[WAEC]: MATHEMATICS ANSWERSS

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

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

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

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

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

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

(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

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

(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

(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°

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

(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

KEEP REFRESHING THIS PAGE FOR MORE ANSWERS

SUBSCRIBE NOW

@N1000 NIGERIA

@GHS30 GHANA

WhatsApp SAPIENTIA:08079678734 OR +233577054983

KEEP REFRESHING THIS PAGE FOR FREE ANSWERS ARE LOADING………..

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

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

SUBSCRIBE NOW @N1000 NIGERIA@GHS30 GHANA SAPIENTIA:08079678734 OR +233577054983

 

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)
🔰 SAPIENTIA:08079678734 OR +233577054983🔰
(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

SUBSCRIBE NOW 

@N1000 NIGERIA

@GHS30 GHANA

🔰 SAPIENTIA:08079678734 OR +233577054983🔰

 

Be the first to comment

Leave a Reply

Your email address will not be published.


*