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? 2017 WAEC GCE MATHEMATICS ANSWERS?
100% VERIFIED MATHEMATICS OBJ ANSWERS
1–10 CBACDBBDCA
11–20 ACCBDADBDC
21–30 BCADDABDAC
31–40 DACBBDACCD
41–50 ACCBDABBCD
100% VERIFIED MATHEMATICS THEORY ANSWERS
PLS NOTE THIS SYMBOLS
^ means Raise to power
* means multiplication
sqrt or root means Sqr root
/ means division or divide
degree means degree sign that looks like small zero
tita means tita sign that looks like crossed-zero
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SWAPPED ANSWERS
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9. CLICK HERE
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2. CLICK HERE
3. CLICK HERE
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1 & 3. CLICK HERE
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11. CLICK HERE
TYPED ANSWERS
(1a)
[2whole1/2+1whole3/4/1whole2/5)]/[2whole1/4)-1whole1/2)]
[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/3√2
PR=3√2*1/√3
=3√2/√3*√3/√3
=3√6/3=√6
From sin45/1*√6/x
x=√6*√2
=√12
=√4*3
=2√3
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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=N2,000,000
(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=√9
h=9.49m
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(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
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(4)
OSP+OSR=90
44+OSR=90
OSR=90-44=46
But OSR+RSP=180
46+RQP=180
RQP=180-46
RQP=134°
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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
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(7a)
Log6y+2log6x= 3
Log6y+log6x²= 3
Log6(x²y)= 3
X²y = 6³
X²y = 216
Y= 216/x²
(7b)
(I). Percentage who liked football and volleyball but not boxing = 20
(II)
Percentage who liked exactly two sports = 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%
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9a) Diagram
(9b)
a= 360-330{ =030°
b=a=030°{alternate angles}
Φ = 090°+030°
= 120degree
(i) find distance AC wing.
|AC|²= |BC|²+|AB|²–2|AB||BC|Costita
|AC|²= 300²+100²–2(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 β, applying the first rule
300/Sinβ = 360.56/Sin120
Sinβ = 300sin120/360.56 = 259.81/360.56
= 0.7206
β = sin inverse(0.7206)
= 46°
Therefore the bearing of the plane from A
=330–046
=284°.
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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
====COMPLETED====
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