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Neco real Mathematics Obj and theory Now available




Maths - Type A
1-10: DBBADBABEE
11-20: CABBEEBBBC
21-30: BCDEBCBCDB
31-40: AEBCDAEDBD
41-50: ABADCEBAED
51-60: BEBDBEDEDA

Maths - TYPE B
1-10: CABBAEBACC
11-20: AEBCDAEDDB
21-30: DEBDBEDEDC
31-40: BBBABBABED
41-50: ECDEBBDBBC
51-60: ADADCECAED

Maths - TYPE C
1-10: BEBDAEDEDA
11-20: EDAEBBDBAB
21-30: ADADDCABEA
31-40: AABBCAEDDB
41-50: DBDBABBBCC
51-60: BBBAEBABEE

Maths - Type D
1-10: ABADCEBAED
11-20: BBBAEBABEE
21-30: AEBCDAEDDB
31-40: DCDEBCBCDB
41-50: BEBDBEDEDA
51-60: CABBEEBABC




(3a.)
13/4 +X/Y = 4/3
X/Y=4/5-13/16
X/Y=64-65/(16)(5)
X/Y= -1/80
X:Y = -1:80
(3b.)
cost each machines
= 25,000/12
= 2083.3naira.
i. cost of machines
= 2083.3 *10
=20833.30
ii. number of such machines
= 175,000/2083.3
=84 machines
(4)
(i) draw a venn diagram
U = universal set =40
mangoes (M) = 15
Pineapples(P)= 21
(ii)
let X reps the no. that do not like mangoes and pineapples
ii. 9+6+15+X = 40
30+X =40
X = 40-30
X =10
(III)
. % of class of mangoes only
= 9/40 * 100 = 22.5%
(6)
(i)
using pythagoras
OQ^2 = OR^2 + QR^2
15^2 = 6^2 + QP^2
QR^2 = 225 - 36
QR^2 = 189 , QR = SQR OF 189
QR = 13.75cm
lenght of chord = PR + QR
= 13.75 + 13.75
= 27.50cm
(II)
Lenght of chord (PQ)
= 2r sin(tita/2) , r = 15cm
27.50 = 2*15*sin(tita/2)
sin(tita/2) = 27.50/30
sin(tita/2) = 0.9165
tita/2 = sin^-1(0.9165)
tita/2 = 66.42
tita = 66.42 *2
tita = 133degree
(iii)
perimeter = lenght of arc + lenght of chord
lenght of arc = tita/360 *2(pile)r
= 133/360 * 2 *22/7 *15
= 34.83cm.
perimeter (P) = 34.83 + 27.50
= 62.30cm
= 62.3cm.

(8)
tabulate . please note that the way you interprete logarith table is diff
No:
10900,root3 10900, 392.9, (392.8)^3, 39.436, 12.03. (12.03)^3, 39.436*(12.03)^3, (root3(10900)*(392.8)^3)/((39.436*12.03)^3), antilog 1.96*10^4
Log:
4.03744, 4.0374/3= 1.3458, 2.594.2, 2.5942*(3)= 7.7826, 1.3458, 7.7826, 1.5959, 1.08003,1.0803*3=3.2409, 1.5959*3=3.2409, 1.5959+3.2409=4.38368, 9.1284, 4.8368, 9.1284+ 4.8368=4.2916
(5)
a)244basen=1022base4
2*n^2+4*n^1+4*n^0=1*4^3+0+4^2+2*4^1+2*4^0
2n^2+4n+4=64+0+8+2
2n^2+4n+4=74
2n^2+4n+4-74=0
2n^2+4n-70=0
n^2+2n-35=0
n^2-5n+7n-35=0
n(n-5)+7(n-5)=0
n+7=0 or n-5=0
n=-7 or n=5
b)Using similar triangle
AZ/XZ=KZ/YZ
8/XZ=11/24
XZ=(8*24)/11
=17.45cm


(1)
(a)
peugeout sector= 180/480*(360)
=135 degree
datsun sector= 108/480*(360)
ford sector= 72/480*360
=54 degree
toyota sector= 56/480*(360)
=48 degree
(b)
Pr(toyota)= 42/480
Pr(volkswagen)= 48/480
Pr(toyota or volkswagen)= (42/480)+ 48/480
=90/480= 9/48
=3/16
(2)
(a) given (a^2-4b^2)
=(a^2-4b^2)/(a^2-3ab-2ab+6b^2)=(a^2-4b^2)/(a(a-3b)-2b(a-3b)
=a^2-(2b)^2/(a-2b)(a-3b))= (a-2b)(a+2b)/(a-2b)(a-3b)
=a+2b/(a-3b)
(2b)
forth term(f4)=-3
ninth term(f9)=123
fn=a+(n-1)d
f4=a+3d
f9=a+8d
a+3d=-3, eqn (1)
a+8d=12, eq(2)
subtract eq(1) from eq(2)
a-a+8d-3d=12-(-3)
5d=15
d=15/5=3
using eq(1)
a+3d=-3
a+3(3)=-3
a+9=-3, a=-3-9
(i) common difference(d)=3
(ii)sum of the first seven terms
sn=n/2(2a+(n-1)d)
s7=7/2

(9)
Total surface area(A)=88cm^2
Height (H)=7cm
i)A=2*pie*r*h+2*pie*r^2
A=2*pie*r(h+r)
88=2*22/7*r(7+r)
88=6.29*r(7+r)
7r+r^2=88/6.29
7r+r^2=14
r^2+7r-14=0
using almighty formular
r=-7+_root(7^2-4*1*-14)/2
r=-7+_root(49+56)/2
=-7+_root(105)/2
=-7+_10.25/2
r=-7-10.2/2 or-7+10.25/2
r=-8.625 or 1.625
Diameter(d)=2r
2*1.625
=3.625
ii)Volume of the solid(v)
=pie*r^2*h
=22/7*(1.625)^2*7
=406.66/7
=58.09cm^3
b) =74degrees
(11)
Sack A
Blue(b)=10
Red(r)=6
total=16
Sack B
Blue(b)=9
Red(r)=12
total=21
pr(drawing b from sack A)=10/16
pr(drawing the second ball b from sack A)=9/15
pr(r from sack A)=6/16
pr(r from sack A)=5/15
pr(b from sack B)9/21
pr(second b from sack B)=8/20
pr(all the four crayons are blue)
=(10/16*9/15)*(9/21*8/20)
=9/240*72/420
=9/24*18/105
=3/8*18/105
=27/420
=9/140

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Posted By naijabams.com On 07:28 Fri, 04 Jul 2014

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