Question Library

To calculate the mean of grouped data, - First step to determine the midpoint (x) of each interval or class. 0 - 4 ►2 5 - 9 ► 7 10 - 14 ►12 These midpoints must then be multiplied by the frequencies of the corresponding classes: 2 X 2 = 4 1 X 7 = 7 2 X 12 = 24 Mean = ( 24 + 7 + 4) \div ( 2 + 1 + 2 ) : Mean = 7

Pr(Ebou did not hit the target) = 1 - \(\frac{3}{4}\) = \(\frac{1}{4}\) = Pr(E'), Pr(Omar did not hit the target) = 1 - \(\frac{4}{5}\) = \(\frac{1}{5}\) = Pr(O') Hence the probability that only Kebba will hit the target = Pr(K) x Pr(E') x Pr(O') = \(\frac{2}{3} \times \frac{1}{4} \times \frac{1}{5}\) = \(\frac{1}{30}\)

In the diagram, \(\frac{16}{12} = \frac{VX}{6}\) (similar \(\Delta\)s) VX = \(\frac{16 \times 6}{12}\) = 8cm

Let Tom's present age be x. Then x = 25 - n If Tom is 5 years younger than Bade, then Bade's present age is x + 5 = 25 - n + 5 = (30 - n)

\(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3}}\) = \(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\) = \(\frac{\sqrt{2} \times \sqrt{3} + \sqrt{3} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\) = \(\frac{\sqrt{6} + 3}{3}\) = \(\frac{3 + \sqrt{6}}{3}\) = Hence, (m + n) = 1 + \(\frac{1}{3}\) = 1\(\frac{1}{3}\)

In the diagram, QOR + 2m(vertically opposite angles) So, m + 90^\circ + 2m = 180^\circ (angles on str. line) 3m = 180^\circ - 90^\circ 3m = 90^\circ m = \(\frac{90^o}{3}\) = 30^\circ substituting 30^\circ for m in 2m + 4n = 200^\circ gives 2 x 30^\circ + 4n = 200^\circ 60^\circ + 4n = 200^\circ 4n = 200^\circ - 60^\circ = 140^\circ n = \(\frac{140^\circ}{4}\) = 35^\circ

T = \(\sqrt{\frac{TK - H}{K - H}}\) Taking the square of both sides, give T2 = \(\frac{TK - H}{K - H}\) T2(K - H) = TK - H T2K - T2H = TK - H T2K - TK = T2H - H K(T2 - T) = H(T2 - 1) K = \(\frac{H(T^2 - 1)}{T^2 - T}\)

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

Diagonal |AC| = (2x + 1)cm In the diagram,area of \(\Delta\)ABC is \(\frac{110}{2}\) = \(\frac{1}{2}\) x |AC| x |HB| 55 = \(\frac{1}{2}\) x (2x + 1) x 10 55 = (2x + 1)5 55 = 10x + 5 55 - 5 = 10x 50 = 10x x = \(\frac{50}{10}\) = 5.0

\(\frac{3x - y}{xy} - \frac{2x + 3y}{2xy} + \frac{1}{2}\) = \(\frac{2(3x - y) - 1(2x + 3y) + xy}{2xy}\) = \(\frac{6x - 2y - 2x - 3y + xy}{2xy}\) = \(\frac{4x - 5y + xy}{2xy}\)

Let x represent the entire farmland then, \(\frac{2}{5}\)x + \(\frac{1}{3}\)[x - \(\frac{2}{3}x\)] + M = x Where M represents the part of the farmland used for growing maize, continuing \(\frac{2}{5}\)x + \(\frac{1}{3}\)x [1 - \(\frac{2}{3}x\)] + M = x \(\frac{2}{5}x + \frac{1}{3}\)x [\(\frac{3}{5}\)] + M = x \(\frac{2}{5}\)x + \(\frac{1x}{5}\) + M = x \(\frac{3x}{5} + M = x\) M = x - \(\frac{2}{5}\)x = x[1 - \(\frac{3}{5}\)] = x[\(\frac{2}{5}\)] = \(\frac{2x}{5}\) Hence the part of the land used for growing maize is \(\frac{2}{5}\)

R \(\alpha\) D2 R = D2K R = 4 Litres when D = 15cm thus; 4 = 152k 4 = 225k k = \(\frac{4}{225}\) This gives R = \(\frac{4D^2}{225}\) Where R = 9litres equation gives 9 = \(\frac{4D^2}{225}\) 9 x 225 = 4d2 D2 = \(\frac{9 \times 225}{4}\) D = \(\sqrt{9 \times 225}{4}\) = \(\frac{3 \times 15}{2}\) = 22\(\frac{1}{2}\)km

Cost price CP of the 100 oranges = \(\frac{100}{5}\) x N40.00 selling price SP of the 100 oranges = \(\frac{100}{20}\) x N120 = N600.00 so, profit or loss per cent = \(\frac{SP - CP}{CP}\) x 100% = \(\frac{600 - 800}{800}\) x 100% = \(\frac{-200}{800}\) x 100% Hence, loss per cent = 25%

The shaded portion represents the elements that are in Q only, but not in P or R. In set notation, this is: Q ∩ P' ∩ R'

\(\frac{1}{4}\)p + 3p = 10...(1) 2p - \(\frac{1}{3}\)q = 7...(2) Multiply equation (2) by 3 to clear fraction 3 x 2p - 3 x \(\frac{1}{3}\)q = 3 x 7 6p - q = 21 6p - 21 = q....(3) substituting 6p - 21 for q in (1) \(\frac{1}{4}\)p + 3(6p - 21) = 10...(4) Multiply equation (4) by 4 to clear fraction 4 x \(\frac{1}{4}\)p + 4 x 3(6p - 21) = 4 x 10 p + 12(6p - 21) = 40 p + 72p - 252 73p = 292 p = \(\frac{292}{73}\) = 4