Neco mathematics obj and theory answers verified by Bmasterz

Please,invite your friends and share this page to your NECO friends.

   

NECO MATHS OBJ VERIFIED BY BMASTERZ

1-10=CCAADDEECE

11-20=EEBABAADCC

21-30=CBEEDABDDD

31-40=BCCCDDCADE

41-50=BDCDDCDECD

51-60=BCDCAEDCDD

☆BMASTERZ☆

.MATHEMATICS-Answers
Use This No1 pls is more acurate
1)
tabulate
Number-
5932
*6141
num
3679
*3113
den
Log
3.7732
3.7882
7.5612
3.5657
3.4931
7.0588
anti log
7.5612
7.0588/0.5024/3=0.1675
anti log
=1471*10
================================
3a)
P=100I/RT=100*30,000/4*3
P=N250,000
================================
5)
tabulate
Mark- 21-25,26-30,31-35,36-40,41-45,46-50
F- 6,8,12,4,6,4=40
X- 23,28,33,38,43,48
FX- 138,224,396,152,258,192=1360
Class Boundaries- 20.5-20.5, 25.5-30.5,35.5-40.50,40.5-45.5
5i)
MEAN
(X)=EFX/EF=1360/40=34
MODE
L1+[Fm-Fa/2fm-fa-fb]c
=30.5+[12-8/2(12)-8-4]5
=30.5+[4/12]5
=30.5+1.67
=32.17

 6i)
 7+8+x = 47
 15+ 2x =47
 2x = 47- 15
 2x/2 = 32/2

 divide both side by 2
 x=16
 6ii)
 tran only;:
 7+8 + 3 + y=30
 18 +y =30
 y=30 -18 =12
 :. Tran only =12 iii) at least two: 7+ 16+ 3 +8 =34

 iv)16 + 7 +12 +16 +8 + 3 + Q=95
 62+ Q =95
 Q= 95 -62
 Q=33 :. Car only = 33

(1)
TABULATE
No|Log|
5932|3.7732|
*6141|3.7882|
Num|7.5612|7.5612
3679|3.5657|
*3113|3.4931|
Den|7.0588|7.0588
| |0.5024/3=0.1675
Antilog:-1.471*10^0
=1.471

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



(2)
a = 3, b = 20, C = -7
Sum is
(A/1 + 1/B + B/1 + 1/ A) Note A means Alpha and B means beta
A^2 B + B^2 + A + B / AB
= -7/3 (-20/3) + (-20/3) = 140/9 - 20/3 / -7/3
= 80/9 x -3/7 = -80/21
Product:
-7/3 + 2 + 1/-7/3 = -7/3 + 2/1 - 3/7
= -49+42-9/21
= -16/21
x^2 - (sum)x + product= 0
x^2 - 80/21x + (-16/21) = 0
Final Answer : 21x^2 - 80x - 16 = 0

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

(3a)
P=100I/RT=100*30,000/4*3
P=N250,000


(3b)
#7000 to Francs
#1 = 8 francs
.:.#7000 = ? france
CROSS MULTIPLY
8/1 * 7000 = 56,000 franc
Spending 49,400 franc
56, 000- 49,400 = 6600
Converting 6 ,600 franc to # AT 10 franc TO 1#
10 franc —– 1#
6600 FRANCS —? #
= 6,600/10 X 1
=#660

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


(4i )
difference in latitude = 36 degree + 36 degree = 72degree
distance travelled =@/ 360* 2pier
= 72 / 360* 2* 22 / 77* 64
= 20275200 / 2520
= 8045 . 71
= 8050 km ( 3 s.f )

(4ii)
given that speed = 800 km / hr
distance travelled = 804571 km
time = distance travelled / speed
= 8045 . 71/ 800
= 10 .057
= 10 hrs ( to the nearest hour )

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


(5)
Tabulate
Mark - 21- 25, 26- 30, 31- 35,36 -40 ,41 -45 ,46- 50
F - 6 ,8, 12,4 ,6, 4= 40
X - 23,28 ,33 ,38, 43, 48
FX - 138, 224, 396, 152, 258, 192= 1360
Class Boundaries- 20 .5 -20 .5, 25 .5 -30 .5, 35. 5- 40. 50,40 .5 -45 .5

