## Waec 2019 Mathematics Obj And Theory Now Available May/June Expo

**Waec 2019 Mathematics Obj And Theory Now Available May/June Expo**

*I PITY THOSE THAT COPY WRONG ANSWERS FROM WRONG WEBSITE !!!*

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*MATHS ANSWERS UNLOCKING PLEASE WAIT=====*

*MATHS OBJ UNLOCKING PLEASE WAIT…..*

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Maths Obj

1-10 BABDBADCBC

11-20 ACBCBBAACD

21-30 DCCBCCABBA

31-40 AABBDCDDCC

41-50 BBCDCCBCCD

Completed✅

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2019 Waec General Maths Special Symbols

* means multiplication

/ means division

^ means raise to power

= means equality

θ means theta

β means beta

α means alpha

Ø means empty set

μ means population mean

π means pi constant

√ means square root

ANSWERS….

Question 1.

110ₓ = 40₅

change all to base ten.

110ₓ = 40₅

1x²+1x ¹+0x⁰ = 45¹+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

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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

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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

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(4)

Since <PQR = <PRS = 90°

Using Pythagoras theorem

|PR|² = |PQ|² + |QR|²

|PR|² = 3² + 4²

|PR|² = 9 + 16

|PR|² = 25 PR = √25

|PR| = 5cm

Considering PRS

|PS|² = |PR|²+|SR|²

13² = 5² + |SR|²

169 = 25 + |SR|²

|SR|² = 169 – 25

|SR|² = 144

|SR| = √144 = 12cm

Hence the area of the quadrilateral = Area of triangle PQR + area of PRS

= 1/2bh + 1/2bh

= 1/2×4×3 + 1/2×12×5

= 6+30 = 36cm

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(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

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Question 6.

Draw the digram

6 ai) F = m ₁ m₂/d^2

Given F = 20N, m=25kg, m₂ = 10kg and d = 5m

20 = k(25)(10)/5^2

250k = 500

k = 500/250 = 2

=> Expression is: F = m ₁ m₂/d^2

6aii) making d subject,

d = √m ₁ m₂/f

d = √27.54/30

d = √60/30 = √2

d = √2m or 1.41m

6b) 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 = 68

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(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

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(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

(12a)

Given : siny = 8/17

Draw the right angle

From Pythagorean triple, third side is 15

Draw the right angle triangle

tan y = 8/15

tan y/1+2tany = 8/15/1+2(8/15) = 8/15/1+16/15

tany /1+2tan y = 8/31

(12b)

Amount shared = #300,000

Otobo’s share = #60,000

Ade’s share = 5/12 × #(300,000-60,000)

= 5/12 × #240,000

=#100,000

Adeobi’s share = #300,000 – (#60,000 + #100,000)

= 300,000 – 160,000

=#140,000

Ratio : Otobo : Ade : Adeola

60,000 : 100,000 : 140,000

60 : 100 : 140

6 : 10 : 14

3 : 5 : 7

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