Mathematics neco answers

[6/4, 12:26 AM] Alaba: 1a)

(2x+1)/(3-4x)=2/3

3(2x+1)=2(3-4x)

6x+3=6-8x

6x+8x=6-3

14x/14=3/14

x=3/14

1bi)

E=MV^2/2

2E/M =MV^2/M

V^2=2E/M

V=sqr2E/M

1bii)

Vsqr2E/M

Vsqr2*64/2

Vsqr64

V=8

[6/4, 12:28 AM] Alaba: 2 a )

number of sides =12

radius of circle =10 cm

area =?

n ×© 2= 360

12 © 2= 360

© 2 = 360 / 12 =30 °

© 1 + © 2= 180 – 30

© 1 = 150

When © 1 and © 2 are interior and exterior angle

of a polygon A sector has are .

Area of sector =© / 360 × rot 8 ^2

= 150 / 360 ×22 / 7× 100 / 1

A =130 – 95 cm^2

2 b )

1 /2 ( 2x + 1 ) – 2 /5 ( x – 2 )= 3

2 x +1/ 3 – 2 x – 4/ 3= 3

10 x +5 – 6x + 12 / 15 =3 / 1

Cross multiple

4 x +17 =45

4 x /4 =28 / 4

x = 7.

3 )

Apply 5m rule to find C P

t / sin T = P /sin P

t / sin 110 = 6/ sin 40

t =6 * 0. 9396 / 0. 6428

= 56376 / 6. 6428

= 87704

= 877 km

4 )

Total Fruit = 80 + 60 = 140

( a)

( i ) Pr one of each fruit is picked

( 79 / 140 * 60 / 139 ) + ( 59 / 140 * 80 / 139 )

= 4740 / 19, 460 + 4720 / 19 , 460

= 9460 / 19, 460 = 0.486

4 aii )

Pr one type of fruit is picked

( 79 / 140 * 78 / 139 ) + ( 39 / 140 * 5 p/ 139 )

= 6162 / 19, 460 + 3422 / 19 , 460

= 9584 / 19 , 460 = 0.492

4 b )

5 X / 8 – 1 / 6 ≤ X / 3 + 7 /24

Multiply through by 24 i : e

15 X – 4≤ 8X + 7

15 X – 8X ≤ 7 + 4

7 X = 11

X ≤ 11 / 7 ===> X ≤ 1 4 / 9

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

5 a )

3 /X + 2 – 6 /3 X – 1

3 ( 3 X – 1) – 6 ( X + 2) / ( X + 2) ( 3X – 1)

9 X – 3 – 6X – 12 /( X + 2) ( 3X – 1)

3 X – 15 / ( X + 2) ( 3X – 1)

5 b )

C .I= P [1 +r / 100 ]^

= 25000 [ 1+ 12/ 100 ]^ 3

= 25000 [ 1+ 0. 12 ]^ 3

= 25000 * 1 . 4049

= 35122 . 50

= N 35 ,122 . 50

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

6 a )

X +- 3/ 2

X =2/ 3 or X =2

( X + 3/ 2)^ 2 or ( X – 2 )

( X + 3/ 2) ( X – 2 )

X ( X – 2) + 3 / 2 ( X – 2)

X ^2 – 2X + 3 X / 2 – 3

2 X ^2 – 4X + 3X – 6

2 X ^2 – X – 6

6 b )

h /h +8 = 6 / 10

10 h = 6 h + 48

h = 12

H = h + 8

H = 12 + 8

H = 20

Volume = 1 /3 A . h

= 1 / 3 ( 10 * 10 ) * 20 – 432 / 3

= 200 / 3 – 432 /3

= 1568 / 3

= 522 . 67 cm3

7 a )

titan / 360 ×2 pie r cos t

d = 55 / 360 ×2 ×22 / 7×640 cos 4

d = 55 × 44 × 6400 cos 4/ 2520

d = 55 × 44 × 6400 ×0496 / 2520

d = 15 , 449 . 28 / 2520

d = 6130. 67

d ~ 6130 km .

ii ) distance along gent circle

D = tita / 360 ×2 pie r

D = 55 / 360 × 2/ 7× 22 /7 × 6400/ 1

D = 55 × 44 ×6400 / 2520

D = 15 , 488 ×6400 / 2520

D = 6144 .03

D =~ 6146 km .

7 b )

Length of sector tita/ 360 × 2 pie r

L= 120 / 360 ×2 /1 × 22 / 7× 42/ 1

L= 120 ×44 ×42 / 2520

L= 221760 / 2520

L= 88cm

L= 2pie r

Where r is the radius of circumference

88 =2 × 22 / 7× r

88 ×7 = 44 r

R =88 ×7 /44

R =616 / 44

R =14 cm.

Curved surface area

= pie rc

A =22 / 7 ×14 × 42 / 1

A =22 ×14 ×4 ^2 /7

A =12936 /7

A =1848 cm^ 2.

8 a )

X =60 / t — – – – – – – – – > ( i )

Y = 180 / t – – – – – — – – – > ( ii )

T 1=60 / X

T 2=100 /Y

T 1+T 2= 5

60 / X + 180 / Y = 300 – — – – – – – – – > ( i )

180 / X + 60 / Y = 260 – – – – – – – – — > ( ii )

Let P =1 / X

2 = 1 / Y

60 p + 180 Q = 300

180 p + 6Q = 200

P + 3 Q =5

9 P + 3Q = 13

Substract ( i ) from ( ii )

8 p = 8

P = 8 / 8 ÷ P = 1

Subtract P into ( i )

P + 3 Q =5

1 + 3 Q = 5

3 Q =5 – 1

3 Q =4

Q = 4/ 3

P = 1 ÷ 1 = 1/ X ÷ X = 1

4 /3 = 1/ Y ÷ Y = 3/ 4

8 b )

2001 – – – – – — – – 25 , 700

2002 – – – – – — – – 15 / 100 X 25, 700 + 25,

700 = 29 , 555

Amount of tax in 2002

= 29 , 555 * 12 .5 /100

= N 3694 .375

= N 3690

8 c )

Log 25

Log 16 25 / 100

Log 16 2/ 4

Log 4 6 – 1

– 1/ 2 Log 4^4

– 1/ 2

9 ai)

W= K +C / 2

24 = k + C / 16

384 = 16 K + C – — – – – – ( i )

18 = K + C / 4

72 + 4K + C – – – – – – – — ( ii )

16 K + C = 384

4 K + C = 72

Substact ( ii ) from ( i )

12 k / 12 = 312 /12

K = 26

Substract K into ( i )

16 k + C = 384

C = 384 – 416

C = – 32

W= k +C / t 2

W= 26- 32 / t 2

( 9aii )

When W= – 46, t =?

