2024/2025 Waec SSCE Further Mathematics Expo Questions and Answers

Answers Loading….. ….. Keep Refreshing this Page. FURTHER MATHS ANSWERS (1) 5^4^(3x/4 -1) + 5^3(x-1)/5^(3x – 2) 5^3x/4 – 4 + 5^3x – 3/5^3x – 2 5^3x – 4 + 5^3x – 3/5^3x – 2 (5^3x ÷ 5^4) + 5^3x ÷ 5^3÷5^3x – 2 5^3x/625 + 5^3x/125÷5^3x- 2 5^3x + 5(5^3x) ÷ 5^3x-3 5^3x + 5(5^3x)/625 ÷ 5^3x/25 Let 5^3x = y y + 5(y)/625 ÷ y/25 y + 5y/625 × 25/y =6y/625 × 25/y =6/25 1(b) F(x+2)-6x^2+5x-8 f(5 X+2=5z X=5-2 X=3 6x^2 +5x-8 =6(3)^2 + 5(3)-8 =54+15-8 =6! (2a) Using y2 – y1/x2 – x1 Where y2 = 7, y1 = -5, X2 = -2, and X1 = 7 7 – -5/-2 – 7 =7+5/-9 =12/-9 =4/-3 Coordinate points : -4/3(3 : 2) =-12/3 : -8/3 = -4 : -2⅔ X = (-4, 2⅔) (2b) 2/1-√2 – 2/2+√2 =2(2+√2)-2(1-√2)/(1-√2)(2+√2) =4+2√2 – 2+2√2/2+√2-2√2 – 2 =2 + 4√2/-√2 =(2+4√2)(-√2)/-√2(-√2) = -2√2 – 4(2)/2 = -8 – 2√2/2 = -4 – √2 (3) Sn = A/2[2n+(n-1)d] Where Sn = 165 a = -3, d = 2 165 = A/2[2(-3)+(n-1)2] 165 = n[-6+2n-2]/2 165×2 = n[2n – 8] 330 = 2n² – 8n 2n² – 8n – 330 = 0 n²-4n-165 = 0 Using -b±√b²-4ac/2a 4±√-4²-4(1)(-165)/2(1) 4±√16 + 660/2 4±√676 = 4±26/2 4+26/2 = 30/2 = 15 terms (4) Draw the right angled triangle Using Pythagoras theorem Third side = √(p+q)² – (p-q)² =√(p+q+p-q)(p+q-p+q) Difference of two squares. =√(2p)(2q) =√4pq Adjacent side = 2√pq Tanx = opp/adj = p – q/2√pq 1 – tan²X = 1-(p-q)²/4pq =(4pq)-(p²-2pq+q²)/4pq = -p²+6pq-q²/4pq = -(p² – 6pq + q²)/4pq (5) Draw the diagram Using cosine law Cos∅ = 16²+10²-14²/2(16)(10) Cos∅ = 256 + 100 – 196/320 Cos∅ = 160/320 Cos∅ = 0.5 ∅ = cos-¹(0.5) ∅ = 60° Angle between 10N and 16N = 180 – ∅ (sum of angles on a straight line) = 180 – 60 =120° (6) Draw the diagram Taking moment about the pivot, (T × 25)=(50×10)+(20×45) 25T = 500 + 900 25T = 1400 T = 1400/25 T = 56N (7) In a tabular form Under class interval: 1-5, 6-10, 11-15, 16-20, 21-25, 26-30 Under class mark (X): 3, 8, 13, 18, 23, 28 Under X-Xbar: -10, -5, 0, 5, 10, 15 Under frequency: 18, 12, 25, 15, 20, 10 Ef = 100 Under f(X – XbarA): -180, -60, 0, 75, 200, 150 f(X – XbarA) = 185 Where xA = 13 Mean = xA + Ef(X – Xbar)/Ef =13 + 185/100 =13 + 1.85 =14.85years PLS NOTE THAT XBAR LOOKS LIKE X WITH MINUS SIGN ON TOP. (10a) given 4x² – px +1 = 0 For real roots: b² – 4ac >0 (-p) ² – 4(4) (1) > 0 p² – 16 > 0 p² >16 p > ± 4 (10bi) Given: (1 +3x)⁶ Using pascal’s triangle: 1, 6, 15, 20, 15, 6, 1 (1)⁶(3x)º + 6(1)⁵ (3x)¹ + 15 (1)⁴ (3x)₂ + 20(1)³ (3x)³ + (15) (1)²(3x)⁴+6(1)¹(3x)⁵ + 1(1)º(3x)⁶ 1 + 6(3x) + 15 (9x²) + 20 (27x³) + 15 (81x⁴) + 6(243 x⁵) 729x⁶ 1 + 18x + 135x² + 540x³ + 1215x⁴ 1458x⁵ + 729x⁶ (ii) (1.03)⁶ = (1 + 3(0.01)] Therefore (1.03)⁶ = 1 + 18(0.01) + 135 (0.01)² + 540(0.01)³ + 1215(0.01)⁴ + 1458 (0.01)⁵ + 729 (0.01)⁶ + 1 + 0.18 + 0.0135 + 0.005 + 0.00001215 + 0.0000001458 + 0.000000000729 = 1.1940523 = 1.194 (4s.f) (12) prob (pass) = 60% = 60/100 =3/5 Prob (fail) 1-3/5 = 2/5 (a) Prob (atleast two failed) = 1 – prob (ome pass) = 1 – 10Ci (3/5)¹ (2/5)⁹ 1 – (10!/9!) (3/5) (2/5)⁹ = 1 – 10 (3/5) (2/5)⁹ = 1 – 10 (3/5) (0.000262144)) = 1 – 0.001572864 = 0.9984 (12b) prob (exastly half passed) = 10C5 (3/5)⁵ (2/5)⁵ = 10!/5!5! (6/25)⁵ = 252 (6/25)⁵ = 252 x 0.0007962624 = 0.2007 (12c) prob (at most two failed) = prob (zero/fail) + prob (one/fail) +prob (two/fail) = 10C (2/5)º (3/5)10 + 10C (2/5)¹ (3/5)⁹ + 10 C2(2/5)² (3/5)⁸ = (3/5)10 + 10 (2/5) (3/5)⁹ + 45 (2/5)²(3/5)⁸ = 0.060466176 + 0.040310784 + 0.120932352 = 0.1673 (14a) Draw the diagram (14b) From the diagram a = v – 20/4 2.5 = v – 20/4 V – 20 = 10 V = 10 + 20 = 30m/s acceleration/retardation = 3/4 2.5/30/T-12 = 3/4 2.5(T – 12)/30 = 3/4 (T – 12) = 30×3/2.5×4 = 90/10 = 9 T – 12 = 9 T = 9 + 12 = 21 t = T – 12 t = 21 – 12 t = 9secs (14c) Total distance of the journey = Area of BCDI + Area of AFEO + Area of DFI = 1/2(12+8)10 + 1/2(9×10) + 1/2(21+21)20 = (20/2)10 + 90/2 +(42/2)20 =10(10) + 45 + 21(20) =100 + 45 + 420 = 565m

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