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

Explanation

There are 3 triangles: \(\bigtriangleup\)MRS, \(\bigtriangleup\) and \(\bigtriangleup\) PRQ

Area PRMS has 2 triangles: \(\bigtriangleup\) MRS and \(\bigtriangleup\) RSP

the fraction = \(\frac{2}{3}\)

Correct Answer: C. 20

Explanation

score 0 = 4 students

score 1 = 2 students

score 2 = 5 students

score 3 = 6 students

score 4 = 3 students

= 20 students

Correct Answer: C. 65%

Correct Answer: A. \(\begin{pmatrix} -8 & -5 \\ 3 & 2 \end{pmatrix}\)

Explanation

Let \(\begin{pmatrix} a & b \\ c & d \end{pmatrix} = T^{-1}\)

\(T . T^{-1} = I\)

\(\begin{pmatrix} -2 & -5 \\ 3 & 8 \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)

\(\implies -2a - 5c = 1\)

\(-2b - 5d = 0 \implies b = \frac{-5d}{2}\)

\(3a + 8c = 0 \implies a = \frac{-8c}{3}\)

\(3b + 8d = 1\)

\(-2(\frac{-8c}{3}) - 5c = \frac{16c}{3} - 5c = \frac{c}{3} = 1 \implies c = 3\)

\(3(\frac{-5d}{2}) + 8d = \frac{-15d}{2} + 8d = \frac{d}{2} = 1 \implies d = 2\)

\(b = \frac{-5 \times 2}{2} = -5\)

\(a = \frac{-8 \times 3}{3} = -8\)

\(\therefore T^{-1} = \begin{pmatrix} -8 & -5 \\ 3 & 2 \end{pmatrix}\)

Correct Answer: D. 14

Explanation

T₁, T₂, T₃

2, -9, -20 .... -141

l = a + (n - 1)d

first term, a = 2

common difference d = T₃ - T₂

= T₂ - T₁

= -20 - (-9) = -9 -2

= -20 + 9

= -9 -2

= -20 + 9

= -11

-11 = -11

d = -1

last term l = -141

-141 = 2 + (\(\cap\) - 1) (-11)

-141 = 2 + (-11 \(\cap\) + 11)

= 2 - 11\(\cap\) + 11

-141 = 13 - 11\(\cap\)

-141 - 13 = -11\(\cap\)

-154 = -11\(\cap\)

\(\cap\) = \(\frac{-154}{-11}\)

\(\cap\) = 14

Correct Answer: C. i and iii

Correct Answer: C. \(\frac{1}{3}\)

Explanation

p = { 1, 2, 3}

Q = {2, 3, 5}

prob(prime number) = prob(1 and 2) or

= prob(2 and 3) or

= prob(3 and 2)

There are three possibilities of the sum being prime

Total possibility = 9

probability(sum being prime) = \(\frac{3}{9}\)

= \(\frac{1}{3}\)

Correct Answer: A. 23.4

Explanation

\(Mean = \frac{\sum fx}{\sum f}\)

= \(\frac{1170}{50} = 23.4\)

Correct Answer: C. t = 2 and 3

Explanation

For collinear points (points on the same line), the slopes are equal for any 2 points on the line.

Given (-1, t - 1), (t, t - 3), (t - 6, 3),

\(slope = \frac{(t-3) - (t-1)}{t - (-1)} = \frac{3 - (t-3)}{(t-6) - t} = \frac{3 - (t-1)}{(t-6) - (-1)}\)

Taking any two of the equations above, solve for t.

\(\frac{t - 3 - t + 1}{t + 1} = \frac{6 -t}{-6}\)

\(12 = (6 - t)(t + 1)\)

\(-t^{2} + 5t + 6 - 12 = 0 \implies t^{2} - 5t + 6 = 0\)

Solving, we have t = 2 and 3.

Explanation

\(xy^{2} + x^{2} y = 4xy\)

Differentiating with respect to x,

\(2xy \frac{\mathrm d y}{\mathrm d x} + y^{2} + x^{2} \frac{\mathrm d y}{\mathrm d x} + 2xy = 4x \frac{\mathrm d y}{\mathrm d x} + 4y\)

\((2xy + x^{2} - 4x) \frac{\mathrm d y}{\mathrm d x} = 4y - y^{2} - 2xy\)

\(\frac{\mathrm d y}{\mathrm d x} = \frac{4y^{2} - y^{2} - 2xy}{2xy + x^{2} - 4x}\)

At (1, 3), Gradient = \(\frac{4(3) - (3)^{2} - 2(1)(3)}{2(1)(3) + 1^{2} - 4(1)}\)

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

= -1