# WAEC 2018 Mathematics Obj And Essay Answer – May/June Expo

## WAEC Mathematics Obj And Essay/Theory Solution Questions and Answer – May/June 2018 Expo Runz.

**SUBSCRIBING BEFORE THE EXAM DAY MAKES YOU SAFER BCOS YOU’LL GET PASSWORD EARLIER ON THAT EXAM DAY.**

PLEASE ALWAYS SUBSCRIBE A DAY BEFORE EACH EXAM.

PLEASE ALWAYS SUBSCRIBE A DAY BEFORE EACH EXAM.

**Maths OBJ:**

1ABBCDDBBAA

11ADAACCDCCC

21DADBABCDAD

31BADADBCACB

41ADDDBDADCA

1ABBCDDBBAA

11ADAACCDCCC

21DADBABCDAD

31BADADBCACB

41ADDDBDADCA

9a)9a)

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

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

**13c)DRAW THE GRAPH**

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)

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 – AΔPTRAΔPTR = 1/2×4×6×sin30=2×6×0.5=6cm²AanglePTQ/AanglePSR = |PT|²/|PS|²6/AanglePSR = 4²/10²6/AanglePSR = 16/10016×AanglePSR = 6×100AanglePSR = 600/16 = 37.5cm2ATQRS = 37.5 – 6=31.5cm2=32cm2**

5a)

m+n+s+p+q/5=12

m+n+s+p+q=60……(1)

Now;

(m+4)+(n-3)+(5+6)+p-2)+(q+8)/5

=(m+n+s+p+q)+(4-3+3+6-2+8)/5

=60+13/5

=73/5

=14.6

5a)

m+n+s+p+q/5=12

m+n+s+p+q=60……(1)

Now;

(m+4)+(n-3)+(5+6)+p-2)+(q+8)/5

=(m+n+s+p+q)+(4-3+3+6-2+8)/5

=60+13/5

=73/5

=14.6

**5b)75% of 500 = 375 peopleNumber of people above 65 years = 500-375=12525% of 500 = 125Number of people below 15 years = 125Number between 15 years and 65 years=500-(125+125)=500-250=250 people**

7a)

X1-X/Y1-Y = X2-X1/Y2-Y1

2-X/5-Y = -4-2/-7-5

2-X/5-Y= -6/-12

-12(2-X)=-6(5-Y)

-24+12X=-30+6Y

6Y-12X=30+24

6Y-12X=-6

6y-12x+6=0

y-2x+1=0

7a)

X1-X/Y1-Y = X2-X1/Y2-Y1

2-X/5-Y = -4-2/-7-5

2-X/5-Y= -6/-12

-12(2-X)=-6(5-Y)

-24+12X=-30+6Y

6Y-12X=30+24

6Y-12X=-6

6y-12x+6=0

y-2x+1=0

**7bi)DRAW THE DIAGRAM7bii)I)p^2=q+r^2-2qrcosPp^2=8^2+5^2-2*8*5*cos90p^2=64+25-0p^2=89p=sqroot(89)p=9.4339kmtherefore |QR|=9.43km(3 sf)II)q/sinQ=p/sinP8/sinQ=9.4339/sin90sinQ=(8*sin90/9.4339sinq=(8*1)/9.4339 =0.8480Q=sin^1(0.8480)=57.99 degreesbut Q=30+ AA=Q-30=57.99-30A=27.99 degreesThe bearing of R from Q=180-A180-27.99=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)

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<2x2x > -4X > -2Therefore; Range is -2 **

10a)

Using Pythagoras theorem from SPQ

|SQ|^2 = 12^2 + 5^2

= 144+25

=169

SQ= sqroot of 169

= 13cm

Sin tita= 5/13 = 0.3846

Tita= Sin^-1(0.3846)

= 22.6degrees

From PRQ

Sin tita= |PR|/12

Sin 22.6 = PR/12

Sin 22.6= PR/12

PR= 12xsin 22.6

PR= 12×0.3843

PR= 4.61cm

10a)

