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WAEC 2019 Mathematics Questions & Answers | Theory/Obj PDF Download of 16th May 2019 is Here

5a) No of red balls = 3 No of green balls = 5 No of blue balls = x Prob.(red ball) = no of total outcome/no of possible outcome Pr(red) = 3/3+5+x = 1/6 3/8+x = 1/6 6(3) = 1(8+x) 18 = 8 + x X = 18 - 8 = 10 Therefore the no of blue ball = 10 (5b) Probability of picking a green ball P(g) = no of green balls/no of possible outcome P(g) = 5/3+5+10 = 5/18 =5/18 (6ai) F α M1M2/d² F = KM1M2/d² Given F = 20N, M1= 25kg, M2 = 10kg and d = 5m 20 = k(25)(10)/5² 250k = 500 k = 500/250 = 2 Expression is F = 2M1M2/d² (6aii) Making d subject d = √2M1M2/F d = √2 ×7.5×4/30 d = √60/30 = √2 d = √2m or 1.41m (6b) Draw the diagram X+X+60+X+80+X+40+X+20 = 540(sum of angles in a Pentagon) 5x + 200 = 540 5x = 540 - 200 5x = 340 X = 340/5 X = 68 SOLVING IN PROGRESS (5a) No of red balls = 3 No of green balls = 5 No of blue balls = x Prob.(red ball) = no of total outcome/no of possible outcome Pr(red) = 3/3+5+x = 1/6 3/8+x = 1/6 6(3) = 1(8+x) 18 = 8 + x X = 18 - 8 = 10 Therefore the no of blue ball = 10 (5b) Probability of picking a green ball P(g) = no of green balls/no of possible outcome P(g) = 5/3+5+10 = 5/18 =5/18 (6ai) F α M1M2/d² F = KM1M2/d² Given F = 20N, M1= 25kg, M2 = 10kg and d = 5m 20 = k(25)(10)/5² 250k = 500 k = 500/250 = 2 Expression is F = 2M1M2/d² (6aii) Making d subject d = √2M1M2/F d = √2 ×7.5×4/30 d = √60/30 = √2 d = √2m or 1.41m (6b) Draw the diagram X+X+60+X+80+X+40+X+20 = 540(sum of angles in a Pentagon) 5x + 200 = 540 5x = 540 - 200 5x = 340 X = 340/5 X = 68 (8a) 1/3x - 1/4(x+2)>_ 3x -1⅓ 1/3x - 1/4(x+2)>_3x - 4/3 Multiply through by the L. C. M(12), we have 4x - 3(x + 2)>_36x - 16 4x - 3x - 6 >_ 36x - 16 -6+16 >_36x + 3x - 4x 10 >_ 35x 35x _< 10 X = 10/35 X = 2/7 (8bi) Draw the triangle |AB|/66 = sin35 |AB| = 66sin35 = 66×0.5736 = 37.8576 Draw the right angled triangle |AD|/|AB| = Tan52 |AD| = 37.8576 × Tan52° = 37.8576 × 1.2799 = 48.45m Height of tower = 48.45m