2018/2019 Waec Wassce Gce Mathematics Answers/Expo/Runz

WAEC GCE 2018 MathematicsAnswer
Waec gce mathematics expo WAEC GCE 2018 MathematicsAnswer

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Free WAEC GCE NOV/DEC Maths Answer below 👇👇👇
 
 
Maths – Obj
1DCBBDCCADB
11ACBACCDADA
21DAAAABCDCB
31ABCDABCBCA
41DCABCDABBB
 
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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
—————————————————————-

(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

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(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
—————————————————————-
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

—————————————————————-

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)
2×7
1! 121 168

= 1+ 121 +168

14

x= 1ltra cand

and 3: -47
and x=-9
—————————————————————-
(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•
—————————————————————-

(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
—————————————————————-
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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