## 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

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 YearP=372500+27937.5+50000=N450437.50I=450437.50*15*1/200I=N33782.81total 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

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)SolutionFrom equation (2)x² – y² = 15(x+y)(x-y) = 15_____(3)Recall x-y = 3Substitute the value of (x-y) in equation (3)3(x + y) = 15Therefore (x +y) =5The value of x+y =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

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}**

4a)

length of chord

=2rsm^θ/2

=2*4sm^9^4/2

=85m47

=8*0.7314

=5.85cm

4a)

length of chord

=2rsm^θ/2

=2*4sm^9^4/2

=85m47

=8*0.7314

=5.85cm

**4b)draw the diagram7cm>7cmarea(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

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

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=52.8kmactual length=x% error=2%but % error =difference in length/actual length2/100=x-52.8/x1/50=x-52.8/xx=50(x-52.8)x=50x-2640x-50x=-2640-49x=-2640x=2640/49x=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

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 = 0y^2/36 – 1/9therefore:y^2/6^2 = 1/3^2y/6 = 1/3y = 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

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 formy-y0/x-x0 = y1-y0/x1-x0=>y-6/x-3 = 8-6/7-3y-6/x-3 = 2/4y-6/x-3 = 1/22y-12 = x-32y-x-12+3=02y-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+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

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

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/TimeTime = Distance/SpeedTotal distance = 6970+2230= 9200kmSpeed = 600km/hTime = 9200/600=15.3hrs**

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