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# NECO Mathematics Objectives And Theory 2017 – Free Expo.

## NECO Mathematics Objectives And Theory 2017 – Free Expo. Looking for NECO Mathematics Objectives and Theory answer Expo you are at the right website.

Academyhint is the site that you can get 100% Correct NECO 2017 Answer and we also posts all exam expo answer earlier than others

This is to Inform all 2017 National Examination Council NECO candidates that are writing Mathematics on Wednesday 14th June by 10:00am – 02:30pm.

### NECO Mathematics Objectives Answer Expo.

1a)
Tabulate
x – 1 ,2 ,3 ,4
1 – 1 ,2 ,3 ,4
2 – 2 _ , 4 , 0 _ ,2 _
3- 3 , 0_ , 3, 0 _
4 – 4 _ , 2_ , 0_ , 4

1 b )
I = PRT /100 , p = N15000 R = 10 % and I= 3 years
A = P + I
where I = 15000 * 10 * 3 / 100 = N4500
A = 4500+ 15000 = N19500
==================================

2 a)
No2

Using sine rule
b/sin120 = 6/sin30
bsin30 = 6sin120
b= 6sin120/sin30
b=6sin30/0.5
b= 12*sin120
b=12*sin120
b= 10.4cm

2 bi )
the diagram is equivalent t triangles .
where
| AX | / | BC | = | BY | / | AC | = | XY | /| YC |
XY = 9 , BY = 7
YC = 18 -7 = 11
9 /11 = 7 / | AC |
9 | AC | = 77
| AC | = 77 /9
| AC | = 8 cm

2 bii)
XY /AB = BY /AC
9 /| AB| = 7 / 8 .6
| AB | = 9 x 8 .6 /7
| AB | = 11 cm
==================================

3)
let the son age be x
man = 5 x
son= x
4 yrs ago; the man age = 5 x – 4
the son age = x – 4
the product of their ages
(5 x – 4 )( x – 4 ) = 448
==================================
4 a)
volume of fuel = cross -sectional area of X depth of fuel rectangular
tank
30, 000litres = 7 . 5* 4 .2 * d m ^ 3
but ; 1000litres = 1 m ^ 3
therefore ; 30( M ^3 ) = 7 .5 *4 . 2* d ( M ^ 3)
30= 31.5 d
==== d = 30/31. 5 = 0 .95( 2 d .p )
4 b)
to fill the tank /volume of fuel needed
= 7 .5 * 4. 2 *1 .2
= 37.8 m ^ 3
= 37, 800 litres
addition fuel = 37, 800- 30, 000
= 7 , 800 litres
therefore , 7 , 800 more litres would be needed

5 a)
sector for building project = 48000 /144000* 360 = 120degree
sector for education = 32, 000 /144000* 360 =80degree
sector for saving = 19200 /144000* 360 = 48degree
sector for maintenance = 12000 / 144000*360 = 30degree
sector for miscellaneous = 7200/144000* 360 =18degree
sector for food items = 360 -( 120 + 80+ 48+30+ 18)
= 360- 296
= 64degree
5 b)
amount spent =144000- [48, 000 + 32000+ 19200 + 12000 +7200] = 144000-118400
= N 25600

6)
/————————— —
/ 41.02 × √0.7124
/ ———— ———— — —
√42.87 × 0.207 × 0.0404

| No | Log |
| 41.02 | 1.6130 | 1.6130
| 0.7124 | T. 8527 ÷ 2 | T.9263/1.5393
| 42.77 | 1.6321 |
| 0.207 | T.3160 |
| 0.0404 | 2. 6064 / |
| | T.5544 | T.5544/1.9849

Antilog = 9658 ≈ 96.58

=================================

7 a)
3^ 2 n + 1 – 4 ( 3 ^ n + 1 )+ 9 = 0
3^ 2 -3 – 4 ( 3^ n – 3) + 9 = 0
(3 ^ n )^2- 3 – 4 (3 ^ n -3 )+ 9 = 0
let 3^ n = p
p ^ 2 -3 – 4 (p -3 ) + 9 = 0
3p ^ 2 /3 – 12 p /3 + 9 / 3 = 0
p ^ 2 – 4 p + 3 = 0
p ^ 2 – 3 p – p + 3 = 0
p ^ 2 p ( p – 3 ) – 1 (p – 3) = 0
(p – 1 ) (p – 3) = 0
p -1 = 0 or p – 3 = 0
p = 1 or 3
Recall 3 ^ n = p
when p = 1
3^ n = 3^ 0
n = 0
when p = 3
3^ n = 3^ 1
n = 1

7 b )
log (x ^ 2 + 4 ) = 2 + logx – log ^ 20
log (x ^ 2 + 4 ) = log ^ 100 = log ^ x – log ^ 20
(x ^ 2 + 4 ) = log ( xx )
x ^ 2 + 4 = 5 x
x ^ 2 -5 x + 4 = 0
x ^ 2 -4 x – x + 4 = 0
x (x -4 ) – 1 ( x -4 ) = 0
(x -1 ) (x -4 ) = 0
x -1 = 0 or x -4 = 0
x = 1 or 4

8 a)
| AD | ^ 2 = 13^2 -5 ^ 2
| AD | ^ 2 = 169- 25
| AD | ^ 2 = 144
| AD | = 12- r
r ^2 = ( 12-r )^2 – 5 ^ 2
r ^2 = ( 12-r ) ( 12- r ) + 25
r ^2 = 144 -24r + 25
r ^2 = 169 -24r
r ^2 + 24r -169 = 0
r ^2 + 24r = 169
r ^2 + 24r + 14^2 = 169 + 14^2
( r + 14)^2= 169 + 196
( r + 14)^2= 365
( r + 14=sqr 365
r + 14=19. 105
r = 19.105 -14
r = 5. 105
r = 5. 1 cm
8 aii )
circumfrenece of a circle = 2 pie R
C = 2 x22/ 2* ( 5 .1 )^2
C = 1144.44/ 7
C = 163 .4914cm
C = 163 .5 cm
8 b)
y2 -y1 / x2 -x1 = y- y1 /x- x1
6 -2 /2 – ( -1 )=y- 2/ x- ( -1 )
4 /2 + 1 = y- 2 /x+ 1
4 /3 = y-2 / x+ 1
3 ( y-2 )=4 ( x+ 1 )
3 y-6 = 4 x+ 4
3 y-4 x= 4 + 6
3 y-4 x= 10
y= 4 x/3 + 10/ 3

9 a)
let Xy represent the two digit number
x-y =5 — — ( i )
3 xy – ( 10x + y)=14
3 xy – 10x – y = 14 – — -( ii )
from eqn ( i )
x= 5 +y
3 y( 5 +y ) -10( 5 + y) -y= 14
15y+ 3 y^2 – 50 – 10y – y =14
3 y^2 + 4 y – 50 = 14
3 y^2 + 4 y – 50 – 14 = 0
3 y^2 + 4 y – 64 = 0
3 y^2 + 12y + 16y – 64 = 0
( 3 y^2 – 12y ) ( + 16y – 64)=0
by
( y-4 ) + 16( y- 4)= 0
( y-4 )=0
ii )
( 3 y+ 16) ( y-4 )=0
3 y+ 16=0 or y- 4 =0
3 y= -16 or y= 4
y= -16/ 3 or y= 4
when y =4!
x= 5 +y
x= 5 +4
x= 9
the no is 94
9 b)
3 -2 x/ 4 + 2 x-

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