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 Engr NASER DAHER 09030302775 (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 www.mrwhyhurry.blospot.com ___________________________________________________ 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° www.mrwhyhurry.blospot.com 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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