– 46 = 26 – 32 / t 2

t 2 = – 32 / -72

t = Sqrt 16 / 36 = 4 /6

= 2 / 3

9 b )

V = Pie r 2 . d = 14 , r = 7cm

1232 = 22 / 7 * 7 ^2 * h

h = 7 * 1232/ 22* 49

h = 8624 /1074

h = 8 cm

11 a)

y ^ 1 =x ^2 ( 3x +1 )^ 2

v = ( 2x + 1 )^ 2

v = m ^2

dm / dx =2

dv / dm =2 m

dy /dx = dv / dm ×dm / dx

= 2 m× 2

= 4 m

dy /dx = 4( 2x + 1)

dy /dx = udv /dx +v whole no . dy / dx

= x ^2 4( xx + 1)^ 2, × 3x

= 4 x ^2 ( 2x +1) + 2x ( 2x + 1)^ 2.

11 b)

[ 3 3 -1 ] [ 1 0 2] [ 3 – 2 3] + 2 [0 – ( – 4) – 3[ 63 ] + – 1 ( – 2)

8 + 9+ 2

= 19 .

11 c )

m= y 2- y 1 /x 2- x 1

y – y 1 = m ( x – x 2)

m= 4- 3/ – 1- 2

m= – 1/ 3

y 1 -y 2=- 1 /3 ( x – x 2)

y – 3= 1/ 3 ( x – 2)

y – 3= – 4/ 3 + 2/ 3

3 y =- x + 1[truncated by WhatsApp]

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8a)

|AD|^2=13^2-5^2

|AD|^2=169-25

|AD|^2=144

AD=sqr144

AD=12CM

|AD|=12-r

r^2=(12-r)^2 – 5^2

r^2=(12-r)(12-r)+25

r^2=144-24r+25

r^2=169-24r

r^2+24r-169=0

r^2+24r=169

r^2+24r+14^2=169+14^2

(r+14)^2=169+196

(r+14)^2=365

(r+14=sqr365

r+14=19.105

r=19.105-14

r=5.105

r=5.1cm

8aii)

circumfrenece of a circle=2pie R

C=2×22/2*(5.1)^2

C=1144.44/7

C=163.4914cm

C=163.5cm

8b)

y2-y1/x2-x1=y-y1/x-x1

6-2/2-(-1)=y-2/x-(-1)

4/2+1 = y-2/x+1

4/3=y-2/x+1

3(y-2)=4(x+1)

3y-6=4x+4

3y-4x=4+6

3y-4x=10

y=4x/3+10/3

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

9a)

let Xy represent the two digit number

x-y=5 —–(i)

3xy – (10x +y)=14

3xy – 10x – y =14 —-(ii)

from eqn (i)

x=5+y

3y(5+y)-10(5+y)-y=14

15y+3y^2 – 50 – 10y – y =14

3y^2 + 4y -50 = 14

3y^2 + 4y -50 – 14 =0

3y^2 + 4y – 64 =0

3y^2 + 12y + 16y – 64 =0

(3y^2 – 12y) (+16y – 64)=0

by

(y-4)+16(y-4)=0

(y-4)=0

9aii)

(3y+16)(y-4)=0

3y+16=0 or y-4=0

3y=-16 or y=4

y=-16/3 or y=4

when y=4

x=5+y

x=5+4

x=9

the no is 94

9b)u

3-2x/4 + 2x-3/3

=3(3-2x)+4(2x-3)/12

=9-6x+8x-12/12

=2x-2/12

=========================2 a )

number of sides =12

radius of circle =10 cm

area =?

n ×© 2= 360

12 © 2= 360

© 2 = 360 / 12 =30 °

© 1 + © 2= 180 – 30

© 1 = 150

When © 1 and © 2 are interior and exterior angle

of a polygon A sector has are .

Area of sector =© / 360 × rot 8 ^2

= 150 / 360 ×22 / 7× 100 / 1

A =130 – 95 cm^2

2 b )

1 /2 ( 2x + 1 ) – 2 /5 ( x – 2 )= 3

2 x +1/ 3 – 2 x – 4/ 3= 3

10 x +5 – 6x + 12 / 15 =3 / 1

Cross multiple

4 x +17 =45

4 x /4 =28 / 4

x = 7.

3 )

Apply 5m rule to find C P

t / sin T = P /sin P

t / sin 110 = 6/ sin 40

t =6 * 0. 9396 / 0. 6428

= 56376 / 6. 6428

= 87704

= 877 km

4 )

Total Fruit = 80 + 60 = 140

( a)

( i ) Pr one of each fruit is picked

( 79 / 140 * 60 / 139 ) + ( 59 / 140 * 80 / 139 )

= 4740 / 19, 460 + 4720 / 19 , 460

= 9460 / 19, 460 = 0.486

4 aii )

Pr one type of fruit is picked

( 79 / 140 * 78 / 139 ) + ( 39 / 140 * 5 p/ 139 )

= 6162 / 19, 460 + 3422 / 19 , 460

= 9584 / 19 , 460 = 0.492

4 b )

5 X / 8 – 1 / 6 ≤ X / 3 + 7 /24

Multiply through by 24 i : e

15 X – 4≤ 8X + 7

15 X – 8X ≤ 7 + 4

7 X = 11

X ≤ 11 / 7 ===> X ≤ 1 4 / 9

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

5 a )

3 /X + 2 – 6 /3 X – 1

3 ( 3 X – 1) – 6 ( X + 2) / ( X + 2) ( 3X – 1)

9 X – 3 – 6X – 12 /( X + 2) ( 3X – 1)

3 X – 15 / ( X + 2) ( 3X – 1)

5 b )

C .I= P [1 +r / 100 ]^

= 25000 [ 1+ 12/ 100 ]^ 3

= 25000 [ 1+ 0. 12 ]^ 3

= 25000 * 1 . 4049

= 35122 . 50

= N 35 ,122 . 50

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

6 a )

X +- 3/ 2

X =2/ 3 or X =2

( X + 3/ 2)^ 2 or ( X – 2 )

( X + 3/ 2) ( X – 2 )

X ( X – 2) + 3 / 2 ( X – 2)

X ^2 – 2X + 3 X / 2 – 3

2 X ^2 – 4X + 3X – 6

2 X ^2 – X – 6

6 b )

h /h +8 = 6 / 10

10 h = 6 h + 48

h = 12

H = h + 8

H = 12 + 8

H = 20

Volume = 1 /3 A . h

= 1 / 3 ( 10 * 10 ) * 20 – 432 / 3

= 200 / 3 – 432 /3

= 1568 / 3

= 522 . 67 cm3

7 a )

titan / 360 ×2 pie r cos t

d = 55 / 360 ×2 ×22 / 7×640 cos 4

d = 55 × 44 × 6400 cos 4/ 2520

d = 55 × 44 × 6400 ×0496 / 2520

d = 15 , 449 . 28 / 2520

d = 6130. 67

d ~ 6130 km .

ii ) distance along gent circle

D = tita / 360 ×2 pie r

D = 55 / 360 × 2/ 7× 22 /7 × 6400/ 1

D = 55 × 44 ×6400 / 2520

D = 15 , 488 ×6400 / 2520

D = 6144 .03

D =~ 6146 km .

