Contents

## NECO MATHEMATICS FREE EXPO 2017.

### MATHEMATICS OBJ ANSWERS.

1 **ecadddceee** 10

11 **ccdcaaceca** 20

21 **dcdeeccded** 30

31 **eaadbeeadc** 40

41 **bbdccdcecb** 50

51 **cadbaecdec** 60

### MATHEMATICS THEORY ANSWERS.

1(a)

Tabulate

x – 1 ,2 ,3 ,4

1 – 1 ,2 ,3 ,4

2 – 2 _ , 4 , 0 _ ,2 _

3- 3 , 0_ , 3, 0 _

4 – 4 _ , 2_ , 0_ , 4

1b)

I = PRT/100, p=N15000 R=10% and I=3years

A = P+ I

where I = 15000*10*3/100=N4500

A=4500+15000 =N19500

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

2a)

using sine rule

b/sin20 = 6/sin30

bsin30 = 6sin120

b 6sin120/sin30

b = 6×0.2511/0.4540

b = 5.7063/0.4540

b = 12.57 ≠ 12.6cm

2bi)

the diagram is euivalent triangles.

where

|AX|/|BC| = |BY|/|AC| = |XY|/|YC|

XY = 9, BY = 7

YC = 18-7=11

9/11 = 7/|AC|

9|AC| = 77

|AC| = 77/9

|AC| = 8cm

2bii)

XY/AB = BY/AC

9/|AB| = 7/8.6

|AB| = 9×8.6/7

|AB| = 11cm

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

2a)

using sine rule

b/sin20 = 6/sin30

bsin30 = 6sin120

b 6sin120/sin30

b = 6×0.2511/0.4540

b = 5.7063/0.4540

b = 12.57 ≠ 12.6cm

2bi)

the diagram is euivalent triangles.

where

|AX|/|BC| = |BY|/|AC| = |XY|/|YC|

XY = 9, BY = 7

YC = 18-7=11

9/11 = 7/|AC|

9|AC| = 77

|AC| = 77/9

|AC| = 8cm

2bii)

XY/AB = BY/AC

9/|AB| = 7/8.6

|AB| = 9×8.6/7

|AB| = 11cm

3a)

let the son age be x

man = 5 x

son= x

4 yrs ago; the man age = 5 x – 4

the son age = x – 4

the product of their ages

(5 x – 4 )( x – 4 ) = 448

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

4 a)

volume of fuel = cross -sectional area of X depth of fuel

rectangular

tank

30, 000litres = 7 . 5* 4 .2 * d m ^ 3

but ; 1000litres = 1 m ^ 3

therefore ; 30( M ^3 ) = 7 .5 *4 . 2* d ( M ^ 3)

30= 31.5 d

==== d = 30/31. 5 = 0 .95( 2 d .p )

4 b)

to fill the tank /volume of fuel needed

= 7 .5 * 4. 2 *1 .2

= 37.8 m ^ 3

= 37, 800 litres

addition fuel = 37, 800- 30, 000

= 7 , 800 litres

therefore , 7 , 800 more litres would be needed

5 a)

sector for building project = 48000 /144000* 360 = 120degree

sector for education = 32, 000 /144000* 360 =80degree

sector for saving = 19200 /144000* 360 = 48degree

sector for maintenance = 12000 / 144000*360 = 30degree

sector for miscellaneous = 7200/144000* 360 =18degree

sector for food items = 360 -( 120 + 80+ 48+30+ 18)

= 360- 296

= 64degree

5 b)

amount spent =144000- [48, 000 + 32000+ 19200 + 12000

+7200]

= 144000-118400

= N 25600

6)

/————————— —

/ 41.02 × √0.7124

/ ———— ———— — —

√42.87 × 0.207 × 0.0404

| No | Log |

| 41.02 | 1.6130 | 1.6130

| 0.7124 | T. 8527 ÷ 2 | T.9263/1.5393

| 42.77 | 1.6321 |

| 0.207 | T.3160 |

| 0.0404 | 2. 6064 / |

| | T.5544 | T.5544/1.9849

Antilog = 9658 ≈ 96.58

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

7 a)

3^ 2 n + 1 – 4 ( 3 ^ n + 1 )+ 9 = 0

3^ 2 -3 – 4 ( 3^ n – 3) + 9 = 0

(3 ^ n )^2- 3 – 4 (3 ^ n -3 )+ 9 = 0

let 3^ n = p

p ^ 2 -3 – 4 (p -3 ) + 9 = 0

3p ^ 2 /3 – 12 p /3 + 9 / 3 = 0

p ^ 2 – 4 p + 3 = 0

p ^ 2 – 3 p – p + 3 = 0

p ^ 2 p ( p – 3 ) – 1 (p – 3) = 0

(p – 1 ) (p – 3) = 0

p -1 = 0 or p – 3 = 0

p = 1 or 3

Recall 3 ^ n = p

when p = 1

3^ n = 3^ 0

n = 0

when p = 3

3^ n = 3^ 1

n = 1

7 b )

log (x ^ 2 + 4 ) = 2 + logx – log ^ 20

log (x ^ 2 + 4 ) = log ^ 100 = log ^ x – log ^ 20

(x ^ 2 + 4 ) = log ( xx )

x ^ 2 + 4 = 5 x

x ^ 2 -5 x + 4 = 0

x ^ 2 -4 x – x + 4 = 0

x (x -4 ) – 1 ( x -4 ) = 0

(x -1 ) (x -4 ) = 0

x -1 = 0 or x -4 = 0

x = 1 or 4

8 a)

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

| AD | ^ 2 = 169- 25

| AD | ^ 2 = 144

AD =sqr 144

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=sqr 365

r + 14=19. 105

r = 19.105 -14

r = 5. 105

r = 5. 1 cm

8 aii )

circumfrenece of a circle = 2 pie R

C = 2 x22/ 2* ( 5 .1 )^2

C = 1144.44/ 7

C = 163 .4914cm

C = 163 .5 cm

8 b)

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 )

3 y-6 = 4 x+ 4

3 y-4 x= 4 + 6

3 y-4 x= 10

y= 4 x/3 + 10/ 3

9 a)

let Xy represent the two digit number

x-y =5 — — ( i )

3 xy – ( 10x + y)=14

3 xy – 10x – y = 14 – — -( ii )

from eqn ( i )

x= 5 +y

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

15y+ 3 y^2 – 50 – 10y – y =14

3 y^2 + 4 y – 50 = 14

3 y^2 + 4 y – 50 – 14 = 0

3 y^2 + 4 y – 64 = 0

3 y^2 + 12y + 16y – 64 = 0

( 3 y^2 – 12y ) ( + 16y – 64)=0

by

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

( y-4 )=0

ii )

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

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

3 y= -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

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