Class 10 CBSE Maths Sample Paper 2026 with Solutions

Step 1: Find the factors of each number:

Step 2: Factors of 12: [12 = 2 x 2 x 3]

Step 3: Factors of 18: [18 = 2 x 3 x 3]

Step 4: Common factors: 2 and 3 [HCF = 2 x 3 = 6]

Step 1: Step 1: [867 = 255 x 3 + 102]

Step 2: Step 2: [255 = 102 x 2 + 51]

Step 3: Step 3: [102 = 51 x 2 + 0]

Step 4: Remainder is 0, so HCF = 51

Step 1: Assume sqrt(2) + sqrt(3) = a/b (rational)

Step 2: Square both sides: [(sqrt(2) + sqrt(3))^2 = a^2/b^2]

Step 3: Expand: [2 + 2sqrt(6) + 3 = a^2/b^2]

Step 4: So: [5 + 2sqrt(6) = a^2/b^2]

Step 5: This gives sqrt(6) = (a^2/b^2 - 5)/2, making sqrt(6) rational -- contradiction!

Step 1: By definition, a number 'k' is called a zero of a polynomial P(x) if P(k) = 0.

Step 2: This means that when 'k' is substituted for 'x' in the polynomial, the entire expression evaluates to zero.

Step 1: Set the polynomial to zero: `6x² - 7x - 3 = 0`.

Step 2: Split the middle term: `6x² - 9x + 2x - 3 = 0`.

Step 3: Factor by grouping: `3x(2x - 3) + 1(2x - 3) = 0`.

Step 4: This gives `(2x - 3)(3x + 1) = 0`. Setting each factor to zero, `2x - 3 = 0` implies `x = 3/2`, and `3x + 1 = 0` implies `x = -1/3`.

Step 1: Perform polynomial long division of (x³ + 4x² + 10x + 12) by (x² + 2x + k).

Step 2: The first term of the quotient is x. Multiplying (x² + 2x + k) by x gives x³ + 2x² + kx. Subtract this from the dividend: (x³ + 4x² + 10x + 12) - (x³ + 2x² + kx) = 2x² + (10-k)x + 12.

Step 3: The next term of the quotient is 2. Multiplying (x² + 2x + k) by 2 gives 2x² + 4x + 2k. Subtract this from the current remainder: (2x² + (10-k)x + 12) - (2x² + 4x + 2k) = (10-k-4)x + (12-2k) = (6-k)x + (12-2k).

Step 4: For (x² + 2x + k) to be a factor, the remainder must be 0. Therefore, (6-k)x + (12-2k) = 0 for all x. This implies 6-k = 0 and 12-2k = 0. Both conditions give k = 6.

Step 1: A solution to a pair of linear equations must satisfy both equations when the values of x and y are substituted into them.

Step 2: Graphically, this solution represents the point where the two lines corresponding to the equations intersect.

Step 1: The given equations are: 3x + y - 1 = 0 and (2k - 1)x + (k - 1)y - (2k + 1) = 0.

Step 2: For no solution, we must have a₁/a₂ = b₁/b₂ ≠ c₁/c₂. So, 3/(2k - 1) = 1/(k - 1).

Step 3: Cross-multiplying, we get 3(k - 1) = 1(2k - 1), which simplifies to 3k - 3 = 2k - 1.

Step 4: Solving for k: 3k - 2k = -1 + 3, so k = 2. We also need to check that 1/(k-1) ≠ -1/-(2k+1) for k=2. 1/(2-1) = 1/1 = 1. And (2k+1)/1 = (2(2)+1)/1 = 5. So 1 ≠ 5, which satisfies the condition.

Step 1: For infinitely many solutions, the ratio of coefficients must be equal: a1/a2 = b1/b2 = c1/c2.

Step 2: From the given equations, a1 = k-3, b1 = 3, c1 = k; and a2 = k, b2 = k, c2 = 12.

Step 3: Set up the ratios: (k-3)/k = 3/k = k/12.

Step 4: From 3/k = k/12, we get k² = 36, so k = ±6. Since (k-3)/k is involved, let's check k=6 and k=-6. If k=6, then (6-3)/6 = 3/6 = 1/2, and 3/6 = 1/2, and 6/12 = 1/2. All ratios are equal. If k=-6, then (-6-3)/(-6) = -9/-6 = 3/2, and 3/(-6) = -1/2. These are not equal, so k=-6 is not a solution. Thus, k=6.

Step 5: The variable 'm' is not present in the given system of equations, hence it is not constrained by these equations. Therefore, k=6 and 'm' can be any real number (arbitrary).

Step 1: Option A: (x + 1)² = 2x + 3. Expanding the left side gives x² + 2x + 1 = 2x + 3. Simplifying, we get x² - 2 = 0, which is a quadratic equation (a=1, b=0, c=-2).

Step 2: Option B: x(x + 2) = x² + 5. Expanding gives x² + 2x = x² + 5. Simplifying, we get 2x - 5 = 0, which is a linear equation.

Step 3: Option C: x³ - 4x² + 5 = 0. This is a cubic equation, not quadratic, as the highest power of x is 3.

Step 4: Option D: (x - 2)(x + 2) = x² - 4. Expanding gives x² - 4 = x² - 4. Simplifying, we get 0 = 0, which is an identity and not an equation in x.