Correct Answer: C. (20, 5)

Correct Answer: D. 5

Explanation

\(P = \begin{pmatrix} 3 & 4 \\ 2 & x \end{pmatrix}; Q = \begin{pmatrix} 1 & 3 \\ -2 & 4 \end{pmatrix}; R = \begin{pmatrix} -5 & 25 \\ -8 & 26 \end{pmatrix}\)

PQ = \(\begin{pmatrix} 3 & 4 \\ 2 & x \end{pmatrix} \begin{pmatrix} 1 & 3 \\ -2 & 4 \end{pmatrix} = \begin{pmatrix} -5 & 25 \\ 2 - 2x & 6 + 4x \end{pmatrix} = R\)

\(\implies 2 - 2x = -8; -2x = -8 - 2 = -10\)

\(6 + 4x = 26 \implies 4x = 26 - 6 = 20\)

\(\implies x = 5\)

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.

Correct Answer: B. 9

Explanation

\(\frac{9*3^{n+1} - 3^{n+2}}{3^{n+1} - 3^{n}}\)

= \(\frac{3^n * 3^n * 3^1 * - 3^2 * 3^2}{3^n * 3^1 - 3^n}\)

= \(\frac{3^n (3^2 * 3^1)}{3^n (3^1 - 1)}\)

= \(\frac{27-9}{3-1}\)

= \(\frac{18}{2}\)

= 9

Correct Answer: A. 1, 3

Explanation

\(f(x) = 3x^{2} - 12x + 12\) and \(f(x) = 3\)

\(\therefore f(x) = 3 = 3x^{2} - 12x + 12 \implies 3x^{2} - 12x + 12 - 3 = 0\)

\(3x^{2} - 12x + 9 = 0; 3x^{2} - 9x - 3x + 9 = 0\)

\(3x(x - 3) - 3(x - 3) = (3x - 3)(x - 3) = 0\)

3x - 3 = 0 or x - 3 = 0

\(x = 1 or 3\)

Correct Answer: D. 5y - 2x - 12 = 0

Explanation

\(Line: 2y + 5x - 6 = 0\)

\(2y = 6 - 5x \implies y = 3 - \frac{5x}{2}\)

Gradient = \(\frac{-5}{2}\)

For the line perpendicular to the given line, Gradient = \(\frac{-1}{\frac{-5}{2}} = \frac{2}{5}\)

The midpoint of P(4, 3) and Q(-6, 1) = \((\frac{-6 + 4}{2}, \frac{3 + 1}{2})\)

= (-1, 2).

Therefore, the line = \(\frac{y - 2}{x + 1} = \frac{2}{5}\)

\(2(x + 1) = 5(y - 2) \implies 5y - 2x - 12 = 0\)

Explanation

(a)

Momentum before collision = \((20 \times 80) + (30 \times 50) = 1600 + 1500\)

= \(3100 kgm/s\)

Momentum after collision = 50v kgm/s

Momentum before collision = Momentum after collision

\(3100 = 50v\)

\(v = \frac{3100}{50} = 62m/s\)

(b)

Momentum before collision = \((20 \times 80) - (30 \times 50) = 1600 - 1500\)

= \(100 kgm/s\)

Momentum after collision = \(50v kgm/s\)

\(\therefore 100 = 50v\)

\(v = 2m/s\)

Explanation

(a)

(b) Assumed mean = 1.1m

\(Mean = A + \frac{\sum fd}{N}\)

= \(1.1 + \frac{0.4}{34}\)

= \(1.1 + 0.0118\)

= \(1.1118\)

\(\approxeq 1.1\) (to 1 d.p)

Explanation

Total number of students =6 boys + 8 girls = 14 students.

Total ways to choose 5 students out of 14:

Total ways = \(^{14}C_5=\frac{14!}{5!*(14 - 5)!}=2002.\)

The number of ways to havemoregirls than boys:

(Selecting 3 girls and 2 boys) or (Selecting 4 girls and 1 boy)

Selecting 3 girls and 2 boys:\(^8C_3\times ^6C_2\)

=(\(\frac{8!}{3!*(8 - 3)!})\times(\frac{6!}{2! * (6 - 2)!})=56\times 15=840.\)

Selecting 4 girls and 1 boy:\(^8C_4\times^6C_1\)

\(=(\frac{8!}{4!*(8 - 4)!})\times(\frac{6!}{1! * (6 - 1)!})=70\times 6=420.\)

Total favorable cases =840+420=1260.

Finally, the probability is given by:

Probability (more girls than boys)

= \(\frac{Total Favorable Cases}{Total Ways}=\frac{1260}{2002}=\frac{90}{143}≈0.629.\)

The probability that more girls than boys are selected is approximately 0.629, or about 62.9%.

Correct Answer: C. -17

Explanation

\(\frac{8x+m}{x^2-3x-4} ≡ \frac{5}{x+1} + \frac{3}{x-4}\)

\(\frac{8x+m}{x^2-3x-4}\) ≡ \(\frac{5(x-1)+ 3(x+4)}{x^2-3x-4}\)

multiplying both sides by x2-3x-4

8x+m ≡ 5(x-4)+3(x+1)

8x + m ≡ 5x - 20 + 3x + 3

8x - 5x - 3x + m = -20 + 3

m = -17