7 b )

Length of sector tita/ 360 × 2 pie r

L= 120 / 360 ×2 /1 × 22 / 7× 42/ 1

L= 120 ×44 ×42 / 2520

L= 221760 / 2520

L= 88cm

L= 2pie r

Where r is the radius of circumference

88 =2 × 22 / 7× r

88 ×7 = 44 r

R =88 ×7 /44

R =616 / 44

R =14 cm.

Curved surface area

= pie rc

A =22 / 7 ×14 × 42 / 1

A =22 ×14 ×4 ^2 /7

A =12936 /7

A =1848 cm^ 2.

8 a )

X =60 / t — – – – – – – – – > ( i )

Y = 180 / t – – – – – — – – – > ( ii )

T 1=60 / X

T 2=100 /Y

T 1+T 2= 5

60 / X + 180 / Y = 300 – — – – – – – – – > ( i )

180 / X + 60 / Y = 260 – – – – – – – – — > ( ii )

Let P =1 / X

2 = 1 / Y

60 p + 180 Q = 300

180 p + 6Q = 200

P + 3 Q =5

9 P + 3Q = 13

Substract ( i ) from ( ii )

8 p = 8

P = 8 / 8 ÷ P = 1

Subtract P into ( i )

P + 3 Q =5

1 + 3 Q = 5

3 Q =5 – 1

3 Q =4

Q = 4/ 3

P = 1 ÷ 1 = 1/ X ÷ X = 1

4 /3 = 1/ Y ÷ Y = 3/ 4

8 b )

2001 – – – – – — – – 25 , 700

2002 – – – – – — – – 15 / 100 X 25, 700 + 25,

700 = 29 , 555

Amount of tax in 2002

= 29 , 555 * 12 .5 /100

= N 3694 .375

= N 3690

8 c )

Log 25

Log 16 25 / 100

Log 16 2/ 4

Log 4 6 – 1

– 1/ 2 Log 4^4

– 1/ 2

9 ai)

W= K +C / 2

24 = k + C / 16

384 = 16 K + C – — – – – – ( i )

18 = K + C / 4

72 + 4K + C – – – – – – – — ( ii )

16 K + C = 384

4 K + C = 72

Substact ( ii ) from ( i )

12 k / 12 = 312 /12

K = 26

Substract K into ( i )

16 k + C = 384

C = 384 – 416

C = – 32

W= k +C / t 2

W= 26- 32 / t 2

( 9aii )

When W= – 46, t =?

– 46 = 26 – 32 / t 2

t 2 = – 32 / -72

t = Sqrt 16 / 36 = 4 /6

= 2 / 3

9 b )

V = Pie r 2 . d = 14 , r = 7cm

1232 = 22 / 7 * 7 ^2 * h

h = 7 * 1232/ 22* 49

h = 8624 /1074

h = 8 cm

11 a)

y ^ 1 =x ^2 ( 3x +1 )^ 2

v = ( 2x + 1 )^ 2

v = m ^2

dm / dx =2

dv / dm =2 m

dy /dx = dv / dm ×dm / dx

= 2 m× 2

= 4 m

dy /dx = 4( 2x + 1)

dy /dx = udv /dx +v whole no . dy / dx

= x ^2 4( xx + 1)^ 2, × 3x

= 4 x ^2 ( 2x +1) + 2x ( 2x + 1)^ 2.

11 b)

[ 3 3 -1 ] [ 1 0 2] [ 3 – 2 3] + 2 [0 – ( – 4) – 3[ 63 ] + – 1 ( – 2)

8 + 9+ 2

= 19 .

11 c )

m= y 2- y 1 /x 2- x 1

y – y 1 = m ( x – x 2)

m= 4- 3/ – 1- 2

m= – 1/ 3

y 1 -y 2=- 1 /3 ( x – x 2)

y – 3= 1/ 3 ( x – 2)

y – 3= – 4/ 3 + 2/ 3

3 y =- x + 1[truncated by WhatsApp]

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8a)

|AD|^2=13^2-5^2

|AD|^2=169-25

|AD|^2=144

AD=sqr144

AD=12CM

|AD|=12-r

r^2=(12-r)^2 – 5^2

r^2=(12-r)(12-r)+25

r^2=144-24r+25

r^2=169-24r

r^2+24r-169=0

r^2+24r=169

r^2+24r+14^2=169+14^2

(r+14)^2=169+196

(r+14)^2=365

(r+14=sqr365

r+14=19.105

r=19.105-14

r=5.105

r=5.1cm

8aii)

circumfrenece of a circle=2pie R

C=2×22/2*(5.1)^2

C=1144.44/7

C=163.4914cm

C=163.5cm

8b)

y2-y1/x2-x1=y-y1/x-x1

6-2/2-(-1)=y-2/x-(-1)

4/2+1 = y-2/x+1

4/3=y-2/x+1

3(y-2)=4(x+1)

3y-6=4x+4

3y-4x=4+6

3y-4x=10

y=4x/3+10/3

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

9a)

let Xy represent the two digit number

x-y=5 —–(i)

3xy – (10x +y)=14

3xy – 10x – y =14 —-(ii)

from eqn (i)

x=5+y

3y(5+y)-10(5+y)-y=14

15y+3y^2 – 50 – 10y – y =14

3y^2 + 4y -50 = 14

3y^2 + 4y -50 – 14 =0

3y^2 + 4y – 64 =0

3y^2 + 12y + 16y – 64 =0

(3y^2 – 12y) (+16y – 64)=0

by

(y-4)+16(y-4)=0

(y-4)=0

9aii)

(3y+16)(y-4)=0

3y+16=0 or y-4=0

3y=-16 or y=4

y=-16/3 or y=4

when y=4

x=5+y

x=5+4

x=9

the no is 94

9b)u

3-2x/4 + 2x-3/3

=3(3-2x)+4(2x-3)/12

=9-6x+8x-12/12

=2x-2/12

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

[6/4, 12:33 AM] Alaba: Neco Mathematics Answers 2018:

Section II [Theory]

Answer ALL questions in Part I

1. Using ruler and a pair of compasses only:

(a) construct a triangle PQRwith /PQ/ = lOcm, ∠QPR = 90° and ∠PQR = 30°;

(b) (i) construct I, the locus of all points equidistant from PR and QR;

(ii) locate M, the point where / intersects with po,

(c) (i) with M as centre and radius MP, draw a circle;

(ii) calculate the area of the circler correct to one decimal place.

