**Waec Gce 2017 Mathematics Obj And Theory Answers – Nov/Dec Expo**- By

**Mr. Maraj**On September 6, 2017 At 11:03 am

MATHS OBJ:

1CCAACBADAB

11BCCCBBBBBB

21DDBDBCCCBA

31BDDDCCCCCD

41ACADADCDDC

12a)

9x+10=28

9x=28-10

9x=18

x=18/9

x=2

-6x+4y=4

-6(2)+4y=4

4y=4+12

4y=16

y=16/4=4

13ai)

x*y=x+y+2xy

2*3=2+3+2*2*3

=5+12=17

(2*3)*5=17*5

17+5+2*17*5

=22+170=192

ii)x*7=x+7+2(x)(7)

=x+7+14x

=15x+7

x*5=x+5+2(x)(5)

=x+5+10x

=11x+5

(x*5)*2=(11x+5)*2

=11x+5+2+2(11x+5)^2

=11x+5+2+44x+20

=55x+27

15x+7=55x+27

15x-55x=22-7

-40x=20

x=-20/40=-1/2

2a)

A=P(1+r/100)^n

A=2,205,000

n=2

r=5%

p=?

2205000=p(1+5/100)^2

2205000m=p(1.05)^2

p=2205000/(1.05)^2=N2,000,000

4)

area of shaded portion=area of sector pqs-area of triangle pqs

=(θ/360*pieR^2)-(1/2r^2sinθ)

(60/360*22/7*14^2)-(1/2*14^2sin60)

102.2667-84.8705

=17.796

=17.8cm^2

7a)

log6Y+2log6x=3

log6y+log6x^2=3

log6(x^y)=3

x^y=6^3

x^y=216

y=216/x^2

9a)

CLICK HERE FOR NO.9a IMAGE

1a)

[2(1/2)+1(3/4)/1(2/5)]/[2(1/4)-1(1/2)]

[5/2+7/4/7/5]/[9/4-3/2]

=[5/2+7/4*5/7]/[9/4-3/2]

[5/2+5/4]/[9/4-3/2]

=[(10+5/4)/(9-6/4)]

=(15/4)/(3/4)

=15/4*4/3

=5

5a)

Daigrams

CLICK HERE FOR NO.1,3,7 IMAGE

3)

2x/1 – 2/5 y = 2y/1

LCM = 5

10x – 2y = 10y

10x = 10y + 2y

10x = 12y

X = 12y/10 ————-1

11)

A={4,5,6,7,8,9,10,11,12}

B={10,11,12,13,14,15,16,17,18,19,20,21,22,23,24}

M={4,8,12,11,6,20,24,28}

a)prob (A)=3/7

b)prob( B)=7-4/7=3/7

c)prob(AuB)=n(AuB)/n(m)

but (AuB)={4,8,12,11,6,20,24,}

prob(AuB)=6/7

MATHS OBJ:

1CCAACBADAB

11BCCCBBBBBB

21DDBDBCCCBA

31BDDDCCCCCD

41ACADADCDDC

12a)

9x+10=28

9x=28-10

9x=18

x=18/9

x=2

-6x+4y=4

-6(2)+4y=4

4y=4+12

4y=16

y=16/4=4

13ai)

x*y=x+y+2xy

2*3=2+3+2*2*3

=5+12=17

(2*3)*5=17*5

17+5+2*17*5

=22+170=192

ii)x*7=x+7+2(x)(7)

=x+7+14x

=15x+7

x*5=x+5+2(x)(5)

=x+5+10x

=11x+5

(x*5)*2=(11x+5)*2

=11x+5+2+2(11x+5)^2

=11x+5+2+44x+20

=55x+27

15x+7=55x+27

15x-55x=22-7

-40x=20

x=-20/40=-1/2

2a)

A=P(1+r/100)^n

A=2,205,000

n=2

r=5%

p=?

