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2017/2018 Waec Gce Wassce Mathematics Questions And Answer Expo/runs Free by Excellentresult(m): 3:53pm On Sep 14, 2018 |
+2348032155450 Maths-Obj 1DCBBDCCADB 11ACBACCAADA 21DAAAABCDCB 31ABCDABCBCA 41DCABCCABBB (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 ___________________ ___________________ ___________ (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 ___________________ _____________ (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° 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 ___________________ ______________ (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 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. ___________________ ______________ ( (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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