**WAEC 2017 Mathematics Objectives And Theory Answer | May June Free Expo.**

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## Waec 2017 Mathematics Objectives Answer

**1-10**: CBBCCACBBC

**11-20:** CACBAAAADB

**21-30:** BBADBCCACB

**31-40:** BBCACBDADA

**41-50:** DDACCCDDCB

## Waec 2017 Mathematics Theory Answer

**NOTE:**

**Where ever u c
^ it means raise to power
/ division
* multiplication
X normal X**

1a)

(y-1)log4^10= ylog16^10

log4^10 (y-1)= log16^y10

4^(y-1)=16y

4^y-1=4^2y

y-1=2y

-1=2y=y

-1=y

y= -y

1b)

let the actual time for 5km/hr be t

for 4km/hr=30mint + t

4km/hr=0.5 + t

distance = 4(0.5+t)

=2*4t

for 5km/hr, time= t

distance =5t

1+4t=5t

t=2hrs

actual distance = 5*2=10km

2a)

2/3(3x-5)-3/5(2x-3)=3/1

L C M =15

10(3x-5)-9(2x-3)=45

30x-50-18x+27=45

30x-18x=45+50-27

12x-23=45

12x=45+23

12x=68

x=68/12

x=34/6

x=17/3

2b)

U’aS=180-(n+88)

=180-n-88=92-n

also, u’TQ=18m

80degree + 92-n+180-m=180degree

80+92+180-n-m=180degree

352-n-m=180degree

-n-m=180-352

-n-m=-172

+(n+m)= +172

m+n=172dgree

3a)

Tan 23.6° = h/50

Cross multiply

Tan 23.6° x h/50

h = 50 tan 23.6°

= 21.844m

3b)

Area of A = 1/2bh

45 = 1/2 x 10 x h

45 = 5h

h = 9cm

Area of
= 1/2 ( 6 + 16)9

= 99cm^2

4a)

T6=37

T6=a+(6-1)d

T6=a+5d

a+5d=37 —–(eq1)

s6=147

sn=n/2(2a+(n-1)d)

147=3(2a+5d)

49=2a+5d

2a+5d=49 —-(eq2)

a+5d=37 —(eq1)

2a+5d=49 —(eq2)

a=12

4b)

S15=15/2(2(12)+14d)

S15=15/2(24+14d)

from(1)

a+5d=37

12+5d=37

5d=37-12

5d=25

d=5

S15 = 15/2(24+14(15)

S15= 15/2(24+70)

S15=15/2*94

S15=15*42

S15=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)=77customers

probability =77/120

=0.642

**SECTION B ANS 5 QUESTIONS ONLY**

10a)

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

10bi)

Considering
/MB/^2. = 12^2 – 9.6^2

/MB/^2 = 51.84

/MB/ = √51.84

/MB/ = 7.2m

From
/LA/^2 = 2.8^2 + 9.6^2

/LA/ ^2 = 100

/LA/ = √100

/LA/ = 10m

10bii)

Let the angle be. θ

From Tanθ = 9.6/2.8

Tan θ = 3.4288

θ = Tan^-1 ( 3.4288)

= 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)

8a)

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

8bi)

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

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

= 2omins

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

**NOTE:** Get yourself prepare for WAEC 2017 Mathematics by reading your notebooks and textbooks.

You can also have interest in checking west Africa examination Council WAEC syllabus

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