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2017 general mathematics obj and theory waec expo

MATHS-ANSWERS[Rated 1st Answers That Will Be Posted[100%] 
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Maths-Obj
1CBBACCCBBA
11ADBBAABCDB
21BCADBCCAAB
Subscribe now

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Note
Where ever u c
^ it means raise to power
/ division
x multiplication
X normal X



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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

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

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°‎

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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 <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
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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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