Question Library

\(\angle\)N + \(\angle\)P + \(\angle\)M = 180 ( sum of angles in \(\triangle\) NPM) 65º + 55º + \(\angle\)M = 180 \(\angle\)M = 180 - 65 - 55 = 60º x + \(\angle\)M = 180(opp interior angles of parallel lines are supplementary) x = 180 - 60 = 120º

Mean(\(\overline{x}\)) = \(\frac{\sum{x}}{n}\) Mean(\(\overline{x}\)) = \(\frac{ 59 + 44 + 53 + 57 + 49 + 40 + 48 + 50}{8}\) Mean(\(\overline{x}\)) = \(\frac{397}{8}\) ≈ 50 kg

Variance(s\(^2\)) = \(\frac{\sum{(x - \overline{x})^2}}{n}\) But mean = 50 (from question 39 same information) x x - \(\overline{x}\) (x - \(\overline{x})^2\) 59 9 81 44 - 6 36 53 3 9 57 7 49 49 - 1 1 40 -10 100 48 - 2 4 50 0 0 \(\sum{(x - \overline{x})^2}\) = 280 Variance(s\(^2\)) = \(\frac{\sum{(x - \overline{x})^2}}{n}\) = \(\frac{280}{8}\) = 35

Opposite sides of a rectangle are equal, then (5x + 3) = (2x + 9) 5x - 2x = 9 - 3 3x = 6 x = \(\frac{6}{3}\) = 2. Put x = 2 in either (5x + 3) or (2x + 9) = 13m Adjacent side (6x - 7) (given) also put x = 2 into it, 6(2) - 7 = 5 Area of rectangle = length x breadth(width) = 13 x 5 = 65 m\(^2\)

Sample space = { 1, 2, 3, 4, 5, 6} Prime numbers = ( 2, 3, and 5) Pr(prime Nos) = \(\frac{\text{number of required outcome}}{\text{total possible outcome}} = \(\frac{3}{6}\) = \(\frac{1}{2}\)

Area of a sector = \(\frac{\theta}{360}\) x \(\pi\) r\(^2\) 51.3 = \(\frac{\theta}{360}\) x \(\frac{22}{7}\) x 7 x 7 \(\theta\) = \(\frac{51.3 \times 360}{22 \times 7}\) = 119.9 ≈ 120º to the nearest whole number.

From the figure above, using SOH CAH TOA Tan\(\alpha\) = \(\frac{\text{opposite}}{\text{adjacent}}\) Tan\(\alpha\) = \(\frac{87}{22}\) = 3.9545 \(\alpha\) = Tan\(^{-1}\)(3.9545) = 76º

\(\angle\)SPT = 35º (alternate segment) From \(\triangle\)TSP 44 + \(\angle\)SPT + \(\angle\)PST = 180º (sum of angles in a triangle) \(\angle\)PST = 180 - 44 - 35 = 101º

Pr(Amaka passes) = \(\frac{3}{7}\), Pr(Bala passes) = \(\frac{4}{9}\). Pr(both Amaka and Bala pass the exam) = \(\frac{3}{7}\) x \(\frac{4}{9}\) = \(\frac{4}{21}\).

If a point lies on a particular line. It will satisfy the equation of the line. Test the points given in the options any that satisfy the equation of the line, is the point. Thus, the point(1, -1) lies on the line 3x - 8y = 11 That is, putting x = 1 and y = -1 into 3x - 8y to see if the result will equal 11 3(1) - 8(-1) = 11.

Range = max. value - min. value max. value = 39, min.value = 15 Range = max. value - min. value = 39 - 15 = 24.

T\(_n\) = a + (n - 1)d T\(_4\) = a + 3d = 16 - - - - - - - - - - - - - - (i) T\(_8\) = a + 7d = 40 - - - - - - - - - - - - - - -(ii) Solving simultaneously using the elimination method(subtract eqn (i) from eqn(ii) 4d = 24 d = \(\frac{24}{4}\) = 6 Thus, the common difference = 6.

\(\frac{y + 2}{8y^2 - 10y + 3}\) not defined when the denominator is equal to zero 8y\(^2\) - 10y + 3 = 0 using factorization method 8y\(^2\) - 6y - 4y + 3 = 0 2y(4y - 3) - 1(4y - 3) = 0 (4y - 3)(2y - 1) = 0 4y - 3 = 0 or 2y - 1 = 0 Hence, y = \(\frac{3}{4}\) or \(\frac{1}{2}\)

Given P = 5, t = 10mins, V = 20 (a) t ∝ \(\frac{\text{V}}{\text{P}}\) t = \(\frac{\text{KV}}{\text{P}}\) K = \(\frac{\text{tP}}{\text{V}}\) = \(\frac{10 \times 5}{20}\) = \(\frac{5}{2}\) Therefore, the relationship between t, P, and V is t = \(\frac{\frac{5}{2}V}{\text{P}}\) = \(\frac{5V}{2P}\) (b) to find t, when V = 50 and P = 2 t = \(\frac{5 \times 50 }{2 \times 2}\) = \(\frac{250}{4}\) = 62.5mins. (c) to find P when V = 40 and t = 20mins P = \(\frac{5V}{2t}\) (simply by switching t for P) P = \(\frac{5 \times 40}{2 \times 20}\) = \(\frac{200}{40}\) = 5 pumps.

Time = \(\frac{\text{distance}}{\text{speed}}\) t\(_1\) = \(\frac{(2x + 13) }{67.5}\) and t\(_2\) = \(\frac{(5x - 20) }{72}\) Total time = t\(_1\) + t\(_2\) = 90mins = 1.5hrs \(\frac{(2x + 13) }{67.5}\) + \(\frac{(5x - 20) }{72}\) = 1.5 multiply through by the LCM of 67.5 and 72 72(2x + 3) + 67.5(5x - 20) = 1.5(67.5)(72) 144x + 936 + 337.5x - 1350 = 7290 144x + 337.5 = 7290 + 1350 - 936 481.5x = 7704 (after collecting like terms) x = \(\frac{7704}{481.5}\) = 16