Monday 4th June 2018
MATHEMATICS OBJ:
1-10 CDAAEABAEC
11-20 AEDDCDCDCC
21-30 CEBDEDCBBC
31-40 CBEEECBDCC
41-50 DBCBCDDBCA
51-60 BCBDCDCCEC

10a)
Obtuse <BOD + Reflex<BOD = 360degrees (angle at a point)
105 + reflex<BOD = 360degrees
Reflex <BOD= 360 - 105
=255°
Now 2w = reflex<BOD(angle at centre = twice angle at circumference)
2w =255°
W = 255/2 =127.5°

Also 2x = obtuse<BOD(angle at centre = twice angle at circumference)
2x = 105°
X = 105/2 = 52.5°
Now EDF = y(base angles of an isosceles triangle)
BED=X=52.5°(angles in the  same segment)
EFD+EDF=BED (sum of interior angles of a triangle equal exterior angle)
Y+y = 52.5°
2y = 52.5°
Y = 52.5°/2
=26.25°

(10b)
Draw the diagram
Opp/adj = TanR
|TB|/|BR| = TanR
100/|BR| = Tan60°
|BR| = 100/tan60
|BR| = 100√3
|BR| = 100√3 * √3/√3

=100√3/3m OR  57.7m

11a)
x+y/2 =11
x+y= 11*2
x+y= 22 ---(1)
x-y= 4 ----(11)
x+y = 22----(1)
-
x-y= 4----(11)
____
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

Correct 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=600100
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

No 4(i)
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

(ii) perimeter of the sector

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

iii) 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?

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

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

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

11a)
x+y/2 =11
x+y= 11*2
x+y= 22 ---(1)
x-y= 4 ----(11)
x+y = 22----(1)
-
x-y= 4----(11)
____________
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

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

(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

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