# WAEC Mathematics Questions and Answers 2022 (100% Theory & Obj)

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.

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### WAEC Mathematics Questions and Answers 2022

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### WAEC Mathematics Questions 2022 and 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

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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