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Waec Gce Maths 2017 Obj and Theory Answer – Nov/Dec Expo

Waec Gce  Maths  2017 Questions & Answers Objectives Theory & Essay  Expo/Runz/Site   ->Complete Waec Gce Mathematics Answer 2017 Expo/Runz ( Maths )



UPDATE... questions is been changed...
inconviniency inconviniency caused

MATHS OBJ:
1–10 CBACDBBDCA

11–20 ACCBDADBDC

21–30 BCADDABDAC

31–40 DACBBDACCD

41–50 ACCBDABBCD

Check before you shade 

✍✍💯💯✔✔.


Maths-Theory
PLS NOTE THIS SYMBOLS
^ means Raise to power
* means multiplication
√ means Sqr root
/ means division or 
divide


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

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



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