2020 Waec Mathematics Questions and Answers Solutions

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Completed.

MATHEMATICS THEORY

(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
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(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
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(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 CLICK HERE FOR THE IMAGE (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 CLICK HERE FOR THE IMAGE (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 (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 [50 + 2(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 CLICK HERE FOR THE IMAGE (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 CLICK HERE FOR THE IMAGE (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 CLICK HERE FOR THE IMAGE (11) CLICK HERE FOR THE IMAGE (12) CLICK HERE FOR THE IMAGE (12a) BCD=ABC=40°(alternate D) DDE=2*BCD( DDE = 2*40 = 80° OD3=OED(base < of I sealed D ODE) ODE + OED + DOE= 180°(sum of < is in D) 2ODE+DOE=180° 2ODE+80°=180 2ODE+180=180 2ODE+100° ODE+100/2=50° (12bi) Digram (12bii) Area of parallelogram = absin =5*7*sin125° =35*sin55° =35*0.8192 =28.67 =28.7cm²(1dp) (12c) Given x=1/2(1-√2) 2x²-2x=2[1/2(1-√2]²-2(1/2(1-√2)} =2[1-2√2+2/4]-(1-√2) =(3-2√2/2)-(1-√2) =3-2√2-2+2√2/2=1/2

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