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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

2b)

80 = n+r ( ext. < equal sum of opp int. < s )

80 = n+r ------------ (i)

M = 80 + 92 - n

M = 172 - n ----------- (ii)

80 + 92 - n + 180 - m = 180°

80 + 92 + 180 - n - m = 180°

352 - n - m = 180

-n - m = 180 - 352

- n - m = - 172°

m + n = 172°

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

3a)

Tan 23.6° = h/50

Cross multiply

Tan 23.6° x h/50

h = 50 tan 23.6°

= 21.844m

aprox. 22m

3b)

Area of 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

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

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

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

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

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

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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