2020 WAEC Expo OBJ & Theory (Questions And Answers) olinerunzWaec



MATHS-OBJ 1CBCDACDCCD 11AADBDACBBC 21BDDABDADAD 31CDACCCCCDA 41BBBCDCACDB Completed. 💪💪💪💪🇳🇬🇳🇬🇳🇬🇳🇬 ================== WAEC 2020 MATHEMATICS THEORY ANSWERS =========================================== (1a) Given A={2,4,6,8,...} B={3,6,9,12,...} C={1,2,3,6} U= {1,2,3,4,5,6,7,8,9,10} A' = {1,3,5,7,9} B' = {1,2,4,5,7,8,10} C' = {4,5,7,8,9,10} A'nB'nC' = {5, 7} (1b) Cost of each premiere ticket = $18.50 At bulk purchase, cost of each = $80.00/50 = $16.00 Amount saved = $18.50 - $16.00 =$2.50 ================= (2ai) P = (rk/Q - ms)⅔ P^3/2 = rk/Q - ms rk/Q = P^3/2 + ms Q= rk/P^3/2 + ms (2aii) When P =3, m=15, s=0.2, k=4 and r=10 Q = rk/p^3/2 + ms = 10(4)/(3)^3/2 + (15)(0.2) = 40/8.196 = 4.88(1dp) (2b) x + 2y/5 = x - 2y Divide both sides by y X/y + 2/5 = x/y - 2 Cross multiply 5(x/y) - 10 = x/y + 2 5(x/y) - x/y = 2 + 10 4x/y = 12 X/y = 3 X : y = 3 : 1 ===================== (3a) Diagram CBD = CDB (base angles an scales D) BCD+CBD+CDB=180° (Sum of < in a D) 2CDB+BCD=180° 2CDB+108°=180° 2CDB=180°-108°=72° CDB=72/2=36° BDE=90°(Angle in semi circle) CDE=CDB+BDE =36°+90 =126 (3b) (Cosx)² - Sinx given (Sinx)² + Cosx Using Pythagoras theory thrid side of triangle y²= 1²+√3 y²= 1+ 3=4 y=√4=2 (Cosx)² - sinx/(sinx)² + cosx (1/2)² - √3/2/ (√3/2)² + 1/2 = 1/4 - √3/2 = 1-2√3/4 3/4+1/2 = 3+2/4 =1-2√3/4 * 4/5 =1-2√3/5 ===================== (4a) Given: r : l = 2 : 5 (ie l = 5/2r) Total surface area of cone =πr² + url 224π = π(r² + r(5/2r)) 224 = r² + 5/2r² 224 = 7/2r² 7r² = 448 r² = 448/7 = 64 r = root 64 = 8.0cm (4b) L = 5/2r = 5/2 × 8 = 20cm Using Pythagoras theorem L² = r² + h² h² = l² - r² h² = 20² - 8² h² = (20 + 8)(20 - 8) h² = 28 × 12 h = root28×12 h = 18.33cm Volume of cone = 1/3πr²h = 1/3 × 22 × 7 × 8² × 18.33 =1229cm³ ======================= (5a) Total income = 32+m+25+40+28+45 =170+m PR(²)=m/170+m = 0.15/1 M=0.15(170+m) M=25.5+0.15m 0.85m/0.85=25.5/0.85 M=30 (5b) Total outcome = 170 + 30 = 200 (5c) PR(even numbers) = 30+40+50/200 =115/200 = 23/40 ========================== === SECTION B ==== ================================ (7a) Click here to view image Using Pythagoras theorem, l²=48² + 14² l²=2304 + 196 l²=2500 l=√2500 l=50m Area of Cone(Curved) =πrl Area of hemisphere=2πr² Total area of structure =πrl + 2πr² =πr(l + 2r) =22/7 * 14 [50 + 2(14)] =22/7 * 14 * 78 =3432cm² ~3430cm² (3 S.F) (b) let the percentage of Musa be x Let the percentage of sesay be y x + y=100 -------------------1 (x - 5)=2(y - 5) x - 5=2y - 10 x - 2y=-5 -------------------2 Equ (1) minus equ (2) y - (-2y)=100 - (-5) 3y=105 y=105/3 y=35 Sesay's present age is 35years ==================================== (8a) Let Ms Maureen's Income = Nx 1/4x = shopping mall 1/3x = at an open market Hence shopping mall and open market = 1/4x + 1/3x = 3x + 4x/12 = 7/12x Hence the remaining amount = X-7/12x = 12x-7x/12 =5x/12 Then 2/5(5x/12) = mechanic workshop = 2x/12 = x/6 Amount left = N225,000 Total expenses = 7/12x + X/6 + 225000 = Nx 7x+2x+2,700,000/12 =Nx 9x + 2,700,000 = 12x 2,700,000 = 12x - 9x 2,700,000/3 = 3x/3 X = N900,000 (ii) Amount spent on open market = 1/3X = 1/3 × 900,000 = N300,000 (8b) T3 = a + 2d = 4m - 2n T9 = a + 8d = 2m - 8n -6d = 4m - 2m - 2n + 8n -6d = 2m + 6n -6d/-6 = 2m+6n/-6 d = -m/3 - n d = -1/3m - n ==================================== (9a) Draw the triangle (9b) (i)Using cosine formulae q² = x² + y² - 2xycosQ q² = 9² + 5² - 2×9×5cos90° q² = 81 + 25 - 90 × 0 q² = 106 q = square root 106 q = 10.30 = 10km/h Distance = 10 × 2 = 20km (ii) Using sine formula y/sin Y = q/sin Q 5/sin Y = 10.30/sin 90° Sin Y = 5 × sin90°/10.30 Sin Y = 5 × 1/10.30 Sin Y = 0.4854 Y = sin‐¹(0.4854), Y = 29.04 Bearing of cyclist X from y = 90° + 19.96° = 109.96° = 110° (9c) Speed = 20/4, average speed = 5km/h *MR KISS & BABIE BOY SOLVERS*💯💯💯.








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