2017 WAEC GCE Mathematics Answers

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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