2015 NECO GCE NOV/DEC
EXAM DAY: SATURDAY
PAPER: MATHEMATICS
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}
Note that the 0 in {0} stands for null that look like tita
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/x = 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 40 degree = 100/ |AT|
|AT | = 100/ cos 40 degree
|AT | = 100/ 0. 766
|AT | = 130. 5m
8b )
In triangle ATP Tan 40 degree = |PT |/100
|PT | = 100 tan 40 degree
=100* 0 . 8391
=83 . 9 m
In trangle PBT ,
tan 55 degree = PT /x tan 55 degree = 83 . 9/ x
x tan 55 degree = 83 . 9
x = 83 . 9/ tan 55 degree
x = 83 . 9 /1 . 4281
x = 58 . 75 m
x = 58 . 8 m
8c )
Distance btwn points A and B
=| AP | +| BP |
=100m + xm
=100m+ 58 . 8 m
=158. 8 m
9a)
Draw the diagram
Radius = 13cm
|OM| = 5cm 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| = 12cm
Hence, the length of the chord = 2|MB|
=2*12cm
=24cm
9b) 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
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
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