1-10: CBCDACDCCD

11-20: AADBDACBBC

21-30: BDDABDCDAD

31-40: CBACCCCCDA

41-50: BBBADCCABA

**MATHS OBJ**1-10: CBCDACDCCD

11-20: AADBDACBBC

21-30: BDDABDCDAD

31-40: CBACCCCCDA

41-50: BBBADCCABA

**COMPLETED**

**2023 WAEC GENERAL MATHEMATICS ANSWERS.**

**NOTE: Waec 2023 general mathematics is not out yet but if you need the questions and answers kindly subscribe with EXAMLORD now. Our WhatsApp contact is 08063608463.**

**ESSAY**

__(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)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)
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__

__======================================================__

__(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__

**COMPLETED**

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