2020/2021 WAEC/NECO Answers: WAEC 2021 - MATHS ANSWERS



+234 8108515604[email protected]
+234 703070473508138274394

Notice
Our 2021 NECO Expo Subscription for answers supply ends by Sunday Night. Note: we don't accept calls.

CRK-OBj
1DABDBABCDA
11DDCADBCCDD
21ACADBBCBDC
31BDDCBDBBDB
41BCCBDDCACB

This is actually the WAEC 2019 - MATHS ANSWERSAnswers as questions sent to our subscribers on WHATSAPP, if you are looking for 2021 WAEC/NECO Answers midnight before the exam, try contacting us.



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


(1)

110â‚“ = 40â‚…

change all to base ten.

110â‚“ = 40â‚…


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


name

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

(2)

(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




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

(3a)

Total ratio =3+5+8=16

Ali's share =3/16 * 420,000/1

=210,000

Son's of Ali and Yusuf's share = N78,750+ N210,000

= N288,750

 

(3b)

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

21(2-3)x=25(x-1)

21* 2-3x=25(x-1)

2-3x+1=25(x-1)

 

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

-3x+1=5x-5

-5x-3x= -5-1

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

x=3/4

name

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

(4)
Since 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
name

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


(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
name

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

(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
name

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

(8a) 
1/3x - 1/4(x+2)>_ 3x -1â…“
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
name



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


(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%
name

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

(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
name

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

(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
name



How do I getWAEC 2021 - MATHS ANSWERSBEFORE THE EXAM?
. You can get 2020/2021 Exam answers directly from examplaza.com, visit HERE NOW
Tags:

FAQ

WHATSAPP GROUP: Related Posts

2021 Neco Specimen | 2021 Neco 2021 NECO SPECIMEN | 2021 NECO SPECIMENS PRACTICAL ON PHYSICS


2021 NECO SCIENCE SPECIMENS

Get all your questions answered about the 0ne-time-only NECO exams for free and get a schedule of events. We will also help you with mock runs, tips on how best to prepare for the exam and even what to wear during the whole process


BEST NECO EXPO 2021 QUESTIONS AND ANSWERS | Neco Expo

Are you attempting the 2021 National Examinations Council (NECO) Senior Secondary School Examination (SSCE)? If yes, you can download or copy the timetable for the exam from this page.


2021 NECO OFFICIAL TIMETABLE

2021 NECO Science Specimens: 2021 Complete NECO Specimens. 2021 NECO Specimen | 2021 SPECIMEN for NECO


2021 NECO Specimen | 2021 SPECIMEN for NECO

Best Waec Expo runz site 2021, ARE YOU LOOKING FOR A PERFECT SOURCE FOR YOUR EXAM SUCCESS IN waec?, we provides real waec questions and answers.


2021 WAEC: BEST WAEC EXPO 2021 QUESTIONS AND ANSWERS

2021 Waec Science Specimens: 2021 Complete WAEC Specimens. 2021 Waec Specimen | 2021 SPECIMEN for WAEC


2021 Waec Specimen | 2021 SPECIMEN for WAEC

2021 NABTEB: NABTEB TIMETABLE FOR 2021/2021 ACADEMIC YEAR PDF DOWNLOAD. 2021 NABTEB TIMETABLE: 2021/2022 NABTEB TIMETABLE Free Download


2021 NABTEB: NABTEB TIMETABLE FOR 2021/2021 ACADEMIC YEAR PDF DOWNLOAD

This is 2021 WAEC Practical Specimens. Are you looking for WAEC specimen 2021? this page is all about 2021 WAEC Practical Specimens: 2021 WAEC Specimen - 2021 WAEC Science Specimens .


2021 WAEC Specimen | 2021 WAEC Science Specimens - WAEC Practical Specimens
Joseph Michael said: I used examplaza for waec in 2018 and I got 6B... view all reviews