Verified Neco Gce Mathematics Objectives & Theory Answers 2017

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Verified Neco Gce Mathematics Objectives & Theory Answers 2017 

Here is verified Neco Gce Mathematics Objectives & Theory Answers 2017

Neco GCE mathemtics Objectives

Maths Obj Answers
01-10 ACCBBECACA
11-20 BDBDBCCABC
21-30 BBDBAAAECB
31-40 DCCCEAEDCA
41-50 EEDCEDBBCE
51-60 DDCEDEDEDA

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(5a) Pls Draw the Histogram
[VIEW at www.tinyurl.com/necomaths5]
(5b) Pls locate the modal mark estimate on ur histogram drawing
[VIEW at www.tinyurl.com/necomaths5]
(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
5b)Mode = L1 + (i(f1 – f0) / (2f1 – f0 – f2))
Where:
L1 = Lower limit of the modal class
f1 = Frequency of the modal class
f0 = Frequency of the class preceding the modal class
f2 = Frequency of the class succeeding the modal class
i = Class interval
i = 10
L1 = 30.5
f1 = 15
f0 = 12
f2 = 7
Mode = L1 + (i(f1 – f0) / (2f1 – f0 – f2))
Mode = 30.5 + (10(15 – 12) / (2(15) – 12 – 7))
Mode = 30.5 + (10(3) / (30 – 12 – 7))
Mode = 30.5 + (30 / 11)
Mode = 30.5 + 2.73
Mode = 33.23
5ci)5ci)Mean = ∑fm / ∑f
∑fm = 33 + 155 + 306 + 532.5 + 301
∑fm = 1327.5
∑f = 6 + 10 + 12 + 15 + 7
∑f = 50
Mean = ∑fm / ∑f
Mean = 1327.5 / 50
Mean = 26.55
5cii5b)5cii)5c)L ₁  + i(((N / 2) – cf) / f)
Where: 
L ₁  = Lower limit of the median class
f = Frequency of the median class
cf = Cumulative frequency of the class preceding the median class
i = Class interval of the median class
N = Sum of the Frequencies = 49
Data table
x f cf C.B
1-10 6 6 0.5 -10.5
11-20 10 16 10.5 – 20.5
21-30 11 27 20.5 – 30.5
31-40 15 42 30.5 – 40.5
41-50 7 49 40.5 – 50.5
5c)L ₁  + i(((N / 2) – cf) / f)
Where: 
L ₁  = Lower limit of the median class
f = Frequency of the median class
cf = Cumulative frequency of the class preceding the median class
i = Class interval of the median class
N = Sum of the Frequencies = 49
Data table
x f cf C.B
1-10 6 6 0.5 -10.5
11-20 10 16 10.5 – 20.5
21-30 12 28 20.5 – 30.5
31-40 15 43 30.5 – 40.5
41-45 7 50 40.5 – 50.5
Median Class = 20.5 – 30.5
L1 = 20.5
N = 50
cf = 16
i = 10
f = 12
Median = L1 + i(((N / 2) – cf) / f)
Median = 20.5 + 10(((50 / 2) – 16) / 12)
Median = 20.5 + 10((25 – 16) / 12)
Median = 20.5 + 10(9 / 12)
Median = 20.5 + 10(0.75)
Median = 20.5 + 7.5
Median = 28
11a) DISTANCE PW ON A PARRALLEL OF LATITUDE
P(42°N ,48°E) and Q(42°N, 36°W)
θ = 48° + 36°
θ = 84°
operation is y = (θ⁄360) x 2πR
Where: 
y = Distance Along Two Great Circles
θ = Angular Difference
R = Radius of the Earth
PQ= (θ⁄360) x 2πR
pQ= (84⁄360) x (2 x π x 6400)
PQ = 0.233 x 40212.39
PQ= 9382.89Km.
11b) DISTANCE WE ALONG THE LINE OF LATITUDE
θ = 42° + 22°
θ = 64°
Therefore, your angular difference = 64°
QR = (θ⁄360) x 2πR
Where: 
y = Distance Along Two Great Circles
θ = Angular Difference
R = Radius of the Earth
QR= (θ⁄360) x 2πR
QR = (64⁄360) x (2 x π x 6400)
QR = 0.178 x 40212.39
QR = 7148.87Km
12a) Mean = ∑fm / ∑f
x f. m. Fm
1-10 12 5.5 66
11-20 35 15.5 542.5
21-30 21 25.5 535.5
31-40 22 35.5 781
41-50 10 45.5 455
∑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
12b)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
9a)Log₄(x² + 7x + 28) = 2
Solution
Log₄(x² + 7x + 28) = 2
x² + 7x + 28 = 2⁴
x² + 7x + 28 = 16
x² + 7x + 12 =0
((-b ± √(b² – 4ac)) / 2a)
where: 
a = coefficient of x²
b = coefficient of x and
c = constant
((-7 ± √(7² – 4 x 1 x 12)) / (2 x 1))
((-7 ± √(49 – 48)) / 2)
((-7 ± √(1)) / 2)
(-7 ± 1) / 2
(-7 + 1) / 2 or (-7 – 1) / 2
(-6) / 2 or (-8) / 2
-3 or -4
12a mean is 20. After adding the total frequency it will give you 100 den divide by 5
4b)A = πrl + πr²
Where: 
A = Area of the Cone
r = Radius of the Cone
l = Slant Height of the Cone
A = (π x 7 x 7) + (π x 7²)
A = 153.94 + (π x 49)
A = 153.94 + 153.94
A = 307.88cm²
7b)
y²/36 – 1/9 = 0
Multiply through by 36
y² – 4 = 0
y² = 4
y =±2
2b) x + y =3______(1)
x² – y² = 15_____(2)
Solution
From equation (2)
x² – y² = 15
(x+y)(x-y) = 15_____(3)
Recall x-y = 3
Substitute the value of (x-y) in equation (3)
3(x + y) = 15
Therefore (x +y) =5
The value of x+y =5
=====================
(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
========================================
(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
=====================================
(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}
=========================
 (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
=================================
(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 

 (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 
===========================
(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 ========================================= (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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