# 2020 WAEC MATHEMATICS OBJ AND ESSAY QUESTIONS/ANSWERS VERIFIED EXPO

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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

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(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

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(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

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(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

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(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

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**(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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