Using Pythagoras theorem from SPQ

|SQ|^2 = 12^2 + 5^2

= 144+25

=169

SQ= sqroot of 169

= 13cm

Sin tita= 5/13 = 0.3846

Tita= Sin^-1(0.3846)

= 22.6degrees

From PRQ

Sin tita= |PR|/12

Sin 22.6 = PR/12

Sin 22.6= PR/12

PR= 12xsin 22.6

PR= 12×0.3843

PR= 4.61cm

**10bii)Let the height at which m touches the wall= yCos x^degrees= 8/10= 0.8x^degrees= Cos^-1(0.8)= 36.87degreesSin x^degrees = y/12Sin 36.87= y/12y= 12xsin36.87y= 12×0.60000y= 7.2m**

CLICK HERE FOR NO.2 nd 8 co. ANSWER

CLICK HERE FOR NO.9 ANSWER

6a)

Draw the Venn diagram

Let the number of cars with faults in brakes only be x

CLICK HERE FOR NO.2 nd 8 co. ANSWER

CLICK HERE FOR NO.9 ANSWER

6a)

Draw the Venn diagram

Let the number of cars with faults in brakes only be x

**6b) Number that passed = 60% × 240 = 144Number that failed =240 – 144 = 96Therefore; 28+2x+x+14+6+6-x+8 = 962x + 62 = 962x = 96 – 622x = 34X = 34/2X = 17i) faulty brakes cars = 8+6+x+6-x= 8+6+6=20ii) only one fault = 28+x+2x=28+3x=28+3(19)=28+51= 79**

2)

Given that y = 2pxˆ² – p² x – 14

AT (3, 10)

10 = 2p(3)² – p² (3) – 14

10 = 18p – 3p² – 14

3p² – 18p + 24 = 0

p² – 6p + 8 = 0

using factor method,

p² – 2p -4p + 8 = 0

p(p-2) – 4(p-2) = 0

(p-4)(p-2) = 0

p-4 = 0 or p-4 = 0

p= 4 or p =2

2)

Given that y = 2pxˆ² – p² x – 14

AT (3, 10)

10 = 2p(3)² – p² (3) – 14

10 = 18p – 3p² – 14

3p² – 18p + 24 = 0

p² – 6p + 8 = 0

using factor method,

p² – 2p -4p + 8 = 0

p(p-2) – 4(p-2) = 0

(p-4)(p-2) = 0

p-4 = 0 or p-4 = 0

p= 4 or p =2

**2b)The lines must be solved simultenously3y – 2x = 21 ——- (1)4y + 5x = 5 ——-(2)using elimination method,(4) 3y – 2x = 21(30 4y + 5x = 512y – 8y = 84 ——— (3)12y + 15x = 15 ——-(4)equ (4) minus equ(3)23x = -69x = -69/23x = -3Put this into equation (1)3y -2(-3) = 213y = 6 = 213y = 21 -63y = 15y =15/3y = 5coordinates of Q is (-3, 5)**

3a)

The diagonal = 10.2m and 9.3cm

Using Pythagoras theory

Ac² = 10.2² + 9-3²

Ac² = 104.04 + 86.49

Ac² = 190.53

Ac² = √190.53

Ac² = 13.80

3a)

The diagonal = 10.2m and 9.3cm

Using Pythagoras theory

Ac² = 10.2² + 9-3²

Ac² = 104.04 + 86.49

Ac² = 190.53

Ac² = √190.53

Ac² = 13.80

**3b)DRAW THE DIAGRAMUsing Pythagoras theory5² = 3² + x²x² = 5² – 3²X²= 25 – 9X² = √16X= 4cmCosX = adjacent/hyp= 4/5Tan X = opp/adj. = 3/45cos x – 4tan x5(4/5)- 4(3/4)20/5 – 12/44-3= 1**

4ai)

sum of angle in a D =180degree

xdegree + 90degree + 180degree – (3x+15)=180degree

xdegree + 90degree + 180degree – 3x+15=180degree

-2x=180degree – 255

+2x/2=+75/2

x=37.5

4aii)

<RsQ =180 – (3x+15)

<RsQ =180-(3*37.5+15)

