# 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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