# Waec 2024 general mathematics questions and answers

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Subject: General mathematics ESSAY & OBJ.

Date: 2nd — 2024

Time: 9:30am-12:00pm(ESSAY)

Time: 3:00pm-4:30pm(OBJ)

## OBJECTIVES:

1-10: ABBDBBBCAC
11-20: ABDDCCCBDD
31-40: DCBCABBCCD
41-50: DBABAAACBB

COMPLETED

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OBJECTIVE QUESTIONS

ESSAY:

MATHS TYPE FORMAT
(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)
Gradient= y2 -y1/x2 -x1
= -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

No1.

No2

No3

5a) m=tan30°= 1/√3 and n=tan45°=1

=> (m-n)/mn= (1/√3 -1)/(1/√3 ×1)

=> √3(1/√3 -1)=> 1-√3.
5b) let the probability that the a woman wear glasses be Pr(g) and that of wearing wristwatch be Pr(w)

=> Pr(g)=15/20 and Pr(w)=10/20

=> Pr(gnw)=Pr(g)×Pr(w) => 15/20 ×10/20=150/400=15/40=3/8

5a) m=tan30°= 1/√3 and n=tan45°=1

=> (m-n)/mn= (1/√3 -1)/(1/√3 ×1)

=> √3(1/√3 -1)=> 1-√3.
5b) let the probability that the a woman wear glasses be Pr(g) and that of wearing wristwatch be Pr(w)

=> Pr(g)=15/20 and Pr(w)=10/20

=> Pr(gnw)=Pr(g)×Pr(w) => 15/20 ×10/20=150/400=15/40=3/8

No6:

13a) Area=2πr²+2πrh

=> Area=2πr(r+h)

=> r+h=Area/(2πr)

=> h=(Area/(2πr))-r

given that area=209cm² and r=7cm and π=22/7

=> h=(209)/(2×22/7×7) -7

=> h=(209/44) -7