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

NECO GCE Mathematics Obj And Essay/Theory Solution Questions and Answer – NOV/DEC 2017 Expo Runz.

MATHS OBJ:
1.CEDADDBEDB
11.EECECCACED
21.CBDDBEEBCD
31.ACEBDBDDAB
41.BDDBCCABCB
51.AAACADDCBA
PART I
ANSWER ALL(1-5) QUESTIONS - completed
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______(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

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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 br="">S:{all even number <14 br="">T:{all even nus<-20divisible 3="" br="" by="">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)
length of chord
=2rsm^θ/2
=2*4sm^9^4/2
=85m47
=8*0.7314
=5.85cm

4b)
draw the diagram
7cm>7cm
area(A)=θper^2/360-1/2r^2 8mθ
=r^2/2{22/180-8mθ}
=49/2[90/180*22/7-8m90]
=49/2{1/2*22/77-1}
=49/2{11/7-1}=44/2{1.57-1}
=49/2{0.57}=13.97cm^2

---

5a/5b)
CLICK HERE FOR THE SOLUTION
5c)
mean(x̅) = Ʃfx/Ʃf = 535.5/43
=12.5
median = 20 students from the histogram

---

PART II
ANSWER 5 QUESTIONS - completed
6a)
Draw
Let n(FnBnV)=x
Number of student that play football
only
=90-[35-x+x+60-x]
=90-(95-x)
=90-95+x
=x-5
basketball only
=95-[35-x+x+25-x]
=95-60+x
=x+35
volleyball only
=105-[25-x+x+60-x]
=105-[85-x]
=105-85+x
=x+20
but n(E) = x-5+x+35+x+20+35-x+25-x+60-x
+x
180=x+170
x=180-170
x=10 student
6ai)
number of students who played football only=x-5
=10-5
=5 student
6aii)
Number of student who play exactly two games
=35-x+25-x+60-x
=120-3x
=120-3(10)
=120-30
=90 student

6b)
measured length=52.8km
actual length=x
% error=2%
but % error =difference in length/actual length
2/100=x-52.8/x
1/50=x-52.8/x
x=50(x-52.8)
x=50x-2640
x-50x=-2640
-49x=-2640
x=2640/49
x=53.9km

---

7a)
y = x^3-6x^2+9x-5, the value of x when gradient dy/dx = 0
therefore:
dy/dx = 3x^2-12x+9
3x^2-12x+9=0
3x^2-3x-9x+9=0
3x(x-3)-9(x-1)=0
(3x-9)(x-1)=0
x=3/9 or x=1
x = 3,1

7b)
y^2/36 - 1/9 = 0
y^2/36 - 1/9
therefore:
y^2/6^2 = 1/3^2
y/6 = 1/3
y = 6/3
=2

---

8a)
volume = 155,232cm^3
(i)curve surfeca area = 2πr^2
(ii)total surface area = 3πr2
(iii)volume = 2/3πr3 = 155232
2πr^3 = 3*155232
r^3 = 3*155232/2π
r^3 = 23284.8/π
r^3 = 232848*7/22
r^3 = 74088
r = 3root74088
r = 41.9cm
r = 42cm

8b)
A(3,6), B(7,8)
using the two point form
y-y0/x-x0 = y1-y0/x1-x0
=>y-6/x-3 = 8-6/7-3
y-6/x-3 = 2/4
y-6/x-3 = 1/2
2y-12 = x-3
2y-x-12+3=0
2y-x-9=0

---

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
y+dy=3(x+dx)^2-4
y+dy=3(x^2+2xDx+Dx^2)-4
dy=3(x^2+2xDx+Dx+Dx^2)-4-y
=3x^2+6xDx+3Dx^2-4-(3x^2-4)
Dy=6xDx+3Dx^2
Dy/Dx=6xDx+3Dx^2/Dx
=6x+3Dx
but as Dx=0
Dy/Dx=dy/dx
dy/dx=6x+3(0)
=6x

---

11a)
Draw The diamgram
-Angular difference between P and Q = 48+36 = 84°
-Angular difference betwwen Q and R = 42-22 = 20°
(a)distance PQ = θ/360 * 2πRcos∝
∝ = 42°
PQ = 84/360 * 3.142*6400cos42
=84*2*3.142*6400*0.7431/360
=2510398.68/360
=6970km

11b)
Distance QR is along the great circle
QR = θ/360*2πR
=20/360*2*3.142*6400
=2*3.142*6400/18
=40217.60/18 = 2234.31
=2230km(3 s.f)

11c)
Speed = Distance/Time
Time = Distance/Speed
Total distance = 6970+2230
= 9200km
Speed = 600km/h
Time = 9200/600
=15.3hrs

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