Waec Mathematics Past Questions Pdf

Waec Mathematics Questions 2021 | Waec 2021 Mathematics Theory and Obj Questions and Answers

All those who are writing waec mathematics questions 2021for the may/June examination, it is good to announce to you that the we have sourced and gotten the required and the needed questions and answers for waec 2021 and this we have presented to you in this post on the waec questions and answers to be written today.

Our waec questions and answers also contains bothwaec 2021 mathematics theory questionsand also the correct2021 waec mathematics questions and answersfor Objective all in this may/junewaec 2021 mathematics question and answer.The questions and answers was sourced from an insider of the west African examination council and was confirmed to be accurate by another insider of the West African examination Council.

The waec mathematics questions 2021 also contain thewaec 2021 mathematics answersto both theory Questions and Objective questions. There is a need to carefully look into thewaec mathematics 2021questions and answers for all those who wants to get their As and make it through in Waec 2021.

The questions are available for May/June and notwaec Gce mathematics questions 2021and we promise to make available for GCE questions once the questions are ready. For you to make good grades in this year's examination, you should endeavor to prepare accurately and confirm every beats of answers provided here by us and one way to do that is through the effective use of thewaec mathematics past questions and answers pdf.

The theory questions is one of thewassce core maths 2021session and candidates should concentrate on having all the theory questions answered because it carries more marks.

WASSCE is one of the core subjects, infact it is the main core subject thst every student should strive to pass in waec May/June and that's why we have made availablewassce 2021 core maths questionsfor you to study before the examination time.

waec questions and answers on mathematics 2021

It is usualyy our nature to make questions and answers available before the hour of the examination and we consider it pertinent to also make thewaec 2021 mathematics question.

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In ourwaec mathematics questions 2021, you will find some questions in thewaec maths objpart as something which may appear difficult, it is you that should confirm the answers to this questions and that is why you should check ourwaec May/June mathematics answer 2021and confirm them by solving appropriately.

The May/June 2021 mathematics questions and answersis not very difficult on line thegce maths questions 2017which gave our solvers a bit of the tough times on how to solve thewaec mathematics questions 2021.

If you have been studying your past questions and answers closely, you will realized that like five theory questions can also be seen in the2014 waec mathematics question and answerthis we can see in some of the questions ofwaec mathematics questions 2021.

We do encourage readers to study activelywaec May\June past questions pdfwhich shows y thewaec questions and answers on mathematics 2021being repeated and sourced largely from thewaec past questions on mathematics.

If you also check thewaec questions and answers on mathematics 2015you will find a lot of questions on the theory aspect of questions waec mathematics 2021.

Format and Structure of waec mathematics questions 2021

There is a known format for waec examination of having both the objective and the theory questions. The format also stands forwaec mathematics questions 2021because waec questions have been on the same format from inception and this we can see in thewaec maths past questions and answers.

Like we have provided questions for Literature, English and Physics Quetions here is also that for mathematics and this we have done for free without asking you to provide us with money or subscribe for it.

For those who are not familiar with the structure, the theory questions consist of Five questions and you are required to answer a particular required number of questions and not to answer all the questions.

The Obj questions do contain about 50 questions and you are also required to answer them all with the OMR sheet of paper provided. The Objective questions may not be very difficult as you may think, but if you studious, then you will encounter little problems in solving the mathematics questions and answers.

In the case where you solve any question and have a different answer please resolve the questions again to make sure it is correct.

According to the timetable, thewaec mathematics questions 2021will start at about 8 :30 with an Essay questions while the objective questions will commence after the Essay Questions at 11:00 hrs. This is almost similar for the timing for English Language

WAEC General Mathematics Questions Objective and Theory

Without beating around the bush, here is the waec Mathematics questions for Objectives which have been gotten from an inner source. Though the questions are not complete, but please visit this page regularly to get the full questions.

OBJ

1-10: ACBCDDCBAA
11-20: CDCABCCCAC
21-30: DADBABCDAD
31-40: BADADBCACB
41-50: ADDDBDADCA

Theory

(1)

On February 28th 2012, value = (100-30/100) * #900,00.00

= 70/100 * #900,00

= #630,000.00

On february 28th 2013, value = (100-22/1000 * #630,00

= 78/100 8 #630,000

= #491,400

On february 28th 2014, value = 78/100 8 #491,400

=383,292

On february 28th 2015, value = 78/100 * #383,292

= #298,967.76

= #299,000

(3a)

Using pythagoras theorem

L^2=5.1^2 + 4.65^2

L^2=26.01 + 21.6225

L^2=47.6325

L=sqroot(47.6325)

L=6.9cm(1 d.p)

Perimeter of rhombus=4

=4*6.9

=27.6cm

(3b)

Sin x=3/5

DRAW THE TRIANGLE

Using pythagoras tripple the third side=4

therefore cosx=4/5

tanx=3/4

therefore 5cosx-4tanx

=5(4/5)-4(3/4)

=4-3

(4ai)

Draw the diagram

ai X + 90 = 3x + 15

90 = 3x – X + 15

90 = 2x + 15

2x + 15 = 90

2x = 90 – 15

X = 75/2

X = 37.5•

(4aii)

(4b)

2N4seven = 15Nnine

Converting both to base 10

2×7+N×7¹+4×7 = 1×9²+5×9¹+N×9

98 + 7N + 4 = 81 + 45 + N

7N + 102 = 126 + N

7N – Ń = 126 – 102

6N = 24

Ń = 24/6

Ń = 4

WE RE-SOLVED NUMBER 6. PLS CHECK.