[Take Π= 22/7].

Answer any FIVE questions in Part II.

Write your answers on the answer booklet provided.

1. (a) Two children shared an amount of money in the ratio 3/4 : 2/5. If the smaller share was GH¢25.00, how much was shared between them?

(b) A box contains 5 red, 3 green and 4 blue balls of the same size. If a boy picks two balls from the box one after the other without replacement, what is the probability that both balls are red?

2. a) In a right angled triangle, sin x = 3/5. Evaluate 5(cos x)³ – 3.

(b) The angle of elevation of the top of a vertical pole from a point 63m east of the base of the pole is 30°. From another point due west of the pole, the angle of elevation of the top is 60°

(i) Draw a sketch diagram to illustrate the information.

(ii) Calculate, correct to three significant figures, the distance of the second point from the base of the pole.

3.

In the diagram above, the points A, B, C and D lie on the circle, centre O. TA and TB are tangents touching the circle at A and B respectively.

AÔB = 132°, AĈD = 59° and AOC is a straight line.

(i) Find A T B.

(ii) Find B D A.

(iii) Find B D C.

(iv) Find OB D.

4. (a) The diagonals of a rhombus are 14em and gem. Calculate, correct to the nearest centimetre, the perimeter of the rhombus.

(b) The cross section of a rectangular tank measures l.2m by 0.9m. It contains water to a depth of O.4m. If a cubical block of side 50em is lowered into the tank, calculate, correct to 2 significant figures, the rise in the water level (in metres).

5. (a) In a class of 50 students, 30 offered History, 15 offered History and Geography while 3 did not offer any of the two subjects.

(i) Represent the information on a Venn diagram.

(ii) Find the number of candidates that offered:

(A) History only;

(B) Geography only.

(b) A trader sold an article at a discount of 8% for N 828.00. If the article was initially marked to gain 25%, find the

(i) cost price of the article;

(ii) discount allowed.

6. (a) The second, fourth and sixth terms of an Arithmetic Progression (A.P.) are x – 1, x + 1 and 7 respectively. Find the:

(i)common difference;

(ii)first term;

(iii)value of x.

(b) A spherical bowl of radius r em is one-quarter full when 6 litres of water is poured into it. Calculate, correct to three significant figures, its diameter. [Take π =22/7]

7. P(lat 400N, long 18°W) and Q(lat 400N, long 78OW) are two cities on the surface of the earth. Calculate, the:

(a) radius of the parallel of latitude on which P and Q lie, correct to the nearest 10km;

(b) length of the minor arc PQ, correct to the nearest 100 km;

(c) vertical distance between the centre of the earth and the centre of the small circle on which P and Q lie, correct to the nearest km.

[Take π = 22/7 and radius of the earth= 6400 km]

8. The table below gives the distribution of marks for 360 candidates who sat for an examination.

[6/4, 9:59 AM] Alaba: 7a)

A) T3=6 & T7 =30

I)common difference using Tn=a+(n-1)d

In the 3rd term; n =3

=>T3=a+(3-1)d=6

=>a+2d=6 equation (1)

In the 7th term ;n =7

=> T7=a+(7-1)d=30

=>a + 8d=30 equation (1) & (2)

Simultaneous

A+2d =6 equation (1)

A+ 8d=30 equation (2)

0+(1-6d) =-24

=> -6d= -24

=>d = -24/-6 =4

II) first term put d=4 into equation

DIAMOND 💎 AND SURE B👌💪

[6/4, 10:22 AM] Alaba: No 4(i)

length of Arc of the sector

Titter= 72?, r = 14cm

L= titter / 360 x 2 pie r

==> L= 72/360 x 2 x 22/7 x 14

=44352/2520 = 17.6cm

(ii) perimeter of the sector

Perimeter = titter/360 x 2 pie r + 2r = 17.6 +(2x14) =17.6+28= 45.6cm

(iii) Area of the sector

Area = Titter/360 x pie r? =

72/360 x 22/7 x (14)? = 72 x 22 x 196/2520

Area= 310464/2520 = 123.2cm?

Dis one correct and images correct

[6/4, 10:22 AM] Alaba: 1a)

Log 10(20*-10)-log10(*+3)=log105

(20*-10/*+3)=log10 =5

20*-10/*+3=5

5(*+3)=20*-10

5*+15=20*-10

15+10=20*-5*

25=15*

*=25/15

*=5/3=1 2/3

1b)

Discount percent =15%

Discount amount =#600

Actual amount paid on the article =?

Original amount on the article =*

15%*=#600

15/100* =600

15*=600*100

15*=60000

*=60000/15

*=#4,000

Therefore actual amount paid on the article

=#4,000-#600

=#3,400

Actual amount paid on the article =#3,400

[6/4, 10:22 AM] Alaba: : 7a)

A) T3=6 & T7 =30

I)common difference using Tn=a+(n-1)d

In the 3rd term; n =3

=>T3=a+(3-1)d=6

=>a+2d=6 equation (1)

In the 7th term ;n =7

=> T7=a+(7-1)d=30

=>a + 8d=30 equation (1) & (2)

Simultaneous

A+2d =6 equation (1)

A+ 8d=30 equation (2)

0+(1-6d) =-24

=> -6d= -24

=>d = -24/-6 =4

II) first term put d=4 into equation

[1b

Discount percent =15%

Discount amount =#600

Actual amount paid on the article =?

Original amount on the article =*

15%*=#600

15/100* =600

15*=600*100

15*=60000

*=60000/15

*=#4,000

Therefore actual amount paid on the article

=#4,000-#600

=#3,400

Actual amount paid on the article =#3,400

[6/4, 09:45] london boy: 1a)

Log 10(20*-10)-log10(*+3)=log105

(20*-10/*+3)=log10 =5

20*-10/*+3=5

5(*+3)=20*-10

5*+15=20*-10

15+10=20*-5*

25=15*

*=25/15

*=5/3=1 2/3

No 4(i)

length of Arc of the sector

Titter= 72?, r = 14cm

L= titter / 360 x 2 pie r

==> L= 72/360 x 2 x 22/7 x 14

=44352/2520 = 17.6cm

(ii) perimeter of the sector

Perimeter = titter/360 x 2 pie r + 2r = 17.6 +(2x14) =17.6+28= 45.6cm

(iii) Area of the sector

Area = Titter/360 x pie r? =

72/360 x 22/7 x (14)? = 72 x 22 x 196/2520

Area= 310464/2520 = 123.2cm?

