Neco 2017 Mathematics Obj And Theory Answers – May/June Expo

NECO MATHEMATICS ANSWERS

Verified-Maths-Obj
1CDAAEDEECE
11EEBABAABCC
21BBCEDABECD
31BCCCDDCADE
41BDCBDCCECD
51BCDCBEDCDB
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
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2)
a = 3, b = 20, C = -7
Sum is
(α/1 + 1/β + β/1 + 1/ α) Note α means Alpha
α^2 β + β^2 + α + β / α β
= -7/3 (-20/3) + (-20/3) = 140/9 – 20/3 / -7/3 (Note / means
Divided)
= 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
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3a)
P=100I/RT=100*30,000/4*3
P=N250,000
3b)
#7000 T0 FRANCS
8FRANS= #1
?FRANCS =7000
CROSS MULTIPLY
=#56,000 FRANCS
SPENDING 49,400 FRANCS
56,000-49,400 = 6600
CONVERTING 6,600 FRANCS TO # AT 10 FRANCS TO 1#
10 FRANCS—– 1#
6600 FRANCS—?#
=6,600/10 X1
4i)
difference in latitude=36degree + 36 degree =72degree
distance travelled=@/360*2pir
=72/360*2*22/77*64
=20275200/2520
=8045.71
=8050km(35.f)
4ii)
given that speed =800km/hr
distance travelled =804571km
time=distance travelled/speed
=8045.71/800
=10.057
=10hrs(to the nearest hour)
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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
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SECTION B

ANSWER 5 QUESTIONS
6i)
7+8+x*x=47
15+2x=47
2x=47-15
2x/2 =32/2
x=16
6ii)
Tram only
7+8+3+y =30
18+y =30
y=30-18=12
Tram only =12
6iii)
at least two
7+16+3+8
=34//
6iv)
16+7+12+16+8+3+9=95
62+9=95
9=95-62
9=33// or car only =33

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8ai)
y=(4x+9)^3
let u =4x+9, y=U^3
du/dx=4, dy/du=3U^2
dy/dx=dy/du * du/dx = 4*3u^2=12u^2
recall= U=4x+9
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
=4x(x^2+4)
in other words
dy/dx=Vdu/dx + udv/dx
=(x^2+4)^2 9 (3x-2)^2+(3x-2)^3 4x(3x-2)
=(x^2+4) (3x-2)^2 (9x^2+36+12x^2-8x)
=(x^2+4) (3x-2)^2 (21x^2-8x+36)
8b)
y-y1=m(x-x1)
y-6=2(x-2)
y-6=2x-4
y=2x-4+6=2x+2
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9)
TYPING IT OUT
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10)
TYPING IT OUT
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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)
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1ci)
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.

12a)
from economics
3x+2y-12=90
3x+2y=90+12
3x+2y=102 —–(1)
sum of angle in a circle=360degree
5x+y-9+3x+2y-12+3y
+4x+21=360
5x+3x+4x+y+2y+3y-9-12+21=360
12x+6y=360 —–(2)
solving equation 1and2
simulteanously
13x+2y=102
12x+6y=360
using elimination method
3x+2y=102*4
12x+8y=408
12x+6y=360
2y/2=48/2
y=24
substitute y=24 into
equation (1) for x
3x + 2y =102
3x+2*24=102
3x+48=102
3x=102-48
3x/3=54/3
x=18
x+y=18+24
x+y=42
12b)
candidate that registerd for literature
=4x+21/360 * 1200
=4*18+21/360 * 1200
=72+21/360 * 1200
93/360*1200 =111600/360
=310
candidate that registered for government
=3y/360 * 1200
=3*24/360 * 1200
=72/360 * 1200 =86400/360
=240
310 – 240 = 70
the candidate that registered for litereature are 70, more than candidate that registered for government