6
Total number of cars on road worthiness = 240
60% passed ie 60/100×240/1 = 144cars.
Number that failed = 240-144 = 96cars

6a) draw the Venn diagram
C = clutch
B = brakes
S = steering

(b) From the diagram above
E = 28+12+8+6+x+6+2x
96=60+3x
96-60=3x
36/3 = 3x/3
Therefore X = 12

(i) The no of cars that had faulty brakes
=12+8+6+x (Since X = 12)
=12+8+6+12 = 38

(ii) Only one fault = (no of clutch only) + (no of brakes only) + (no of steering only)
=28+x+12x = 28+12+24
=64cars

(7a)

(y-y1)/(x-x1)=(y2-y1)/(x2-x1)

(y-5)/(x-2)=(-7-5)/(-4-2)

(y-5)/(x-2)=-12/-6

(y-5)/(x-2)=2

Cross multiply

y-5=2(x-2)

y-5=2x-4

2x-y-4+5=0

2x-y+1=0

(7bi)

DRAW THE DIAGRAM

(7bii)

(I)

p^2=q+r^2-2qrcosP

p^2=8^2+5^2-2*8*5*cos90

p^2=64+25-0

p^2=89

p=sqroot(89)

p=9.4339km

therefore |QR|=9.43km(3 sf)

(II)

q/sinQ=p/sinP

8/sinQ=9.4339/sin90

sinQ=(8*sin90/9.4339

sinq=(8*1)/9.4339 =0.8480

Q=sin^1(0.8480)=57.99 degrees

but Q=30+ A

A=Q-30

=57.99-30

A=27.99 degrees

The bearing of R from Q

=180-A

180-27.99

=155.01

=>152 degrees

(8a)

Cost price for Lami= #300.00

Profit made by lami = x%

Ie selling price for lami=(100+x/100)×#300

=#3(100+x)

=#(300+3x)

Bola's cost price = #3(100+x)

Profit made by bola =x%

Selling price for bola =(100+x/100)×#3(100+x)

=#3/100(100+x)²

James cost price =#3/100(100+x)²=300+(6x+3/4)

expanding;

3/100(10000+200+x²) = 300+3/4+6x

3(10000+200x+x²)=30000+75+600x

30000+600x+3x²=30000+75+600x

3x²=75

X² = 75/3

X² = 25

X = square root 25

X = 5

(8b)

3x-2<10+x<2+5x

3x-2<10+x & 10+x<2+5x

3x-x<10+2 & 10-2<5x-x

2x<12 8<4x

X<12/2 4x>8

X<6 x>8/4

X>2

Also; 3x-2<2+5x

-4<2x

2x > -4

X > -2

Therefore; Range is -2

(9a)
Draw the diagram
Angles PTR and PSR are similar
|PT|/|PS| = |TQ|/|SR|
In angle PTR
|TQ|²=|PT|²+|PQ|²-2|PT||PQ|cos30degrees
=4²+6²-2×4×6×cos30
=16+36-48×0.8660
=52-41.568
=10.432
|TQ|=√10.432 =3.22cm
4/10 = 3.22/|SR|
4|SR| = 10×3.22
|SR| = 32.2/4
|SR| = 8.05cn
Approximately 8cm(to the nearest whole number)

(9b)
Atqrs = AΔPSR – AΔPTR
AΔPTR = 1/2×4×6×sin30
=2×6×0.5
=6cm²
AanglePTQ/AanglePSR = |PT|²/|PS|²
6/AanglePSR = 4²/10²
6/AanglePSR = 16/100
16×AanglePSR = 6×100
AanglePSR = 600/16 = 37.5cm2
ATQRS = 37.5 – 6
=31.5cm2

=32cm2

(13a)
Frequency=16+x+y
16+x+y=30
x+y=30-16
x+y=14–(eqi)
(900+30x+50y)/30=52
900+30x+50y=52*30
30x+50y=1560-900
30x+50y=660
divide through by 10
3x+5y=66–(eqii)
From (i)
x+y=14
x=14-y–(eqiii)
sub for x in eqii
3(14-y) +5y=66
42-3y+5y=66
2y=66-42
y=24/2
y=12
feom eqiii
x=14-12
x=2

(13b)
TABULATE
Class interval:1-10,11-20,21-30,41,50,51-60,61-70,71-80,81-90
Freq:1,1,2,5,12,1,4,3,1
Class boundary:0.5-10.5,10.5-20.5,20.5-30.5,30.5-40.5,40.5-50.5,50.5-60.5,60.5-70.5,70.5-80.5,80.5-90.5

waec mathematics questions 2021

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