TS SSC Mathematics 2026 Question Paper with Solutions PDF

Step 1: Understanding the Concept:

The expression \( \log_{b} a = x \) is equivalent to the exponential form \( b^x = a \). To solve this, we express both the base and the number in terms of a common base, which is 2.


Step 2: Key Formula or Approach:

Use the property \( \log_{b} a = x \Rightarrow b^x = a \).


Step 3: Detailed Explanation:

Let \( \log_{\sqrt{2}} 64 = x \).
Converting to exponential form: \[ (\sqrt{2})^x = 64 \]
Since \( \sqrt{2} = 2^{1/2} \) and \( 64 = 2^6 \), we can write: \[ (2^{1/2})^x = 2^6 \] \[ 2^{x/2} = 2^6 \]
Equating the exponents since the bases are equal: \[ \frac{x}{2} = 6 \] \[ x = 12 \]


Step 4: Final Answer:

The value of \( \log_{\sqrt{2}} 64 \) is 12. Quick Tip: Always convert roots to fractional exponents (e.g., \( \sqrt{x} = x^{1/2} \)) to make equating bases easier in logarithmic equations.

Step 1: Understanding the Concept:

A system of linear equations is consistent if it has at least one solution. If the lines are parallel (same slope but different intercepts), they never intersect, meaning the system is inconsistent.


Step 2: Key Formula or Approach:

For two equations \( a_1x + b_1y = c_1 \) and \( a_2x + b_2y = c_2 \), the condition for inconsistency is: \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]


Step 3: Detailed Explanation:

Given equations: \( 2x - 3y = 8 \) and \( 4x - 6y = 11 \).
Here, \( a_1 = 2, b_1 = -3, c_1 = 8 \) and \( a_2 = 4, b_2 = -6, c_2 = 11 \).
Calculating the ratios: \[ \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} \] \[ \frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2} \] \[ \frac{c_1}{c_2} = \frac{8}{11} \]
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the lines are parallel.


Step 4: Final Answer:

The pair of equations is inconsistent. Quick Tip: Graphically, inconsistent equations represent parallel lines. If the x and y coefficients are proportional but the constants are not, the lines will never meet.

Step 1: Understanding the Concept:

This is a right-angled triangle problem where the tree represents the perpendicular, the ground represents the base, and the line of sight represents the hypotenuse.


Step 2: Detailed Explanation:

To draw the diagram:
1. Draw a vertical line segment \( AB \) representing the tree, with \( B \) as the foot.
2. Draw a horizontal line segment \( BC \) representing the distance on the ground, where \( BC = 30 m \).
3. Draw a line segment \( AC \) representing the line of sight from the observer at \( C \) to the top of the tree at \( A \).
4. Mark the angle of elevation \( \angle ACB = 60^\circ \).
5. Mark the right angle \( \angle ABC = 90^\circ \).


Step 3: Final Answer:

The diagram is a right-angled triangle \( ABC \) where \( BC = 30 m \) and \( \angle C = 60^\circ \). Quick Tip: The angle of elevation is always measured from the horizontal ground looking upwards to the object.

Step 1: Understanding the Concept:

We can express any trigonometric ratio in terms of another by using fundamental identities like the quotient identity and the Pythagorean identity.


Step 2: Key Formula or Approach:

Quotient identity: \( \cot \theta = \frac{\cos \theta}{\sin \theta} \)
Pythagorean identity: \( \sin^2 \theta + \cos^2 \theta = 1 \Rightarrow \cos \theta = \sqrt{1 - \sin^2 \theta} \)


Step 3: Detailed Explanation:

Start with the definition: \[ \cot \theta = \frac{\cos \theta}{\sin \theta} \]
Substitute \( \cos \theta \) with \( \sqrt{1 - \sin^2 \theta} \): \[ \cot \theta = \frac{\sqrt{1 - \sin^2 \theta}}{\sin \theta} \]


Step 4: Final Answer:

The expression is \( \cot \theta = \frac{\sqrt{1 - \sin^2 \theta}}{\sin \theta} \). Quick Tip: If \( \theta \) is in a specific quadrant, remember to consider the positive or negative sign of the square root based on the ASTC rule.

Step 1: Understanding the Concept:

According to the Basic Proportionality Theorem (Thales' Theorem), if a line is drawn parallel to one side of a triangle to intersect the other two sides, then the other two sides are divided in the same ratio.


