### Verified Neco Gce Mathematics Objectives & Theory Answers 2017

Here is verified Neco Gce Mathematics Objectives & Theory Answers 2017

Neco GCE mathemtics Objectives

**Maths Obj Answers01-10 ACCBBECACA11-20 BDBDBCCABC21-30 BBDBAAAECB31-40 DCCCEAEDCA41-50 EEDCEDBBCE51-60 DDCEDEDEDA**

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(5a) Pls Draw the Histogram

[VIEW at www.tinyurl.com/necomaths5]

[VIEW at www.tinyurl.com/necomaths5]

(5b) Pls locate the modal mark estimate on ur histogram drawing

[VIEW at www.tinyurl.com/necomaths5]

[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

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

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)

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

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

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

11b) DISTANCE WE ALONG THE LINE OF LATITUDE

θ = 42° + 22°

θ = 64°

Therefore, your angular difference = 64°

QR = (θ⁄360) x 2πR

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

12a) Mean = ∑fm / ∑f

x f. m. Fm

1-10 12 5.5 66

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

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

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

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

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

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

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)

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

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

(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}

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)