# 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

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

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

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}

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

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

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

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

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

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

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

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

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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