New radius = r/2

Number of spheres = (Volume of big sphere) / (Volume of small sphere)

⇒ = r³ / (r/2)³
⇒ = r³ / (r³/8)
⇒ = 8

Final Answer: Number of spheres = 8

05 15 PM IST - 27 Mar'26

Mensuration Solved Question - 2

The radius of a cylinder is doubled while its height remains the same. How does its curved surface area change?

CSA of cylinder = 2πrh

New radius = 2r

New CSA = 2π(2r)h = 4πrh

Original CSA = 2πrh

Ratio = 4πrh : 2πrh = 2 : 1

Final Answer: Curved surface area becomes twice (double)

05 00 PM IST - 27 Mar'26

Mensuration Solved Question - 1

The ratio of radius and slant height of a Right circular cone is 7:25. If its curved surface area is 550 cm², then find its radius.

Let radius = 7x, slant height = 25x

Curved Surface Area (CSA) = πrl

⇒ 550 = π × (7x) × (25x)
⇒ 550 = π × 175x²

Using π = 22/7:
⇒ 550 = (22/7) × 175x²
⇒ 550 = 22 × 25x²
⇒ 550 = 550x²

⇒ x² = 1 ⇒ x = 1

Radius = 7x = 7 cm

Final Answer: Radius = 7 cm

04 45 PM IST - 27 Mar'26

Trigonometry Practice Question - 4

Prove that

sinθ / (1 + cosθ) + (1 + cosθ) / sinθ = 2 cosecθ

04 30 PM IST - 27 Mar'26

Trigonometry Practice Question - 3

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

04 15 PM IST - 27 Mar'26

Trigonometry Practice Question - 2

Practice the question given below:

04 00 PM IST - 27 Mar'26

Trigonometry Practice Question - 1

“An observer standing at a distance of 50 metres from the foot of a tower observes its top at an angle of elevation of 45°.” Draw a suitable diagram for this situation.

03 45 PM IST - 27 Mar'26

Trigonometry Solved Question - 4

Here is the solution of the question:

03 30 PM IST - 27 Mar'26

Trigonometry Solved Question - 3

Is it right to say that Cos(60° + 30°) = Cos 60° Cos 30° - Sin 60°. Sin 30°

Using identity:
cos(A + B) = cos A cos B − sin A sin B

LHS:
cos(60° + 30°) = cos 90° = 0

RHS:
cos 60° cos 30° − sin 60° sin 30°
= (1/2)(√3/2) − (√3/2)(1/2)
= √3/4 − √3/4 = 0

LHS = RHS

Final Answer: Yes, the statement is correct.

03 15 PM IST - 27 Mar'26

Trigonometry Solved Question - 2

Express ‘sin θ’ in terms of ‘tan θ’.

We know:
tan θ = sin θ / cos θ

Also,
sin²θ + cos²θ = 1

Divide by cos²θ:
tan²θ + 1 = 1 / cos²θ

⇒ cos θ = 1 / √(1 + tan²θ)

Now,
sin θ = tan θ × cos θ

⇒ sin θ = tan θ / √(1 + tan²θ)

Final Answer: sin θ = tan θ / √(1 + tan²θ)

03 00 PM IST - 27 Mar'26

Trigonometry Solved Question - 1

A flagpole stands vertically on the ground. From a point which is 15 metres away, the angle of elevation of the top of the tower is 45°. Draw a suitable diagram for the given data.

Let:
AB = height of the flagpole
BC = distance from the pole = 15 m
AC = line of sight

∠ACB = 45° (angle of elevation)

Diagram description:

• Draw a vertical line AB (flagpole)
• Draw a horizontal line BC (ground)
• Join A to C
• Mark ∠ACB = 45°
• Angle at B is 90° (vertical pole on ground)

Final Answer: A right-angled triangle with vertical height AB, horizontal base BC = 15 m, and angle of elevation at point C = 45°.

