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Re-arranging them 2, 3, 5, 7, 8, 9, 12, 14 Mean = P and median = Q P = \(\frac{2 + 2 + 5 + 7 + 8 + 9 + 12 + 14}{8}\) \(\frac{60}{8}\) = 7.5 median (Q) = \(\frac{7 + 8}{2}\) \(\frac{15}{2}\) = 7.5 P + Q = 7.5 + 7.5 = 15

CF (6) = 5 + 6 + 8 + 10 + 6 + 3 = 38

A rectangle, a square and a rhombus are the only parallelogram ∴ Probability of a parallelogram = \(\frac{3}{5}\)

(n-2)\(^2\) = 25 n - 2 = 5 n = 5 + 2 = 7

From the diagram, POQ is a diameter; o is the centre of the circle and TP is a tangent where RTP = 30o RTP = RQP = x(at circumference = made by tangent outside the circle) i.e x = 30o

h1 = \(\frac{h_1 + h_2}{1.5}\) = \(\frac{h_1}{h_2} = 2\) \(\frac{A_{\bigtriangleup PQR}}{A_{TP(QRST)}} = \frac{\frac{1}{2} \times 2 \times h_1}{\frac{1}{2} \times h_2 (2 + 3)}\) \(\frac{2}{5} \times \frac{h_1}{h_2} = \frac{2}{5} \times 2\) = \(\frac{4}{5}\)

Sum of interior angles of any polygon is (2n - 4) right angle; n angles of the nonagon = 9 where 3 are equal and 6 other angles = 1110o ( 2 x 9 - 4)90o = (18 - 4)90o = 14 x 90o = 1260o 9 angles = 12600, 6 angles = 1110o Remaining 3 angles = 1260o - 1110o = 150o size of one of the3 angles \(\frac{150}{3}\) = 50o

Z = X = Y = 180o .........(i) 2Z + Y + Y = 190 = 2Z + 2Y = 180 Z + Y = 90o........(ii) hence x + (z + y) = 180 x + 90o = 180o x = 180o - 90o = 90o

QR is a given line and P is a given point. The construction is the perpendicular from a given point P to a given line QR

From 40 to 50 = these number are prime i.e. 41, 43, 47 prob. of selecting a prime No. is \(\frac{3}{11}\)

Evaluate (212)3 - (121)3 + (222)3 (221)3+ (222)3 1211 121 ------ (1020)3

Let x represent Musa's total mark when he scores 57 in biology and Let Y represent Musa's total mark when he now scored 75 in biology, if he scored 75 in biology his new total mark will be \(\frac{Y}{4}\) = 60, y = 4 x 60 = 240 To get his total mark when he scored 57, subtract 57 from 75 to give 18, then subtract this 18 from the new total mark(ie. 240) = 240 - 18 = 222

48 = 24 x 3, 64 = 26, 80 = 24 x 5 L.C.M = 26 x 3 x 5 H.C.F = 24 \(\frac{2^6 \times 3 \times 5}{2^4}\) = 22 x 3 x 5 = 4 x 3 x 5 = 12 x 5 = 60

\(252 = 2^{2} \times 3^{2} \times 7\) \(\therefore\) To make 252 a perfect square, the smalest number to multiply with = 7. \(252 \times 7 = 1764 = 42^{2}\)

\(\frac{2}{3}\) = \(\frac{2}{3}\) = \(\frac{2}{3}\) x \(\frac{6}{5}\) = \(\frac{4}{5}\) reciprocal of \(\frac{4}{5}\) = \(\frac{1}{\frac{4}{5}}\) = \(\frac{5}{4}\)