**UPCOMING NEXT**: WAEC 2019 Economics Obj And Essay Answer Now Available

Mathematics OBJ:

01-10: CABDDADCBC

11-20: BCBCBBAACD

21-30: DCCBCCABBA

31-40: AABDDCDDCC

41-50: BBCDCCBCCD

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Question 1.

110ₓ = 40₅

change all to base ten.

110ₓ = 40₅

1x²+1x ¹+0x⁰ = 4×5¹+0×5⁰

x² + x + 0 = 20

x² + x – 20 = 0

Solve quadratically

x² + 5x – 4x – 20 = 0

x(x+5) – 4(x+5) = 0

(x-4)(x+5) = 0

Either x-4 = 0 or x + 5 = 0

x = -5 or 4

therefore, x= 4

1b)

15/√75 + √108 + √432

= 15/√253 + √363 + √144×3

= 15/5√3 + 6√3 + 12√3

= 3/√3 + 18√3

= 3√3/3 = 18√3

= √3 + 18√3

=19√3

Question 2.

M = -3-7/2+2 = m = (y₂-y₁/-10)/4

m = -5/2

Equation line = y – y₁ = m(x-x₁)

= y – 7 = -5/2(x+2)

= 2y – 14 = -5(x+20)

= 2y – 14 = -5x – 10

5x + 2y – 14 + 10 = 0

5x + 2y – 4 = 0

2b) 5b -a /8b + 3a = 1/5

5b – a = 1 * 3

8b + 3a = 5

15a – 3b = 3

8b + 3a = 5/23b = 8

b = 8/23 = 0.35

using equation (1)

5b – a = 1

5(0.35) – a = 1

1.75 – a = 1

-a = 1 – 1.75

-a = -0.75 (- divide -)

a = 0.75

a/b = 0.75/0.35

a/b = 2.14

Question 3.

Ali : Musah : Yusif = ₦420,000

3 : 5 : 8

Sum of ratio shared;

3 + 5 + 8 = 16

therefore, Ali share = 3/16 * ₦420,000

= ₦78,750

Musah share = 5/16 * ₦420,000

= ₦131,250

Yusif share = 8/16 * ₦420,000

= ₦210,000

therefore, sum of Ali + Yusif = ₦78,750 + ₦210,000

=₦288,750

3b)

Solve: 2(1/8)^x = 32^x-1

2^1 x (8^-1) = 32^x-1

2^1 x (8^-1) = 32^x-1

2^1 x (2^3(-1))^3 = 2^5(x-1)

2^1 x (2^-3)^x = 2^5x-5

2^1 x 2^-3x = 2^5x-5

2^1+(-3x) = 2^5x-5

2^1-3x = 2^5x-5 (2 cancel 2)

1-3x = 5x -5

-3x – 5x = -5 -1

-8x/-8 = -6/-8

x = 3/4

========

(4)

Since (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

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