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2013/2014 NECO/SSCE GCE MATHEMATICS Obj/Theory Answers



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Topic: 2013/2014 NECO/SSCE GCE MATHEMATICS Obj/Theory Answers

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imageGinger4me(,)- PLS NOTE THESE PAGE WE CREATE WE ARE NOT POSTING IT HERE ONLINE GO AND SUBCRIBE,ONLY PASSWORD LINK USERS GET WERE WE POST THE ANSWER BELIVE ME WE ARE NOT POSTING IT FREE CALL Ginger4me for info 08135193104
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Reply 2 on :

imageOkoj1(Male,40)- Verified MATHS OBJ:
1-10: BBACBECBDE
11-20: CBAAABCDBB
21-30: EBDABCCDBD
31-40: DCDACCBBAB
41-50: EEBCEBBCDA
51-60: CCCECECDBE

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Reply 3 on :

imageOkoj1(Male,40)-
NECO GCE 2013 MATHEMATICS THEORY
Section A
(1a)
U=32, n(p)=16, n(G)=18 and n
(P^!union G^!)
16-x+x+18-x+3=32
34+3-x=32
37-x=32 -x=32-37
-x=-5
x=5
(1b)
((9/4+(3/2))/((49/8)-11/4)=
((18+12)/8)/((196-88)/32) =(30/8)*(32/108)
=(30*4)/108=30/27=10/9
=1 whole no 1/9

(2a)
rx=99cm,ry=101cm
hx=hy=2.1m
Vx=pie(rx sqr)hx
=pie(0.99)^2*2.1
Vx=2.006 pie m^3
Vx=2.06 pie M^3
Vy=pie (ry)^2hy
Vy=pie (1.10)^2*2.1
Vy=2.14 pie m^3
their difference is
V=vy-vx
v=2.14pie-2.06pie
v=0.08pie=0.08*3.142
v=0.258m^3
(2b)
(a^2-2a)/a-(a^2-a^3)/a^2-(a)
a-2-(1-a)-a
a-2-(1-a)-a
a-2-1+a-a
=a-3

(3a)
tan25 degreee=TP/XP
let tita be the angle of
elevation
tan tita=TP/120
tan 25 degree= TP/150, TP=tan 25 degree*150
TP=69.95m=70m
tan tita=70/120, tan
tita=0.583
tita=tan^-1(0.583)=30.24
dgree. the angle is 30 degree
(3b)
let TP be the distance ,
Tp=70cm

(4a)
(i) Log 27(base 10)= log 3^
3(base 10)
=log 10(base 3) +log 3(base10) + log 10 (base 3)
(4b).
(3x+2y)(3X+2y) -((2y-3x) (2y-3x))
9x^2 + 6xy+6xy+4y^2-
(4y^2-6xy-6xy+9x^2)
9x^2+12xy+4y^2 -4y^2 +6xy
- 9x^2
12xy+12xy= 24xy.

(5)
n=8,
Xbar 8=N55000
Xbar= (summation x)/n, 55000= (summation x)/8
summation x= 55000*8
summation x= N440000
total(summationx)= 440,000+25,0000
=465,000
x bar of 10=9summation x)/n, n=10
x bar of 10= N46,500
(ii)
10% increase of N46,500
10/100*(46,500)
=N4650
=25,100*4650
= N116,250,000

(Q6a)
number 16|17|18|19|20|21|22|23
binary equivalant 10000|10001|
10010|10011|10100|10101|10110|
10111
(i)prob(last digit zero) =2/8=1/4
(ii)prob(at least two ones)=7/8
(iii)prob(threeones and
zeros)=6/8=3/4
(iv)prob (last digit one)=4/8=1/2
(6b)t2=-1and t4=8+t2
(i)t2=a+d=-1........(1)
t4=8+a+d
a+3d=8++d
2d=d=8/2=4
from (1) a+4=-1 ;a=-5
(ii)t6=a+5d=-5+5(4)
t6=-5+20=15
(iii)S12=n/2{+(9-1)d}
S12=12/2(-5+(12-1)4)
=6(-5+ 44) =6(39)
S12=234

(7a)
AB=50km, AC=90km
(i)
Bc sqr= AC sqr+ BA sqr- 2(AC)(BA)cos125 degree
BC sqr= 90 sqr+ 50 sqr- 2(90)(50)cos 125 degree
BC sqr= 8100 +2500-9000x-0.5736
Bc sqr=10600+5162.4
BC sqr= 15762.4
BC= sqr root(15762.4)
(7aii)
using sine rule
AC/sin tita= 125.5/sin125.5 degree
90/sin tita=125.5/0.8192
sin tita= (90*0.8192)/125.5
sin tita=0.5874
tita= sin^-1(0.5874)
tita=35.97 degree
tita= gamma+15
35.97=gamma+15
gamma=35.97-15
Bearing of C from B is
=90 degree+ 90 degree+ 90 degree +20.97
=290.97 degree
(7b)
area of base =L*B
=11*9=99cm^2
volume of pyramid(v)
(1/3)*area of base*height
=1/3*(99)*16
=528cm^3

(8a)
P=2*A
A=P(1+1/r)^n
A=P(1+1/0.04)^n
A=P(1+1/0.04)^n
A=P(1+25)^n
1/2=(26)^n, Log(0.5)=Log(26)^n
n=log 0.5/(log 26)
=4.7
n=5
8b)
(8/3)*(4/5)/(43/12)*(-2/3)
=(32/15)/(43/1*(-1/5)
=(-32*6)/(5*43*5)=(-192)/1075
(8c)
(2^2x)-9*2x+4=0
(2^x)^2-9*2^x+4=0
Let 2^x=p
2p^2-9p+4=4
2p^2-p-8p+4=0
(2p^2-p)-(8p-4)=0
2p-1=0
2p=1
P=1/2
Or p-4=0
P=4
When p=1/2
2x=2^-'
X=(-1)
When p=4
2x=2^2
X=2

(9a) log2-log25/100+log32
=log(2/25/100)*32
log2/1*100/25*32
Log8[2*4*32]
Log8[8*32] =log8[8]+log8[32]
1+log8[32]
(9b)
2Sn= n(2a+(n-1)d)
2Sn =n (2a+nd-d)
2sn= 2an + n^2d - nd
2sn - 2an = n^2d- nd
2sn - 2an/n = nd -d
d(n-1) = 2sn - 2an/n
d = 2sn - 2an/n(n-1)
n= 10
d = 2*115-2x-4*10/10(10-1)
d = 230 + 80 / 90
d = 310/90 = 31/9 = 3.4

10a)
((16)^3/4*(4)1/2)/(8)^5/3
=((2^4)^3/4*(2^2)^1/2)/((2^3)^5/3
=2^3*2^1/2^5
=2^4/2^5
b) M(32 degree N,45 degree W) and N(32 degree N,30 degree E)
i) Latitude(alpha)=32 degree
Longitude(tita)=45+30
=75 degree
i) r=Rcos alpha
r=6400*cos32 degree
r=5427.5km
r=5430km
ii)d=tita/360*2pie R
d=75/360*2*(22/7)*6400*cos 32 degree
d=(75*2*2286400*0.8480)/(360*7)
d=17909760/2520
d=7107.0km
d=7110km
iii)speed(v)=distance/time
=7110/9
=789.7km/hr
=790km/hr

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imageOkoj1(Male,40)-
==>MISSION ACCOMPLISH====

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