Step 2: Key Formula or Approach:
\[ \frac{AD}{DB} = \frac{AE}{EC} \]


Step 3: Detailed Explanation:

Substitute the given values into the ratio: \[ \frac{x}{x^2 - 1} = \frac{x + 2}{x} \]
Cross-multiply: \[ x^2 = (x + 2)(x^2 - 1) \] \[ x^2 = x^3 - x + 2x^2 - 2 \]
Rearrange the equation: \[ x^3 + x^2 - x - 2 = 0 \]
By testing values (Trial and Error), let \( x = 2 \): \[ (2)^3 + (2)^2 - (2) - 2 = 8 + 4 - 2 - 2 = 8 \neq 0 \]
(Note: If the problem intended \( DB = x-1 \), then \( x^2 = x^2+x-2 \Rightarrow x=2 \). Given the specific cubic equation resulting from \( x^2-1 \), \( x \approx 1.2 \) is a root, but typically in these textbook problems, a simpler linear expression for \( DB \) is intended).


Step 4: Final Answer:

The value of \( x \) is found by solving the BPT ratio \( \frac{x}{x^2-1} = \frac{x+2}{x} \). Quick Tip: Always ensure your final value for \( x \) makes all side lengths positive. Length cannot be zero or negative.

Step 1: Understanding the Concept:

In probability, a "Sure event" (or certain event) is an event that is guaranteed to happen, having a probability of 1. An "Impossible event" is an event that can never occur, having a probability of 0.


Step 2: Detailed Explanation:

1. Sure Event: Consider rolling a standard six-sided die once. The event "getting a number less than 7" is a sure event because every possible outcome (1, 2, 3, 4, 5, 6) satisfies this condition.

2. Impossible Event: Consider the same six-sided die. The event "getting the number 8" is an impossible event because the maximum value on the die is 6. There is no outcome in the sample space that matches 8.


Step 3: Final Answer:

An example of a Sure event is rolling a number between 1 and 6 on a standard die, and an example of an Impossible event is rolling a 7 on the same die. Quick Tip: Probability (\(P\)) always ranges from \(0 \le P \le 1\). If \(P(E) = 1\), the event is sure. If \(P(E) = 0\), the event is impossible.

Step 1: Understanding the Concept:

Venn diagrams represent sets geometrically. Operations like union (\( \cup \)), intersection (\( \cap \)), and difference (\( - \)) correspond to specific regions: the total area, the overlap, and the unique areas of the circles respectively.


Step 2: Detailed Explanation:

Based on the visual regions of a standard Venn diagram:

1. \( A \cup B \): Identify all elements present in circle A, circle B, or both. This represents the total collection of unique items.

2. \( A \cap B \): Identify the elements located in the overlapping "football" shaped region. These are elements common to both sets.

3. \( A - B \): Identify the elements in circle A that are NOT in the intersection. This is the "A-only" region.

4. \( B - A \): Identify the elements in circle B that are NOT in the intersection. This is the "B-only" region.


Step 3: Final Answer:

The sets are derived by listing the elements found in the specific shaded or bounded regions as defined by set algebra. Quick Tip: To remember set difference, think of \( A - B \) as "Subtracting B from A." You simply take the whole circle A and cut out anything that belongs to B.

Step 1: Understanding the Concept:

An Arithmetic Progression (A.P.) is a sequence where each term is obtained by adding a fixed number (common difference) to the preceding term. The general term formula calculates any value in the sequence based on its position.


Step 2: Key Formula or Approach:

The formula for the \( n^{th} \) term is: \[ a_n = a + (n - 1)d \]


Step 3: Detailed Explanation:

The variables in the formula represent the following:

1. \( a_n \): The value of the term at the \( n^{th} \) position.

2. \( a \): The starting value or the first term of the progression.

3. \( n \): The position of the term (e.g., for the 10th term, \( n = 10 \)).

4. \( d \): The common difference between any two consecutive terms.


Step 4: Final Answer:

The \( n^{th} \) term of an A.P. is given by \( a_n = a + (n - 1)d \). Quick Tip: The reason we use \( (n-1) \) is that for the first term, we have added the difference zero times. For the second term, we add it once, for the third twice, and so on.

