WAEC Mathematics 2021 Answers – WAEC mathematics Questions 2021, free verified objective and theory answers (Expo) for West Africa examination council (May/June) 2021. WAEC mathematics objective and theory answers For you to have good WAEC result in Mathematics examination as well as repeated questions for free in this post. You will also understand how WAEC mathematics questions are set and how to answer them. The West African Examinations Council is an examination board established by law to determine the examinations required in the public interest in the English-speaking West African countries, to conduct the examinations and to award certificates comparable to those of equivalent examining authorities internationally. **CLICK TO SEE WAEC MATHEMATICS EXAM 2021 QUESTIONS & ANSWERS.**

**WAEC Mathematics Questions and Answers 2022 (Theory and Objective)**

**WAEC mathematics exam date and time is**

- Paper III: Objective – General Mathematics – 10:00am – 11:45am –
- Paper II: Essay – General Mathematics – 12:00noon – 2:30pm

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**2021 WAEC Mathematics questions and answers**

Table of Contents

**Today’s Mathematics Essay**

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**(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

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)

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 + r ²

20² = h² + 8²

400 – 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

=======================================

### WAEC Mathematics Objectives Answers Answers 2021

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**WAEC Mathematics Objectives Answers** (2021 Answers)

**WAEC Mathematics Questions and Answers 2021**

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(1b)

Cost of each premiere ticket = $18.50

At bulk purchase, cost of each = $80.00/50 = $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)

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 + r ²

20² = h² + 8²

400 – 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

=======================================

(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

=======================================

(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

=======================================

### More WAEC Mathematics 2021

(9c)

Speed = 20/4, average speed = 5km/h

(12a)

BCD=ABC=40°(alternate D)

DDE=2*BCD(<at centre = twice < at circle)

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

WAEC Mathematics Theory Answers 2021

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- WAEC Mathematics Questions and Answers 2022 (Theory and Objective)

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