02 45 PM IST - 27 Mar'26

Trigonometry Heights & Distances (Basic Relation)

Main Formula

tan θ = Height / Distance

Using sin and cos (when needed)

sin θ = Height / Slant height

cos θ = Base / Slant height

02 30 PM IST - 27 Mar'26

Trigonometry Important Formulas - 2

Pythagorean Identities

sin²θ + cos²θ = 1

1 + tan²θ = sec²θ

1 + cot²θ = cosec²θ

Complementary Angles

sin(90° − θ) = cos θ

cos(90° − θ) = sin θ

tan(90° − θ) = cot θ

02 15 PM IST - 27 Mar'26

Trigonometry Important Formulas - 1

Basic Trigonometric Ratios (Right Triangle)
sin θ = Perpendicular / Hypotenuse
cos θ = Base / Hypotenuse
tan θ = Perpendicular / Base

Reciprocal Identities
sin θ = 1 / cosec θ
cos θ = 1 / sec θ
tan θ = 1 / cot θ

02 00 PM IST - 27 Mar'26

Trigonometry Values at Standard Angles

The respective values at angles are given below:

01 45 PM IST - 27 Mar'26

Geometry Practice Question - 4

In ΔABC, DE ∥ BC.

If AD = x - 2, DB = 5, AE = 2x - 1, EC = 2x + 5, then find the value of x.

01 30 PM IST - 27 Mar'26

Geometry Practice Question - 3

Draw a circle of radius 3 cm. Construct a pair of tangents to the circle from an external point that is at a distance of 8 cm from the centre of the circle.

01 15 PM IST - 27 Mar'26

Geometry Practice Question - 2

Construct a triangle ABC with AB = 5.6 cm, BC = 7.2 cm and CA = 4.8 cm. Construct another triangle similar to ΔABC whose sides are 5/3 times the corresponding sides of ΔABC.

01 00 PM IST - 27 Mar'26

Geometry Practice Question - 1

Show that the triangle with vertices A(-4, 2), B(2, -4) and C(12, 6) forms a Right angled triangle.

12 45 PM IST - 27 Mar'26

Geometry Solved Question - 4

In a triangle ABC, D is a point on BC such that AD is perpendicular to BC. If BD = DC, what can you say about triangle ABC?

Given:
AD ⟂ BC
BD = DC

So, D is the midpoint of BC

Since AD is perpendicular and passes through midpoint,
AD is the perpendicular bisector of BC

Therefore,
AB = AC

So, triangle ABC is isosceles

Final Answer: Triangle ABC is an isosceles triangle (AB = AC)

12 30 PM IST - 27 Mar'26

Geometry Solved Question - 3

In a circle of radius 7 cm, find the length of a chord that is at a distance of 5 cm from the centre.

Radius (r) = 7 cm
Distance from centre to chord (d) = 5 cm

Perpendicular from centre bisects the chord

Using Pythagoras:
Half chord length = √(r² − d²)
= √(7² − 5²)
= √(49 − 25)
= √24 = 2√6

Full chord length = 2 × 2√6 = 4√6 cm

Final Answer: Length of chord = 4√6 cm

12 15 PM IST - 27 Mar'26

Geometry Solved Question - 2

Determine x so that 2 is the slope of the line through P (2, 5) and Q (x, 3).

Slope formula:
m = (y₂ − y₁) / (x₂ − x₁)

Given slope = 2

⇒ 2 = (3 − 5) / (x − 2)
⇒ 2 = (−2) / (x − 2)

Cross multiply:
2(x − 2) = −2
⇒ 2x − 4 = −2
⇒ 2x = 2
⇒ x = 1

Final Answer: x = 1

11 45 AM IST - 27 Mar'26

Algebra Practice Question - 4

If x² + y² = 27xy, then show that 2 log (x - y) = 2 log 5 + log x + log y.

11 30 AM IST - 27 Mar'26

Algebra Practice Question - 3

Draw the graph of the quadratic polynomial y = x² − 7x − 12 and find the zeroes of the polynomial from the graph.

11 15 AM IST - 27 Mar'26

Algebra Practice Question - 2

In an arithmetic progression, if 4 times the fourth term is equal to 8 times the eighth term, then prove that the twelfth term of the progression is zero.

