**WAEC 2017 Mathematics Theory and Obj Answers**

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MATHS OBJ:

1-10: CBBCCACBBC

11-20: CACBAAAADB

21-30: BBADBCCACB

31-40: BBCACBDADA

41-50: DDACCCDDCB

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

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

3a)

DRAW THE DIAGRAM

tanx=50/x

tan66.4=50/x

x=50/2.289

x=21.844m

x=22m

3b)

DRAW THE DIAGRAM

Area of=1/2*10*h=45

=>5h=45

h=45/5

h=9cm

Area of parallelogram=L*h

=9*6

=54cm^2

Therefore the area of QTUS=54 45

=99cm

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

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

(8a)

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 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

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)

9ai)

DRAW THE DIAGRAM

9aii)

RTS=?

PTS=180-65(Sum of9b)

Therefore tita=180-52-32-70(sum of < in a triangle)

tita=20degrees

Therefore LWYZ=20 degrees

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Comments: 41Publics: 581Registration: 13-02-2017
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