Waec 2017 General Mathematics Obj and Theory Complete Answer Now Available

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Maths-Obj

1-10: CBBDCADBBC

11-20: CACBAAACAB

21-30: BBACBCCAAB

31-40: CBCACBDADB

41-50: DBDCCDBDAB

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SECTION A ANS ALL QUESTIONS ================================

1 a) (y -1 ) log 4 ^ 10 = ylog 16 ^ 10

log 4 ^ 10 ( y -1 )= log 16 ^ y 10

4 ^ ( y -1 )= 16 y

4 ^ y -1 = 4 ^ 2 y

y- 1 = 2 y

-1 = 2 y= y

-1 = y

y= -y

1 b )

let the actual time for 5 km / hr be t

for 4 km /hr = 30 mint + t

4 km /hr =0 . 5 + t

distance = 4 (0 . 5 + t )

= 2 * 4 t

for 5 km /hr , time = t

distance =5 t

1 + 4 t = 5 t

t= 2 hrs

actual distance = 5 * 2 = 10 km

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

2a)

2/3 (3x – 5) – 3/5 (2x – 3) = 3

(15) x 2/3 (3x – 5) – 3/5 (15) (2x – 3) = 3(15)

10(3x -5) – 9(2x – 3) = 45

30x – 50 – 18x + 27 = 45.

12x – 23 = 45

12x = 45+23

12x = 68

X = 68/12

X = 5.67

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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

aprox. 22m

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4 a)

T 6 =37

T 6 =a + ( 6 – 1 )d

T 6 =a + 5 d

a + 5 d = 37 – — -( eq1 )

s 6 = 147

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

147 = 3 (2 a + 5 d )

49 = 2 a + 5 d

2 a+ 5 d = 49 — — (eq 2 )

a + 5 d = 37 – –( eq1 )

2 a+ 5 d = 49 — -( eq2 )

a =12

4 b ) S 15 = 15 /2 ( 2 (12 )+ 14 d )

S 15 = 15 /2 ( 24 + 14 d )

from(1 )

a + 5 d = 37

12 + 5 d = 37

5 d =37 -12

5 d =25

d =5

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

S 15 = 15 / 2 (24 + 70 )

S 15 = 15 /2 * 94

S 15 = 15 * 42

S 15 = 630

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5a) Let bag=B

Shoe= S

U=120

n(BnS)=45, n(s)=x+11, n(b)=x

n(SnB’)=x+11-45

=x-34

n(BnS’) = 45

5b)

Y – 45 + 45 + Y – 34 = 120

2Y – 34 = 120

2Y = 120 – 34

2Y = 154

Y = 154/2

Y = 77

11+x=77+11

= 88

Therefor 88 bought shoes costumer

5c)

n(bag)= 77 customers

Pr. =77/120

================================ SECTION B ANS 5 QUESTIONS ONLY

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

8) 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

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

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

11a)

8 students finished

12 tanks in 2/3 (60) mins

= 40 mins

4 student wil finish

X tanks in 1/3 (60)min

= 20mins

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

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

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]

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)

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