Waec 2017 Mathematics Obj And Theory Answers – May/June Expo -Now available

Maths OBJ:
1CBBACCCBBA
11ADBBAABCDB
21BCADBCCAAB
31CDCACDBACB
41BDABCDDCAD

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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)

12a)
3y^2-5y+2=0
y^2 – 5/3y + 2/3=0
y^2-5/3y=-2/3
y^2-5/3y+(^-5/6)^2=(-^5/6)^2-2/3
(y-5/6)^2=25/36-2/3
(y-5/6)^2=25/-24/36
(y-5/6)^2=1/36
(y-5/6)=+sqr1/36
y=5/6+1/6
y=5+1/6 or 5-1/6
y=6/6 or 2/3
y=1 or 2/3
12b)
given
M N = [2,3 1,4] hence
[1,4 2,3] * [m,n x,y] =[2,3 1,4] [m+2n, x*2y] [4m+3n, 4x+5y] = [2,3, 1,4] therefore
m+2n=2——(i)
4m+3n=3——(ii)
from ——(i)
m=2-2n
4(2-2n)+3n=3
8-8n+3n=3
8-5n=3
8-3=5n
5=5n
n=1
hence
m=2-2(1)
M=0
also
x+2y=1——(i)
4x+3y=4——(ii)
from ——(iii)
x=1-2y
4(1-2y)+3y=4
4-8y+3y=4
y=0
therefore x=1-2(0)
x=1
this N=[i i]

11a)
8 students finished
12 tanks in 2/3 (60) mins
= 40 mins
4 student wil finish
X tanks in 1/3 (60)min
= 20mins
X = 4x20x12/8×40
= 3tanks
11b)
L(AB) = 200m |ON| = 12cm
r2 = (AN)2 + (ON )2
r2 = (10)2 +(12)2
r2 = 100 + 144
r2 = 244
r = Sqr 244
r = 15.6CM
11bii)
L(AB) = 2r sin 0/2
20 = 2 (15.6) sin 0/2
20 = 31.2 sin 0/2
sin0/2 = 20/31.2
sin0/2 = 0.6410
0/2 = sin -1 (0.6410)
0/2 = 39.87
0 = 2 (39.87)
0 = 79.74
= 79.7′ (1 d.p )
11bii)
p 2r + 0/360 x 2TTr
= 2 (15.6 ) + 79.7/360 x 2x 3 x42x15.6
=31.2 + 21.7
= 52.9 cm

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4b )
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
5d = 37 – 12
5d = 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 )
draw
U =20
B = y – 45
S = y – 34
B =bag
S =shoe
let n ( B)=y
n( S )=y + 11
for bag only y – 45
for shoe only y – 11 – 45 =y – 34
5b )
y – 45 + 45 +y – 34 = 120
2y – 34 = 120
2y = 154
y = 154/ 2
y = 77
number of customers who bought shoe = y +11
77 + 11 = 88
5c )
n( bag)=77 customers
probability = 77 /120
=0 . 642

SECTION B ANS 5 QUESTIONS ONLY

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

10 a )

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 – 2 sin 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

10 bi)

Considering < LMB / MB / ^ 2 . = 12 ^ 2 - 9. 6 ^ 2 / MB / ^ 2 = 51 . 84 / MB / = √ 51 . 84 / MB / = 7 . 2 m From < AML / LA / ^ 2 = 2 . 8 ^ 2 + 9. 6 ^ 2 / LA / ^ 2 = 100 / LA / = √ 100 / LA / = 10 m 10 bii) Let the angle be . θ From < AML Tanθ = 9 . 6 / 2 . 8 Tan θ = 3 . 4288 θ = Tan ^ - 1 ( 3 . 4288 ) = 73 . 74 ° If you want to Get the Answers Direct To Your Phone As Sms, Kindly Send MTN Recharge Card worth of N1000

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