Waec Questions And Answers On Mathematics 2017

Waec MATHEMATICS ANSWER
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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= -1

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

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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

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3a)

Tan 23.6° = h/50

Cross multiply

Tan 23.6° x h/50

h = 50 tan 23.6°

= 21.844m

3b)

Area of <TRU = 45cm^2 (Note: This ^ means Raise to power)

A = 1/2bh

45 = 1/2 x 10 x h

45 = 5h

h = 9cm

Area of < QTUS = 1/2 ( QT + US)h

= 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)=77 customers

probability = 77 /120

=0 . 642

SECTION B ANS 5 QUESTIONS ONLY

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

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

10 a )

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 – 2 sin 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

10 bi)

Considering < LMB

/ MB / ^ 2 . = 12 ^ 2 – 9. 6 ^ 2

/ MB / ^ 2 = 51 . 84

/ MB / = √ 51 . 84

/ MB / = 7 . 2 m

From < AML

/ LA / ^ 2 = 2 . 8 ^ 2 + 9. 6 ^ 2

/ LA / ^ 2 = 100

/ LA / = √ 100

/ LA / = 10 m

10 bii)

Let the angle be . θ

From < AML

Tanθ = 9 . 6 / 2 . 8

Tan θ = 3 . 4288

θ = Tan ^ – 1 ( 3 . 4288 )

= 73 . 74 °

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 x 12 /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 bii )

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

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

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)