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Author Topic: NECO GCE 2013/2014 General Mathematics Paper I & II Objective and Essay Questions And Answers  (Read 6749 times)

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Offline MEGAX

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NECO GCE 2013/2014 Confirmed General Mathematics Paper I & II (Objective and Essay) Questions And Answers

NOTE: The Below Answers was posted @ EarlyFace Examination Room for Only Our Subscribers before The Exam start. You must first Register Or Login as an EarlyFace member to be able to gain access any of the Links in Order to get this 2013/2014 Real NECO GCE Questions And Answers. Or pay For it..

LINK1: http://www.earlyface.com.ng/index.php?action=ezportal;sa=page;p=4

LINK2: http://www.earlyface.com.ng/index.php?action=ezportal;sa=page;p=10


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


ESSAY/THEORY:
Quote
Note that
* means mulplication
/ means division
^ means Riase to power.
* means multiplication
tita  you should know dat.
sqr rut/ squar rut means square root



(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(base 10) + 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(summation x)= 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(three ones and zeros)=6/8=3/4

(6b)
given; F2=-1
a+d=-1
F4=8+F2
a+3d=8+a+d
a-a+3d-d=8
2d=8
d=8/2
substitute d=4 into eqn (1)
a+d=-1, a=-1-4
a+4=-1
a=-5
(i) first term(a)=-5
(ii)
sixth term(F6)=a+5d
F6=-5+5(4)
F6=-5+5(4)
F6=-5+20
F6=15
(iii)
S=n/2(2a+(n-1)d),n=12
S=12/2(2*(-5)+(12-1)4)
s=6(-10+44)
s=6(34)=204
sum of the first twelve terms is 204

(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)
(((-10/3)*(-6/5))/((13/12)*(-4/3))*(-11/6)
=(4/(-43/9))*(-11/6)
=4*(-9/43)*(-11/6)
=4*-6/43*(-11/2)=(2*66)/43
=132/43=3 whole number 3/43

(8c)
(2^x)^2- 9*2^x+4=0
let P=2^x
((P^2)*2)-9*p+4=0
2p^2-(9p)+4=0
2(P sqr) -8p+4=0
2p(p-4)-1(p-4)=0
(2p-1)(p-4)=0
2p-1=0 or p-4=0
2p=1 0r p=4
p=1/2 or p=4

(9a)
log (2/0.25)(base 8 ) + log 32 (base 8 )
log 8 (base 8 ) + log 32 (base 8 )
= log (8*32)(base 8 )
= log 256 (base 8 )
- log 2^8 (base 2^3) = 1/3 *8log2^2
= 8/3log 2(base 2)= 8/3 *1 =8/3
log 2 (base 2) = 1.

(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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