WAEC GCE 2018 Mathematics (Obj & Essay) Answer - Nov/Dec Expo
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**Free WAEC GCE NOV/DEC Maths Answer below ðŸ‘‡ðŸ‘‡ðŸ‘‡**

**Maths - Obj**

**1DCBBDCCADB**

**11ACBACCDADA**

**21DAAAABCDCB**

**31ABCDABCBCA**

**41DCABCDABBB**

**====GOODLUCK====**

Mr Spy - 8th, September 2018

Mr Spy - 8th, September 2018

==> 1a)

sn=n/2{2a+(n-1)d}

n=10, sn=130

130=10/2{2a+(10-1)d}

130=5(2a+ad)

130=10a+45d---------(eq1)

Tn=a+(n-1)d

T5=3*a=3a,n=5

3a=a+(5-1)d

3a=a+4d

3a-a-4d=0

2a-4d=0----------(eq2)

Solve equation 1 and 2 simultaneously

10a+45d=130*1

2a-4d=0*5

10a+45d=130

10a-20d=0

65d/65=130/65

d=2

(1b)

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

(1c)

L = a+(n-1)d

28 = 4+(? - 1)2

28 = 4+2n - 2

2n = 28 - 2

2n = 26

2n/2 = 26/2

n = 13

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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

5-x >1

5-1>x

4>x

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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30+3 = 52-3

2x+3)(3-) = (2-3)

3* (3-2) 43(3-2) = 5% (-)-361)

6x-23 +9 325I -5x3x3

62-3% -2x +9= 52-63+3

32- +9=562373

5x tor-FOC-36+3-9-0

obh-979.20 heyn

OL---|(-1)-64x+v-fe)

2x7

1! 121 168

= 1+ 121 +168

14

x= 1ltra cand

and 3: -47

and x=-9

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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

(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

(6b)

Draw the diagram

Using ?TOP

tan 28 = H/OP

OP = H/tan28

Then for ?ROP

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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WAEC GCE 2018 Mathematics (Obj & Essay) Answer - Nov/Dec Expo
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