## 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

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

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