Correct Answer: D. SSA

Correct Answer: A. 19.50

Correct Answer: A. 0.5

Explanation

y = k\(\sqrt{x}\) when y = 1, x = 4

k = \(\frac{y}{\sqrt{x}}\)

= \(\frac{1}{\sqrt{4}}\)

= \(\frac{1}{2}\)

= 0.5

Correct Answer: C. 3

Explanation

3x \(\equiv\) 4(mod 5)

In modulo 5, multiples of 5 that give solution to the given equation are 5, 20, 35, 50,... but 5 will yield the leaast value of x.

Thus; 3x = 4 + 5 = 9

x = \(\frac{9}{3}\)

x = 3

Explanation

(a) \(\frac{15.05 \times \sqrt{0.00695}}{6.95 \times 10^{2}}\)

\(\therefore \frac{15.05 \times \sqrt{0.00695}}{6.95 \times 10^{2}} \approxeq 0.00181\) (3 sig. figs)

(b) No of boys = 2, No of girls = 3

\(\therefore\) Total students = 5

(i) P(both are boys) = \(\frac{2}{5} \times \frac{1}{4} = \frac{1}{10}\)

(ii) P(both are of different sexes) = P(first a boy, then a girl) or P(first a girl, then a boy)

= \(\frac{2}{5} \times \frac{3}{4} + \frac{3}{5} \times \frac{2}{4} \)

= \(\frac{6}{20} + \frac{6}{20}\)

= \(\frac{12}{20} = \frac{3}{5}\)

Explanation

\(\overline{PQ}\) = \(\begin{pmatrix} 1 \\ 3 \end{pmatrix}\) - \(\begin{pmatrix} 4 \\ -5 \end{pmatrix}\) = \(\begin{pmatrix} -3 \\ 8 \end{pmatrix}\)

\(\overline{MR}\) = \(\begin{pmatrix} -5 \\ 2 \end{pmatrix}\) - \(\begin{pmatrix} x \\ y \end{pmatrix}\) = \(\begin{pmatrix} -5 & -x \\ 2 & - y \end{pmatrix}\)

\(\begin{pmatrix} -3 \\ 8 \end{pmatrix}\) = \(\begin{pmatrix} -5 & -x \\ 2 & - y \end{pmatrix}\)

⇒ -5 - x = -3; x = -2

⇒ 2 - y = 8; y = -6

The coordinate of M (-2,-6)

(b) \(\overline{PM}\) = \(\overline{OM}\) - \(\overline{OP}\)

= \(\begin{pmatrix} -2 \\ -6 \end{pmatrix}\) - \(\begin{pmatrix} 4 \\ -5 \end{pmatrix}\) = \(\begin{pmatrix} -6 \\ -1 \end{pmatrix}\)

\(\overline{PQ}\) = \(\overline{OQ}\) - \(\overline{OP}\)

= \(\begin{pmatrix} 1 \\ 3 \end{pmatrix}\) - \(\begin{pmatrix} 4 \\ -5 \end{pmatrix}\) = \(\begin{pmatrix} -3 \\ 8 \end{pmatrix}\)

|\(\overline{PM}\)| = √(-6\(^2\) + [-1\(^2\)]) = √37

|\(\overline{PQ}\)| = √(-3\(^2\) + 8\(^2\)) = √73

cosØ = \(\frac{PM.PQ}{|\overline{PM}||\overline{PQ}|}\)

cosØ = \(\frac{[-6i - j].[-3i+8j]}{√37 . √73}\) = \(\frac{10}{√2710}\)

⇒ Ø = cos\(^{-1}\) \(\frac{10}{√2710}\)

: Ø = 78.91° ≈ 79°

Correct Answer: D. - 5

Explanation

2x + 3y = 10 .......x2

3x - 2y = -11 ........x3

4x + 6y = 20

9x - 6y = -33

\(\overline{\frac{13x}{13} = \frac{-13}{13}}\), x = 1

from

2x + 3y = 10

2(-1) + 3y = 10

-2 + 3y = 10

3y = 10 + 2

\(\frac{3y}{3} = \frac{12}{3}\), y = 4

x - y = -1 - 4

= -5

Explanation

\(|TQ|^{2} = 4^{2} + 6^{2} - 2 \times 4 \times 6 \times \cos 30°\)

= \(16 + 36 - 48 \times 0.8660\)

= \(52 - 41.568 = 10.432\)

\(\therefore |TQ| = \sqrt{10.432} \approxeq 3.23 cm\)

(a) By the rules of similar triangles,

\(\frac{|PT|}{|TQ|} = \frac{|PS|}{|SR|}\)

\(\frac{4}{3.23} = \frac{10}{|SR|}\)

\(|SR| = \frac{3.23 \times 10}{4}\)

= \(8.075 cm \approxeq 8 cm\)

(b) Using sine rule,

\(\frac{|PQ|}{\sin \alpha} = \frac{|TQ|}{\sin 30}\)

\(\frac{6}{\sin \alpha} = \frac{3.23}{\sin 30}\)

\(\sin \alpha = \frac{6 \sin 30}{3.23}\)

sin \(\alpha\) = 0.9288

From the diagram, \(\alpha\) = \(\beta\)

\(\sin \alpha = \sin \beta = 0.9288\)

\(\sin \beta = \frac{h}{6}\)

h = \(6 \sin \beta\)

Hence, area of quadrilateral

TQRS = \(\frac{1}{2} (TQ + SR) \times h\)

= \(\frac{1}{2} (3.23 + 8.075) \times 6\sin \beta\)

= \(\frac{1}{2} (11.305) (6 \times 0.9288)\)

= 31.501 cm\(^2\)

\(\approxeq\) 32 cm\(^2\) (to the nearest whole number)

Correct Answer: B. 3

Explanation

23 x 54 = 1242

In mod 7;

1242

Correct Answer: A. 45^o, 225^o