WAEC

# Waec General Mathematics 2022 Questions And Answers

The Waec mathematics 2022 questions and answers have been released. The West African Examination Council (WAEC) General Mathematics essay and objective paper will now be written on Thursday, 2nd June 2022 as scheduled.

The Waec mathematics 2022 Questions will be a general subjects which will be written by all candidates of all departments both Commercials, Science, Art and Technicals.

• #### NECO Registration Deadline: Procedure and Process

The WAEC Maths 2 (Essay) paper will start at 8:30 am and will last for 2hrs 30mins while the Waec Mathematics 1 (Objective) exam will commence at 2:00 pm and will last for 1hr 30mins.

In this post, you will be seeing the WAEC Maths questions and answers 2022 for candidates that will participate in the examination from past questions.

The questions and answers below on WAEC Mathematics have been provided to assist candidates to understand the required standards expected in WAEC Mathematics Examination.

There is nothing like Waec mathematics expo 2022 online yet for now till the day approach.

### Waec General Mathematics 2022 Questions And Answers   (2a)

y= (Pr/m – P²r) -³/²

Multiply both index by -⅔

y-⅔= (Pr/m – P²r)-³/²*-⅔

y-⅔/1 = Pr/m – P²r/1

my-⅔ = Pr – mP²r

my-⅔/P-mP² = r (P-mP²)/P-mP²

m/(P-mP²)y⅔ = r

(2b)

y= -8, m=1, P=3

r = m/(P-mP²)y⅔

r = 1/(3-1*3²)* -8⅔

r = 1/(3-9)*(³√-8)²

r = 1/-6*(-2)² = 1/-6*4 = 1/-24

:. r = -1/24

(3)

Perimeter of minor segment= AB + AB

AB = 2AM = 2rSin36°

= 2*24.5*Sin36°

= 28.8m

AB = 72°/360° * 2*22/7* 24.5

= 72/360* 154 = 30.8m

:. Perimeter= 30.8+28.8

= 57.6m

(4a)

2(2y+10) = 5x-35 (angle at centre = twice at circle)

4y+20= 5x-35

5x = 4y+55

x = 4/5y +11……….(1)

(2y+10)+(2x+40) = 180 (supplementary angle)

2x+2y+50 = 180

x+y = 65……..(2)

Put (1) into (2)

4/5y +11+y =65

9/5y = 54

y = 5/9*54 = 30

From (2) x=65-y = 65-30= 35

(4b)

<ABC = 360 – (5x-35+40+2x+40)

= 360 – (175-35+40+70+40)

= 360-290

= 70°

(5a)

Given m= tan30° = 1/√3

n = tan45° = 1

m -n/mn= 1/√3 -1/(1/√3)(1)

= √3(1/√3 -1)

= 1 – √3

(5b)

15-x+x+10-x =20

25-x = 20

x= 25-20 = 5

PnB(both) = 5/20 = 1/4

(6a)

On x-axis: 2cm to 2units

On y-axis: 2cm to 10units

(6b)

(x+2)(x-4) = 0

x²-4x+2x-8 = 0

x²-2x-8= 0

mx² + nx + r = y

:. m=1, n= -2, r = -8

(6c)

= -27 -5/-5 -3

= -32/-8 = 4

(6d)

(x+2)(x-4) > 0

x+2<0 or x-4<0

x< -2 or x<4

And x-4>0

x+2>0 or x-4>0

x> -2 or x>4

:. x> -2

(10a)

[TABULATE]

Age X;3|| 4|| 5|| 6|| 7|| 8|| 9|| 10||

Number of children F; 2 || 6 ||5 || x=4 || 6 || 9 || 8 || 5||

Fx; 6 || 24 || 25 || 6x=24 || 42 || 72 || 72 || 50||

x – x̄; -4 || – 3 || – 2 || – 1 || 0 || 1 || 2 || 3||

(x – x̄)²; 16|| 9 || 4 || 1 || 0 || 1 || 4 || 9 ||

f(x – x̄)²; 32 || 54 || 20 || 4 || 0 || 9 || 32 || 45||

Σf = x + 41

Σfx = 6x + 291

Σf(x – x̄)² = 196

Mean x̄ = Σf/Σf

x̄ = 7 = (6x + 291)/(x +41)

7x + 287 = 6x + 291

7x – 6x = 291 – 287

x = 4

(10b)

S.D = √[(Σf(x – x̄)²]/Σf

S.D = √(196/x+41)

S.D = √(196/45)

S.D = 2.1

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