WAEC 2021 Answers For Mathematics Maths

2021 WAEC Maths Mathematics

Question And Answer General. Maths 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 +104x/y=12X/y=3X: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) Total Surface Area = 224πcm² r:l = 2:5 r/l = 2/5 Cross multiply 2l/2 = 5r/2 L = 5r / 2 Total surface = πrl + πr² = πr (l + r) 24π/π = πr (5r/2 + r )/ π 224 = 5r²/2 + r²/1 L.c.m = 2 448 = 5r² + 2r² 448 / 7= 7r²/7 r² = 64 r = √64 = 8cm L = 5*8/2 = 20cm (4b) Volume = 1/2πr²h = 1/3 * 22/7 * 8 * 8 * 18.33 = 1228.98cm³ L² =h²+r²20²=h²+8²400–64=h² h² = 336 h = √ 336 h = 18.33cm (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 (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 +104x/y=12X/y=3X:y=3:1(7a) Diagram 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 =22/7 * 14 * 78 =3432cm² ~3430cm² (3 S.F) (7b) 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–2nT9=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 (9c) Speed = 20/4, average speed = 5km/h (12a) BCD=ABC=40°(alternate D) DDE=2*BCD(