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### Author Topic: WAEC GENERAL MATHEMATICS EASY AND OBJECTIVE 2021/2022 ANSWERS  (Read 1728 times)

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#### NORTOLA ##### WAEC GENERAL MATHEMATICS EASY AND OBJECTIVE 2021/2022 ANSWERS
« on: September 19, 2021, 12:01:44 pm »

#### NORTOLA ##### Re: WAEC GENERAL MATHEMATICS EASY AND OBJECTIVE 2021/2022 ANSWERS
« Reply #1 on: September 20, 2021, 08:26:41 am »

#### NORTOLA ##### Re: WAEC GENERAL MATHEMATICS EASY AND OBJECTIVE 2021/2022 ANSWERS
« Reply #2 on: September 20, 2021, 10:17:47 am »
(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 = 224πcm²
r:l = 2:5
r/l = 2/5
Cross multiply
2l/2 = 5r/2
L = 5r / 2
Total surface = πrl + πr²
= πr (l + r)
24π/π = πr (5r/2 + r )/ π
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 = 1/2πr²h
= 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) =πrl
Area of hemisphere=2πr²
Total area of structure =πrl + 2πr²
=πr(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

(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

#### NORTOLA ##### Re: WAEC GENERAL MATHEMATICS EASY AND OBJECTIVE 2021/2022 ANSWERS
« Reply #3 on: September 20, 2021, 11:04:43 am »
Because of pictures and other things used this website and pin to write the general mathematics examking pin is 4822

#### Donlife ##### Re: WAEC GENERAL MATHEMATICS EASY AND OBJECTIVE 2021/2022 ANSWERS
« Reply #4 on: September 20, 2021, 01:02:10 pm »
This is wrong

#### manospace9

• Young Earlyfacian
• Posts: 3
• Vote for me: +0/-0
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• HomeTown: mbaise ##### Re: WAEC GENERAL MATHEMATICS EASY AND OBJECTIVE 2021/2022 ANSWERS
« Reply #5 on: September 20, 2021, 01:07:31 pm »

#### menegbobakel ##### Re: WAEC GENERAL MATHEMATICS EASY AND OBJECTIVE 2021/2022 ANSWERS
« Reply #6 on: September 20, 2021, 02:17:40 pm »
Totally wrong

#### passino ##### Re: WAEC GENERAL MATHEMATICS EASY AND OBJECTIVE 2021/2022 ANSWERS
« Reply #7 on: September 20, 2021, 02:19:50 pm »
Passino

#### menegbobakel ##### Re: WAEC GENERAL MATHEMATICS EASY AND OBJECTIVE 2021/2022 ANSWERS
« Reply #8 on: September 20, 2021, 02:25:40 pm »
MATHS OBJ

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