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*2017 WAEC CERTIFIED MATHEMATICS OBJECTIVES*💯💯💯✅✅✅✅

*1-10 CBBACCCBBA*

*11-20 ADBBAABCDB*

*21-30 BCBDBCCAAB*

*31-40 CAAACDBACB*

*41-50 BDABAADCAD*
================================
 *SECTION A ANS ALL QUESTIONS*
 ================================

 1 a)
 (y -1 ) log 4 ^ 10 = ylog 16 ^ 10
 log 4 ^ 10 ( y -1 )= log 16 ^ y 10
 4 ^ ( y -1 )= 16 y
 4 ^ y -1 = 4 ^ 2 y
 y- 1 = 2 y
 -1 = 2 y= y
 -1 = y
 y= -y

 1 b )
 let the actual time for 5 km / hr be t
 for 4 km /hr = 30 mint + t
 4 km /hr =0 . 5 + t
 distance = 4 (0 . 5 + t )
 = 2 * 4 t
 for 5 km /hr , time = t
 distance =5 t
 1 + 4 t = 5 t
 t= 2 hrs
 actual distance = 5 * 2 = 10 km

 ================================

 2a) 
 2/3 (3x - 5) - 3/5 (2x - 3) = 3
 (15) x 2/3 (3x - 5) - 3/5 (15) (2x - 3) = 3(15)
 10(3x -5) - 9(2x - 3) = 45
 30x - 50 - 18x + 27 = 45. 
 12x - 23 = 45
 12x = 45+23
 12x = 68
 X = 68/12
 X = 5.67

 2b)
 80 = n+r ( ext. < equal sum of opp int. <s )
  80 = n+r ------------ (i)
  <UQT = 180 - 88 - n = 92 - n
  M = 80 + 92 - n
  M = 172 - n ----------- (ii)
 80 + 92 - n + 180 - m = 180°‎
80 + 92 + 180 - n - m = 180°‎
352 - n - m = 180
  -n - m = 180 - 352‎
 - n - m = - 172°‎
   m + n = 172°‎

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

3a) ‎
Tan 23.6° = h/50
  Cross multiply
 Tan 23.6° x h/50
 h = 50 tan 23.6°
  = 21.844m‎
 ‎aprox. 22m
3b) 
 Area of <TRU = 45cm^2 (Note: This ^ means Raise to power)
 A = 1/2bh
  45 = 1/2 x 10 x h
  45 = 5h
  h = 9cm
 Area of < QTUS = 1/2 ( QT + US)h
  = 1/2 ( 6 + 16)9
  = 99cm^2

 ================================

 4 a)
 T 6 =37
 T 6 =a + ( 6 - 1 )d
 T 6 =a + 5 d
 a + 5 d = 37 - --- -( eq1 )
 s 6 = 147
 sn = n / 2 (2 a + (n -1 ) d )
 147 = 3 (2 a + 5 d )
 49 = 2 a + 5 d
 2 a+ 5 d = 49 -- -- (eq 2 )
 a + 5 d = 37 - --( eq1 )
 2 a+ 5 d = 49 -- -( eq2 )
 a =12

 4 b )
 S 15 = 15 /2 ( 2 (12 )+ 14 d )
 S 15 = 15 /2 ( 24 + 14 d )
 from(1 )
 a + 5 d = 37
 12 + 5 d = 37
 5 d =37 -12
 5 d =25
 d =5
 S 15 = 15 /2 ( 24 + 14 (15 )
 S 15 = 15 / 2 (24 + 70 )
 S 15 = 15 /2 * 94
 S 15 = 15 * 42
 S 15 = 630

 ================================

 5a)
 Let bag=B
 Shoe= S
 U=120
 n(BnS)=45, n(s)=x+11, n(b)=x
 n(SnB')=x+11-45
 =x-34
 n(BnS') = 45

 5b)
 Y - 45 + 45 + Y - 34 = 120
 2Y - 34 = 120
 2Y = 120 - 34
 2Y = 154
 Y = 154/2
 Y = 77
 11+x=77+11
 = 88
 Therefor 88 bought shoes costumer

 5c)
 n(bag)= 77 customers
 Pr. =77/120


 ================================
 *SECTION B ANS 5 QUESTIONS ONLY*
 ================================

8) 
In Table Form / Tabular form
X = 1,2,3,4,5
F = m+2, m-1, 2m-3, m+5, 3m-4 = 8m - 1‎
Fx = m+2, 2m-2, 6m-9, 4m+20, 15m-20 = 28m - 9‎

But x̄ ( this symbol (x̄) means X bar) 
= 75/23‎
ΣFx / Σf = 75/23 = 28m - 9/8m-1
  75/23 = 28m - 9/8m - 1
Cross multiply 
75(8m-1) = 23(28m-9)
600m - 75 = 644m - 207
 -75 + 207 = 644m - 600m
132 = 44m
M = 3

8b)
In tabular form
X = 1,2,3,4,5
F = 5,2,3,8,5
Cum Freq= 5,7,10,18,23

Q1 = (N+1/4) = (23+1/4)
 = 6

Q3 = (3N + 1/4) = (3*23+1/4)
 = 18

Inter quarter range = Q3 - Q1
 =. 18-6
 = 12

8bii)
Pr. (at least 4 mark)
= 8+3+2+5/23
 = 18/23‎



 10) 
 Sin x = 5/13
 Using pythagoras rule
  M^2 = 13^2 - 5^2 (^ means Raise to power)
  M^2 = 169 - 25
  M ^2 = 144
  M = √144
  M = 12
 Hence: 
 Cos x - 2sin x / 2tan x
 12/13 - 2(5/13) / 2(5/12)
‎= 12/13 - 10/23 / 5/6
  FIND LCM
  = 12 - 10/13 / 5/6
  =  12/65

 10b) 
 Draw a triangle LACB
 in triangle LCB
 Hyp^2 = Opp^2 + Adj^2
 12^2 = 9.6^2 + |CB|^2
 144 = 92.16 = |CB|^2
 144 – 92.16 = |CB|^2
 51.84 = |CB|^2
 therefore, |CB| = √51.84
 |CB| = 7.2m
 |AC| + |CB| =|AB|
 |AC| + 7.2m = 10m
 |AC| = 10m – 7.2m

 |AC| = 2.8m
 In triangle LCA
 Hyp^2 = Opp^2 + Adj^2
 |LA|^2 = |AC|^2 + |LC|^2

 |LA|^2 = 2.8^2 + 9.6^2

 |LA|^2 = 7.84 + 92.16

 |LA|^2 =100

 |LA| = √100

 |LA| = 10m

 10bii) 
 in triangle LCA
 sinθ = Opp/Hyp
 sinθ = |LC|/|LA|
 sinθ = 9.6/10
 sinθ = 0.96
θ = sin^-1 (0.96)
θ = 73.74

 ===============================

 13ai)
 given
 x(*)y=x+y/2
 i)3(*)2/5=3+2/5/2
 =(15+2/5)*1/2
 =17/5*1/2
 =17/10= 1,7/10

 13aii)
 8(*)y=8^1/4
 =8+y/2 =33/4
 32+4y=66
 4y=66-32
 4y=34
 y=34/4
 y=17/2
 y=8^1/2

 13b)
 given DABC
 AB=(^-4/6) and AC =(3/^-8)
 so AP =1/2(^-4/6)
 AP=(^-2/3)
 hence
 CP = CA + AP
 CP= -(3/^8)+(^-2/3)
 CP = (^-5/11)

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