Correct Answer: B. \(\frac{3}{4}\)

Explanation

A = [2, 4, 6, 8}

B = {2, 3, 5, 9}

Pr = (Prime in B) = \(\frac{3}{4}\)

Correct Answer: B. 2:1

Explanation

From the diagram, XYZQ is a parallelogram. Thus, |YZ| = |XQ| = |PX|; \(\bigtriangleup\)PXY, Let the area of XYZQ = A₁, the area of \(\bigtriangleup\)PXY

= Area of \(\bigtriangleup\)YZR = A₂

Area of \(\bigtriangleup\)PQR = A = A₁ + 2A₂

But from similarity of triangles

\(\frac{\text{Area of PQR}}{\text{Area of PXY}} = (\frac{PQ}{PX})^2 = (\frac{QR}{XY})^2\)

\(\frac{A}{A_2} = (\frac{2}{1})^2 = \frac{2}{1}\)

A = 4A₂ But, A = A₁ + 2A

A₁ = 4A₂ - 2A₂

A₁ = 2A₂

\(\frac{A_1}{A_2}\) = 2

A₁:A₂ = 2:1

Area of XYZQ:Area of \(\bigtriangleup\)YZR = 2:1

Correct Answer: D. zero

Explanation

\(\begin{array}{c|c} x & 2 & 3 & 5 & 9 \\ \hline 2 & 4 & 6 & 10 & 18 \\ \hline 4 & 8 & 12 & 20 & 36 \\ \hline 6 & 12 & 18 & 30 & 54 \\ \hline 8 & 16 & 24 & 40 & 72 \end{array}\)

Note: A {horizontal}

B {vertical}

Pr (Odd Product) = \(\frac{0}{16}\)

= 0

Correct Answer: D. \(\begin{pmatrix} -4 & 3 \\ -2 & 9 \end{pmatrix}\)

Explanation

\(\begin{pmatrix} 1 & -2 \\ 3 & 4 \end{pmatrix} \(\begin{pmatrix} -2 & 3 \\ 1 & 0 \end{pmatrix}\)

=\(\begin{pmatrix} (1 \times -2) + (-2 \times 1) & (1 \times 3) + (-2 \times 0) \\ (3 \times -2) + (4 \times 1) & (3 \times 3) + (4 \times 0) \end{pmatrix}\)

= \(\begin{pmatrix} -4 & 3 \\ -2 & 9 \end{pmatrix}\)

Correct Answer: B. 52cm

Explanation

For a rhombus, all the sides are equal and diagonals bisect each other at 90^o. Hence, the triangles formed are congruent: (under RHS).

Thus: \(\bar{PS}^2\) = 5² + 12²

\(\bar{PS} = \sqrt{25 + 144}\)

= \(\sqrt{169}\)

\(\bar{PS}\) = 13cm

perimeter = 4 x length of a side

= 4 x 13cm

= 52cm

Correct Answer: B. -1

Explanation

p - 2g + 1 = g + 3p.........(1)

p - 2 = 0 .........(2)

From (2), p = 2; put p = 2 into (1);

2 - 2g + 1 = g + 3(2)

3 - 2g = g + 6

-2g - g = 6 - 3

-3g = 3

g = \(\frac{3}{-3}\)

g = -1

Correct Answer: C. \(\frac{1}{8}\)

Explanation

\(\begin{array}{c|c} - & 2 & 3 & 5 & 9 \\ \hline 2 & 0 & 1 & 3 & 7 \\ \hline 4 & 2 & 1 & 1 & 5\\ \hline 6 & 4 & 3 & 1 & 3 \\ \hline 8 & 6 &5 & 3 & 1 \end{array}\)

Note: A {horizontal}

B {vertical}

Pr(Difference of 6 or 7) = \(\frac{2}{16} = \frac{1}{8}\)

Correct Answer: C. \(f(x) = x^{3} - 3x^{2} + x + 2\)

Explanation

\(f ' (x) = 3x^{2} - 6x + 1\)

\(f(x) = \int (3x^{2} - 6x + 1) \mathrm {d} x\)

= \(x^{3} - 3x^{2} + x + c\)

\(f(3) = 5 = 3^{3} - 3(3^{2}) + 3 + c\)

\(27 - 27 + 3 + c = 5 \implies 3 + c = 5\)

\(c = 2\)

\(f(x) = x^{3} - 3x^{2} + x + 2\)

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

Explanation

For perpendicular vectors, their dot product = 0.

\((4i - kj). (3i + 8j) = 12 - 8k = 0\)

\(8k = 12 \implies k = \frac{3}{2}\)

Explanation

Given :

\(v = 50kmh^{-1} ; s = 20 km ; a = 6kmh^{-2} ; u = ?\)

\(v^{2} = u^{2} + 2as\)

\(50^{2} = u^{2} + 2(6)(20)\)

\(2500 = u^{2} + 240 \implies u^{2} = 2500 - 240 = 2260\)

\(u = \sqrt{2260} = 47.54 kmh^{-1}\)

To find time, we use

\(v = u + at \)

\(50 = 47.54 + 6t\)

\(t = \frac{50 - 47.54}{6} = 0.41h\)

\(\approxeq 0.4hour\)