**MATHS OBJ**1-10: CBCDACDCCD

11-20: AADBDACBBC

21-30: BDDABDCDAD

31-40: CBACCCCCDA

41-50: BBBADCCABA

**COMPLETED**

**2023 WAEC GENERAL MATHEMATICS ANSWERS.**

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**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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