WAEC 2019 MATHEMATICS EXPO

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(4)
Since <PQR = <PRS = 90°
Using Pythagoras theorem 
|PR|² = |PQ|² + |QR|²
|PR|² = 3² + 4²
|PR|² = 9 + 16
|PR|² = 25     PR = √25
|PR| = 5cm
Considering PRS
|PS|² = |PR|²+|SR|²
13² = 5² + |SR|²
169 = 25 + |SR|²
|SR|² = 169 - 25
|SR|² = 144
|SR| = √144 = 12cm

Hence the area of the quadrilateral = Area of triangle PQR + area of PRS
= 1/2bh + 1/2bh
= 1/2×4×3 + 1/2×12×5

= 6+30 = 36cm


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
name

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(6ai)
F α M1M2/d²
F = KM1M2/d²
Given F = 20N, M1= 25kg, M2 = 10kg and d = 5m
20 = k(25)(10)/5²
250k = 500
k = 500/250 = 2
Expression is 
F = 2M1M2/d²

(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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10)
130kg of tomatoes for #52,000 
Half of the tomatoes 
130/2 = 65kg sold at 30%
Profit = #52,000/2 = 26,000
#26,000 = 100%
X = 130%
X = 26000 × 130/100
= #33,800

Then 65kg was then sold at reduction of 12% per kg
Recall that the initial cost price = 52000/130
=400kg
65kg sold at = 33,000/65
=#520/kg
Then for 12% reduction 
520

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