[6/4, 10:49 AM] Alaba: 1CDAAEABAEC

11ACDDCDDDAC

21CEADEDCBBC

31CBEEECBDC-

41DBCBCBDBA

51 BCBDCDB

Still solving ✍🏻✍🏻

[6/4, 12:26 AM] Alaba: 1a)

(2x+1)/(3-4x)=2/3

3(2x+1)=2(3-4x)

6x+3=6-8x

6x+8x=6-3

14x/14=3/14

x=3/14

1bi)

E=MV^2/2

2E/M =MV^2/M

V^2=2E/M

V=sqr2E/M

1bii)

Vsqr2E/M

Vsqr2*64/2

Vsqr64

V=8

[6/4, 12:28 AM] Alaba: 2 a )

number of sides =12

radius of circle =10 cm

area =?

n ×© 2= 360

12 © 2= 360

© 2 = 360 / 12 =30 °

© 1 + © 2= 180 – 30

© 1 = 150

When © 1 and © 2 are interior and exterior angle

of a polygon A sector has are .

Area of sector =© / 360 × rot 8 ^2

= 150 / 360 ×22 / 7× 100 / 1

A =130 – 95 cm^2

2 b )

1 /2 ( 2x + 1 ) – 2 /5 ( x – 2 )= 3

2 x +1/ 3 – 2 x – 4/ 3= 3

10 x +5 – 6x + 12 / 15 =3 / 1

Cross multiple

4 x +17 =45

4 x /4 =28 / 4

x = 7.

3 )

Apply 5m rule to find C P

t / sin T = P /sin P

t / sin 110 = 6/ sin 40

t =6 * 0. 9396 / 0. 6428

= 56376 / 6. 6428

= 87704

= 877 km

4 )

Total Fruit = 80 + 60 = 140

( a)

( i ) Pr one of each fruit is picked

( 79 / 140 * 60 / 139 ) + ( 59 / 140 * 80 / 139 )

= 4740 / 19, 460 + 4720 / 19 , 460

= 9460 / 19, 460 = 0.486

4 aii )

Pr one type of fruit is picked

( 79 / 140 * 78 / 139 ) + ( 39 / 140 * 5 p/ 139 )

= 6162 / 19, 460 + 3422 / 19 , 460

= 9584 / 19 , 460 = 0.492

4 b )

5 X / 8 – 1 / 6 ≤ X / 3 + 7 /24

Multiply through by 24 i : e

15 X – 4≤ 8X + 7

15 X – 8X ≤ 7 + 4

7 X = 11

X ≤ 11 / 7 ===> X ≤ 1 4 / 9

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

5 a )

3 /X + 2 – 6 /3 X – 1

3 ( 3 X – 1) – 6 ( X + 2) / ( X + 2) ( 3X – 1)

9 X – 3 – 6X – 12 /( X + 2) ( 3X – 1)

3 X – 15 / ( X + 2) ( 3X – 1)

5 b )

C .I= P [1 +r / 100 ]^

= 25000 [ 1+ 12/ 100 ]^ 3

= 25000 [ 1+ 0. 12 ]^ 3

= 25000 * 1 . 4049

= 35122 . 50

= N 35 ,122 . 50

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

6 a )

X +- 3/ 2

X =2/ 3 or X =2

( X + 3/ 2)^ 2 or ( X – 2 )

( X + 3/ 2) ( X – 2 )

X ( X – 2) + 3 / 2 ( X – 2)

X ^2 – 2X + 3 X / 2 – 3

2 X ^2 – 4X + 3X – 6

2 X ^2 – X – 6

6 b )

h /h +8 = 6 / 10

10 h = 6 h + 48

h = 12

H = h + 8

H = 12 + 8

H = 20

Volume = 1 /3 A . h

= 1 / 3 ( 10 * 10 ) * 20 – 432 / 3

= 200 / 3 – 432 /3

= 1568 / 3

= 522 . 67 cm3

7 a )

titan / 360 ×2 pie r cos t

d = 55 / 360 ×2 ×22 / 7×640 cos 4

d = 55 × 44 × 6400 cos 4/ 2520

d = 55 × 44 × 6400 ×0496 / 2520

d = 15 , 449 . 28 / 2520

d = 6130. 67

d ~ 6130 km .

ii ) distance along gent circle

D = tita / 360 ×2 pie r

D = 55 / 360 × 2/ 7× 22 /7 × 6400/ 1

D = 55 × 44 ×6400 / 2520

D = 15 , 488 ×6400 / 2520

D = 6144 .03

D =~ 6146 km .

7 b )

Length of sector tita/ 360 × 2 pie r

L= 120 / 360 ×2 /1 × 22 / 7× 42/ 1

L= 120 ×44 ×42 / 2520

L= 221760 / 2520

L= 88cm

L= 2pie r

Where r is the radius of circumference

88 =2 × 22 / 7× r

88 ×7 = 44 r

R =88 ×7 /44

R =616 / 44

R =14 cm.

Curved surface area

= pie rc

A =22 / 7 ×14 × 42 / 1

A =22 ×14 ×4 ^2 /7

A =12936 /7

A =1848 cm^ 2.

8 a )

X =60 / t — – – – – – – – – > ( i )

Y = 180 / t – – – – – — – – – > ( ii )

T 1=60 / X

T 2=100 /Y

T 1+T 2= 5

60 / X + 180 / Y = 300 – — – – – – – – – > ( i )

180 / X + 60 / Y = 260 – – – – – – – – — > ( ii )

Let P =1 / X

2 = 1 / Y

60 p + 180 Q = 300

180 p + 6Q = 200

P + 3 Q =5

9 P + 3Q = 13

Substract ( i ) from ( ii )

8 p = 8

P = 8 / 8 ÷ P = 1

Subtract P into ( i )

P + 3 Q =5

1 + 3 Q = 5

3 Q =5 – 1

3 Q =4

Q = 4/ 3

P = 1 ÷ 1 = 1/ X ÷ X = 1

4 /3 = 1/ Y ÷ Y = 3/ 4

8 b )

2001 – – – – – — – – 25 , 700

2002 – – – – – — – – 15 / 100 X 25, 700 + 25,

700 = 29 , 555

Amount of tax in 2002

= 29 , 555 * 12 .5 /100

= N 3694 .375

= N 3690

8 c )

Log 25

Log 16 25 / 100

Log 16 2/ 4

Log 4 6 – 1

– 1/ 2 Log 4^4

– 1/ 2

9 ai)

W= K +C / 2

24 = k + C / 16

384 = 16 K + C – — – – – – ( i )

18 = K + C / 4

72 + 4K + C – – – – – – – — ( ii )

16 K + C = 384

4 K + C = 72

Substact ( ii ) from ( i )

12 k / 12 = 312 /12

K = 26

Substract K into ( i )

16 k + C = 384

C = 384 – 416

C = – 32

W= k +C / t 2

W= 26- 32 / t 2

( 9aii )

When W= – 46, t =?

