#WAEC 2021: Mathematics Practical Questions & Answers | WAEC Maths Expo 2021 100%

WAEC 2021: Mathematics Science Practical Questions & Answers | WAEC Maths Verified Expo 2021

This page is for those in search of the WAEC 2021: Mathematics Science Practical Questions & Answers to prepare for a successful examination in 2021.

Below are Maths waec answers and answer 2021 for Ghana, Nigeria and other west African countries 

(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)

loading…

(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

(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

(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

Make money writing on TDPel Media

-6d = 2m + 6n
-6d/-6 = 2m+6n/-6
d = -m/3 – n

d = -1/3m – n

(9c)
Speed = 20/4, average speed = 5km/h

(12a)
BCD=ABC=40°(alternate D)

DDE=2*BCD(

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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