Explanation

(a) Let S = sample space = integers from 5 to 25 inclusive.

F = multiple of 5 ; T = multiple of 3.

S = {5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25}

F = {5, 10, 15, 20, 25}; T = {6, 9, 12, 15, 18, 21, 24}

P(F or T) = \(\frac{5}{21} + \frac{7}{21} = \frac{12}{21} = \frac{4}{7}\)

(b) Let B = blue balls and R = red balls

n(B) = 4 ; n(R) = 6.

(i) P(both are red) = \(\frac{6}{10} \times \frac{6}{10} = \frac{36}{100}\)

= \(\frac{9}{25}\)

(ii) P(both are same colour) = P(both are red) or P(both are blue)

= \(\frac{9}{25} + (\frac{2}{5} \times \frac{2}{5})\)

= \(\frac{9}{25} + \frac{4}{25}\)

= \(\frac{13}{25}\)

Explanation

(a) \(\frac{\frac{1}{2} of \frac{1}{4} \div \frac{1}{3}}{\frac{1}{6} - \frac{3}{4} + \frac{1}{2}}\)

Numerator - \(\frac{1}{2} \times \frac{1}{4} \div \frac{1}{3} = (\frac{1}{2} \times \frac{1}{4}) \div \frac{1}{3}\)

\(\frac{1}{8} \times 3 = \frac{3}{8}\)

Denominator - \(\frac{1}{6} - \frac{3}{4} + \frac{1}{2} = \frac{1}{6} - (\frac{3}{4} - \frac{1}{2})\)

\(\frac{1}{6} - (\frac{3 - 2}{4}) = \frac{1}{6} - \frac{1}{4}\)

\(\frac{2 - 3}{12} = \frac{-1}{12}\)

\(\therefore \frac{\frac{1}{2} \times \frac{1}{4} \div \frac{1}{3}}{\frac{1}{6} - \frac{3}{4} + \frac{1}{2}} = \frac{3}{8} \div \frac{-1}{12}\)

\(\frac{3}{8} \times -12 = \frac{-9}{2}\)

= \(-4.5\)

(b) \(\sqrt{x} = 10^{\bar{1}.6741}}\)

Squaring both sides, we have

\(x = (10^{\bar{1}.6741}})^{2}\)

\(x = 10^{\bar{1}.3482}\) (checking antilogarithm)

\(x = 0.2229\)

Correct Answer: C. \(\frac{1}{2}\)

Explanation

\(log_{y}\frac{1}{8} = 3 \implies y^{3} = \frac{1}{8}\) (Laws of logarithm)

\(y^{3} = \frac{1}{2^{3}} = (\frac{1}{2})^{3}\)

Equating both sides, we have

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

Correct Answer: A. 156

Correct Answer: A. 1

Explanation

\((x+3) \propto y\\

?x + 3 = ky\hspace{1mm}when\hspace{1mm}x = 3, y = 12\\

3+3 = 12k\\

? k = \frac{1}{2}\\

\Longrightarrow x + 3 = \frac{1}{2}y\) to find x when y = 8

\(x + 3 = \frac{1}{2}\times 8\\

x=4-3\\

x = 1\)

Correct Answer: A. -5

Correct Answer: D. x\(^2\) - y\(^2\) + 3x - 5y = 2

Correct Answer: D. HA 539KST

Correct Answer: C. 4

Explanation

Given the equation of the circle \(x^{2} + y^{2} - 8x - 2y + 1 = 0\).

The equation of a circle is given as \((x - a)^{2} + (y - b)^{2} = r^{2}\)

Expanding, we have \(x^{2} - 2ax + a^{2} + y^{2} - 2by + b^{2} = r^{2} \equiv x^{2} - 2ax + y^{2} - 2by = r^{2} - a^{2} - b^{2}\)

Comparing the RHS of the equation above with the equation rewritten as \(x^{2} + y^{2} - 8x - 2y = -1\), we have

\(-2a = -8; -2b = -2 \implies a = 4, b = 1\)

\(\therefore r^{2} - 4^{2} - 1^{2} = -1 \implies r^{2} = -1 + 16 + 1 = 16\)

\(r = \sqrt{16} = 4\)

Correct Answer: B. \(\begin{pmatrix} -1 & -6 \\ 1 & 18 \end{pmatrix}\)

Explanation

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

= \(\begin{pmatrix} 9 & 6 \\ -3 & 12 \end{pmatrix}\)

2N = 2 \(\begin{pmatrix} 5 & 6 \\ -2 & -3 \end{pmatrix}\)

= \(\begin{pmatrix} 10 & 12 \\ -4 & -6 \end{pmatrix}\)

(3M - 2N) = \(\begin{pmatrix} 9 & 6 \\ -3 & 12 \end{pmatrix}\) - \(\begin{pmatrix} 10 & 12 \\ -4 & -6 \end{pmatrix}\)