Step 1: Understanding the Concept:

Probability measures the likelihood of an event. It is calculated as the ratio of favorable outcomes to the total number of equally likely outcomes in the sample space.


Step 2: Key Formula or Approach:
\[ P(E) = \frac{n(E)}{n(S)} \]


Step 3: Detailed Explanation:

Total number of balls in the bag: \( 5 + 3 + 2 = 10 \).

1. Probability of a red ball: There are 5 red balls. \[ P(red) = \frac{5}{10} = \frac{1}{2} \]
2. Probability of not getting a white ball: This includes red and black balls. Total favorable outcomes = \( 5 + 3 = 8 \). \[ P(not white) = \frac{8}{10} = \frac{4}{5} \]


Step 4: Final Answer:

The probability of a red ball is \( 1/2 \) and the probability of not getting a white ball is \( 4/5 \). Quick Tip: "Not getting a white ball" is the same as \( 1 - P(white) \). Since \( P(white) = 2/10 = 1/5 \), the result is \( 1 - 1/5 = 4/5 \).

Step 1: Understanding the Concept:

This identity is proven by rationalizing the square root expression and using the reciprocal property of the terms.


Step 2: Detailed Explanation:

First, rationalize \( P \): \[ P = \sqrt{\frac{1 + \sin \theta}{1 - \sin \theta} \times \frac{1 + \sin \theta}{1 + \sin \theta}} = \sqrt{\frac{(1 + \sin \theta)^2}{1 - \sin^2 \theta}} = \frac{1 + \sin \theta}{\cos \theta} = \sec \theta + \tan \theta \]
Since \( \frac{1}{P} \) is the reciprocal: \[ \frac{1}{P} = \frac{\cos \theta}{1 + \sin \theta} = \frac{\cos \theta(1 - \sin \theta)}{1 - \sin^2 \theta} = \frac{\cos \theta(1 - \sin \theta)}{\cos^2 \theta} = \sec \theta - \tan \theta \]
Adding \( P \) and \( \frac{1}{P} \): \[ (\sec \theta + \tan \theta) + (\sec \theta - \tan \theta) = 2 \sec \theta \]


Step 3: Final Answer:

The sum \( P + \frac{1}{P} \) simplifies to \( 2 \sec \theta \). Quick Tip: Whenever you see \( \sqrt{1 \pm \sin \theta} \), rationalizing often removes the square root and simplifies the term into basic secant and tangent functions.

Step 1: Understanding the Concept:

The relationship between roots and coefficients (\( \alpha + \beta = -b/a \) and \( \alpha\beta = c/a \)) allows us to derive conditions for specific root ratios.


Step 2: Detailed Explanation:

Let the roots be \( \alpha \) and \( 2\alpha \).

1. Sum of roots: \( \alpha + 2\alpha = 3\alpha = -\frac{b}{a} \Rightarrow \alpha = -\frac{b}{3a} \).

2. Product of roots: \( \alpha \cdot 2\alpha = 2\alpha^2 = \frac{c}{a} \).

Substitute the value of \( \alpha \) into the product equation: \[ 2 \left( -\frac{b}{3a} \right)^2 = \frac{c}{a} \] \[ 2 \left( \frac{b^2}{9a^2} \right) = \frac{c}{a} \] \[ \frac{2b^2}{9a^2} = \frac{c}{a} \Rightarrow 2b^2 = 9ac \]


Step 3: Final Answer:

If one root is twice the other, the condition \( 2b^2 = 9ac \) is satisfied. Quick Tip: For any ratio \( m:n \), the condition is \( (m+n)^2 ac = mnb^2 \). Here, \( m=1, n=2 \), so \( (1+2)^2 ac = (1)(2)b^2 \Rightarrow 9ac = 2b^2 \).

Step 1: Understanding the Concept:

When liquid is transferred, the total volume remains constant. The number of containers is found by dividing the total volume of the tank by the volume of a single container.


Step 2: Key Formula or Approach:

Volume of Hemisphere = \( \frac{2}{3}\pi r^3 \); Volume of Cylinder = \( \pi r^2 h \).


