**MATHEMATICS OBJ**:

1-10 CDAAEABAEC

11-20 AEDDCDCDCC

21-30 CEBDEDCBBC

31-40 CBEEECBDCC

41-50 DBCBCDDBCA

51-60 BCBDCDCCEC.

**Mathematics Theory**

1a)

Log 10(20*-10)-log10(*+3)=log105

(20*-10/*+3)=log10 =5

20*-10/*+3=5

5(*+3)=20*-10

5*+15=20*-10

15+10=20*-5*

25=15*

*=25/15

*=5/3=1 2/3

1b)

Discount percent =15%

Discount amount =#600

Actual amount paid on the article =?

Original amount on the article =*

15%*=#600

15/100* =600

15*=600*100

15*=60000

*=60000/15

*=#4,000

Therefore actual amount paid on the article

=#4,000-#600

=#3,400

Actual amount paid on the article =#3,400

(2a)

(X^2 Y^-3 Z)^3/4/X^-1 Y^4 Z^5

= (X^2)^3/4/X^-1 * (Y^-3)^3/4/Y^4 * Z^3/4/

Z^5

= X^3/2/X^-1 * Y^-9/4/Y^4 * Z^3/4/Z^5

=X^3/2+1 * Y^-9/4-4 * Z^3/4-5 =X^5/2 * Y^-25/4 * Z^-17/4

=X^10/4 * Y^-25/4 * Z^-17/4

=(X^10/Y^25 Z^17)^1/4

(2b)

√2/k + √2 = 1/k – √2

Multiply both sides by (k+√2)(k-√2)

√2(k-√2) = k+√2

√2k-√2 = k+√2

√2k-k = 2+√2 K(√2 -1) = 2+√2

K = 2+√2/√2-1

K = -(2+√2)/1-√2

Rationalizing

K = -(2+√2) * 1+√2/1-√2

K = -(2+√2)(1+√2)/1 – 2 K = (2+√2)(1+√2)

K = 2+2√2 + √2+2

K = 4+3√2

3)

V = Mg√1 – r²

Square both sides

V² = m²g²(1-r²)

V²/m²g² = 1-r²

r² = 1 – v²/m²g²

r = √1-(v/mg)²

If v = 15, m = 20, and g = 10

r = √1 – (15/20*10)²

r = √1 – (0.075)²

r= √(1.075)(0.925)

r = √0.994375

r = 0.9972

4i)

length of Arc of the sector

Titter= 72?, r = 14cm

L= titter / 360 x 2 pie r

==> L= 72/360 x 2 x 22/7 x 14

=44352/2520 = 17.6cm

4ii) perimeter of the sector

Perimeter = titter/360 x 2 pie r + 2r = 17.6 +(2×14) =17.6+28= 45.6cm

4iii) Area of the sector

Area = Titter/360 x pie r? =

72/360 x 22/7 x (14)? = 72 x 22 x 196/2520

Area= 310464/2520 = 123.2cm

6

8)

x=a+by(eqi)

when y=5 and x=19

19=a+5b(eqii)

when y=10 and x=34

34=a+10b(eqiii)

solving eqii and eqiii

a+10b=34

a+5b=19

=>5b=15

b=15/5=3

putting b=3 in eqii

19=a+5(3)

19=a+15

a=19-15

a=4

(8i)

Putting a=4 and b=3 in eqi

x=4+3y

This is the relationship between xand y

(8ii)

When y=7

x=4+3(7)

x=4+21

x=25

7a)

A) T3=6 & T7 =30

I)common difference using Tn=a+(n-1)d

In the 3rd term; n =3

=>T3=a+(3-1)d=6

=>a+2d=6 equation (1)

In the 7th term ;n =7

=> T7=a+(7-1)d=30

=>a + 8d=30 equation (1) & (2)

Simultaneous

A+2d =6 equation (1)

A+ 8d=30 equation (2)

0+(1-6d) =-24

=> -6d= -24

=>d = -24/-6 =4

II) first term put d=4 into equation

x+y/2 =11

x+y= 11*2

x+y= 22 —eq(i)

x-y= 4 —-eq(ii)

x+y = 22—-(i)

–

x-y= 4—-(ii)

____________

2y = 18

y= 18/2

y=9

Substitute y=9 in equ 1

x+9=22

x=22-9

x=13

x=13, y=9

x+y= 13+9= 22

Sum of the two number

11b)

(6x + 3) dx

(6x + 3)dx

(6x +3)^6 – (6x + 3)^1

(6 x + 3)^5

(7776x^5 + 243)

38,880x/6 + 243

6480 x^6 + 243x

9(720x^6 + 27x)

11c)

y = x² + 5x – 3 (x = 2)

y = 2² + 5(2) – 3

y = 4 + 10 – 3

y = 14 – 3

y = 11

Gradient of the curve = 11

5a)

Mode = mass with highest frequency = 35kg

Median is the 18th mass

= 40kg.

5b)

IN A TABULAR FORM

Under Masses(x kg)

30,35,40,45,50,55

Under frequency(f)

5,9,7,6,4,4

Ef = 35

Under X-A

-10, -5, 0, 5, 10, 15

Under F(X-A)

-50, -45, 0, 30, 40, 60

Ef(X – A) = 35

Mean = A + (Ef(X – A)/Ef)

= 40 + 35/35

= 40 + 1

= 41kg

(12a)

Pr of Abu to pass = 3/7

Pr of Abu to fail = 1 – 3/7 = 7-3/7 = 4/7

Pr of kuranku to pass = 5/9

Pr of kuranku to fail = 1 – 5/9 = 9 – 5/9 = 4/9

Pr of musa to pass = 12/13

Pr of musa to fail = 1 – 12/13 = 13 – 12/13 = 1/13

Pr of only one of them passing is

=(3/7*4/9*1/13)+(5/9*4/7*1/13)+(12/13*4/

7*4/9)

=12/819+ 20/819 + 192/819

=12+20+192/819 = 224/819

= 32/117

(12b)

10Red + 8green + 7blue = 25

(i)

pr of different colour is

Prof(RG)+(RB)+(GB)+(BG)+(BR) +(GR)

=

(10/25*8/24)+(10/25*7/24)+(8/25*7/

24)+(7/25*8/24)+(7/25

=80/100 + 70/600 + 56/600 + 56/600 + 70/600 + 80/600

= 80+70+56+56+70+80/600

= 412/800 = 103/200

(ii)

pr of atleast one must be

=Pr[RB+BR+GB+BG+BB]

= (10/25*7/24)+(7/25*10/24)+(8/25*7/24)+ (7/25*8/24)

+ (7/25*7/24)

=70/600+70/600+56/600+56/600+49/600

=70+70+56+56+49

/600

=301/600

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