Every Book Back multiple-choice question from Algebra (10th Standard Mathematics, Samacheer Kalvi) — each with the correct option highlighted and a clear, worked explanation. Free to read in English and Tamil.
Q1
A system of three linear equations in three variables is inconsistent if their planes
- A. intersect only at a point
- B. intersect in a line
- C. coincide with each other
- D. do not intersectCorrect
Explanation. A system of linear equations has no solution if there is no common point of intersection among the planes. Therefore, when the planes do not intersect, the system is inconsistent.
Q2
The solution of the system \(x + y - 3z = -6\), \(-y + z = 1\), \(3z = 9\) is
- A. \(x = 1, y = 2, z = 3\)Correct
- B. \(x = -1, y = 2, z = 3\)
- C. \(x = -1, y = -2, z = 3\)
- D. \(x = 1, y = -2, z = 3\)
Explanation. Solving \(3z = 9\) gives \(z = 3\). Substituting \(z = 3\) into \(-y + z = 1\) yields \(y = 2\). Substituting both values into \(x + y - 3z = -6\) gives \(x = 1\).
Q3
If \((x - 6)\) is the HCF of \(x^2 - 2x - 24\) and \(x^2 - kx - 6\), then the value of \(k\) is
- A. 3
- B. 5Correct
- C. 6
- D. 8
Explanation. Since \((x - 6)\) is a common factor, substituting \(x = 6\) into the polynomial \(x^2 - kx - 6\) must equal zero. This gives \(36 - 6k - 6 = 0\), which simplifies to \(k = 5\).
Q4
\(\frac{3y - 3}{y} \div \frac{7y - 7}{3y^2}\) is
- A. \(\frac{9y}{7}\)Correct
- B. \(\frac{9y^3}{21y - 21}\)
- C. \(\frac{21y^2 - 42y + 21}{3y^3}\)
- D. \(\frac{7(y^2 - 2y + 1)}{3y^2}\)
Explanation. Simplifying the expressions, we get \(\frac{3(y-1)}{y} \times \frac{3y^2}{7(y-1)}\). Cancelling out the common factor \((y-1)\) and simplifying the \(y\) terms gives the final result of \(\frac{9y}{7}\).
Q5
\(y^2 + \frac{1}{y^2}\) is not equal to
- A. \(\frac{y^4 + 1}{y^2}\)
- B. \(\left(y + \frac{1}{y}\right)^2\)Correct
- C. \(\left(y - \frac{1}{y}\right)^2 + 2\)
- D. \(\left(y + \frac{1}{y}\right)^2 - 2\)
Explanation. Expanding the expression \(\left(y + \frac{1}{y}\right)^2\) yields \(y^2 + \frac{1}{y^2} + 2\), which is not equal to the original expression. All other options simplify correctly to \(y^2 + \frac{1}{y^2}\).
Q6
\(\frac{x}{x^2 - 25} - \frac{8}{x^2 + 6x + 5}\) gives
- A. \(\frac{x^2 - 7x + 40}{(x - 5)(x + 5)}\)
- B. \(\frac{x^2 + 7x + 40}{(x - 5)(x + 5)(x + 1)}\)
- C. \(\frac{x^2 - 7x + 40}{(x^2 - 25)(x + 1)}\)Correct
- D. \(\frac{x^2 + 10}{(x^2 - 25)(x + 1)}\)
Explanation. Expressing with a common denominator \((x^2 - 25)(x + 1)\), the numerator becomes \(x(x + 1) - 8(x - 5)\). This simplifies to \(x^2 - 7x + 40\), giving the correct fraction.
Q7
The square root of \(\frac{256x^8 y^4 z^{10}}{25x^6 y^6 z^6}\) is equal to
- A. \(\frac{16}{5}\left|\frac{x^2 z^4}{y}\right|\)
- B. \(16\left|\frac{y^2}{x^2 z^4}\right|\)
- C. \(\frac{16}{5}\left|\frac{y}{x z^2}\right|\)
- D. \(\frac{16}{5}\left|\frac{x z^2}{y}\right|\)Correct
Explanation. Simplifying inside the fraction gives \(\frac{256 x^2 z^4}{25 y^2}\). Taking the square root of the coefficients and variables yields \(\frac{16}{5}\left|\frac{x z^2}{y}\right|\).
Q8
Which of the following should be added to make \(x^4 + 64\) a perfect square?
- A. \(4x^2\)
- B. \(16x^2\)Correct
- C. \(8x^2\)
- D. \(-8x^2\)
Explanation. By completing the square, \(x^4 + 64 = (x^2)^2 + 8^2\). To form the perfect square \((x^2 + 8)^2\), we must add the middle term \(2(x^2)(8) = 16x^2\).
Q9
The solution of \((2x - 1)^2 = 9\) is equal to
- A. \(-1\)
- B. 2
- C. \(-1, 2\)Correct
- D. None of these
Explanation. Taking the square root on both sides gives \(2x - 1 = \pm 3\). Solving \(2x - 1 = 3\) gives \(x = 2\), and solving \(2x - 1 = -3\) gives \(x = -1\).
Q10
The values of \(a\) and \(b\) if \(4x^4 - 24x^3 + 76x^2 + ax + b\) is a perfect square are
- A. \(100, 120\)
- B. \(10, 12\)
- C. \(-120, 100\)Correct
- D. \(12, 10\)
Explanation. Expanding \((2x^2 - 6x + 10)^2\) yields \(4x^4 - 24x^3 + 76x^2 - 120x + 100\). Comparing coefficients gives \(a = -120\) and \(b = 100\).