– 46 = 26 – 32 / t 2

t 2 = – 32 / -72

t = Sqrt 16 / 36 = 4 /6

= 2 / 3

9 b )

V = Pie r 2 . d = 14 , r = 7cm

1232 = 22 / 7 * 7 ^2 * h

h = 7 * 1232/ 22* 49

h = 8624 /1074

h = 8 cm

11 a)

y ^ 1 =x ^2 ( 3x +1 )^ 2

v = ( 2x + 1 )^ 2

v = m ^2

dm / dx =2

dv / dm =2 m

dy /dx = dv / dm ×dm / dx

= 2 m× 2

= 4 m

dy /dx = 4( 2x + 1)

dy /dx = udv /dx +v whole no . dy / dx

= x ^2 4( xx + 1)^ 2, × 3x

= 4 x ^2 ( 2x +1) + 2x ( 2x + 1)^ 2.

11 b)

[ 3 3 -1 ] [ 1 0 2] [ 3 – 2 3] + 2 [0 – ( – 4) – 3[ 63 ] + – 1 ( – 2)

8 + 9+ 2

= 19 .

11 c )

m= y 2- y 1 /x 2- x 1

y – y 1 = m ( x – x 2)

m= 4- 3/ – 1- 2

m= – 1/ 3

y 1 -y 2=- 1 /3 ( x – x 2)

y – 3= 1/ 3 ( x – 2)

y – 3= – 4/ 3 + 2/ 3

3 y =- x + 1[truncated by WhatsApp]

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8a)

|AD|^2=13^2-5^2

|AD|^2=169-25

|AD|^2=144

AD=sqr144

AD=12CM

|AD|=12-r

r^2=(12-r)^2 – 5^2

r^2=(12-r)(12-r)+25

r^2=144-24r+25

r^2=169-24r

r^2+24r-169=0

r^2+24r=169

r^2+24r+14^2=169+14^2

(r+14)^2=169+196

(r+14)^2=365

(r+14=sqr365

r+14=19.105

r=19.105-14

r=5.105

r=5.1cm

8aii)

circumfrenece of a circle=2pie R

C=2×22/2*(5.1)^2

C=1144.44/7

C=163.4914cm

C=163.5cm

8b)

y2-y1/x2-x1=y-y1/x-x1

6-2/2-(-1)=y-2/x-(-1)

4/2+1 = y-2/x+1

4/3=y-2/x+1

3(y-2)=4(x+1)

3y-6=4x+4

3y-4x=4+6

3y-4x=10

y=4x/3+10/3

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

9a)

let Xy represent the two digit number

x-y=5 —–(i)

3xy – (10x +y)=14

3xy – 10x – y =14 —-(ii)

from eqn (i)

x=5+y

3y(5+y)-10(5+y)-y=14

15y+3y^2 – 50 – 10y – y =14

3y^2 + 4y -50 = 14

3y^2 + 4y -50 – 14 =0

3y^2 + 4y – 64 =0

3y^2 + 12y + 16y – 64 =0

(3y^2 – 12y) (+16y – 64)=0

by

(y-4)+16(y-4)=0

(y-4)=0

9aii)

(3y+16)(y-4)=0

3y+16=0 or y-4=0

3y=-16 or y=4

y=-16/3 or y=4

when y=4

x=5+y

x=5+4

x=9

the no is 94

9b)u

3-2x/4 + 2x-3/3

=3(3-2x)+4(2x-3)/12

=9-6x+8x-12/12

=2x-2/12

=========================2 a )

number of sides =12

radius of circle =10 cm

area =?

n ×© 2= 360

12 © 2= 360

© 2 = 360 / 12 =30 °

© 1 + © 2= 180 – 30

© 1 = 150

When © 1 and © 2 are interior and exterior angle

of a polygon A sector has are .

Area of sector =© / 360 × rot 8 ^2

= 150 / 360 ×22 / 7× 100 / 1

A =130 – 95 cm^2

2 b )

1 /2 ( 2x + 1 ) – 2 /5 ( x – 2 )= 3

2 x +1/ 3 – 2 x – 4/ 3= 3

10 x +5 – 6x + 12 / 15 =3 / 1

Cross multiple

4 x +17 =45

4 x /4 =28 / 4

x = 7.

3 )

Apply 5m rule to find C P

t / sin T = P /sin P

t / sin 110 = 6/ sin 40

t =6 * 0. 9396 / 0. 6428

= 56376 / 6. 6428

= 87704

= 877 km

4 )

Total Fruit = 80 + 60 = 140

( a)

( i ) Pr one of each fruit is picked

( 79 / 140 * 60 / 139 ) + ( 59 / 140 * 80 / 139 )

= 4740 / 19, 460 + 4720 / 19 , 460

= 9460 / 19, 460 = 0.486

4 aii )

Pr one type of fruit is picked

( 79 / 140 * 78 / 139 ) + ( 39 / 140 * 5 p/ 139 )

= 6162 / 19, 460 + 3422 / 19 , 460

= 9584 / 19 , 460 = 0.492

4 b )

5 X / 8 – 1 / 6 ≤ X / 3 + 7 /24

Multiply through by 24 i : e

15 X – 4≤ 8X + 7

15 X – 8X ≤ 7 + 4

7 X = 11

X ≤ 11 / 7 ===> X ≤ 1 4 / 9

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

5 a )

3 /X + 2 – 6 /3 X – 1

3 ( 3 X – 1) – 6 ( X + 2) / ( X + 2) ( 3X – 1)

9 X – 3 – 6X – 12 /( X + 2) ( 3X – 1)

3 X – 15 / ( X + 2) ( 3X – 1)

5 b )

C .I= P [1 +r / 100 ]^

= 25000 [ 1+ 12/ 100 ]^ 3

= 25000 [ 1+ 0. 12 ]^ 3

= 25000 * 1 . 4049

= 35122 . 50

= N 35 ,122 . 50

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

6 a )

X +- 3/ 2

X =2/ 3 or X =2

( X + 3/ 2)^ 2 or ( X – 2 )

( X + 3/ 2) ( X – 2 )

X ( X – 2) + 3 / 2 ( X – 2)

X ^2 – 2X + 3 X / 2 – 3

2 X ^2 – 4X + 3X – 6

2 X ^2 – X – 6

6 b )

h /h +8 = 6 / 10

10 h = 6 h + 48

h = 12

H = h + 8

H = 12 + 8

H = 20

Volume = 1 /3 A . h

= 1 / 3 ( 10 * 10 ) * 20 – 432 / 3

= 200 / 3 – 432 /3

= 1568 / 3

= 522 . 67 cm3

7 a )

titan / 360 ×2 pie r cos t

d = 55 / 360 ×2 ×22 / 7×640 cos 4

d = 55 × 44 × 6400 cos 4/ 2520

d = 55 × 44 × 6400 ×0496 / 2520

d = 15 , 449 . 28 / 2520

d = 6130. 67

d ~ 6130 km .

ii ) distance along gent circle

D = tita / 360 ×2 pie r

D = 55 / 360 × 2/ 7× 22 /7 × 6400/ 1

D = 55 × 44 ×6400 / 2520

D = 15 , 488 ×6400 / 2520

D = 6144 .03

D =~ 6146 km .