Step 3: Detailed Explanation:

1. Volume of the hemispherical tank (\( r = 3 \)): \[ V_H = \frac{2}{3} \pi (3)^3 = \frac{2}{3} \pi (27) = 18\pi m^3 \]
2. Volume of one cylindrical container (\( r = 1, h = 2 \)): \[ V_C = \pi (1)^2 (2) = 2\pi m^3 \]
3. Number of cylinders: \[ n = \frac{V_H}{V_C} = \frac{18\pi}{2\pi} = 9 \]


Step 4: Final Answer:

The total number of cylinders needed is 9. Quick Tip: Keep \( \pi \) as a symbol during your calculations. It will almost always cancel out in problems involving ratios of volumes, saving you tedious multiplication.

Step 1: Understanding the Concept:

To prove a number is irrational, we typically use the method of contradiction. We assume the number is rational (expressible as \( p/q \)) and show that this lead to a logical inconsistency, such as equating a rational number to a known irrational number like \( \sqrt{5} \) or \( \sqrt{7} \).


Step 2: Detailed Explanation:

Assume \( 3\sqrt{5} + 2\sqrt{7} = r \), where \( r \) is a rational number.

Isolate one radical: \( 3\sqrt{5} = r - 2\sqrt{7} \).

Squaring both sides: \[ (3\sqrt{5})^2 = (r - 2\sqrt{7})^2 \] \[ 45 = r^2 - 4r\sqrt{7} + 28 \]
Rearrange to isolate the radical \( \sqrt{7} \): \[ 4r\sqrt{7} = r^2 + 28 - 45 \] \[ 4r\sqrt{7} = r^2 - 17 \] \[ \sqrt{7} = \frac{r^2 - 17}{4r} \]
Since \( r \) is rational, \( \frac{r^2 - 17}{4r} \) must also be rational. However, \( \sqrt{7} \) is a known irrational number. A rational number cannot equal an irrational number.


Step 3: Final Answer:

Our assumption that the number is rational is false. Therefore, \( 3\sqrt{5} + 2\sqrt{7} \) is irrational. Quick Tip: When squaring an expression like \( (a-b)^2 \), don't forget the middle term \( -2ab \). This term is what allows you to isolate the radical.

Step 1: Understanding the Concept:

The graph of a quadratic polynomial is a parabola. The "zeroes" of the polynomial are the x-coordinates of the points where the graph intersects the x-axis (where \( y = 0 \)).


Step 2: Detailed Explanation:

To draw the graph, calculate points \( (x, y) \):

- If \( x = -2, y = (-2)^2 - 3(-2) - 4 = 4 + 6 - 4 = 6 \)

- If \( x = -1, y = (-1)^2 - 3(-1) - 4 = 1 + 3 - 4 = 0 \)

- If \( x = 0, y = (0)^2 - 3(0) - 4 = -4 \)

- If \( x = 1, y = (1)^2 - 3(1) - 4 = -6 \)

- If \( x = 2, y = (2)^2 - 3(2) - 4 = -6 \)

- If \( x = 4, y = (4)^2 - 3(4) - 4 = 16 - 12 - 4 = 0 \)

Plotting these points results in a U-shaped parabola opening upwards. The curve crosses the x-axis at \( (-1, 0) \) and \( (4, 0) \).


Step 3: Final Answer:

From the graph, the zeroes of the polynomial are \( -1 \) and \( 4 \). Quick Tip: The vertex of a parabola \( ax^2 + bx + c \) is always at \( x = -b/2a \). For this graph, the vertex is at \( x = 1.5 \), which helps in making the drawing symmetrical.

Step 1: Understanding the Concept:

Trisection means dividing a line segment into three equal parts. This requires two points, \( P \) and \( Q \). Point \( P \) divides the segment in a \( 1:2 \) ratio, and point \( Q \) divides it in a \( 2:1 \) ratio.


Step 2: Key Formula or Approach:

Section Formula: \( \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) \).


Step 3: Detailed Explanation:

Let \( A = (-4, 3) \) and \( B = (5, 9) \).

1. Point P (1:2 ratio): \[ x = \frac{1(5) + 2(-4)}{1+2} = \frac{5 - 8}{3} = -1 \] \[ y = \frac{1(9) + 2(3)}{1+2} = \frac{9 + 6}{3} = 5 \]
So, \( P = (-1, 5) \).

2. Point Q (2:1 ratio): \[ x = \frac{2(5) + 1(-4)}{2+1} = \frac{10 - 4}{3} = 2 \] \[ y = \frac{2(9) + 1(3)}{2+1} = \frac{18 + 3}{3} = 7 \]
So, \( Q = (2, 7) \).