11 00 AM IST - 27 Mar'26

Algebra Practice Question - 1

Construct a Quadratic equation having the roots log₁₀ 8 and log₁₀ 100.

10 45 AM IST - 27 Mar'26

Algebra Solved Question - 4

Is the pair of linear equations 3x − 5y + 7 = 0 and 6x − 10y + 13 = 0 inconsistent? Justify your answer.

Equation 1: 3x − 5y + 7 = 0
Equation 2: 6x − 10y + 13 = 0

Compare ratios:

a₁/a₂ = 3/6 = 1/2
b₁/b₂ = (−5)/(−10) = 1/2
c₁/c₂ = 7/13

Since:
a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Therefore, lines are parallel and do not intersect.

Conclusion: The system is inconsistent (no solution). 

10 30 AM IST - 27 Mar'26

Algebra Solved Question - 3

If the pair of linear equations :

(3k + 1)x + 3y - 2 = 0

and

(k² + 1)x + (k - 2)y - 5 = 0 has no solutions, then find the value of k.

If the pair of linear equations:
(3k + 1)x + 3y − 2 = 0
(k² + 1)x + (k − 2)y − 5 = 0
has no solutions, find the value of k.

Solution:
For no solution:
a₁/a₂ = b₁/b₂ ≠ c₁/c₂

So,
(3k + 1)/(k² + 1) = 3/(k − 2)

Cross multiply:
(3k + 1)(k − 2) = 3(k² + 1)

⇒ 3k² − 6k + k − 2 = 3k² + 3
⇒ 3k² − 5k − 2 = 3k² + 3

Cancel 3k²:
−5k − 2 = 3

⇒ −5k = 5
⇒ k = −1

Check condition:
c₁/c₂ = −2/−5 = 2/5

a₁/a₂ = −2/2 = −1

So, a₁/a₂ ≠ c₁/c₂ 

Final Answer: k = −1

10 15 AM IST - 27 Mar'26

Algebra Solved Question - 2

In an arithmetic progression, the first term is 1, the last term is 20, and the sum of all the terms is 399; then find the number of terms in the progression.

First term (a) = 1
Last term (l) = 20
Sum (Sₙ) = 399

Formula:
Sₙ = n/2 (a + l)

⇒ 399 = n/2 (1 + 20)
⇒ 399 = n/2 × 21
⇒ 399 = 21n/2

⇒ 798 = 21n
⇒ n = 798 / 21 = 38

Final Answer: Number of terms = 38

10 00 AM IST - 27 Mar'26

Algebra Solved Question - 1

Write a Quadratic equation whose roots are the values of sin 30° and cos 60°.

sin 30° = 1/2
cos 60° = 1/2

So, roots are: α = 1/2, β = 1/2

Sum of roots = α + β = 1/2 + 1/2 = 1
Product of roots = αβ = (1/2)(1/2) = 1/4

Quadratic equation:
x² − (sum)x + product = 0

⇒ x² − x + 1/4 = 0

Multiplying by 4 to remove fraction:
4x² − 4x + 1 = 0

Final Answer: 4x² − 4x + 1 = 0

09 45 AM IST - 27 Mar'26

Algebra Important Formulas - 4

Laws of Logarithms
log(ab) = log a + log b
log(a/b) = log a − log b
log(aⁿ) = n log a

Relation from Given Roots
If roots are α, β, then the equation is:
x² − (α + β)x + αβ = 0

09 30 AM IST - 27 Mar'26

Algebra Important Formulas - 3

Sum of n Terms of AP
Sₙ = n/2 [2a + (n − 1)d]
or
Sₙ = n/2 (a + l)

Basic Identities
(a + b)² = a² + 2ab + b²
(a − b)² = a² − 2ab + b²
(a² − b²) = (a − b)(a + b)

