Waec 2020 General Mathematics Essay & Obj Question and Answers Now Available


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WAEC 2020 GENERAL MATHEMATICS OBJ & THEORY ANSWERS

MATHS OBJ

MATHS-OBJ
1 -10-CBCDACDCCD
11 – 20-AADBDACBBC
21-30-BDDABDADAD
31-40-CDACCCCCDA
41-50-BBBCDCACDB


COMPLETED
==================================

(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)

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

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

 

(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

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

 

(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

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

 


(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





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

(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)
Diagram

(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


Solving/Typing…..[Keep Refreshing]


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