*2017 WAEC CERTIFIED MATHEMATICS OBJECTIVES*💯💯💯✅✅✅✅
*1-10 CBBACCCBBA*
*11-20 ADBBAABCDB*
*21-30 BCBDBCCAAB*
*31-40 CAAACDBACB*
*41-50 BDABAADCAD*
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*SECTION A ANS ALL QUESTIONS*
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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
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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.
80 = n+r ------------ (i)
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°
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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
A = 1/2bh
45 = 1/2 x 10 x h
45 = 5h
h = 9cm
Area of
= 1/2 ( 6 + 16)9
= 99cm^2
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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
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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
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*SECTION B ANS 5 QUESTIONS ONLY*
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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
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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)
