# NECO GCE 2017 Mathematics Obj And Essay Solution Answer – Nov/Dec Expo

NECO GCE Mathematics Expo 2017, NECO GCE Mathematics Runs 2017, NECO GCE Mathematics Paper II (Essay) Questions 2017, NECO GCE Mathematics Paper III (Objectives) Questions and Answers 2017.

Exam Time: Saturday 11th Nov. 2017
General Mathematics Paper III (Objective)
11.00 am – 12.45 pm
General Mathematics Paper II (Essay)
1.00 pm – 3.30 pm
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STUDY THESE SIGNS BELOW.
^ Means Raise to power, so just carry the next number after u see ^ to the top.
You should know how to write CHANGE IN Y OR X OR ANYTHING. “CHANGE IN” LOOKS LIKE A SMALL TRIANGLE.
You should know degree sign, so write the sign wen u see Degree
* means x ie multiply
You should know the tita sign, replce TITA with the sign anywhr u see it.
Learn to draw a Frequncy table or any table eg
X: 10, 20, 30
F: 2, 2.7, 4
This means that 10,20,30 is under x in a table likewise 2,2.7,4 are under F in a table.

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MATHEMATICS OBJ:
11-20: EECECCACED
21-30: CBDDBEEBCD
31-40: ACEBDBDDAB
41-50: BDDBCCABCB
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NECO GCE MATHEMATICS OBJ AND THEORY FROM EXAMHOT.COM
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(1)
P=N300,000
R=7^1/2%
T=3yrs
At the end of year 1
I=PRI/100=300000*15*1/100*2
I=N22500
2nd Year
P=300000+22500+50000
=N372500
I=372500*15*1/200=N27937.50
3rd Year
P=372500+27937.5+50000
=N450437.50
I=450437.50*15*1/200
I=N33782.81
total saving of 3years
=450437.5+33782.81+50000
=N534220.31

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(2a)
T=thickness
P=Pages
T*P
T=KP
T=3
P=900
3=900k
k=3/900=1/300
T=P/300
WHERE T=45CM
P=?
45=P/300
P=300*45
P=13500pages

(2b)
X-Y=3 & x^2-Y^2=15
(x+y)(x-y)=15
x+y=15/x-y
=15/3
=5

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(3a)
n(y)=40
n(c)=35
n(b)=26
n(CnB)=x
[Drawing]
40=35-x+x+26-x
40=61-x
x=61-40=21
21 student offer both

(3b)
U:{all positive <-20}
S:{all even number <14}
T:{all even nus<-20divisible by 3}
U={1,2,3,,,20}
S={2,4,6,8,10,12}
T={6,12,18}
SUT={2,4,6,8,10,12,18}

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(4a)
DRAW THE DIAGRAM
Consider triangle AON
Sin tita=opp/hyp
Sin47=AN/4
0.7314=|AN|/4
|AN|=0.7314*4=2.9256cm
but chord AB=2|AN| since |AN|=|BN|
Therefore the length of the chord
=2*2.9256
=5.8512cm
=5.85cm(2 d.p)

(4b)
DRAW THE DIAGRAM
A sector-A triangle
=tita/360(pier^2)-1/2sinr^2sintita
=90/360*22/7*7^2-(1/2*7^2*sin90)
=(11*7)/2-(1/2*49*1)
=72/2-49/2
28/2
=14cm^2

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(5c)
Draw a table
|1-10, 11-20, 21-30, 31-40, 41-50|
X|5.5, 15.5, 25.5, 35.5, 45.5
F|6,10,12,15,17
Fx |33,155,306,532.5,318.5
Efx = 1345
Ef = 50
Mean X = Efx/Ef = 1345/50
= 26.9
Median = (N +1/2)th
= (51/2)th
= 25.5

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(7a)
Y = x³ – 6x² + 9x – 5
The gradient is zero means that the derivative is equal to zero
dy/dx = 3x² – 12x + 9 = 0
3x² – 12x + 9 = 0
3x² – 9x – 3x + 9 = 0
(3x² – 9x) – (3x + 9) = 0
3x(x – 3) -3(x – 3) = 0
(3x – 3) ( X – 3) = 0
3x – 3 = 0 or X – 3 = 0
3x = 3 or X = 3
X= 3/3 or X = 3
X = 1 or X = 3

(7b)
y²/36 – 1/9 = 0
Multiply through by 36
y² – 4 = 0
y² = 4
y = ±2

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(8ai)
Volume (v) =⅔πr²
V = 155.232cm³,
π = 3.142
V = ⅔πr²
155.232 = 2/3 × 3.142 × r²
= 155.232 = 3.142×2×2×r²/3
Cross multiply
3.142×2×r² = 3×155.232
6.284r² = 465.698
Divide both sides by 6.284
6.284r²/6.284 = 465.698/6.284
r² = 74.108²
r = √74.108
r = 8.61cm
Curved surface Area = 2πr²
=2×3.142×(8.61)²
=2×3.142×74.1321
Therefore; curved surface Area = 465.8461cm²

(8aii)
Total surface Area = 3πr²
= 3×3.142×(8.61)²
= 698.7692cm²

(8b)
(3, 6) and (7, 8)
X1 = 3, X2 = 6, Y1 = 7, Y2 = 8
Slope m = y2 – y1/x2 – x1
8 – 6/7 – 3 = 2/4 = 1/2
Y – Y1 = m(x – x1)
Y – 6 = 1/2(x – 3)
2(y – 6) = (X – 3)
2y – 12 = X – 3
2y = X – 3 + 12
2y = X + 9
Y = x/2 + 9/2
Therefore; y = 1/2x + 9/2x