=180-(112.5 + 15)

=180 – 127.5

<RsQ= 52.5degree

4ai)

sum of angle in a D =180degree

xdegree + 90degree + 180degree – (3x+15)=180degree

xdegree + 90degree + 180degree – 3x+15=180degree

-2x=180degree – 255

+2x/2=+75/2

x=37.5

4aii)

<RsQ =180 – (3x+15)

<RsQ =180-(3*37.5+15)

=180-(112.5 + 15)

=180 – 127.5

<RsQ= 52.5degree

**4b)2N4seven =15Nnine2*7^2+N*7^1+4*7degree =1*9^2 + 5*9^1+N*9degree9*49+N*7+4*1=1*81+5*9+N*198+7N+4=81+45+N7N+102=126+N7N-N=126-1026N/6 =24/6N=4**

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

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

CLICK HERE FOR NO.3 AND 4 ANSWERS

CLICK HERE FOR NO.3 AND 4 ANSWERS

CLICK HERE FOR NO.1 ANSWER

CLICK HERE FOR NO.1 ANSWER

CLICK HERE FOR NO.7 TABLE

CLICK HERE FOR NO.7 TABLE

*042tvseries.Com send all exam expo answer earlier than others**WAEC ssce 2018 Mathematics expo answers, Ssce 2018 Mathematics objective and essay Expo answers now available, nov/dec Mathematics theory Answers for 2018, Mathematics obj and theory Questions WAEC Ssce Expo Answers Here, Free wassce Mathematics objective, essay & theory correct Expo 2018 answers Runz WAEC Mathematics, WAEC Ssce 2018 Free Mathematics essay, obj and theory Questions and Answers Expo for WAEC Mathematics Paper 1 and Paper 2 objective and essay Questions and Answers Expo Now Posted, Free Mathematics objective ans for internal ssce, see WAEC ssce expo Mathematics obj and theory, essay questions for free, loadedexpo correct wassce Economics ans, sure WAEC ssce Mathematics objective answers runz runs expo free real and correct WAEC ssce 2018 nov/dec Mathematics free expo runz runs ans answers online for free see free answers online, 2018 Verified WAEC Free Ssce Mathematics Obj and theory Answers have been posted REAL WAEC SSCE Mathematics QUESTIONS & ANSWERS Direct To Phone Number AS Text message, RE: 2018 WASSCE Mathematics (Expo) WAEC 2018 certified runs Mathematics questions and answers WAEC SSCE November/December 2018 Mathematics THEORY / ESSAY / OBJ QUESTION AND EXPO I need WAEC ssce free expo site, how can I get WAEC Mathematics answer for free, WAEC 2018 Mathematics essay, Objective And Theory Question and Answer Now Posted, Real 2018/2019 nov/dec wassce WAEC Mathematics objective, essay and theory 100% correct expo questions and answers runz chokes runs, verified 2018 WAEC Mathematics theory, essay and obj WAEC Mathematics expo website, WAEC ssce Mathematics live cheats, sure November/December WAEC SSCE 2018 Mathematics original objective, theory and essay questions and answer now available here, 100% WAEC ssce Mathematics expo, 2018 wassce Mathematics obj, essay and theory runz now posted, correct WAEC ssce Mathematics objective, essay (theory) answer now ready, free 2018 wassce Mathematics obj, theory and essay best expo site WAEC ssce Mathematics questions and answers, password for certified WAEC ssce Mathematics essay theory obj expo, how to get/ where to get wassce Mathematics theory, essay and objective expo answer direct to my phone as text message, trusted WAEC ssce Mathematics questions and answer, i need WAEC Ssce nov/dec Mathematics essay theory objective obj Expo 2018 answers on whatsapp, gidifans Mathematics answer, solutionclass Mathematics answer, 042tvseries Mathematics answer, codedexam WAEC Mathematics answer, wapbaze Mathematics answer, wapextra Mathematics expo, Martinlibrary Mathematics expo, examgists WAEC Mathematics expo, WAEC Mathematics objectives answer, naija WAEC Mathematics expo free answer*

**GOODLUCK!!!**