Q11
If the roots of the equation \(q^2 x^2 + p^2 x + r^2 = 0\) are the squares of the roots of the equation \(qx^2 + px + r = 0\), then \(q, p, r\) are in
- A. A.P
- B. G.PCorrect
- C. Both A.P and G.P
- D. none of these
Explanation. Let the roots of the first be \(\alpha, \beta\) and the second be \(\alpha^2, \beta^2\). Using sum and product relations gives \(p^2 = qr\), which means \(q, p, r\) are in G.P.
Q12
Graph of a linear equation is a
- A. straight lineCorrect
- B. circle
- C. parabola
- D. hyperbola
Explanation. A linear equation of the form \(ax + by + c = 0\) always represents a straight line on the Cartesian coordinate plane.
Q13
The number of points of intersection of the quadratic polynomial \(x^2 + 4x + 4\) with the \(X\)-axis is
- A. 0
- B. 1Correct
- C. 0 or 1
- D. 2
Explanation. Since \(x^2 + 4x + 4 = (x + 2)^2\), the equation has real and equal roots. Therefore, the graph touches the \(X\)-axis at exactly one point, \(x = -2\).
Q14
For the given matrix \(A = \begin{pmatrix} 1 & 3 & 5 & 7 \\ 2 & 4 & 6 & 8 \\ 9 & 11 & 13 & 15 \end{pmatrix}\) the order of the matrix \(A^T\) is
- A. \(2 \times 3\)
- B. \(3 \times 2\)
- C. \(3 \times 4\)
- D. \(4 \times 3\)Correct
Explanation. The matrix \(A\) has \(3\) rows and \(4\) columns, making its order \(3 \times 4\). Transposing interchanged rows and columns, resulting in an order of \(4 \times 3\).
Q15
If \(A\) is a \(2 \times 3\) matrix and \(B\) is a \(3 \times 4\) matrix, how many columns does \(AB\) have
- A. 3
- B. 4Correct
- C. 2
- D. 5
Explanation. The product of an \(m \times n\) matrix and an \(n \times p\) matrix is an \(m \times p\) matrix. Here, the product matrix \(AB\) has order \(2 \times 4\), meaning it has 4 columns.
Q16
If a number of columns and rows are not equal in a matrix then it is said to be a
- A. diagonal matrix
- B. rectangular matrixCorrect
- C. square matrix
- D. identity matrix
Explanation. A rectangular matrix is defined as a matrix in which the number of rows and columns are not equal.
Q17
Transpose of a column matrix is
- A. unit matrix
- B. diagonal matrix
- C. column matrix
- D. row matrixCorrect
Explanation. A column matrix has order m x 1. Taking its transpose swaps the dimensions to 1 x m, which is a row matrix.
Q18
Find the matrix \(X\) if \(2X + \begin{bmatrix} 1 & 3 \\ 5 & 7 \end{bmatrix} = \begin{bmatrix} 5 & 7 \\ 9 & 5 \end{bmatrix}\)
- A. \(\begin{bmatrix} -2 & -2 \\ 2 & -1 \end{bmatrix}\)
- B. \(\begin{bmatrix} 2 & 2 \\ 2 & -1 \end{bmatrix}\)Correct
- C. \(\begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix}\)
- D. \(\begin{bmatrix} 2 & 1 \\ 2 & 2 \end{bmatrix}\)
Explanation. Subtracting the matrix from the right-hand side gives \(2X = \begin{bmatrix} 4 & 4 \\ 4 & -2 \end{bmatrix}\). Dividing each element by 2 yields \(X = \begin{bmatrix} 2 & 2 \\ 2 & -1 \end{bmatrix}\).
Q19
Which of the following can be calculated from the given matrices \(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix}\) and \(B = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}\)?\n(i) \(A^2\)\n(ii) \(B^2\)\n(iii) \(AB\)\n(iv) \(BA\)
- A. (i) and (ii) only
- B. (ii) and (iii) only
- C. (ii) and (iv) onlyCorrect
- D. all of these
Explanation. Since \(A\) has order 3x2 and \(B\) has order 3x3, only square matrices can be squared (\(B^2\)), and product is defined only if columns of the first equal rows of the second (\(BA\)).
Q20
If \(A = \begin{bmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{bmatrix}\), \(B = \begin{bmatrix} 1 & 0 \\ 2 & -1 \\ 0 & 2 \end{bmatrix}\) and \(C = \begin{bmatrix} 0 & 1 \\ -2 & 5 \end{bmatrix}\), which of the following statements are correct?\n(i) \(AB + C = \begin{bmatrix} 5 & 5 \\ 5 & 5 \end{bmatrix}\)\n(ii) \(BC = \begin{bmatrix} 0 & 1 \\ 2 & -3 \\ -4 & 10 \end{bmatrix}\)\n(iii) \(BA + C = \begin{bmatrix} 2 & 5 \\ 3 & 0 \end{bmatrix}\)\n(iv) \((AB)C = \begin{bmatrix} -8 & 20 \\ -8 & 13 \end{bmatrix}\)
- A. (i) and (ii) onlyCorrect
- B. (ii) and (iii) only
- C. (iii) and (iv) only
- D. all of these
Explanation. Multiplying \(A\) and \(B\) gives \(AB = \begin{bmatrix} 5 & 4 \\ 7 & 0 \end{bmatrix}\), so statement (i) holds. Multiplying \(B\) and \(C\) gives \(BC = \begin{bmatrix} 0 & 1 \\ 2 & -3 \\ -4 & 10 \end{bmatrix}\), so statement (ii) holds.
