**Monday 17th August 2020**

General mathematics(core) 2(Essay) – 9:30am – 12:00noon (2hrs 30mins)

General mathematics(core) 2(objective) – 3:00pm – 4:30pm

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KEEP REFRESHING THIS PAGE IN EVERY 5 MINUTES STARTING FROM 1HR TO THE EXAM

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**MATHS OBJ
1-10: CBCDACDCCD
11-20: AADBDACBBC
21-30: BDDABDCDAD
31-40: CBACCCCCDA
41-50: BBBCDCACDB**

**COMPLETED**

WAEC Mathematics Questions and Answers 2022. WAEC Maths Expo for Theory & Objective (OBJ) PDF: verified & correct expo Solved Solutions, 2022 WAEC MATHEMATICS ANSWERS (MAY/JUNE). 2022 WAEC EXAM Mathematics Questions and Answers

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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
=$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)
Draw the diagram
CBD = CDB(Base angles of an issoceles triangle)
BCD + CBD + CDB = 180°(sum of angles in a triangle)
2CDB + BCD = 180°
2CDB + 108° = 180°
2CDB = 180° - 108° =72°
CDB = 72/2 = 36°
BDE = 90°(angle in a semi-circle)
CDE = CDB + BDE
= 36° + 90°
= 126°
(3b)
(CosX)² - SinX/(SinX)²+ CosX
Using Pythagoras theorem, third side of triangle
y² = 1² + |3²
y² = 1 + 3 = 4
y = square root e = 2
Cos X = 1/2(adj/hyp)
Sin X = root 3/2(opp/hyp)
(CosX)² - SinX/(SinX)² + CosX
= (1/2)² - root3/2 / (root3/2)² + 1/2
= 1/4 - root3/2 / 3/4 + 1/2
= 1 - 2root3/4 / 3+2/4
= 1-2root3/5
================
(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
===============
(5a)
Prob(2) = no of 2s/Total outcomes
0.15 = m/32+m+25+40+28+45
0.15 = m/m + 170
m = 0.15m + 25.5
m - 0.15m = 25.5
0.85m = 25.5
m = 25.5/0.85 = 30
(5b)
Number of times dice was rolled = m + 170
= 30 + 70
= 200
(5c)
Prob(even number) = no of even numbers/Total outcome
= m+40+45/200
=30+40+45/200
=115/200
= 23/40 = 0.575
===============
(7a)
Total surface area = url + 2πr²
=πr(l + 2r)
Draw the diagram
From pythagoras theorem
Hyp² = Adj² + Opp²
L² = 14² + 48²
L² = 196 + 2304
L² = 2500
L = /2500 = 50m
=πr(L + 2r)
= 22/7 ×14(50 + 2(14))
= 44(50 + 28)
= 3432m²
Total surface area = 3432m²
~3430m²(to 3s.f)
(7b)
Five years ago,
Let Musa's age = x
Let Sesay's age = y
X - 5 = 2(Y - 5)
X - 5 = 2y - 10
X - 2y = 5 - 10
X - 2y = -5 ..... (1)
-X + y = 100 ..... (2)
-3y = -105
Subtracting eqn 2 from 1
-3y/3 = -105/-3
y = 35
Sesay's present age = 35 years
===============
(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
======================
(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
=====================
(11a)
(11b)
Given 8y+4x=24
8y=-4x + 24
y=4/8x + 24/8
y=-1/2x +3
Gradient = -1/2
Using m = y-y/x-x¹ and given (x¹=-8) (y¹=12)
-1/2=y-12/x+8
2(y-12)=-x-8
2y-24=-x-8
2y+x=24-8
2y+x=16
=====================
(12)
=================**

WAEC Mathematics Questions and Answers 2022. WAEC Maths Expo for Theory & Objective (OBJ) PDF: verified & correct expo Solved Solutions, 2022 WAEC MATHEMATICS ANSWERS (MAY/JUNE). 2022 WAEC EXAM Mathematics Questions and Answers

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