Step 4: Final Answer:

The points of trisection are \( (-1, 5) \) and \( (2, 7) \). Quick Tip: Once you find the first trisection point \( P \), the second point \( Q \) is actually the midpoint of \( PB \). This can save you from doing the full section formula twice!

Step 1: Understanding the Concept:

To find the mean of grouped data, we first identify the class mark (\(x_i\)) for each interval, which is the average of the lower and upper limits. Then, we use the Direct Method formula: \( Mean (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i} \).


Step 2: Detailed Explanation:

Let's calculate the class marks (\(x_i\)) and the products (\(f_i x_i\)):

0-10: \(x_i = 5, f_i x_i = 3 \times 5 = 15\)
10-20: \(x_i = 15, f_i x_i = 6 \times 15 = 90\)
20-30: \(x_i = 25, f_i x_i = 7 \times 25 = 175\)
30-40: \(x_i = 35, f_i x_i = 12 \times 35 = 420\)
40-50: \(x_i = 45, f_i x_i = 9 \times 45 = 405\)
50-60: \(x_i = 55, f_i x_i = 5 \times 55 = 275\)
60-70: \(x_i = 65, f_i x_i = 6 \times 65 = 390\)
70-80: \(x_i = 75, f_i x_i = 2 \times 75 = 150\)

Sum of frequencies (\(\sum f_i\)) = \(3+6+7+12+9+5+6+2 = 50\).

Sum of products (\(\sum f_i x_i\)) = \(15+90+175+420+405+275+390+150 = 1920\).
\[ Mean = \frac{1920}{50} = 38.4 \]


Step 3: Final Answer:

The mean of the given data is 38.4. Quick Tip: If the numbers for \(x_i\) are large, you can use the Assumed Mean Method (\(A + \frac{\sum f_i d_i}{\sum f_i}\)) to make the calculations simpler and faster.

Step 1: Understanding the Concept:

To construct a similar triangle with a scale factor \( k > 1 \), the new triangle will be larger than the original. We use a ray and equal divisions to project the larger sides.


Step 2: Detailed Explanation:

1. Base Triangle: Draw \( \triangle ABC \) with \( AB = 8 cm \), \( BC = 6 cm \), and \( AC = 5 cm \).

2. Ray: Draw a ray \( AX \) making an acute angle with \( AB \).

3. Divisions: Mark 5 equal points (\( A_1 \) to \( A_5 \)) on \( AX \) (since 5 is the larger number in the ratio \( 5/3 \)).

4. Join and Parallel: Join \( A_3 \) (the denominator) to \( B \). Draw a line through \( A_5 \) parallel to \( A_3B \) to intersect the extended line \( AB \) at \( B' \).

5. Final Side: Draw a line through \( B' \) parallel to \( BC \) to intersect the extended line \( AC \) at \( C' \).


Step 3: Final Answer:
\( \triangle AB'C' \) is the required triangle similar to \( \triangle ABC \) with scale factor \( 5/3 \). Quick Tip: Always join the denominator of the ratio (in this case, 3) to the end of the base of your original triangle.

Step 1: Understanding the Concept:

The area of a sector can be expressed in terms of the arc length (\(l\)) and the radius (\(r\)) using the formula: \( Area = \frac{1}{2} l r \).


Step 2: Detailed Explanation:

Let the radius be \(r\). According to the problem, the arc length \(l = r + 7\).

Given Area = 147 cm².

Using the formula \( Area = \frac{1}{2} l r \): \[ 147 = \frac{1}{2} (r + 7)r \] \[ 294 = r^2 + 7r \] \[ r^2 + 7r - 294 = 0 \]
Factorizing the quadratic equation: \[ r^2 + 21r - 14r - 294 = 0 \] \[ r(r + 21) - 14(r + 21) = 0 \] \[ (r - 14)(r + 21) = 0 \]
Since radius cannot be negative, \( r = 14 cm \).

Arc length \( l = r + 7 = 14 + 7 = 21 cm \).


Step 3: Final Answer:

The radius is 14 cm and the arc length is 21 cm. Quick Tip: The formula \( A = \frac{1}{2}lr \) is very similar to the area of a triangle (\( \frac{1}{2}bh \)), where the arc length acts as the base and the radius as the height.