# WAEC 2020/2021 Mathematics Essay /Theory and Obj) Question and answers is out

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General Mathematics/Mathematics (Core) 1 (Objective)*

_3:00 pm – 4:30 pm_

Mathematics Obj is out....

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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)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)

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 = 224Ï€cm²

r:l = 2:5

r/l = 2/5

Cross multiply

2l/2 = 5r/2

L = 5r / 2

Total surface = Ï€rl + Ï€r²

= Ï€r (l + r)

24Ï€/Ï€ = Ï€r (5r/2 + r )/ Ï€

224 = 5r²/2 + r²/1

L.c.m = 2

448 = 5r² + 2r²

448 / 7= 7r²/7

r² = 64

r = √64 = 8cn

L = 5*8/2 = 20cm

(4b)

Volume = 1/2Ï€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

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

=== SECTION B ====

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

(7a) Click here to view image

Using Pythagoras theorem, l²=48² + 14²

l²=2304 + 196

l²=2500

l=√2500

l=50m

Area of Cone(Curved) =Ï€rl

Area of hemisphere=2Ï€r²

Total area of structure =Ï€rl + 2Ï€r²

=Ï€r(l + 2r)

=22/7 * 14 [50 + 2(14)]

=22/7 * 14 * 78

=3432cm²

~3430cm² (3 S.F)

(b)

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

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

(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

(7a)

Total surface area = url + 2Ï€r²

=Ï€r(l + 2r)

Draw the diagram

From pythagoras theorem

Hyp² = Adj² + Opp²

L² = 14² + 48²

L² = 196 + 2304

L² = 2500

L = /2500 = 50m

=Ï€r(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

(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

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