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

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

mean = \(\bar{x}\) = \(\frac{\sum x}{n}\) = \(\frac{5 + 3 + 0 + 7 + 2 + 1}{6}\) \(\frac{18}{6}\) = 3 \(\begin{array}{c|c} x & x - \bar{x} & |x - \bar{x}|\\ \hline 5 & -2 & 2 \\ \hline 3 & 0 & 0 \\ \hline 0 & -3 & 3 \\ \hline 7 & 4 & 4 \\ \hline 2 & -1 & 1 \\ \hline 1 & -2 & 2 \\ \hline & & & \sum|x - \bar{x}| \\ \hline & & & 12 \end{array}\) hence, MD = \(\frac{\sum|x - \bar{x}}{n}|\) \(\frac{12}{6}\) = 2

Area of floor = 3.5m x 2.2m = 7.7m2 Area of carpet = (3.5 - 0.5 - 0.5)m x (2.2 - 0.5 - 0.5)m 2.5m x 1.2m = 3m2 hence, area of margine = area of floor - area of carpet = (7.7 - 3)m2 = 4.7m2

The diagram given simple illustrates that a pentagon contains three \(\Delta\)s. Two angles which are in the ratio 2:3 will have actual values 2xo, 3xo respectively. Thus 2xo + 3xo + 3 x 60 = 3x sum of angles of \(\Delta\)a i.e. 5xo + 180o = 3 x 180o 5xo + 180o = 540o 5xo = 540o - 180o 5xo = 360o xo = \(\frac{360^o}{5}\) = 72o Hence, the smaller of the two angles is 2 x 72o = 144o

Two word mathematics is first rewritten as mathematics, since every letter of mathematics appears once in the english alphabet. Hence the probability that letter selected is from the word mathematics is prob. (matheics) = \(\frac{8}{26}\) = \(\frac{4}{13}\)

In the diagram |AM| = |MB| - \(\frac{|AB|}{2}\) = \(\frac{16\sqrt{3}}{2}\)cm = 8\(\sqrt{3}\)cm in \(\Delta\) AMO, r2 = |AM|Z + |MO|2 r2 = (8\(\sqrt{3}\))2 + 102 = 64 x 3 + 100 = 192 + 100 = 292 r = \(\sqrt{292}\) 17.088cm 17cm

In \(\Delta\) XYZ, 2m + 2n + 68o = 180o 2(m + n) + 68o = 180o...(1) in \(\Delta\) XOZ, m + n + q = 180o ...(2) (m + n) = 180o - q...(3) substituting 180o - q for (m + n) in (1) gives 2(180o - q) + 68o = 180o 360o - 2q = 180o - 68o 360o - 2q = 112o 360o - 112o = 2q 248o = 2q q = \(\frac{248^o}{2}\) = 124o hence, o

(a) \(3\frac{4}{9} \div (5\frac{1}{3} - 2\frac{3}{4}) + 5\frac{9}{10}\) \(\frac{31}{9} \div (\frac{16}{3} - \frac{11}{4}) + \frac{59}{10}\) \(\frac{31}{9} \div (\frac{64 - 33}{12}) + \frac{59}{10}\) \((\frac{31}{9} \div \frac{31}{12}) + \frac{59}{10}\) \((\frac{31}{9} \times \frac{12}{31}) + \frac{59}{10}\) \(\frac{12}{9} + \frac{59}{10} = \frac{120 + 531}{90}\) \(\frac{651}{90} = \frac{217}{30}\). (b) + 1 3 5 2 3 5 7 3 4 6 8 4 5 7 9 Let E be the event of the sum being greater than 3 and less than 7 and S be the total sample space. n(E) = 4; and n(S) = 9. P(E) = \(\frac{n(E)}{n(S)} = \frac{4}{9}\)

(a) \(4 + \frac{3}{4}(x + 2) \leq \frac{3}{8}x + 1\) Multiply through by the LCM of 4 and 8, to clear fractions : \(8 \times 4 + 8 \times \frac{3}{4}(x + 2) \leq 8 \times \frac{3}{8}x + 8 \) \(32 + 6(x + 2) \leq 3x + 8 \) \(32 + 6x + 12 \leq 3x + 8 \) \(44 + 6x \leq 3x + 8\) \(6x - 3x \leq 8 - 44 \implies 3x \leq -36\) \(x \leq - 12\). (b) Area \((A_{1})\), of rectangle PQRS = \(20(20 + x) cm^{2}\) Area \((A_{2})\), of square = \(x \times x = x^{2} cm^{2}\) \(\implies 20(20 + x) - x^{2} = 484\) \(400 + 20x - x^{2} = 484\) \(x^{2} - 20x + 484 - 400 = x^{2} - 20x + 84 = 0\) \(x^{2} - 14x - 6x + 84 = 0\) \(x( x - 14) - 6(x - 14) = 0 \) \((x - 14)(x - 6) = 0\) \(\therefore \text{x = 6 cm or 14 cm}\)

(a) Let the diagram below represent a section of the polygon SEE DIAGRAM ABOVE Also, let : \(i\) represent the size of an interior angle; \(e\) represent the size of an exterior angle. Then \(\frac{i}{e} = \frac{5}{2}\) \(\implies i = \frac{5}{2}e\) But \(i + e = 180^\circ\) (sum of angles on a straight line) Substitute \(\frac{5}{2}e\) for \(i\) in the equation \(\frac{5}{2}e + e = 180^\circ\) \(\frac{7}{2}e = 180^\circ\) \(e = \frac{180^\circ \times 2}{7} = \frac{360^\circ}{7}\) Number of sides of the polygon = \(\frac{360^\circ}{\text{size of one exterior angle}}\) = \(360^\circ \div \frac{360^\circ}{7}\) \(360^\circ \times \frac{7}{360^\circ} = 7 sides\). (b) In the diagram above, < PSR = 68^\circ (interior angle of a cyclic quad = opp exterior angle) < PSQ = 68^\circ - 40^\circ = 28^\circ < PRQ = 28^\circ (angles in same segment) < SRQ = 74^\circ (interior angle of a cyclic quad = opp exterior angle) hence, < PRS = 74^\circ - 28^\circ = 46^\circ