# Waec 2019 Mathematics Obj & Essay Answer Now Available.

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

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

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

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

=========================

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 ========================

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 ========================

(11ai) ar² = 1/4 ......(1) ar^5= 1/32 .....(2) Divide eqn (2) by eqn(1) ar^5/ar² = 1/32÷1/4 r³ = 1/32 × 4/1 r³= 1/8 r³ = 2-³ r = 2-¹ r = 1/2 Common ratio = 1/2 Put this into eqn (1) a(1/2)² = 1/4 a(1/4) = 1/4 a = (1/4)/(1/4) = 1 First term, a = 1

(11aii) Seventh term, T7 = ar^6 =(1)(1/2)^6 =1/64 (1b) Given : X = 2 and X = -3 (X - 2)(X + 3) = 0 X² + 3x - 2x - 6 , 0 X² + x - 6 = 0 Comparing with ax²+bx+c = 0 a = 1 b = 1 C = -6

## No comments