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(9a)
logbase4(x^2+7x+28)=2
logx^2+7x+28=4^2
logx^2+7x+28=log16
x^2+7x+28=16
x^2+7x+28-16=0
x^2+7×12=0
x^2+3x+4x+12=0
x(x+3)+4(x+3)=0
(x+3)(x+4)=0
x=-3 or x=-4

(9b)
y=3x^2-4
Let the minimum increment in x and y be x+changex and y+changey respectively
y+changey=3(x+changex)^2-4
y+chnagey=3(x+changex)(x+changex)-4
y+changey=3(x^2+2xchangex+changex^2)-4
y+changey=3x^2=6xchangex+3(changex^2)-4
subtract y from both sides
y+changey-y=3x^2+6xchangex+3(chngex^2)-4-y
changey=3x^2=6xchangex+3(changex^2)-4-3x^2+4
changey=6xchnagex+3(chnagex^2)
Divide through by change x
changex/changey=(6xchangex/changex) +3(changex^2)/changex
changey/changex=6x+3changex
lim(changey/changex)=lim(6x+3changex)
dy/dx=6x+3(0)
dy/dx=6x
at x=6*2/3
=2*2
therefore x=4

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(10a)
Using Cosine rule
(i) yxz
Cos yxz = 6² + 9² – 11²/2 × 6 ×9
Cos yxz = 36 + 181 – 121/108
Cos yxz = 117 – 121
Cos yxz = -4/108
Cos yxz = 0.0370
yxz = cos-1(-0.0370)

(ii) ym
Using similar triangle
Considering triangle xym & xyz
9/9 + 3 = ym/11
9/12 = ym/11
12(ym) = 11 × 9
12ym = 99
Ym = 99/12
(ym) = 8.25cm

(10bi)
Hence to get the distance of A from C
|AC|² = AB+AC²-2ABACCosB
|AC|² = 110²+120²-2×110×120cos169
|AC|² = 12100+14400-26400cos169
|AC|² = 26500 -26400cos169
|AC|² = 26500 – 26400(-0.981)
|AC|² = 26500+25914.24
|AC|² = 52414.24
AC = √52414.24 =
AC = 228.9km

(10bii)
The bearing of C from A
Using sine Rule
120/sin tita = 228.9/sin169
Sintita = 120sin169/228.9 = 120(0.1908)/228.9
Tita = sin-1 0.1000
Tita = 5.74
The bearing of C from the starting point = 109 + 5.74
= 114.74 degree

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(11a)
DRAW THE DIAGRAM
Angular difference between p and q
=36+48=84degrees
Angular difference between q and r
=42-22=20degrees
r=RcosLAT
r=6400*cos42km
r=6400*0.74314km
r=4756.1km
therefore |PQ| tita/360*2pie r
|pQ|=84/360*2*3.142*4756.13
|PQ|=2510550.112/360km
|PQ|=6973.75km
|PQ|=6970km(3 s.f)

(11b)
QR along the line of longitude
QR=tita/360 *2pieR
=20/360*2*3.142*6400km
=20108.8/9
|QR|=2234.31km
=2230km (3 s.f)

(11c)
Speed=Distance/time
Time=distance/speed
T=(6973.75+2234.31)km/600km/h
T=9208.06/600 hrs
=9207.631/600
=15.346hrs
=15.3hrs (3 s.f)
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(12i)
Mean = ∑fm / ∑f
Draw the Table
x: 1-10, 11-20, 21-30, 31-40, 41-50
f: 12, 35, 21, 22, 10
m: 5.5, 15.5, 25.5 , 35.5, 45.5
Fm: 66, 542.5, 535.5, 781, 455

∑fm = 66 + 542.5 + 535.5 + 781 + 455
∑fm = 2380
∑f = 12 + 35 + 21 + 22 + 10
∑f = 100
Mean = ∑fm / ∑f
Mean = 2380 / 100
Mean = 23.8

(12ii)
Mean Deviation = ∑f|m – Mean| / N
Where:
m = Mid value of the grouped data
N = Sum of the frequencies
Mean = ∑fm / N
N = 12 + 35 + 21 + 22 + 10
N = 100
∑fm = 66 + 542.5 + 535.5 + 781 + 455
∑fm = 2380
Mean = 2380 / 100
Mean = 23.8
Mean Deviation = ∑f|m – Mean| / N
∑f|m – Mean| = 219.60 + 290.5 + 35.69+ 257.4 + 217
∑f|m – Mean| = 1020.19
Mean Deviation = 1020.19 / 100
Mean Deviation = 10.20

(12iii)
Standard Deviation = √(∑f|m – Mean|² / N)
Where:
m = Mid value of each grouped data
N = Sum of the frequencies
Mean = ∑fm / N
N = 12 + 35 + 21 + 22 + 10
N = 100
∑fm = 66 + 542.5 + 535.5 + 781 + 455
∑fm = 2380
Mean = 2380 / 100
Mean = 23.8
∑f|m – Mean|² = 4018.68 + 2411.15 + 60.69 + 3011.58 + 4708.9
∑f|m – Mean|² = 14211
Standard Deviation = √(∑f|m – Mean|² / N)
Standard Deviation = √(14211 / 100)
Standard Deviation = √(142.11)
Standard Deviation = 11.92

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COMPLETED
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