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(4)
Since 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

(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

(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| = 37.8576 × Tan52° = 37.8576 × 1.2799 = 48.45m

(10)

(11ai)
ar² = 1/4 ……(1)
ar^5= 1/32 …..(2)
Divide eqn (2) by eqn(1)
ar^5/ar² = 1/32÷1/4
r³ = 1/32 × 4/1
r³= 1/8
r³ = 2-³
r = 2-¹
r = 1/2
Common ratio = 1/2
Put this into eqn (1)
a(1/2)² = 1/4
a(1/4) = 1/4
a = (1/4)/(1/4) = 1
First term, a = 1
(11aii)
Seventh term, T7 = ar^6
=(1)(1/2)^6
=1/64
(11b)
Given : X = 2 and X = -3
(X – 2)(X + 3) = 0
X² + 3x – 2x – 6 , 0
X² + x – 6 = 0
Comparing with ax²+bx+c = 0
a = 1
b = 1
C = -6

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

MATHS OBJ
1-10: CABD-ACCBC
11-20: -CBC
21-30:
31-40:
41-50:

43) C, 44)D, 45) B,46) C, 47) A,17)A, 18)A, 19)/B

23) B, 24) B, 48) C, 49) C, 50) A, 16) C, 20)D

Use this very accurate.
Maths-Obj