2018 WAEC GCE MATH powered by Engr NASER DAHER
WAEC GCE MATHEMATICS THEORY AND OBJECTIVES (2018
Our work is to DELIVER COMFIRMED AND VERIFIED
WAEC
JAMB
NECO
NABTEB
BECE
ANSWERS
_OBJ LOADING_
(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
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(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
(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
(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
(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°
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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
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