**WAEC 2019 Mathematics Expo Answer – Free Maths WAEC Expo 2019 May/June Runz**

2019 WAEC

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**MATHS OBJ**=================

Maths Obj

**1-10 CABDBADCBC**

11-20 ACBCBBAACD

21-30 DCCBCCABBA

31-40 AABBDCDDCC

41-50 BBCDCCBCCD

11-20 ACBCBBAACD

21-30 DCCBCCABBA

31-40 AABBDCDDCC

41-50 BBCDCCBCCD

hsudfgc

(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

(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

(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

(11b)

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

SOURCE:

**May/June 2019 WAEC Mathematics Expo Answer – Free Maths WAEC Expo 2019 May/June Runz**

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