# 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.

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**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

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 GRAPH13a)

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)

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

=32cm29a)

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

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 people

Number of people above 65 years = 500-375

=125

25% of 500 = 125

Number of people below 15 years = 125

Number between 15 years and 65 years

=500-(125+125)

=500-250

=250 people5a)

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 people

Number of people above 65 years = 500-375

=125

25% of 500 = 125

Number of people below 15 years = 125

Number 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

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 degrees7a)

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 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 -28a)

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

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= 12x0.3843

PR= 4.61cm

10bii)

Let the height at which m touches the wall= y

Cos x^degrees= 8/10= 0.8

x^degrees= Cos^-1(0.8)

= 36.87degrees

Sin x^degrees = y/12

Sin 36.87= y/12

y= 12xsin36.87

y= 12x0.60000

y= 7.2m10a)

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= 12x0.3843

PR= 4.61cm

10bii)

Let the height at which m touches the wall= y

Cos x^degrees= 8/10= 0.8

x^degrees= Cos^-1(0.8)

= 36.87degrees

Sin x^degrees = y/12

Sin 36.87= y/12

y= 12xsin36.87

y= 12x0.60000

y= 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

6b) Number that passed = 60% × 240 = 144

Number that failed =

240 - 144 = 96

Therefore; 28+2x+x+14+6+6-x+8 = 96

2x + 62 = 96

2x = 96 - 62

2x = 34

X = 34/2

X = 17

i) faulty brakes cars = 8+6+x+6-x

= 8+6+6

=20

ii) only one fault = 28+x+2x

=28+3x

=28+3(19)

=28+51

= 79CLICK 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 = 144

Number that failed =

240 - 144 = 96

Therefore; 28+2x+x+14+6+6-x+8 = 96

2x + 62 = 96

2x = 96 - 62

2x = 34

X = 34/2

X = 17

i) faulty brakes cars = 8+6+x+6-x

= 8+6+6

=20

ii) 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

2b)

The lines must be solved simultenously

3y – 2x = 21 ——- (1)

4y + 5x = 5 ——-(2)

using elimination method,

(4) 3y – 2x = 21

(30 4y + 5x = 5

12y – 8y = 84 ——— (3)

12y + 15x = 15 ——-(4)

equ (4) minus equ(3)

23x = -69

x = -69/23

x = -3

Put this into equation (1)

3y -2(-3) = 21

3y = 6 = 21

3y = 21 -6

3y = 15

y =15/3

y = 5

coordinates of Q is (-3, 5)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 simultenously

3y – 2x = 21 ——- (1)

4y + 5x = 5 ——-(2)

using elimination method,

(4) 3y – 2x = 21

(30 4y + 5x = 5

12y – 8y = 84 ——— (3)

12y + 15x = 15 ——-(4)

equ (4) minus equ(3)

23x = -69

x = -69/23

x = -3

Put this into equation (1)

3y -2(-3) = 21

3y = 6 = 21

3y = 21 -6

3y = 15

y =15/3

y = 5

coordinates 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

3b)

DRAW THE DIAGRAM

Using Pythagoras theory

5² = 3² + x²

x² = 5² - 3²

X²= 25 - 9

X² = √16

X= 4cm

CosX = adjacent/hyp

= 4/5

Tan X = opp/adj. = 3/4

5cos x - 4tan x

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

20/5 - 12/4

4-3= 13a)

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 DIAGRAM

Using Pythagoras theory

5² = 3² + x²

x² = 5² - 3²

X²= 25 - 9

X² = √16

X= 4cm

CosX = adjacent/hyp

= 4/5

Tan X = opp/adj. = 3/4

5cos x - 4tan x

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

20/5 - 12/4

4-3= 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

4b)

2N4seven =15Nnine

2*7^2+N*7^1+4*7degree =1*9^2 + 5*9^1+N*9degree

9*49+N*7+4*1=1*81+5*9+N*1

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

7N+102=126+N

7N-N=126-102

6N/6 =24/6

N=44ai)

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 =15Nnine

2*7^2+N*7^1+4*7degree =1*9^2 + 5*9^1+N*9degree

9*49+N*7+4*1=1*81+5*9+N*1

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

7N+102=126+N

7N-N=126-102

6N/6 =24/6

N=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

= #298,967.761)

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 ANSWERSCLICK HERE FOR NO.3 AND 4 ANSWERS

CLICK HERE FOR NO.1 ANSWER

CLICK HERE FOR NO.3 ANSWER

CLICK HERE FOR NO.3b ANSWER

CLICK HERE FOR NO.4 ANSWER

CLICK HERE FOR NO.6a ANSWERCLICK HERE FOR NO.1 ANSWER

CLICK HERE FOR NO.3 ANSWER

CLICK HERE FOR NO.3b ANSWER

CLICK HERE FOR NO.4 ANSWER

CLICK HERE FOR NO.6a ANSWER

CLICK HERE FOR NO.7 TABLECLICK HERE FOR NO.7 TABLE

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