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

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**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 **

**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

-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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