2205000=p(1+5/100)^2

2205000m=p(1.05)^2

p=2205000/(1.05)^2=N2,000,000

2b)

area=30cm^2 h/b=3/2

area=1/2bh

b=2h/3

30=1/2*2h/3*h

2h^2=30*2*3

h^2=30*2*3/2

h^2=90

h=√90=9.49m

4)

area of shaded portion=area of sector pqs-area of triangle pqs

=(θ/360*pieR^2)-(1/2r^2sinθ)

(60/360*22/7*14^2)-(1/2*14^2sin60)

102.2667-84.8705

=17.796

=17.8cm^2

7a)

log6Y+2log6x=3

log6y+log6x^2=3

log6(x^y)=3

x^y=6^3

x^y=216

y=216/x^2

7bi)

CLICK HERE FOR No.7bi IMAGE

7bii)

percentage who liked football and volleyball but not boxing=20

ii)percentage who liked exactly two part=40+20+10=70

iii)summing

5+5+10+40+20+10+5+x=100%

x+95=100

x=100-95=5%

percentage who like none=5%

9a)

CLICK HERE FOR NO.9a IMAGE

9b)

|CA|^2=300^2+100^2-2(300)(100)cos 120degree

=9000+10000-60000(-cos60)

=100000+30000

KA|^2=130000

KA|=√130000

KA|=360.56KM

9bii)

Using sine rule

|AB|/sinC=|AC|/sin120

100/sinC=360.56

sinC=100sin60/360.56

sinC=0.2402

C=sin^-1(0.2402)=13.90degree

bearing=270+13.90=283.9degree = 284degree

1a)

[2(1/2)+1(3/4)/1(2/5)]/[2(1/4)-1(1/2)]

[5/2+7/4/7/5]/[9/4-3/2]

=[5/2+7/4*5/7]/[9/4-3/2]

[5/2+5/4]/[9/4-3/2]

=[(10+5/4)/(9-6/4)]

=(15/4)/(3/4)

=15/4*4/3

=5

1b)

From tan60/1*PR/3root2

PR=3root2*1/root3

=3root2/root3*root3/root3

=3root6/3=root6

From sin45/1*root6/x

x=root6*root2

=root12

=root4*3

=2root3

5a)

Daigrams

5b)

Median = ((N + 1) / 2)th item

Where:

N = 31

40, 40, 40, 40, 40, 40, 40, 41, 41, 41, 41, 42, 42, 42, 42, 42, 42, 43, 43, 44, 44, 44, 44, 45, 45, 46, 46, 46, 46, 46, 46

Data table

F 7 4 6 2 4 2 6

X 40 41 42 43 44 45 46

Median = ((31 + 1) / 2)th item

Median = (32 / 2)th item

Median = 16th item

Arranging the discrete data in ascending order, We go get:

40, 40, 40, 40, 40, 40, 40, 41, 41, 41, 41, 42, 42, 42, 42, 42, 42, 43, 43, 44, 44, 44, 44, 45, 45, 46, 46, 46, 46, 46, 46

Median = 42

CLICK HERE FOR NO.1,3,7 IMAGE

3)

2x/1 – 2/5 y = 2y/1

LCM = 5

10x – 2y = 10y

10x = 10y + 2y

10x = 12y

X = 12y/10 ————-1

2x – 2/5y + 2 1/4x + 1/2y + 2y = 180

LCM = 20

40x – 8y + 45x + 10y + 40y = 3600

40x + 45x + 10y – 8y + 50y = 3600

95x + 42y = 3600

95 (6y/5) + 42y =3600

19(6y) + 42y = 3600

114y + 42y = 3600

156y = 3600

Y = 3600/156

Y = 23.08

Sub for y in eq 1

X = 6(23.08)/5

X = 27.69

11)

A={4,5,6,7,8,9,10,11,12}

B={10,11,12,13,14,15,16,17,18,19,20,21,22,23,24}

M={4,8,12,11,6,20,24,28}

a)prob (A)=3/7

b)prob( B)=7-4/7=3/7

c)prob(AuB)=n(AuB)/n(m)

but (AuB)={4,8,12,11,6,20,24,}

prob(AuB)=6/7

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