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%

(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

(ii)
PQ+QR+RS
=8i+6j+(-4i+3j)+(-12i-9j)
=8i+6j-4i+3j-12i-9j
= -8i

Take Our Free Runz: 2017 WAEC GCE Mathematics Answers will be posted here for free but 60 minutes after exam starts. Take Our Paid Runz: With N400 we will send you the password to view all the answers 3 hours before exam starts. To make payment, simply buy MTN Recharge card of N400 and send it to 09067385575 . You can also do transfer to the number.

How to send the card: Type the card pin, your subjects & phone number and send to 09067385575. You can also call or message the above number for more details.

Why we are the best? We give you 24/7 customer support. We give you another subject for free if you didn’t get the answers for the one you paid for. We offer the cheapest Runz (N400 per Subject, N2000 for 5-9 Subjects and N4000 for all our runz). Our answers are real and genuine. We also give free Runz to candidates who cannot afford our paid Runz. Where else on the internet can you get this kind of service? Subscribe now and enjoy while it lasts!

Refresh this page 60 minutes after exam starts for our Free answers.