Class 9 CBSE Maths Sample Paper 2026 with Solutions

Step 1: Real numbers are composed of both rational and irrational numbers. Therefore, every irrational number is a real number.

Step 2: Option B is false because rational numbers are also real numbers.

Step 3: Option C is false because fractions like 1/2 are rational but not integers.

Step 4: Option D is false because negative integers (e.g., -3) are integers but not whole numbers.

Step 1: Option A: √100 = 10, which is a rational number (can be written as 10/1).

Step 2: Option B: 0.121212... is a non-terminating recurring decimal, which can be expressed as a rational number (12/99).

Step 3: Option C: 3.14 is a terminating decimal, which can be expressed as a rational number (314/100).

Step 4: Option D: √7 cannot be simplified to an integer or a fraction, and its decimal expansion is non-terminating and non-recurring. Therefore, √7 is an irrational number.

Step 1: Rationalize x: x = [(√3 + 1) / (√3 - 1)] × [(√3 + 1) / (√3 + 1)] = (√3 + 1)² / (3 - 1) = (3 + 1 + 2√3) / 2 = (4 + 2√3) / 2 = 2 + √3.

Step 2: Rationalize y: y = [(√3 - 1) / (√3 + 1)] × [(√3 - 1) / (√3 - 1)] = (√3 - 1)² / (3 - 1) = (3 + 1 - 2√3) / 2 = (4 - 2√3) / 2 = 2 - √3.

Step 3: Now, calculate x² and y²: x² = (2 + √3)² = 4 + 3 + 4√3 = 7 + 4√3. y² = (2 - √3)² = 4 + 3 - 4√3 = 7 - 4√3.

Step 4: Finally, sum them: x² + y² = (7 + 4√3) + (7 - 4√3) = 14.

Step 1: The terms are $5x^3$, $4x^2$, and $-7$.

Step 2: The highest power of $x$ is 3. [\text{Degree} = 3]

Step 1: Factor Theorem: $(x-a)$ is a factor of $p(x)$ iff $p(a) = 0$.

Step 2: Since $(x-2)$ is a factor, $p(2) = 0$.

Step 1: $p(1) = 1 - 6 + 11 - 6 = 0$, so $(x-1)$ is a factor.

Step 2: Dividing: $x^3 - 6x^2 + 11x - 6 = (x-1)(x^2 - 5x + 6)$

Step 3: $x^2 - 5x + 6 = (x-2)(x-3)$

Step 4: Final: [(x-1)(x-2)(x-3)]

Step 1: The x-coordinate of a point refers to its horizontal position and is formally called the abscissa.

Step 2: The y-coordinate of a point refers to its vertical position and is formally called the ordinate.

Step 3: Therefore, the x-coordinate is the abscissa, and the y-coordinate is the ordinate.

Step 1: For P(-7, 10), the x-coordinate (abscissa) is -7 and the y-coordinate (ordinate) is 10. Thus, options A and B are false.

Step 2: A point with a negative x-coordinate and a positive y-coordinate (-,+) lies in the Second Quadrant, not the Third. So option C is false.

Step 3: The distance of a point (x, y) from the Y-axis is given by the absolute value of its x-coordinate, |x|. For P(-7, 10), the distance from the Y-axis is |-7| = 7 units. So option D is true.

Step 1: Let the coordinates of point P be (x, y). Since P lies in the third quadrant, both x and y are negative.

Step 2: The perpendicular distance from the y-axis is |x|, and from the x-axis is |y|. Given |x| = 3|y|.

Step 3: Also, the sum of the absolute values of its coordinates is 16, so |x| + |y| = 16.

Step 4: Substitute |x| = 3|y| into the second equation: 3|y| + |y| = 16, which simplifies to 4|y| = 16. Thus, |y| = 4.

Step 5: Now find |x|: |x| = 3 × 4 = 12. Since P is in the third quadrant, x = -12 and y = -4. So, P is (-12, -4).

Step 1: A linear equation in two variables can be written in the form ax + by + c = 0, where a, b, and c are real numbers, and a and b are not both zero.

Step 2: The degree of each variable (x and y) must be 1. Option A, 2x + 3y = 5 (or 2x + 3y - 5 = 0), fits this definition as both x and y have a degree of 1.

Step 3: Options B (x² + y = 7) has x with degree 2, option C (xy = 4) has a product of variables making it non-linear, and option D (3x = 9) is a linear equation in only one variable.