**WAEC MATHEMATICS QUESTIONS AND ANSWERS A’1 ASSURANCE**

Exam Time: Thursday 16th May, 2019

Mathematics 2 (Essay) – 09:30am – 12:00pm

Mathematics 1 (Objective) – 3:00pm – 4:30pm

**MATHEMATICS ANSWERS**

PLEASE MAKE SURE YOU WRITE WITH CLEAR AND GOOD WRITING

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**SAPIENTIA MATHS OBJ**

1-10: CABDBADCBC

11-20: BCBCBBAACD

21-30: BCCBCCABBA

31-40: AABDDCDDCC

41-50: BBCDCCACCD

No 1

(1a)

110x = 40s

Converting both sides to base 10, we have;

(1xX²)+(1xX¹)+(0×X^0) = (4×5¹) + (0×5^0)

X² + X + 0 = 20 + 0

X² + X – 20 = 0

X² + 5x – 4x – 20 = 0

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

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

Since X must be positive

X – 4 = 0

X = 4

(1b)

15/√75 + √108 + √432

= 15/√25×3 + √36×3 + √144×3

=15/5√3 + 6√3 + 12√3

= 3/√3 + 18√3

=3√3/3 + 18√3

= √3 + 18√3

= 19√3

-1a

1+4x/2 – 5+2x/7 < x-2

The LCM

7(1-4x)-2(5+2x)/14 < x-2/1

Cross and multiply

7-28x-10-4x+1<(x-2)

7-28x-10-4x<14x-28

CLT

7-10+28<14x+4x+28x

25<46x

Divide both sides by x

25/46<46x/46

X<25/46

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

(3b)

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

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

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

2^1X2^-3x=2^5(x-1)

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

1-3x=5x-5

-3x-5x=-5-1

-8x=-6

x=-6/-8

x=3/4

==≠======================

(2a)

The equation of the line through the points

A(-2,7) and B(2,-3)

Using the equation Y=mx+b

Where m=slope of the gradient, b=the intercept at the vertical axis

Hence slope=Change in Y/Change in X

=>Y2-Y1/X2-X1=Y-y1/X-x1

=(-3-7)/(2–2)=Y-7/X+2

=-10/4=(Y-7)/(X+2)

=4(y-2)=-10(x+2)

4y-28=-10x-20

4y=-10x+8

5x+2y=4

(2b)

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

5(5b-a)=1(8b+3a)

25b-5a=8b+3a

25b-8b=3a+5a

17b=8a

Therefore a/b=17/8

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

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

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No 4

(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 = 3No of green balls = 5No of blue balls = xProb.(red ball) = no of total outcome/no of possible outcome Pr(red) = 3/3+5+x = 1/63/8+x = 1/66(3) = 1(8+x)18 = 8 + xX = 18 – 8 = 10Therefore 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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(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

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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

(8bii)

|AC|/66 = Cos35°

|AC| = 66 x cos35°

= 66 x 0.8192

= 54.0672

Tan = 41.86°

Angle of elevation of top of tower from c = 41.85°

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(10)

130kg of tomatoes for #52,000

Half of the tomatoes

130/2 = 65kg sold at 30%

Profit = #52,000/2 = 26,000

#26,000 = 100%

X = 130%

X = 26000 × 130/100

= #33,800

Then 65kg was then sold at reduction of 12% per kg

Recall that the initial cost price = 52000/130

=400kg

65kg sold at = 33,000/65

=#520/kg

Then for 12% reduction

520 × 88/100 = 457.6/kg

(a)

The new selling price per kg = #457.6/kg

(b) 65kg – 5kg = 60kg

(60kg×457.6kg)+33,800

= #61,256.00

#profit = selling price /cost price × 1000/1

=61256/52000×100/1= 117.8

= 17.8%

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

(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

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

(13a)

RT^0=RS^0=90° (radius o targets)

T^0S=2TU^0 (angle at centre = 2 time angle at circumstance)

TOs=2*68°=136°

Now RT^0+RS^0+T^0S+SRT^0=360°(sum of angle in a quadrilateral)

90° + 90° + 136° + x = 360°

X+316°=360°

X=360°-316

X=44°

(13b)

Let tank B hold x litres

; Tank A hold (x+600)literally

3(x-100)=(x+600-100)

3x-300=x+500

3x-x=500+300

2x=800

X=800/2=400

Tank B holds 400 litres

Tank A holds (400+600)=1000litres

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**MATHEMATICS COMPUTER SYMBOLS**

WARNING: Please Study the above Symbols, If you Don’t want to write Rubbish in you MATHEMATICS Exam.

NOTE: WE WILL TRY TO MAKE OUR ANSWERS AVAILABLE IN PICTURE FORMAT, but still study the above symbols incase!!!

Know that this is a key subject Maths sub Now

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MATHS COMPUTER

SYMBOLS (1) / means division

or divide

➗

Examples:

2/2 means 2/2

2 whole no 3/4 means

2¾ M=wp+(1/3)yp^2

means m=wp + ¾y 2

(2) Log means

logarithm

Example: (8) Log5

(base means Log8 5

(3) * means

multiplication

Example: 2*2 means

2×2

(4) ^ means Raise to power

Examples:

3^(-1) means 3 -1

5^(2) means 5 2

30cm^(-2) means

30cm -2 (5) Tita means θ

Example: Tan tita

means tan θ

(6) Pie means π

Example: Pie R sqr h

means πr 2 h (7) Base means

subscript

Example: 2 Base 3

means A 3

(8) Sqr root means √

Examples: √3-√6+2√3 means

Root 3-root6+2root 3

√(¾) means sqr root

(3/4)

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(9) bar means a dash

on top a number of letter

Example: X bar

means

(10) —(1) means

equation 1

(11) Proportional means ∝

Example: R is

proportional to 3/4rut

m means R

∝ ¾

(12) Alpha means α (13) Beta means β

(14) Gamma means γ

(15) cube root means

∛

(16) Mu means μ

(17) Rho means ρ (18) Delta means δ

(19) Sigma means σ

(20) Tau means τ

(21) Ohm means Ω

(22) Lamda means λ

(23) Omega means ω (24) Intersection

means ∩

Example: B

intersection C means

B∩C

(25) Union means U Example: B union C

means BUC

(26) Factorial means !

(27) complements in

sets means ‘

(28) (aq),(g),(l),(s) is used in chemistry to

show

the state

of matter in equations

(29) Equivalent to

means ≡ (30) Not equal to

means ≠ (31) Quotation means “ ”

(32) Less or equal to

means ≤

(33) Greater or equal to

≥ Example: (3/4)>=(1/2)

means ¾ ≥ ½

(34) Plus or minus (or

+-) means ±

Example: (3/4)<=(1/2)

means ¾ ≤ ½

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