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MATHEMATICS NECO GCE 2015 FREE EXPO ANSWERS




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Saturday 7th Nov. 2015
GENERAL MATHEMATICS (OBJ)
11.00AM-12.45.00PM
MATHEMATICS (ESSAY)
1.00PM-3.30PM
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MATHS OBJ:
1-10: BCDCECECAB
11-20: EECBBDDDDD
21-30: BCADECDDDA
31-40: EDEDCEDCED
41-50: CBEACBEBCC
51-60: CDACBBBDBA


(1a)
3/5(4x+3) = 3/4(3x-1)
3(4x+3)/5 = 3(3x-1)/4
12x+9/5 = 9x-3/4
5(9x-3) = 4(12x+9)
45x-15=48x+36
45x-48x=36+15
-3x=51
x=-51/3
x = -17

(1b)
x/sin60 degree = 10/sin30 degree
x = 10sin60 degree/sin 30 degree = 10 * 0.866/0.5
x = 8.66/0.5
=17.32cm
Height = 17.32cm


2a)
Draw the angle.
Tan 62degree = x/25
x=25 tan 62degree =25*1.8807
=47.02m
:. Height of the tower = 47.02m
2b)
U = {a,b,c,d,e,f,g}
A = {b,d,g} B = {a,c,e}
AnB = {0}
PLS NOTE {0},the zero is file, it looks like
tita inside{}.



(5)
DRAW A TABLE:
x: 13,28,34,45,51,69
f: 1,1,1,1,1,1 = 6
fx: 13,28,34,45,51,69 = 240
(x-xbar): -27,-12,-6,5,11,29
(x-xbar)^2: 729,144,36,25,121,841 =
1896

(i) Mean = Efx = 240/6 = 40

(ii) Standard deviation
=root (x-xbar)^2/N
=root 1896/40
=root 47.4
=6.88



(8)
Draw the triangle
(8a)
In triangle APT,
Cos 40degree = 100/|AT|
|AT| = 100/cos 40degree
|AT| = 100/0.766
|AT| = 130.5m

(8b)
In triangle ATP
Tan 40degree = |PT|/100
|PT| = 100 tan 40degree
=100*0.8391
=83.9m
In trangle PBT,
tan 55degree = PT/x
tan 55degree = 83.9/x
x tan 55degree = 83.9
x = 83.9/tan 55degree
x=83.9/1.4281
x=58.75m
x=58.8m

(8c)
Distance btwn points A and B
=|AP|+|BP|
=100m + xm
=100m+58.8m
=158.8m



12a)
y=(x^3+1)^4=U^4
Let U=x^3+1
dy/dx=du/dx*dy/du
du/dx=3x^2
dy/du=4u^3
dy/dx=3x^2*4(x^3+1)^3
=12x^2(x^3+1)^3

12bi)
Tabulate
x-0-1-2-3-4-5-6
0-0-0-0-0-0-0-0
1-0-1-2-3-4-5-6
2-0-2-4-6-1-3-5
3-0-3-6-2-5-1-4
4-0-4-1-5-2-6-3
5-0-5-3-1-6-4-2
6-0-6-5-4-3-2-1

12bii)
2sn=n[2a+(n-1)d]
2sn=n[2a+nd-d]
2sn-2an=dn^2-dn
2(sn-an)=d(n^2-n)
d=2(sn-an)/n^2-n
d=2(sn-an)/n(n-1)

11ai)
First term=2
Common ratio=6/2=3
Number of terms required=6
Un=ar^n-1
U6=2*(3)^6-1
=2*3^5
=2*243
=486

11aii)
Radius=6cm
pie=22/7
Volume=4752cm^3
V=Pie*r^2*h
4752=22/7*6^2*h
4752*7=22*36*h
33264=792h
h=33264/792
h=42cm
therefore height of the cylinder is 42cm

11bi)
Let x=-2 or x=3/5
x-2=0 or 5x=3
(x-2)(5x-3)=0
5x^2-3x-10x+6=0
5x^2-13x+6=0

11bii)
2x^2+5x-12=0
2x^2+8x-3x-12=0
(2x^2+8x)-(3x+12)=0
2x(x+4)-3(x+4)=0
(2x-3)(x+4)=0
(
Draw the triangle



(9 a )
Draw the diagram
Radius = 13 cm
|OM | = 5 cm
In triangle MOB ,
|OB| ^ 2 = | OM| ^ 2 + | MB |^ 2
13 ^ 2 = 5 ^ + |MB |^ 2
|MB |^ 2 = 13 ^ 2 - 5 ^ 2
|MB |^ 2 = 169 - 25
|MB |^ 2 = 144
|MB | = root 144
|MB | = 12 cm
Hence, the length of the chord = 2 | MB |
=2 * 12 cm
=24 cm

(9 b )
Green balls = 6
Blue balls = 10
Total balls = 16
Prob (that the balls are of different colors )
= Prb (GB ) + Prb (BG )
=(6 / 16 * 10 / 15 ) + (10 / 16 * 6 / 15 )
=(3 / 8 * 2 / 3 ) + (2 / 8 * 3 / 3 )
=(1 / 4 ) + (1 / 4 )
=1 / 4 + 1 / 4
=1 +1 / 4
=2 / 4
=1 / 2

9c)
x & 1/y^2
x = k/y^2
12 = k/(1/2)^2
k = 12*1/4
k = 12*1/4 = 3

.:. x = 3/y^2
when y = 3/4
x = 3/(3/4)^2
=3/9/16
=3*16/9
=16/3 =5whole number 1/3











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Posted By KellyChi On 11:25 Sat, 07 Nov 2015

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