WAEC 2017 Mathematics Question And Answer – May/June Exam Expo - Examexpo.com.ng – 2019 Waec Expo, Neco 2019 Expo, Jamb 2019 Expo.

Waec 2017 General Mathematics Obj and Theory Complete Answer Now Available

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Maths-Obj
1-10: CBBDCADBBC
11-20: CACBAAACAB
21-30: BBACBCCAAB
31-40: CBCACBDADB
41-50: DBDCCDBDAB
================================

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

===============================
3a)
Tan 23.6° = h/50
Cross multiply
Tan 23.6° x h/50
h = 50 tan 23.6°
= 21.844m
aprox. 22m

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

=======================================
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
=========================
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]
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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