Correct Answer: C. \(A = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}\)

Explanation

\(B.B^{-1} = 1\), let \(B^{-1} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\)

\(\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\)\(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\) = \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)

Multiplying \(B \times B^{-1}\), we have the following equations:

\(2a+5c = 1......... (1)\); \(a+3c = 0 ........(2)\)

\(2b+5d = 0 ..........(3)\); \(b+3d = 1..........(4)\)

Solving the equations simultaneously, we have

\(a = 3; b = -5; c = -1; d = 2 \implies B^{-1} = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}\)

Correct Answer: D. 1000₃

Correct Answer: A. N16,200

Explanation

(a) \(T_{n} = a + (n - 1)d\) (For an AP series)

\(T_{6} = 37 = a + (6 - 1) d = a + 5d\)

\(a + 5d = 37 .... (1)\)

\(S_{n} = \frac{n}{2}(2a + (n - 1) d)\)

or

\(S_{n} = \frac{n}{2}(a + l)\) where a = first term; and l = last term.

Since, we are given the sum of the first 6 terms and also given the sixth term which is the last term in the sum, we can use the second formula where the sixth term is the last term in this case.

\(S_{6} = 147 = \frac{6}{2} (a + 37)\)

\(147 = 3(a + 37) \implies 147 = 3a + 111 \)

\(3a = 147 - 111 = 36 \implies a = 12\)

(b) From equation (1) above,

\(a + 5d = 37\)

\(12 + 5d = 37\)

\(5d = 37 - 12 = 25\)

\(d = 5\)

\(S_{15} = \frac{15}{2} (2(12) + (15 - 1)(5))\)

\(\frac{15}{2} (24 + 70) = \frac{15}{2} (94)\)

= \(15 \times 47\)

= \(705\).

Correct Answer: B. 6\(\frac{2}{3}\)

Correct Answer: A. (360 - x- y - z)°

Correct Answer: B. \(\frac{-3}{2}\)

Explanation

\(\frac{log8}{log\frac{1}{4}}=\frac{log2^3}{log2^{-2}}=\frac{3log2}{-2log2}=-\frac{3}{2}\)

Correct Answer: C. 1.5%

Correct Answer: A. \(\frac{2x}{x - 3}, x \neq 3\)

Explanation

From the rules of binary operation, \(x * e = x\)

\(\implies x * e = 3x + 3e - xe = x\)

\(3e - xe = x - 3x = -2x\)

\(e = \frac{2x}{x - 3}, x \neq 3\)

Correct Answer: C. 60 cm²