WAEC GCE 2017 Mathematics Obj And Theory Answers-Nov/Dec Expo NOW AVAILABLE!!!!!!!!

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=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

===========================

• 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
• ===========================
• 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
• ==========================
• 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%
• -----------------------------------------
• 9a) Diagram
• 
  
• (9b)
• a= 360-330{ =030 degree
• b=a=030degree{alternate angles}
• Φ = 090degree+030degre
• = 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)
• = 46degree
• Therefore the bearing of the plane from A
• =330–046
• =284degrees.
• 
  
• 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
• =============================
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