Correct Answer: A. The diagonals are equal

Explanation

(a) \(2x^{2} + 7x - 11 = 0\)

Using the quadratic formula,

\(a = 2, b = 7, c = -11\)

\(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)

\(x = \frac{-7 \pm \sqrt{7^{2} - 4(2)(-11)}}{2(2)}\)

\(x = \frac{-7 \pm \sqrt{49 + 88}}{4}\)

\(x = \frac{-7 \pm 11.705}{4}\)

\(x = \frac{-7 + 11.705}{4} ; \frac{-7 - 11.705}{4}\)

\(x = \frac{4.705}{4} = 1.17625\) or \(x = \frac{-18.705}{4} = -4.67625\)

\(x \approxeq 1.18\) or \(x \approxeq -4.68\)

(b) \(\frac{2}{x} + \frac{1}{y} = 3\)

\(\frac{1}{x} - \frac{5}{y} = 7\)

Substituting using \(P = \frac{1}{x}; Q = \frac{1}{y}\), we have

\(2P + Q = 3 ... (i)\)

\(P - 5Q = 7 ... (ii)\)

Multiply (ii) by 2, we have \(2P - 10Q = 14 ... (iii)\)

(iii) - (i) : \(-10Q - Q = 14 - 3 \implies -11Q = 11\)

\(Q = -1\)

Substitute Q = -1 in (i),

\(2P - 1 = 3 \implies 2P = 3 + 1 = 4\)

\(\implies P = 2\)

P = 2; Q = -1

But \(P = \frac{1}{x} \implies 2 = \frac{1}{x}\)

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

Also, \(Q = \frac{1}{y} \implies -1 = \frac{1}{y}\)

\(y = -1\)

\(\therefore x = \frac{1}{2} ; y = -1\)

Correct Answer: D. 80

Explanation

Sum of ∠s in a pentagon = (n - 2)180 = 540°

x° + 2x° + x° + 60° + x° + 10° + x° - 10° = 540°

6x° + 60° = 540°; x = 80°

Explanation

(a)

(i) Nobody takes Economics and Chemistry \(n(E \cap C) = \null \).

(ii) 4 pupils take Economics and Government; \(n(E \cap G) = 4\)

(b) (i) Suppose \(n(C \cap G) = x\)

\(22 = 8 + 4 + 8 - 4 - x + x + 7 - x\)

\(22 = 23 - x \implies x = 1\)

Only 1 person does Chemistry and Government.

(ii) Number taking only Government = \(8 - (4 + x)\)

\(8 - (4 + 1) = 3\)

Correct Answer: D. y = 5

Explanation

\(\frac{y + 2}{y^2 - 3y - 10}\)

\(y^2 - 3y - 10 = 0 \implies y^2 - 5y + 2y - 10 = 0\)

\(y(y - 5) + 2(y - 5) = 0\)

\((y - 5)(y + 2) = 0\)

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

\(\therefore\) At y = 5, the expression \(\frac{y + 2}{y^2 - 3y - 10}\) is undefined.

Correct Answer: D. 8

Explanation

Let x be the exterior angle; interior = 3x; but x + 3x = 180^o

4x = 180^o; x = 45

= \(\frac{360}{45}\) = 8

Correct Answer: A. 5850

Explanation

(i) \(2\frac{2}{3} - (2\frac{1}{2} - 1\frac{4}{5})\)

= \(2\frac{2}{3} - (\frac{5}{2} - \frac{9}{5})\)

= \(\frac{8}{3} - (\frac{25 - 18}{10})\)

= \(\frac{8}{3} - \frac{7}{10}\)

= \(\frac{24 - 21}{30}\)

= \(\frac{3}{30} = \frac{1}{10}\)

(ii) \(\frac{3.25 - 1.64}{2.47 - 2.01}\)

= \(\frac{1.61}{0.46}\)

= \(\frac{7}{2}\)

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

Correct Answer: B. 4

Explanation

x = 5, \(\frac{\sum {fx}}{\sum {f}}\) =

5= \(\frac{\sum {fx}}{30}\)

\(\sum {fx}\) = 5 x 30 = 150; 150 - 34 = 116

\(\sum {f}\) = 29

\(\sum {fx}\) = 116

= \(\frac{116}{29}\) = 4

Explanation

Let T and B be the top and base of the control tower respectively; P, the point of pilot's observation and PS, the line of observation.

(a) Shortest distance is |PB|.

\(\therefore \cos 33° = \frac{8.5}{PB}\)

\(PB = \frac{8.5}{\cos 33} = \frac{8.5}{0.8387}\)

\(|PB| = 10.135km \approxeq 10.1km\)

(b) Height of the control tower = |SB| - |ST|

\(\therefore \tan 33° = \frac{|SB|}{8.5 km} \implies |SB| = 8.5 \tan 33°\)

\(|SB| = 5.52 km\)

\(\tan 30° = \frac{|ST|}{8.5 km} \implies |ST| = 8.5 \tan 30°\)

\(|ST| = 4.907 km\)

\(\therefore\) The height of the tower = 5.52 km - 4.907 km = 0.613 km.