Correct Answer: D. 4:5

Explanation

(1 - \(\frac{20}{100}\)) = cash payment fraction= \(\frac{4}{5}\)

The ratio of the cash payment to the marked price = 4:5

Explanation

(a) Total surface area of a sphere = \(4\pi r^{2}\)

Let the TSA of the smaller sphere be \(S_{1}\) with radius r and the bigger sphere be \(S_{2}\) with radius R.

\(\frac{S_{1}}{S_{2}} = \frac{9}{49}\)

\(\frac{9}{49} = \frac{4\pi r^{2}}{4\pi R^{2}}\)

\(\implies \frac{9}{49} = \frac{12^{2}}{R^{2}}\)

\(\frac{49 \times 144}{9} = R^{2}\)

\(R = \sqrt{49 \times 16} = 28 cm\)

Volume of bigger sphere = \(\frac{4}{3} \pi r^{3}\)

= \(\frac{4}{3} \times \frac{22}{7} \times 28 \times 28 \times 28\)

= \(\frac{275,968}{3}\)

= \(91989.33 cm^{3}\)

\(\approxeq 91989 cm^{3}\) (nearest whole number).

(b)

\(|ZX|^{2} = |YZ|^{2} + |YX|^{2} - 2|YZ||YX| \cos Y\)

= \(5^{2} + 3^{2} - 2(5)(3) \cos 135\)

= \(25 + 9 - 30(-0.7071)\)

= \(34 + 21.213\)

\(|ZX|^{2} = 55.213\)

\(|ZX| = 7.43 km \approxeq 7 km\)

(ii) Using sine rule,

\(\frac{\sin \theta}{5} = \frac{\sin 135}{7.43}\)

\(\sin \theta = \frac{5 \times \sin 135}{7.43}\)

= \(\frac{3.5355}{7.43}\)

\(\sin \theta = 0.4758\)

\(\theta = \sin^{-1} (0.4758)\)

= \(28.41° \approxeq 28°\) (to the nearest degree).

Bearing of Z from X = 270° + 28.41° = 298.41°

\(\approxeq\) 298°.

Explanation

nC\(_4\) = \(\frac{n!}{[n-4]! 4!}\) → \(\frac{n[n-1][n-2][n-3][n-4]!}{[n-4]!4!}\)

nC\(_5\) = \(\frac{n!}{[n-5]! 5!}\) → \(\frac{n[n-1][n-2][n-3][n-4][n-5]!}{[n-5]!5!}\)

and nC\(_6\) = \(\frac{n!}{[n-6]! 6!}\) → \(\frac{n[n-1][n-2][n-3][n-4][n-5][n-6]!}{[n-6]!6!]}\)

d = U2 - U1 = U3 - U2

nC\(_5\) - nC\(_4\) = \(\frac{n(n-1)(n-2)(n-3)(n-9)}{5!}\)

nC\(_6\) - nC\(_5\) = \(\frac{n(n-1)(n-2)(n-3)(n-4)(n-11)}{5! 6}\)

nC\(_5\) - nC\(_4\) = nC\(_6\) - nC\(_5\)

\(\frac{n(n-1)(n-2)(n-3)(n-9)}{5!}\) = \(\frac{n(n-1)(n-2)(n-3)(n-4)(n-11)}{5! 6}\)

divide both sides by n(n-1)(n-2)(n-3)

\(\frac{n-9}{5!}\) = \(\frac{[n-4][n-11]}{5! 6}\)

multiply both sides by 5! * 6

6(n-9) = (n-4)(n-11)

6n - 54 = n\(^2\) -11n - 4n + 44

6n - 54 = n\(^2\) - 15n + 44

n\(^2\) - 21n + 98 = 0

n\(^2\) - 7n - 14n + 98 = 0

n(n - 7) -14(n - 7) = 0

(n-7)(n-14) = 0

n = 7 or 14

ii.

d = U2 - U1

when n = 7

d = 7C\(_5\) - 7C\(_4\)

d = \(\frac{7*6*5!}{2! * 5!}\) - \(\frac{7*6*5*4!}{3! * 4!}\)

d = 21 - 35

d = -14

when n = 14

d = 14C\(_5\) - 14C\(_4\)

d = \(\frac{14*13*12*11*10*9!}{9! * 5!}\) - \(\frac{14*13*12*11*10!}{10! * 4!}\)

d = 2002 - 1001

d = 1001

Explanation

(a) \(\frac{1}{1 - \cos \theta} + \frac{1}{1 + \cos \theta}\)

= \(\frac{(1 + \cos \theta) + (1 - \cos \theta)}{1 - \cos^{2} \theta}\)

= \(\frac{2}{1 - \cos^{2} \theta}\)

= \(\frac{2}{\sin^{2} \theta}\)

(b) \(y = x^{2} (x - 3)\)

\(\frac{\mathrm d y}{\mathrm d x} = x^{2} + 2x(x - 3)\)

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

At stationary point, \(\frac{\mathrm d y}{\mathrm d x} = 0\)

\(3x^{2} - 6x = 0\)

\(3x(x - 2) = 0 \implies x = \text{0 or 2}\)

If x = 0, \(y = 0^{2}(0 - 3) = 0\)

If x = 2, \(y = 2^{2}(2 - 3) = -4\)

The points are (0, 0) and (2, -4).

Gradient = \(\frac{-4 - 0}{2 - 0} = \frac{-4}{2} = -2\)

Equation : \(\frac{y - (-4)}{x - 2} = -2\)

\(\frac{y + 4}{x - 2} = -2 \implies y + 4 = -2(x - 2)\)

\(y + 4 = 4 - 2x \implies y + 2x = 0\)

\(y = -2x\)

Distance : \(\sqrt{(2 - 0)^{2} + (-4 - 0)^{2}}\)

= \(\sqrt{4 + 16} = \sqrt{20}\)

= \(2\sqrt{5} units\)

Correct Answer: B. -2.5

Correct Answer: A. \(\frac{5}{4}\sqrt{2}\)

Explanation

\(\frac{10}{\sqrt{32}}=\frac{10}{\sqrt{16\times 2}} = \frac{10}{4\sqrt{2}}\\

\frac{10\times \sqrt{2}}{4\sqrt{2}\times\sqrt{2}}=\frac{10\sqrt{2}}{4\times 2}= \frac{5}{4}\sqrt{2}\)

Correct Answer: C. Mean

Correct Answer: B. 1.75

Explanation

(a) \(625^{\frac{3}{8}} \times 5^{\frac{1}{2}} \div 25\)

= \((5^{4})^{\frac{3}{8}} \times 5^{\frac{1}{2}} \div 5^{2}\)

= \(5^{\frac{3}{2}} \times 5^{\frac{1}{2}} \div 5^{2}\)

= \(5^{\frac{3}{2} + \frac{1}{2} - 2}\)

= \(5^{0} = 1\)

(b)(i) \(\tan (\theta + 25)° = 5.145\)

\(\theta + 25 = \tan^{-1} (5.145)\)

\(\implies \theta + 25 = 79°\)

\(\theta = 79° - 25° = 54°\)

(ii) \(5 \cos \theta - 1 = 0\)

\(\implies 5 \cos \theta = 1\)

\(\cos \theta = \frac{1}{5} = 0.2\)

\(\theta = \cos^{-1} (0.2) = 78.463°\)

\(\approxeq 78.5°\) (to one d.p)

Correct Answer: D. 1.54 x 10⁻⁵