Free Waec 2017 Mathematics Obj And Theory Answers – May/June Expo

Free Waec 2017 Mathematics Obj And Theory Answers – May/June Expo



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Verified Maths 2017 OBJ

1-10 CBBCCACBBC
11-20 CACBAAACBB
21-30 BBADBCCACB
31-40 CBCACBDADB
41-50 DDACCCDDCB

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(1a)
(y-1)log4=ylog16
log4^(y-1)=log16^y
log4^(y-1)=log4^2*y
y-1=2y
y=-1

(1b)
Distance =speed*time
let the time of the walk be x hours
=>4*(x+0.5)=5x
Because 30 minutes 0.5hours
4x+2=5x
5x-4x=2
x=2hours
=>5*2
=10km

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(2a)
2/3(3x-5)-3/5(2x-3)=3
2/3(3x-5)-3/5(2x-3)=3
lcm=15
=>(6x-10)/3-(6x-9)/9=3
Multiply both sides by 15
15(6x-10)/3-15(6x-9)5=15(3)
30x-50-3(6x-9)=45
30x-50-18x+27=45
12x-23=45
12x=45+23
x=68/12
x=5(2/3)

(2b)
=180-n-88=92-n
Also UTQ=180cm--(sum of 80+92-n+180-m=180(sum of 352-n-m=180
-n-m=180-352
-n-m=-172
-(n+m)=-172
m+n=172degrees.

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(4a)
Tnth=a+(n-1)d
Snth=n/2[2a+(n-1)d]
T6th=37
S6th=147
T6th=>37=a+(6-1)d
=>37=a+5d-----(i)
S6th=>147=6/2[2a+96-1)d
=>147=3(2a+5d)
147=6a+15d------(ii)
Therefore a+5d=37----(i)
6a+15d=147-----(ii)
divide equation ii by 3
2a+5d=49--(ii)
multiply eq i by ii
=>a+10d=74
a+5d=49
5d/5=25/5
therefore d=5
substitute for d in eq i
a+5d=37
a+5(5)=37
a=37-25
a=12
The first term=12

(4b)
Sn=n/2[2a+(n-1)d]
S15=15/2[2(12)+(15-1)5]
=15/2(24+(14*5)
=7.5(24+70)
=705
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(5a)
Let bag =B
shoe=S
U=120
n(BnS)=45,
n(S)=x+11
n(B)=x
n(SnB^1)=x+11-45
=x-34
n(BnS^1)=x-45

(5b)
DRAW THE VENN DIAGRAM
x-45+45+x-34=120
2x-34=120
2x=120+34
2x/2=154/2
x=77
=11+x
=11+77=88
Therefore 88customers bought shoes

(5c)
Pr(bag)=n(B)/U=77/120

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(8)
Tabulate:
?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 x bar
= 75/23?
EFx / Ef = 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?

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9ai)
DRAW THE DIAGRAM


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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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11 a )
8 students finished
12 tanks in 2 /3 ( 60 ) mins
= 40 mins
4 student wil finish
X tanks in 1 / 3 (60 )min
= 20 mins
X = 4 x 20 x12 /8 x 40
= 3 tanks

11 b )
L (AB ) = 200 m |ON | = 12 cm
r 2 = (AN )2 + (ON ) 2
r 2 = (10 ) 2 +( 12 )2
r 2 = 100 + 144
r 2 = 244
r = Sqr 244
r = 15 .6 CM

11 bii )
L (AB ) = 2 r sin 0 / 2
20 = 2 (15 . 6 ) sin 0 /2
20 = 31 .2 sin 0 /2
sin 0 / 2 = 20 / 31 .2
sin 0 / 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 )

11 biii )
p 2 r + 0 / 360 x 2 TTr
= 2 ( 15 .6 ) + 79 .7 / 360 x 2 x 3 x 42 x 15 .6
= 31 .2 + 21 .7
= 52 . 9 cm

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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]

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(13ai)
given
x(*)y=x+y/2
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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