Factorisation Identity
x² − (α + β)x + αβ = 0

09 15 AM IST - 27 Mar'26

Algebra Important Formulas - 2

Sum and Product of Roots
For ax² + bx + c = 0
α + β = −b/a
αβ = c/a

Standard Form of Quadratic Equation
ax² + bx + c = 0 (a ≠ 0)

nth Term of AP
aₙ = a + (n − 1)d

09 00 AM IST - 27 Mar'26

Algebra Important Formulas - 1

Quadratic Formula
x = (−b ± √(b² − 4ac)) / 2a

Nature of Roots (Discriminant)
D = b² − 4ac
If D > 0 → Real & distinct
If D = 0 → Real & equal
If D < 0 → Imaginary

08 45 AM IST - 27 Mar'26

Presentation Hacks for Extra Marks

• Even if your final answer is wrong, you can get 80-90% of the marks if your steps are logical. Write the formula used in a box on the right side of the step.
• For every long answer, start with:
  • Given: (List the values provided in the question)
  • To Find: (What is the objective?)
  • Formula: (State it clearly)
• Use a sharp pencil for Geometry diagrams.
• Label your axes in Graphs and always mention the scale.
• Underline the final answer with a double line and include the Units.

08 30 AM IST - 27 Mar'26

Topper's Attempt Strategy

After talking to multiple toppers of the exam, we found a very similar and interesting pattern in how they attempted the paper. 

1. Use the initial reading time to categorise questions:
   • Type A (Instant): Questions you know by heart.
   • Type B (Needs Thought): Questions where you know the method but need careful calculation.
   • Type C (Tricky): Higher-order thinking questions.
2. Start with Section III (6-mark questions) while your mind is fresh and your energy is high. These carry the most weight. Then move to Section II, Section I, and finally the Objective Bits (Part B).
3. Aim to finish the paper 20 minutes before the bell. Use this time only to recalculate values and check if you copied the numbers correctly from the question paper.

08 15 AM IST - 27 Mar'26

Expert Preparation Tips for 2026

To move from a standard score to a 100/100, you can follow these prep tips.

• Create a one-page cheat sheet for every chapter. For Mensuration, group formulas by shape (Cylinder, Cone, Sphere) to avoid confusion between Total Surface Area (TSA) and Lateral Surface Area (LSA).
• Solve the last 5 years of TS SSC papers. You’ll notice that Statistics (Mean, Median, Mode) and Probability follow very predictable patterns.

08 12 AM IST - 27 Mar'26

Geometry Solved Question - 1

AOB is the diameter of a circle with centre ‘O’ and AC is a tangent to the circle at A. If ∠BOC = 130°, then find ∠ACO.

Since AC is a tangent at A, radius OA ⟂ AC
⇒ ∠OAC = 90°

In triangle AOC:
OA = OC (radii of the circle)
So, triangle AOC is isosceles

Given: ∠BOC = 130°
Since AOB is a straight line,
∠AOC = 180° − 130° = 50°

In triangle AOC:
∠OAC + ∠ACO + ∠AOC = 180°

⇒ 90° + ∠ACO + 50° = 180°
⇒ ∠ACO = 180° − 140° = 40°

Final Answer: ∠ACO = 40°

08 00 AM IST - 27 Mar'26

Chapter-Wise Expected Weightage

The expected chapter-wise weightage for the TS Class 10th maths exam is given below:

07 45 AM IST - 27 Mar'26

TS SSC Maths Exam 2026 MS & Blueprint

The marking scheme for the maths exam tomorrow will be:

07 30 AM IST - 27 Mar'26

Exam Pattern Details

The exam will be conducted for 80 marks and will be divided into Part A and Part B. The question paper will include objective questions, short answers, and long answer type questions.

07 15 AM IST - 27 Mar'26

Admit Card Reminder!

Students must carry their SSC board exam hall ticket 2026 to the exam centre without fail. Make sure to check your exam centre details in advance and keep your admit card ready.

07 00 AM IST - 27 Mar'26

TG SSC Maths Exam 2026 Date & Timings

The maths exam will be held tomorrow, March 28, 2026. The exam will take place in a single morning shift from 9:30 AM to 12:30 PM, with an extra 15 minutes given to read the question paper.