7 b )

Length of sector tita/ 360 × 2 pie r

L= 120 / 360 ×2 /1 × 22 / 7× 42/ 1

L= 120 ×44 ×42 / 2520

L= 221760 / 2520

L= 88cm

L= 2pie r

Where r is the radius of circumference

88 =2 × 22 / 7× r

88 ×7 = 44 r

R =88 ×7 /44

R =616 / 44

R =14 cm.

Curved surface area

= pie rc

A =22 / 7 ×14 × 42 / 1

A =22 ×14 ×4 ^2 /7

A =12936 /7

A =1848 cm^ 2.

8 a )

X =60 / t — – – – – – – – – > ( i )

Y = 180 / t – – – – – — – – – > ( ii )

T 1=60 / X

T 2=100 /Y

T 1+T 2= 5

60 / X + 180 / Y = 300 – — – – – – – – – > ( i )

180 / X + 60 / Y = 260 – – – – – – – – — > ( ii )

Let P =1 / X

2 = 1 / Y

60 p + 180 Q = 300

180 p + 6Q = 200

P + 3 Q =5

9 P + 3Q = 13

Substract ( i ) from ( ii )

8 p = 8

P = 8 / 8 ÷ P = 1

Subtract P into ( i )

P + 3 Q =5

1 + 3 Q = 5

3 Q =5 – 1

3 Q =4

Q = 4/ 3

P = 1 ÷ 1 = 1/ X ÷ X = 1

4 /3 = 1/ Y ÷ Y = 3/ 4

8 b )

2001 – – – – – — – – 25 , 700

2002 – – – – – — – – 15 / 100 X 25, 700 + 25,

700 = 29 , 555

Amount of tax in 2002

= 29 , 555 * 12 .5 /100

= N 3694 .375

= N 3690

8 c )

Log 25

Log 16 25 / 100

Log 16 2/ 4

Log 4 6 – 1

– 1/ 2 Log 4^4

– 1/ 2

9 ai)

W= K +C / 2

24 = k + C / 16

384 = 16 K + C – — – – – – ( i )

18 = K + C / 4

72 + 4K + C – – – – – – – — ( ii )

16 K + C = 384

4 K + C = 72

Substact ( ii ) from ( i )

12 k / 12 = 312 /12

K = 26

Substract K into ( i )

16 k + C = 384

C = 384 – 416

C = – 32

W= k +C / t 2

W= 26- 32 / t 2

( 9aii )

When W= – 46, t =?

– 46 = 26 – 32 / t 2

t 2 = – 32 / -72

t = Sqrt 16 / 36 = 4 /6

= 2 / 3

9 b )

V = Pie r 2 . d = 14 , r = 7cm

1232 = 22 / 7 * 7 ^2 * h

h = 7 * 1232/ 22* 49

h = 8624 /1074

h = 8 cm

11 a)

y ^ 1 =x ^2 ( 3x +1 )^ 2

v = ( 2x + 1 )^ 2

v = m ^2

dm / dx =2

dv / dm =2 m

dy /dx = dv / dm ×dm / dx

= 2 m× 2

= 4 m

dy /dx = 4( 2x + 1)

dy /dx = udv /dx +v whole no . dy / dx

= x ^2 4( xx + 1)^ 2, × 3x

= 4 x ^2 ( 2x +1) + 2x ( 2x + 1)^ 2.

11 b)

[ 3 3 -1 ] [ 1 0 2] [ 3 – 2 3] + 2 [0 – ( – 4) – 3[ 63 ] + – 1 ( – 2)

8 + 9+ 2

= 19 .

11 c )

m= y 2- y 1 /x 2- x 1

y – y 1 = m ( x – x 2)

m= 4- 3/ – 1- 2

m= – 1/ 3

y 1 -y 2=- 1 /3 ( x – x 2)

y – 3= 1/ 3 ( x – 2)

y – 3= – 4/ 3 + 2/ 3

3 y =- x + 1[truncated by WhatsApp]

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8a)

|AD|^2=13^2-5^2

|AD|^2=169-25

|AD|^2=144

AD=sqr144

AD=12CM

|AD|=12-r

r^2=(12-r)^2 – 5^2

r^2=(12-r)(12-r)+25

r^2=144-24r+25

r^2=169-24r

r^2+24r-169=0

r^2+24r=169

r^2+24r+14^2=169+14^2

(r+14)^2=169+196

(r+14)^2=365

(r+14=sqr365

r+14=19.105

r=19.105-14

r=5.105

r=5.1cm

8aii)

circumfrenece of a circle=2pie R

C=2×22/2*(5.1)^2

C=1144.44/7

C=163.4914cm

C=163.5cm

8b)

y2-y1/x2-x1=y-y1/x-x1

6-2/2-(-1)=y-2/x-(-1)

4/2+1 = y-2/x+1

4/3=y-2/x+1

3(y-2)=4(x+1)

3y-6=4x+4

3y-4x=4+6

3y-4x=10

y=4x/3+10/3

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

9a)

let Xy represent the two digit number

x-y=5 —–(i)

3xy – (10x +y)=14

3xy – 10x – y =14 —-(ii)

from eqn (i)

x=5+y

3y(5+y)-10(5+y)-y=14

15y+3y^2 – 50 – 10y – y =14

3y^2 + 4y -50 = 14

3y^2 + 4y -50 – 14 =0

3y^2 + 4y – 64 =0

3y^2 + 12y + 16y – 64 =0

(3y^2 – 12y) (+16y – 64)=0

by

(y-4)+16(y-4)=0

(y-4)=0

9aii)

(3y+16)(y-4)=0

3y+16=0 or y-4=0

3y=-16 or y=4

y=-16/3 or y=4

when y=4

x=5+y

x=5+4

x=9

the no is 94

9b)u

3-2x/4 + 2x-3/3

=3(3-2x)+4(2x-3)/12

=9-6x+8x-12/12

=2x-2/12

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

[6/4, 12:33 AM] Alaba: Neco Mathematics Answers 2018:

Section II [Theory]

Answer ALL questions in Part I

1. Using ruler and a pair of compasses only:

(a) construct a triangle PQRwith /PQ/ = lOcm, ∠QPR = 90° and ∠PQR = 30°;

(b) (i) construct I, the locus of all points equidistant from PR and QR;

(ii) locate M, the point where / intersects with po,

(c) (i) with M as centre and radius MP, draw a circle;

(ii) calculate the area of the circler correct to one decimal place.

