2019 WAEC Mathematics Question & Answer

MATHS OBJ:

1-10: CABDBADCBC
11-20: BCBCBBAACD
21-30: BCCBCCABBA
31-40: AABDDCDDCC
41-50: BBCDCCACCD

COMPLETED….

MATHS THEORY:

(1)

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(2)

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(3)

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(4)

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(5a)
No of red balls = 3
No of green balls = 5
No of blue balls = x
Prob.(red ball) = no of total outcome/no of possible outcome
Pr(red) = 3/3+5+x = 1/6
3/8+x = 1/6
6(3) = 1(8+x)
18 = 8 + x
X = 18 – 8 = 10
Therefore the no of blue ball = 10

(5b)
Probability of picking a green ball
P(g) = no of green balls/no of possible outcome
P(g) = 5/3+5+10 = 5/18
=5/18

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(6ai)
F α M1M2/d�
F = KM1M2/d�
Given F = 20N, M1= 25kg, M2 = 10kg and d = 5m
20 = k(25)(10)/5�
250k = 500
k = 500/250 = 2
Expression is
F = 2M1M2/d�

(6aii)
Making d subject
d = √2M1M2/F
d = √2 �7.5�4/30
d = √60/30 = √2
d = √2m or 1.41m

(6b)
Draw the diagram
X+X+60+X+80+X+40+X+20 = 540(sum of angles in a Pentagon)
5x + 200 = 540
5x = 540 – 200
5x = 340
X = 340/5
X = 68

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(8a)
1/3x – 1/4(x+2)>_ 3x -1⅓
1/3x – 1/4(x+2)>_3x – 4/3
Multiply through by the L. C. M(12), we have
4x – 3(x + 2)>_36x – 16
4x – 3x – 6 >_ 36x – 16
-6+16 >_36x + 3x – 4x
10 >_ 35x
35x _< 10
X = 10/35
X = 2/7

(8bi)
Draw the triangle
|AB|/66 = sin35
|AB| = 66sin35 = 66�0.5736 = 37.8576

Draw the right angled triangle
|AD|/|AB| = Tan52
|AD| = 37.8576 � Tan52� = 37.8576 � 1.2799 = 48.45m
Height of tower = 48.45m
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(10)

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(11)

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(12a)
Given : siny = 8/17
Draw the right angle
From Pythagorean triple, third side is 15
Draw the right angle triangle
tan y = 8/15

tan y/1+2tany = 8/15/1+2(8/15) = 8/15/1+16/15

tany /1+2tan y = 8/31

(12b)
Amount shared = #300,000
Otobo’s share = #60,000
Ade’s share = 5/12 � #(300,000-60,000)
= 5/12 � #240,000
=#100,000

Adeobi’s share = #300,000 – (#60,000 + #100,000)
= 300,000 – 160,000
=#140,000

Ratio : Otobo : Ade : Adeola
60,000 : 100,000 : 140,000
60 : 100 : 140
6 : 10 : 14
3 : 5 : 7

COMPLETED….

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