# WAEC Mathematics 2022 Questions and Answers

2022-06-01

Nigeria

**WAEC Mathematics 2022: WAEC Mathematics Questions and Answers 2022 for May/June 2022 is now. WAEC Mathematics Theory and Objective Answers (100% sure) Mathematics 2 Essay verified Free for West African Examinations Council. WAEC Mathematics Questions For you to have good WAEC result in Mathematics 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.

Please Note that the WAEC 2022 Mathematics Questions and Answers and any other WAEC Mathematics 2022 is provided by us for free. We understand that a lot of website charge of collect money from student to provide WAEC exam Mathematics Answers to them. WAEC questions and answers 2022 are provided for free. We will do same during Other Exam like NECO.

**WAEC Mathematics Questions and Answers 2022**

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2022 Mathematics 2 Essay Questions Loading 93.8%

#### WAEC Mathematics OBJECTIVES (OBJ) ANSWERS Loading

**2022 VERIFIED Mathematics OBJ:…Loading**

WAEC Mathematics OBJ Answers:

1-10: CBCADABBCD

11-20: BCBADCCABC

21-30: CDCAAADACD

31-40: CADACBAABC

41-50: ABAADADBBB

### WAEC Mathematics Answers

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We are right now getting things ready for you. The WAEC 2022 Mathematics theory questions and answers will be posted any moment from now. All you need do is to keep refreshing this page until you see the answers.

**WAEC Mathematics 2022: WAEC Mathematics Questions and Answers 2022 for May/June 2022 is now. WAEC Mathematics Theory and Objective Answers (100% sure) Mathematics 2 Essay verified Free for West African Examinations Council. WAEC Mathematics Questions For you to have good WAEC result in Mathematics 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.

Please Note that the WAEC 2022 Mathematics Questions and Answers and any other WAEC Mathematics 2022 is provided by us for free. We understand that a lot of website charge of collect money from student to provide WAEC exam Mathematics Answers to them. WAEC questions and answers 2022 are provided for free. We will do same during Other Exam like NECO.

**WAEC Mathematics Questions and Answers 2022**

WAEC 2022 Mathematics Questions will be posted in this page. Our Team are right now with the question paper. It is under verification and once tthe verification process complete, we will go ahead to upload it here.

**Be at alert! Keep refreshing this page for WAEC Mathematics Theory Questions and Answers**

2022 Mathematics 2 Essay Questions Loading 93.8%

#### WAEC Mathematics OBJECTIVES (OBJ) ANSWERS Loading

**2022 VERIFIED Mathematics OBJ:…Loading**

WAEC Mathematics OBJ Answers:

1-10: CBCADABBCD

11-20: BCBADCCABC

21-30: CDCAAADACD

31-40: CADACBAABC

41-50: ABAADADBBB

### WAEC Mathematics Answers

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Please do not panic and fall into the right hand. 2022 Mathematics Answers will be posted for free here once ready.

We are right now getting things ready for you. The WAEC 2022 Mathematics theory questions and answers will be posted any moment from now. All you need do is to keep refreshing this page until you see the answers.

### WAEC Mathematics Questions 2022 and Answers

**Today’s WAEC Mathematics Answers: Loading…99.1%**

(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)?

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

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

### WAEC Mathematics Questions and Answers 2022 Continues

(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 = 224?cm²

r:l = 2:5

r/l = 2/5

Cross multiply

2l/2 = 5r/2

L = 5r / 2

Total surface = ?rl + ?r²

= ?r (l + r)

24?/? = ?r (5r/2 + r )/ ?

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 = 1/2?r²h

= 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

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

### WAEC Mathematics 2022 Questions and Answers

(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

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

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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)?

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

### WAEC Mathematics 2022 Questions and Answers Continues…

(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 = 224?cm²

r:l = 2:5

r/l = 2/5

Cross multiply

2l/2 = 5r/2

L = 5r / 2

Total surface = ?rl + ?r²

= ?r (l + r)

24?/? = ?r (5r/2 + r )/ ?

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 = 1/2?r²h

= 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) =?rl

Area of hemisphere=2?r²

Total area of structure =?rl + 2?r²

=?r(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

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

### WAEC Mathematics Questions 2022 and Answers

(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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### WAEC Mathematics Questions 2022 and Answers

**Today’s WAEC Mathematics Answers: Loading…99.1%**

(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)?

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

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

### WAEC Mathematics Questions and Answers 2022 Continues

(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 = 224?cm²

r:l = 2:5

r/l = 2/5

Cross multiply

2l/2 = 5r/2

L = 5r / 2

Total surface = ?rl + ?r²

= ?r (l + r)

24?/? = ?r (5r/2 + r )/ ?

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 = 1/2?r²h

= 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

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

### WAEC Mathematics 2022 Questions and Answers

(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

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

**More Answers Loading…. Keep Refreshing**

If you have any questions about the WAEC Mathematics theory & Obj 2022, kindly let us know in the comment box.

**More On WAEC Mathematics Questions and Answers 2022**

We are happy you are here for the 2022 Mathematics answers. The complete solution will be made free in some minutes before the Mathematics examination.

WAEC Mathematics questions 2022, WAEC 2022 Mathematics questions and answers, Mathematics WAEC 2022, WAEC questions 2022/2023, Mathematics 2022,

(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)?

P^3/2 = rk/Q – ms

rk/Q = P^3/2 + ms

Q= rk/P^3/2 + ms

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)

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

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

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

### WAEC Mathematics 2022 Questions and Answers Continues…

(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

(?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

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

Total Surface Area = 224?cm²

r:l = 2:5

r/l = 2/5

Cross multiply

2l/2 = 5r/2

L = 5r / 2

Total surface = ?rl + ?r²

= ?r (l + r)

24?/? = ?r (5r/2 + r )/ ?

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

Volume = 1/2?r²h

= 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

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

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) =?rl

Area of hemisphere=2?r²

Total area of structure =?rl + 2?r²

=?r(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

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

### WAEC Mathematics Questions 2022 and Answers

(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

## Subjects Related To WAEC Mathematics Questions and Answers 2022

We want to hear from you concerning the **WAEC Mathematics Questions and Answers 2022**, kindly use the comment section below to get to us.

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