[Take Π= 22/7].

Answer any FIVE questions in Part II.

Write your answers on the answer booklet provided.

1. (a) Two children shared an amount of money in the ratio 3/4 : 2/5. If the smaller share was GH¢25.00, how much was shared between them?

(b) A box contains 5 red, 3 green and 4 blue balls of the same size. If a boy picks two balls from the box one after the other without replacement, what is the probability that both balls are red?

2. a) In a right angled triangle, sin x = 3/5. Evaluate 5(cos x)³ – 3.

(b) The angle of elevation of the top of a vertical pole from a point 63m east of the base of the pole is 30°. From another point due west of the pole, the angle of elevation of the top is 60°

(i) Draw a sketch diagram to illustrate the information.

(ii) Calculate, correct to three significant figures, the distance of the second point from the base of the pole.

3.

In the diagram above, the points A, B, C and D lie on the circle, centre O. TA and TB are tangents touching the circle at A and B respectively.

AÔB = 132°, AĈD = 59° and AOC is a straight line.

(i) Find A T B.

(ii) Find B D A.

(iii) Find B D C.

(iv) Find OB D.

4. (a) The diagonals of a rhombus are 14em and gem. Calculate, correct to the nearest centimetre, the perimeter of the rhombus.

(b) The cross section of a rectangular tank measures l.2m by 0.9m. It contains water to a depth of O.4m. If a cubical block of side 50em is lowered into the tank, calculate, correct to 2 significant figures, the rise in the water level (in metres).

5. (a) In a class of 50 students, 30 offered History, 15 offered History and Geography while 3 did not offer any of the two subjects.

(i) Represent the information on a Venn diagram.

(ii) Find the number of candidates that offered:

(A) History only;

(B) Geography only.

(b) A trader sold an article at a discount of 8% for N 828.00. If the article was initially marked to gain 25%, find the

(i) cost price of the article;

(ii) discount allowed.

6. (a) The second, fourth and sixth terms of an Arithmetic Progression (A.P.) are x – 1, x + 1 and 7 respectively. Find the:

(i)common difference;

(ii)first term;

(iii)value of x.

(b) A spherical bowl of radius r em is one-quarter full when 6 litres of water is poured into it. Calculate, correct to three significant figures, its diameter. [Take π =22/7]

7. P(lat 400N, long 18°W) and Q(lat 400N, long 78OW) are two cities on the surface of the earth. Calculate, the:

(a) radius of the parallel of latitude on which P and Q lie, correct to the nearest 10km;

(b) length of the minor arc PQ, correct to the nearest 100 km;

(c) vertical distance between the centre of the earth and the centre of the small circle on which P and Q lie, correct to the nearest km.

[Take π = 22/7 and radius of the earth= 6400 km]

8. The table below gives the distribution of marks for 360 candidates who sat for an examination.

[6/4, 9:59 AM] Alaba: 7a)

A) T3=6 & T7 =30

I)common difference using Tn=a+(n-1)d

In the 3rd term; n =3

=>T3=a+(3-1)d=6

=>a+2d=6 equation (1)

In the 7th term ;n =7

=> T7=a+(7-1)d=30

=>a + 8d=30 equation (1) & (2)

Simultaneous

A+2d =6 equation (1)

A+ 8d=30 equation (2)

0+(1-6d) =-24

=> -6d= -24

=>d = -24/-6 =4

II) first term put d=4 into equation

DIAMOND 💎 AND SURE B👌💪

[6/4, 10:22 AM] Alaba: No 4(i)

length of Arc of the sector

Titter= 72?, r = 14cm

L= titter / 360 x 2 pie r

==> L= 72/360 x 2 x 22/7 x 14

=44352/2520 = 17.6cm

(ii) perimeter of the sector

Perimeter = titter/360 x 2 pie r + 2r = 17.6 +(2x14) =17.6+28= 45.6cm

(iii) Area of the sector

Area = Titter/360 x pie r? =

72/360 x 22/7 x (14)? = 72 x 22 x 196/2520

Area= 310464/2520 = 123.2cm?

Dis one correct and images correct

[6/4, 10:22 AM] Alaba: 1a)

Log 10(20*-10)-log10(*+3)=log105

(20*-10/*+3)=log10 =5

20*-10/*+3=5

5(*+3)=20*-10

5*+15=20*-10

15+10=20*-5*

25=15*

*=25/15

*=5/3=1 2/3

1b)

Discount percent =15%

Discount amount =#600

Actual amount paid on the article =?

Original amount on the article =*

15%*=#600

15/100* =600

15*=600*100

15*=60000

*=60000/15

*=#4,000

Therefore actual amount paid on the article

=#4,000-#600

=#3,400

Actual amount paid on the article =#3,400

[6/4, 10:22 AM] Alaba: : 7a)

A) T3=6 & T7 =30

I)common difference using Tn=a+(n-1)d

In the 3rd term; n =3

=>T3=a+(3-1)d=6

=>a+2d=6 equation (1)

In the 7th term ;n =7

=> T7=a+(7-1)d=30

=>a + 8d=30 equation (1) & (2)

Simultaneous

A+2d =6 equation (1)

A+ 8d=30 equation (2)

0+(1-6d) =-24

=> -6d= -24

=>d = -24/-6 =4

II) first term put d=4 into equation

[1b

Discount percent =15%

Discount amount =#600

Actual amount paid on the article =?

Original amount on the article =*

15%*=#600

15/100* =600

15*=600*100

15*=60000

*=60000/15

*=#4,000

Therefore actual amount paid on the article

=#4,000-#600

=#3,400

Actual amount paid on the article =#3,400

[6/4, 09:45] london boy: 1a)

Log 10(20*-10)-log10(*+3)=log105

(20*-10/*+3)=log10 =5

20*-10/*+3=5

5(*+3)=20*-10

5*+15=20*-10

15+10=20*-5*

25=15*

*=25/15

*=5/3=1 2/3

No 4(i)

length of Arc of the sector

Titter= 72?, r = 14cm

L= titter / 360 x 2 pie r

==> L= 72/360 x 2 x 22/7 x 14

=44352/2520 = 17.6cm

(ii) perimeter of the sector

Perimeter = titter/360 x 2 pie r + 2r = 17.6 +(2x14) =17.6+28= 45.6cm

(iii) Area of the sector

Area = Titter/360 x pie r? =

72/360 x 22/7 x (14)? = 72 x 22 x 196/2520

Area= 310464/2520 = 123.2cm?

[6/4, 10:49 AM] Alaba: 1CDAAEABAEC

11ACDDCDDDAC

21CEADEDCBBC

31CBEEECBDC-

41DBCBCBDBA

51 BCBDCDB

Still solving ✍🏻✍🏻

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