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# NECO 2018 MATHEMATICS SUBSCRIPTION

Posted by extralmaster on Jun 01, 2018 in GENERAL DISCUSSIONS » 937 view(s) » 4 Comments

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MATHS OBJ:
1-10 CDAAEABAEC
11-20 AEDDCDCDCC
21-30 CEBDEDCBBC
31-40 CBEEECBDCC
41-50 DBCACDDBCA
15-60 BCBDCDDCCE

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)
Let actual amount be #X
15% of #x = #600
15x/100 = 600
X = (100/15)*600
X = 100*40
X = 4,000
Actual amount = #4,000
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(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)
âˆš2/k + âˆš2 = 1/k - âˆš2
Multiply both sides by (k+âˆš2)(k-âˆš2)
âˆš2(k-âˆš2) = k+âˆš2
âˆš2k-âˆš2 = k+âˆš2
âˆš2k-k = 2+âˆš2
K(âˆš2 -1) = 2+âˆš2
K = 2+âˆš2/âˆš2-1
K = -(2+âˆš2)/1-âˆš2
Rationalizing
K = -(2+âˆš2) * 1+âˆš2/1-âˆš2
K = -(2+âˆš2)(1+âˆš2)/1 - 2
K = (2+âˆš2)(1+âˆš2)
K = 2+2âˆš2 + âˆš2+2
K = 4+3âˆš2
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(3)
V = Mgâˆš1 - rÂ²
Square both sides
VÂ² = mÂ²gÂ²(1-rÂ²)
VÂ²/mÂ²gÂ² = 1-rÂ²
rÂ² = 1 - vÂ²/mÂ²gÂ²
r = âˆš1-(v/mg)Â²
If v = 15, m = 20, and g = 10
r = âˆš1 - (15/20*10)Â²
r = âˆš1 - (0.075)Â²
r= âˆš(1.075)(0.925)
r = âˆš0.994375
r = 0.9972
==============================

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 +(2x14) =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?
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(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
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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
)
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
(7bi)
T3=>ar^2=9/2(eqi)
T6=>ar^5=243/16(eqii)
Dividing eqii by eqi
ar^5/ar^2=243/16 divided by 9/2
r^3=243/16*2/9
r^3=27/8
r^3=3^3/2^3
r=3/2
Putting this into eqi
a(3/2)^2=9/2
a(9/4)=9/2
a=9/2*4/9
a=4/2=2
(7bii)
Common ratio r=3/2 as above
(7biii)
Seventh term=2(3/2)^6
=2(729/64)
=729/32
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(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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(10a)
Obtuse 105 + reflex Reflex =255Â°
Now 2w = reflex 2w =255Â°
W = 255/2 =127.5Â°

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

(10b)
Draw the diagram
|TB|/|BR| = TanR
100/|BR| = Tan60Â°
|BR| = 100/tan60
|BR| = 100âˆš3
|BR| = 100âˆš3 * âˆš3/âˆš3
=100âˆš3/3m OR 57.7m
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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 = xÂ² + 5x - 3 (x = 2)
y = 2Â² + 5(2) - 3
y = 4 + 10 - 3
y = 14 - 3
y = 11
Gradient of the curve = 11

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