WAEC GCE MATHS 2018 OBJ AND THEORY ANSWER – NOV/DEC EXPO

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Mathematics OBJ:


1-10 CCBBDCCADB

11-20 ACBACCDADA

21-30 DAAAABCDCB

31-40 ABCDABCBCA

41-50 DCABCDABBB


  _Completed_ best of luck


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THEORY QUESTION 


 


(1a)

S = ½n[a + L]

Where S = 130

S = n/2[a+a+(n-1)d]

2S = n[2a+(n-1)d]

2(130) = 10[2a + (10 - 1)d]

260 = 10[2a + 9d]

260 = 20a + 90d .....(I)

a + 4d = 3a...... (II) 

4d = 3a - a

4d = 2a

2d = a...... (III) 

Substitute a = 2d into Eqn (I) 

260 = 20(2d) + 90d

260 = 40d + 90d

260 = 130d

260/130 = 130d/130

d = 2

Hence common difference d = 2


(1b) 

First term a = 2d = 2(2) = 4


(1c) 

L = a+(n-1)d

28 = 4+(ń - 1)2

28 = 4+2n - 2

2n = 28 - 2

2n = 26

2n/2 = 26/2

n = 13


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(2a)


Draw the triangle


Using Pythagoras rule


a² = b² + c²

5² = 3² + c²

25 = 9 + c²

c² = 25 - 9

c² = 16

C = √16 = 4, Cos X = 4/5

5(cosx)² - 3 = 5((4/5))² - 3

= 5(4/5 × 4/5) - 3

= 16/5 - 3/1 L. C. M = 5

= 16 - 15/5 = 1/5.


(2b)


Volume of a pyramid

=1/3b×L×h

Volume of a cone = 1/3πr²h

Vp = volume of pyramid

Vc = volume of cone

Hp = height of pyramid

Hc = height of cone

Vp = Vc; Hp = Hc

Vp ± Vc

1/3bLh = 1/3πr²h

42 × 11 = 22/7r²

21 = 1/7 × r²

r² = 21 × 7

r² = 147

r = √147

r = 12 3/25cm



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(4a)

(2y+x) + (6y-2x+1) + 4y = 28 ...(i)

6y-2x + 1 = 4y... (II) 

2y+ x +6y - 2x + 1 +4 = 28 

12y + x +6 -2x + 1 + 4 = 28

12y - x + 1 = 28

12y - x = 29... (III) 

6y-2x + 1 = 4y, 6y-2x - 4y = 1

2y - 2x = -1... (iv)

24y - 2x = 54

2y - 2x = 1


22y/2w = 55/22 y = 2.5

12y - x = 27

12 (2.5) - x = 27

30 x 27 x=3


(b)

2y + x = 2 (2.5) + 3 = 5+3 = 8cm

6y - 2x + 1 = (2.5)-2 (3) + 1

= 15-6 + 1 = 10cm

4y = 4(2.5) = 10cm


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(5a)

5 - X > 1 9 + X >_ 8

5 - 1 > X X >_ 8 - 9

6 > X X >_ -1

Range is

-1_< X < 6 OR 6 >X >_ -1


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(5b) 

PQR + PSR = 180(supplementary angles of a cyclic quad) 

PQR + 56 = 180

PQR = 180 - 56

PQR = 124°

Next, join P to R

QRP = QPR(base angles of an isosceles)

PQR + 2QRP = 180(Sum of angles in a triangle)

124 + 2QRP = 180

2QRP = 56°

QRP = 56/2 = 28°

PRS = 90°(angle in a semi - circle)

= 28 + 90

= 118°




And 9-x>or equals to 8


9-8>or equals to X

1>or equals to x


Therefore 4>x and 1>or equals to x

Final answer.

4>x<or equals to 1


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(6ai)

The profit y = X²/8 + 5x

y = GHc20,000.00

Hence 20,000 = X²/8 + 5x

160,000 = X² + 40x

X² + 40x - 160,000 = 0

Since X is in thousands 

X² + 40x - 160 = 0


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(6aii) 

Using quadratic formula 

X = -b±√b² - 4ac/2a

Where; a = 1, b = 40 & c = -160

X = -40±√40² - 4(1)(-160)/2(1)

X= -40 ±√1600 + 640/2

X = -40 ±√2240/2

X = -40 ± 47.32/2

X = -40±47.32/2

= 7.32/2

X = 3.66

X ≈ 4


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(6b) 

Draw the diagram 

Using ΔTOP

tan 28 = H/OP

OP = H/tan28


Then for ΔROP

tantita = H/2/OP

OP = H/2/tantita

Hence H/tan 28 = H/2/tantita

tantita = H/2 × tan 28/H

tantita = tan28/2

Hence Tita = 28/2 = 14.

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(7a)

2

S(2x³ - 4x + 6)dx

1

= 2x³+¹/3+1 - 4x¹+¹/1+1 + 6x]2

1


=2x^4/4 - 4x²/2 +6x]2, 1

= x^4/2 - 2x² + 6x]2, 1

=(2^4/2 - 2(2²) + 6(2)) - (1^4/2 - 2(1)² + 6(1))

=(8 - 8 + 12) - (1/2 - 2 + 6)

=12 - 4½

= 7½


(7b) 

Given; P^-1 = (-1 1)

(4 -3)

P = (p-1)^-1 = C^T/|p^-¹|

=(-3 -1)

(-4 -1)/3 - 4

=(-3 -1)

(-4 -1)/-1

=(3 1)

(4 1)


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(8)

(I) Draw The Diagram


(II)

V = 1/3 Ah, = x r²

V = 1/3 xr²h

V = 4.158 liters

V = 4.158cm³, V = 1/3 xr²h

4158 = 1/3 x 22/7 x 21 x 21 x h

4158 x 3 = 22 x 63h

h = 4158/21 x 22 = 9cm

h = 9cm


(8b) 

d = 28cm,  r = d/2 = 28/2 = 14cm

V2 = 1/3 xr²h

V2 = 1/3 x 22/7 x 14 x 14 x 9

V2 = 1/2 x 22/1 x 2 X 14 x 3

= 1848cm³

V2 = 1.845 liters


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10a )

area of the farmland = 7200 m ^ 2

length X breath = 7200 m ^ 2.. . ( 1 )

perimeter = 360 m

2length +2breath = 360m . .. ( 2)

LXB = 7200 .. . ( 1 )

2L + 2B = 360 . .. ( 2)

i ) solving eqn ( 2) and ( 1)

L = 7200 / B .. .( 3)

put eqn ( 3) into ( 2)

2( 7200 / B ) +2B = 360

14400 / B + 2B / 1= 360

14400 +2B ^ 2/ B = 360

14400 +2B ^ 2= 360 B

2B ^ 2-360 B + 14400 = 0

B = 120 or 60

the maximum value is 120

hence B = 120 m

L = 7200 / B

L = 7200 / 120

L = 60 M

: . the length is 60m and the breath is 120m

or

the breath is 60 m and the length is 120m


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