WAEC MATHEMATICS OBJECTIVE AND THEORY ANSWER
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11CCBCABABBA
21AACCDCDABC
31BCDCDCBABA
41ACADBBDADD
================ (1a) = 3 4/9 ÷ (5 1/3 - 2 3/4) + 5 9/10 = 31/9 ÷ Lcm (3 4-9/12 ) + 5 9/10 = 31/9 ÷ 3 -5/12 + 5 9/10 = 31/9 ÷ 2 7/12 + 5 9/10 = 31/9 ÷ 31/12 + 5 9/10 = 31/9 x 12/31 + 5 9/10 = 4/3 + 59/10 Find the Lcm = 40+177/30 = 217/30 = 7 7/30 (1b) A { 2,3,4 } B { 1,3,5 } = Possibility are { 1,2 }, {2,3}, {3,1},{3,3}, {3,5},{4,1},{4,3},and {4,5} = Prob! Of sum greater than 3 = 8/9 = Prob of sum less than 7 = 5/9 = Prob of sum greater than 3 and less than 7 = 8/9 x 5/9 = 40/81 ======================= (2a) 4 + 3/4 (x+2) < eqaul to 3/8x + 1 Multiply through by 8 = 8(4) + 3(2)(x+2) < eqaul to 3x + 8(1) = 32 + 6(x+2) < eqaul to 3x + 8 = 32 + 6x + 12 < eqaul to 3x + 8 = 6x - 3x < eqaul to 8 - 44 = 3x < eqaul to -36 = x < eqaul to -36/3 = x < eqaul to -12 (2b) Area of the shaded portion = total area - area of the removed square 484 = 20(20+x) - x^2 484 = 400 + 20x - x^2 X^2 - 20x + 84 = 0 X^2 - 6x - 14x + 84 = 0 X(x-6) - 14 ( x-6 ) = 0 (X-14) , (x-6) = 0 X=14 or 6 Ie x= 6cm or 14cm. ======================= 3a) The ratio of interior angle to. Exterior angle = 6 : 2 interior angle = 5/7 x 180/1 = 900/7 = 128 4/7 Exterior angle = 2/7 x 180/1 = 360/7 = 51 3/7 Number of side = 360/exterior angle = 360 / 51 3/7 = 6.9999 approx!! 7 The number of side of the polygon is 7 ======================= (3b) 74+SPR+40=180(sum of 114degrees + SPR =180 SPR=180-114 SPR=66 degrees. ====================== 4a) Sum of 3(2/3) and 2(1/5) =3(2/3) + 2(1/5) =11/3 + 11/5 =(55+33)/15 =88/15 =5.87 Amount=7-5.87 =1.13 =7-5(13/15) =7-(88/15) =17/15 4b) i)h=6+4cos(15p)degrees h=6+4cos(15*4) h=6+4cos60degrees h=6+(4*0.5) h=6+2 =8m ii)9=6+4cos(15p)degrees 4cos(15p)degrees=3 Cos15p=3/4 =0.75 15p=Cos^-1(0.75) 15p=41.4 P=41.4/15 =2.76 11a) H = mt/d(m+p) find M = cross multiply = mt = hd(m+p) = mt = hdm + hdp = mt - hdm = hdp = m(t-hd) = hdp = m = hdp/t-hd or = m = hdp/t-dh 11b) From Angle WXM < WXM = 90 degree ( angle at the center is twice < at circumference ) < WMX = 180degree - (90 - 48)degrees (sum of < in a triangles) < WMX = 42 < WMX + < XMZ - 180 ( < on a straight lines) < XMZ = 180 - 42 = 138 < WYZ + < XMZ = 180 < WYZ = 42 degree (11c) Operation Table * | 1 | 3 | 5 | 6 1 | 4 | 6 | 1 | 2 3 | 6 | 1 | 3 | 4 5 | 1 | 3 | 5 | 6 6 | 2 | 4 | 6 | 0 I= {5} II= { } ======================= (10a) 2tan60 + cos 30/sin 60 = 2 ( square root 3/1 + square root 3/2 )/ square root 3/2 Open the bracket = 2 square root 3/1 + square root. 3/2 / square root. 3/2 = 4 square root 3 + square root 3/2 x 2/ square root 3 = 8 square root 3 + 2 square root 3/2square root 3 = 2 ( 4square root 3 + square root 3 ) / 2square root 3 = 4 square root 3 + square root 3 / square root 3 x square root 3 / square root 3 = 4(3) + 3 / 3 = 15/3 = 5 (10b) Using < ACE Tan Θ = opp/ adj Tan 41 degree = CE / 1050 CE = 1050 (tan 41) CE = 1050 ( 0.8693 ) CE = 912.751m Using. < BCD Tan 36/1 = CD/1050 CD = 1050 (tan 36) CD = 762.8697m (I) Height of control tower = CE - DE = 912.751 - 762.8697 = 149.8813 Approx! 150m (II) Using < ACE Cos Θ = adj/hyp Cos 41 = 1050/AC AC = 1050/cos 41 AC = 1391.2636 aprox! 1391m The shortest distance = AC ======================= (9a) a= -8 a + 6d : a + 8d 5:8 a+6d/a+8d = 5/8 8a + 48d = 5a + 40d 8a - 5a = 40d - 48d 3a = -8d 3 (-8) = -8d -24 = -8d d= 24/8 d = 3 (9b) 30p + 100m = 2450 ................ (I) 40/100p + 30/100m = 855 .................(II) From equation 1 3p + 10m = 245 ................ (III) From equation 2 4/10p + 3/10m = 855 Multiply through by 10 to give 2 4p + 3m = 8550 ............... (Iv) Multiply (III) by 4 and. (Iv) by 3 12p + 40m = 980 12p + 9m = 25 (12) Vol of. The reservoir = vol of cone + vol of hem = 1/3 pi r^2 h + 2/3 pi r^3 where r =x 333 1/3 = 1/3 pi x^2 6x + 2/3 pi x^3 1000/3 = 2 pi x^3 + 2/3 pi x^3 1000/3 = 2 2/3 pi x^3 1000/3 ÷ 8/3 = pi x^3 1000/3 x 3/8 = pi x^3 7000/176 = x^3 X^3 = 39.77 X= 3sqare root 39.77 X = 3.41 Radius = 3.41m (12bI) Vol of hemsphere = 2/3 pi r^3 Vol = 2/3 x 22/7 x ( 3.41)^3 V = 2/3 x 22/7 x 39.77 V = 1749.88/21 V = 83.33m (12bII) Total surface area of the reservoir = curved surface area of the cone + curve surface area of them. hempsere TSA = 22/7 x R x L + 2 x 22/7 x r^2 L^2 = (20.46)^2 + (3.41)^2 L^2 = 418.61 + 11.63 L^2 = 430.24 L = square root ( 430.24) L= 20.74 TSA = 1555.91/7 + 511.54/7 = 2067.55/7 = 295.36m
