Important Posts

NECO GCE 2017 - MATHS OBJECTIVE AND ESSAY ANSWERS

Get Waec Gce Expo Here

NECO GCE - MATHS ANSWERS

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


==========================
COMPLETED
==========================


PLS STUDY THE MATHS SYMBOLS ON THIS LINK, SO YOU CAN UNDERSTAND THE MATHS WHEN WE POST OR SEND ANSWERS. 
THERE ARE SIGNS/SYMBOLS YOU SHOULD CHANGE WITH THE ORIGINAL SYMBOLS WHEN WRITING ON YOUR PAPER


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.

----------------------------------------------------------------


Maths OBJ:
1-10: CEDADDBEDB
11-20: EECECCACED
21-30: CBDDBEEBCD
31-40: ACEBDBDDAB
41-50: BDDBCCABCB
51-60: AAACADDCBA
=====================



(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}

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

(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
Area of shaded portion
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

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




(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

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


(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+7x12=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|=6980km(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

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


(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






++++++++++++++++++++++++
COMPLETED..
++++++++++++++++++++++++


Get Waec Gce Expo Here
THERE IS LOVE IN SHARING! :

No comments:

Post a Comment

Designed by NAIRACODED MEDIA.