Question Library

In In the diagram given, \(\alpha\) = 22º (vertically opp. angles), \(\alpha\) = \(\beta\) (alternate angles) < PSQ + 113º + \(\beta\) = 180º (sum of angles of a \(\Delta\)) < PSQ + 113º + 22º = 180º < PSQ = 180º - (113 + 22)º 45º

In the diagram above, \(\alpha\) = 2mo (angle at centre = 2 x angle at circumference) \(\alpha\) + 10mo = 360o (angle at circumference) \(\alpha\) + 10mo = 360o(angles round a point) 2mo + 10mo = 360o 12mo = 360o mo = \(\frac{360^o}{12}\) = 30o

By similar triangles, \(\frac{8}{3}\) = \(\frac{8 + 12}{h}\) \(\frac{8}{3} = \frac{20}{h}\) h = \(\frac{3 \times 20}{8}\) = 7.5m

In the diagram above, x1 = 180o - 117o = 63o(opposite angles of a cyclic quad.) x1 = x2 (base angles of isos. \(\Delta\)) x\(_1\) + x\(_2\) + \(\alpha\) = 180º (sum of angles of a \(\Delta\) 63º + 63º + \(\alpha\) = 180º \(\alpha\) = 180º - (63 + 63)º = 54º < PST = 54º (aalternate angles)

In the diagram, < TUQ + 60o(corresp. angles) < QTU = 48o (alternate angles) < QU + 60o + 48o = 180o(sum of angles of a \(\Delta\)) < TQU = 180o - 108o = 72o

In \(\Delta\) UXT, tan\(\alpha\) = \(\frac{1}{\sqrt{3}}\) \(\alpha\) = tan-1(\(\frac{1}{\sqrt{3}}\)) = 30o In \(\Delta\)WXT, tan\(\beta\) \(\frac{\sqrt{3}}{\sqrt{3}}\) = 1 \(\beta\) = tan-1(1) = 45o Hence, o + 45o = 75o

sectoral angle representing food = 360o - (80 + 70 + 90)o = 120o Amount spent on food = \(\frac{\tect{sectoral angle}}{360^o}\) x Le 10,800,000 = Le 3,600,000

\(\begin{array}{c|c} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & 1,1 & 1,2 & 1,3 & 1,4 & 1,5 & 1,6 \\ \hline 2 & 2,1 & 2,2 & 2,3 & 2,4 & 2,5 & 2,6 \\ \hline 3 & 3,1 & 3,2 & 3,3 & 3,4 & 3,5 & 3,6 \\ \hline 4 & 4,1 & 4,2 & 4,3 & 4,4 & 4,5 & 4,6 \\ 5 & 5,1 & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \\ \hline 6 & 6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6\end{array}\) From the table above, event space, n(E) = 6 sample space, n(S) = 36 Hence, probability sum of scores is at least 10, is; \(\frac{n(E)}{n(S)}\) = \(\frac{6}{36}\) = \(\frac{1}{6}\)

First, arrange the marks in order of magnitude; 3, 4, 5, 7, 10, 13, 14, 16 Hence range = 16 - 3 = 13

Log2(3x - 1) = 5 Log2(3x - 1) = Log225 Log2(3x - 1) = Log232 3x - 1 = 32 3x = 32 + 1 = 33 x = \(\frac{33}{3}\) = 11

Volume of sphere = Volume of cylinder i.e. \(\frac{4}{3} \pi r^3 = \pi r^2 h\) \(\frac{4}{3} \pi r^3 = \pi \times 3^2 \times 4\) r3 = \(\frac{\pi \times 9 \times 4 \times 3}{4 \pi}\) r = 3\(\sqrt{27}\) = 3

No expanded working has been attached for this question yet. The verified answer is Option D.

Area of \(\Delta\) QNP = \(\frac{1}{2} \times 9 \times 6 \) = 27cm2 Area of \(\Delta\) QMN = Area of \(\Delta\) QNP = Area of \(\Delta\) PNO (triangles between the same parallels) Hence, area of the trapezium 3 x area of \(\Delta\) QNP = 3 x 27 = 81cm2

Perimeter of a sector = 2r + \(\frac{\theta}{360^o}\) x 2 x \(\frac{22}{7}\) x 21 64 = 2 x 21 + \(\frac{\theta}{360^o}\) x 2 x \(\frac{22}{7}\) x 21 64 = 42 + \(\frac{\theta}{360^o}\) x 44 x 3 64 - 42 = \(\frac{\theta}{360^o}\) x 11 x 3 22 = \(\frac{33\theta}{90}\) \(\theta = \frac{22 \times 30}{11}\) = 60o

From the venn diagram given, M = (a, b, c), N = (c, f, g) U = (a, b, c, d, e, f, g) Thus M' \(\cap\) N = (e, f, g) \(\cap\) (c, f, g) = (f, g)