(5i)
MEAN
( X )= EFX / EF = 1360/ 40 = 34

MODE
L 1+ [Fm - Fa / 2fm - fa -fb ] c
= 30 .5+ [12 -8/ 2 ( 12 ) -8- 4] 5
= 30 .5+ [4/ 12 ]5
= 30 .5+ 1.67
= 32 .17

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


(6)
(i)
7+8+x = 47
15+ 2x =47
2x = 47- 15
2x/2 = 32/2
Divide both side by 2
x=16

(ii)
Train only;:
7+8 + 3 + y=30
18 +y =30
y=30 -18 =12
.:. Train only =12

(iii)
at least two: 7+ 16+ 3 +8 =34

(iv)
16 + 7 +12 +16 +8 + 3 + Q=95
62+ Q =95
Q= 95 -62
Q=33 :. Car only = 33



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


(7a)
a+2d=11---eq1
-(a+8d)=29
-6d/-6=-18/-6
d=3
Sub for d=3 in eq1
a+2(3)=11
a+6=11
a=11-6
a=5

(7b)
L = ar^n-1
729/3 = 3/3 x 3^n-1
243 = 3^n-1
3^5 = 3^-1
5 = n-1
N = 5+1
= 6
Sn = a(r^-1)/r-1 = 3(3^6 - 1)/3-1
Sn = 3( 729 -1)/2
S6 = 3 x 728/2
S6 = 1092

(7c)
3x^3/3 + x | 2,1
(2^3 + 2) - (1^3+1)
(8+2) - (1+1)
= 8‎

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


(8ai)
y=(4x+9)^3
Let u=4x+9
y=u^3
du/dx=4,dy/du=3u^2
dy/dx=dy/du*dy/dx
=4*3u^2
=12u^2
Recall u=4x+9
therefore dy/dx=12(4x+9)^2

(8aii)
y=(3x-2)^3(x^2+4)^2
u=(3x-2)^3, v=(x^2+4)^2
du/dx=3.3(3x-2)^3-1
dv/dx=2x.2(x^2+4)^2-1
du/dx=9(3x-2)^2
dv/dx=4x(x^2+4)
dy/dx=Vdu/dx +Udv/dx
=(x^2+4)^2[9(3x-2)^2+(3x-2)^3][4x(x^2+4)]
=(x^2+4)(3x-2)^2(9x^2+36+12x^2-8x)
=(x^2+4)(3x-2)^2(21x-8x+36)

(8b)
y-y1=m(x-x1)
y-6=2(x-2)
y-6=2x-4
y=2x-4+6
y=2x+2

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


(9a)
y proportional x proportional 1/x
y=kx/1 +k/x
4=2k/1+k/2
4=(4k+k)/2
8=5k
k=8/5
Relationship:
y=kx/1+k/x
y=8x/5+8/5x

(9b)
4(3^x+1)-3^2x=27
4(3^x *3^1)-362x=3^3
let 3^x=p
4(3p)-p^2=27
12p-p2=27
p^2-12p+27=0
p^2-9p-3p+27=0
p(p-9)-3(p-9)=0
(p-9)(p-3)=0
p=9 or p=3
Recall 3^x=p
when p=9
3^x=3^3
x=3
when p=3
3^x=3^1
x=1
therefore x=2 or 1


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


(11a)
Tabulate
X- |-1| |0| |1| |2| |3| |4|
6 |6| |6| |6| |6| |6| |6|
+X |-1| |0| |1| |2| |3| |4|
-X^2 |-1| |0| |-1| |-4| |-9| |16|
Y- |4| |6| |6| |4| |0| |-6|

(11b)
Draw the graph

(11ci)
gradient at x/x
dy/dx =B-A/B-C
=7-4/2-1=3/1=3

(11cii)
roots of equation from the graph
x=-2.2 and x=3

(11ciii)
minimum value of y
=6.4

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

Keep refreshing.more loading...

   

Keep refreshing this page for more answers...