2017 WAEC GCE Mathematics Answers

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Maths OBJ:

1-10 CBACDBBDCA

11-20 ACCBDADBDC

21-30 BCADDABDAC

31-40 DACBBDACCD

41-50 ACCBDABBCD

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

2a)

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

A=2205000

n=2

r=5%

p=?

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

2205000=p(1.05)^2

p=2205000/(1.05)^2

p=N2000000

2b)

Area=30cm^2,

h/b=3/2

Area=1/2bh

b=2h/3

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

2h^2=30*2*3

h^2=30*2*3/2

h^2=9

h=root9

h=9.49m

4)

OSP+OSR=90

44+OSR=90

OSR=90-44=46

But OSR+RSP=180

46+RQP=180

RQP=180-46

RQP=134degree

(5a) Draw a bar chat with the info the table given

(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

(3)

2x/1-2/5y=2y/1(isosceles angle)

LCM = 5

10x-2y=10y

10x=10y+2y

x=12y/10

x=6y/5……(1)

2x-2/55y+2whole1/4c+1/2+2y=180(sum of angle in a triangle)

2x/1-2y/5+(9/4)x+y/2+2y=180

LCM = 20

40x-8y+45x+10y+40y=3600

40x+45x-8y+50y=3600

95x+42y=3600….(2)

Substitute for x in eq…2

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

19(6y)+42y=3600

114y+42y=3600

156y/156=3600/156

y=23.08

Subst for y in equ…1

x=6(23.08)/5

x=27.69

7a)

log6Y+2log6X=3

log6Y+log6X^2=3

log6(X^2)Y=3

(X^2)Y=6^3

(X^2)Y=216

Y=216/X^2

7b)

Let F= Students that liked football

B=Students that liked boxing

V=students that liked volleyball

n(F)=70 n(B)=60 n(V)=45

n(FnB)=45 n(BnV)=15

n(FnV)=25

n(FnVnB)=5

n(FuVuB)^1=x

DRAW THE VENN DIAGRAM

7bii)

I)Students that liked football and volleyball but not boxing

=25-5=20%

II)Exactly two sports

=20+40+10

=70%

III) x+5+5+10+20+40+10+5=100

x+95=100

x=100-95

=5%

(9a) Diagram

(9b)

a= 360-330=30 degree

b=a=30degree{alternate angles}

tita = 90degree+30degre

= 120degree

(i) find distance AC wing.

|AC|^2= |BC|^2+|AB|^2–2|AB||BC|Costita

|AC|^2= 300^2+100^2–2(300)(100)Cos120degree

= 90000+10000–60000×(-0.5)

=100,000+30,000

= 130,000

|AC| = √130,000 = 360.555KM

=360.56Km(2d.p)

(ii)

find beta, applying the first rule

300/Sinbeta = 360.56/Sin120

Sinbeta= 300sin120/360.56 = 259.81/360.56

= 0.7206

beta = sin inverse(0.7206)

= 46degree

Therefore the bearing of the plane from A

=330–46

=284degrees

11)

A=(2,3,4,5,6,7,8,9,10,11,12)

B=10,11,12,13…24)

M=(4,8,12,16,20,24,28)

a)

Pr(MnA)=3/7

b)

Pr(MnB^1)1-PrMnB)

=1-4/7

=3/7

c)

Pr(Mn(AuB)

AuB=(2,3,4,5…14)

Mn(AuB)=(4,8,12,16,20,24

Pr(Mn(AuB)=6/7

12)

MATRIX:

[(3x,2)(5,4)]|[(3,-2)(5,2y)]=[(28,4) (35,22)]
=9x+10=28

9x=28-10

9x=18

x=18/9

x=2

and

-6x+4y=4

-6(2)+4y=4

4y=4+12

4y=16

y=16/y

y=4

(13a)

x* y = X+Y+2xy

(i)

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

= 15x + 7

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

= 11x + 5

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

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

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

= 55x + 27

X*7 = (x*5)*2

15x + 7 = 55x + 27

15x – 55x = 27–7

–40x = 20

X = 20/-40

X = -½

(13b)

PQ=8i+6j

QR=4i+3j

SR=3(PQ)/2 = 3/2(8i+6j)

=12I+9j

(i)

RS = -SR

=-12i-9j

(ii)

PQ+QR+RS

=8i+6j+(-4i+3j)+(-12i-9j)

=8i+6j-4i+3j-12i-9j

= -8i

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