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Mathematics objective
1C 2A 3B 4B 5A 6D 7D 8A 9C 10B 11D12C13B
14A 15C 16D 17A 18A 19C 20A 21A 22B 23C 24D
25D 26C 37 B 28C 29A 30B 31C 32A 33D 34D
35A 36B 37B 38A 39C 40B 41C 42A 43C 44A 45B
46A 47C 48C 49A 50D

10a)
GIVEN: y = 2X^2- 3X – 1
When x = -3
Y= 2(-3)^2- 3(-3) – 1 = 18 + 9 – 1 = 26
When x= -1
Y = 2(-1)^2 – 3(-1)-1 = 2+3-1 = 4
When x = 1
Y = 2(-1)^2 – 3(1)-1 = 2 – 3 – 1 = -2
When x = 2
Y=2(2)^2-3(2)-1= 8 – 6 – 1 =1
When x = 3
Y =2(3)3-3(3)-1 = 18-9-1 = 8
When x = 4
Y = 2(4)^3 – 3(4)-1 = 32-12-1 =19
1a )
6whole 1 / 2 – 3 whole 2/ 5 / 2whole 1 / 2 – 1
whole 3 / 5
25 / 4- 17 / 5 , / 5 / 2- 8/ 5
125 – 68 / , 20 , / 25 – 16,
/ 10
==> 57 / 20 / 9/ 10
=57 / 20 /9 / 10
=57 / 20 x 10 / 9
=57 / 20 x 10 / 9
=57 / 18
= 3whole 1 / 6
1b)
let the number of students be X therefore
total age of students be 15x
15x + 45/x + 1 = 18
15x + 45 = 18 (x + 1)
15x + 45 = 18x + 18
18x – 15x = 45 – 18
3x = 27
X = 27/3
X = 9 students
=======
3a )
c = – 1, y = -3 , z= – 4 and w = -7
x ^2 – y^ 2 / 2 w – z
(-1 )^3 – (-3 )^2 / 2(- 7) -(- 4)
= -1 -9 /+ 4+ 4 = – 10/ – 10
= 1
3b )
< MNQ = 90 degree
< MQN + < NQO = 90 degree
< MNQ + 46 = 90 degree
< MNQ = 90 – 46 degree
< MNQ = 44 degree
Considering < MNQ
y + n + MQN = 180 degree
y + 44 + 44 = 180
y = 180 – 88
Y = 92 degree

4a )
2/ 3(1 -4 x) -1 /2 (5- 3x ) less /equal to 1/ 4(7 +
9x )- 1/ 3
multiply through by 12
8(1 -4 x) -6 (5- 3x ) less /equal to 3(7 + 9x )- 4
8- 32 x- 30 + 18 x less/ equal to 21 + 27 x -4
-32 x + 18 x – 27 x less /equal to 21 -4 + 30 -8
-41 x less /equal to 39
x less/ equal to -39 / 41
4b )
DRAW THE ANGLE
from DTMR /
Tan 65 degree = TR /3
TR = 3Tan 65
= 3 x 22. 1445
From DRMB
Tan 20 = RB / 3
RB = 3Tan 20
= 3 x 0. 3640
= 1. 092 m
H = TR + RB
= 6. 4335 + 1 . 092
= 7. 5255
= 7. 53 m
=========
5)
Area of shed segment = Area of Sector – Area of
Triangle
=Φ / 360 *πr ^2 – (½ abSinc )
=$ 90/ 360 *22 / 7*7 ^2) – ½* 7 *7*sin90 `
=154 / 4 – 42 /2
=(754 – 98 )/4
=56 / 4
=14 cm ^2
Cost of painting it = 14 *750
=N 10500
========
6a )
Taxable income = x
25 /100 x x /1
= 14, 000
25 x= 1400000
X =1400000 / 25
X = 560000
Total income =
56 ,000 / 4 x 5
=70 ,000
6b )
Education = 2 /5
Clothes = 1 /6
Food = 3/ 8
Expenditure = 2 /5 + 1/ 6 + 3 /8
48 + 20 + 45 / 120
= 113 /120 * 36, 000 / 1
Le 39 , 000
Savings annually = Le 21 ,000
To save Le 63 ,000 . 00
To save Le 63 ,000 / 2100 yrs
= 30 years
=================================
8 a )
Drawing
8 bi)
Using Pythagoras theory
| xz| ² =550² 320²
= 302500 102400 =404900
xz=√ 404999
=636 km( 3 s.f )
Total distance = 320 550 636
= 1506km
8 bii )
Using SOHCAHTOA
Tan z = 320 /550
Z = Tan – ¹ ( 320 /550 )
Z = Tan – ¹( 0 . 5818)
Z = 30 °
From the diagram in 8 a
Bearing of X from Z = 55 30
= 085° or N85 °E
9a)
3log10^2 – 2log10^3 = 1 + log(1/x)
log10^(2)^3 – log10^(3)^2 – log10^(1/x) = 1
Log10^( 8 /9 ) / (1 / x) = log10^10
8 / 9 * X / 1 = 10
8x = 9 *10
8x = 90
X = 90/8
= 11.25
9b.
Distance= speed * time
Distance= 3km/h * 3mins
= 3000m/60min * 3min
=50*3 =150min.
9bi.
Circumference= 150m
2π(r+1) = 150
22/1 * 22/7 * (r+1)= 150
r+1= 150*7/44
=1050/44
r+1 = 23.9 ≈ 24 to the nearest whole number.
r+1= 24
r=24-1 = 23min
9bii.
Volume of cylinder=πr²h
Volume= 22/7 * (23)² * 8
22/7 * 529/1 * 8/1
= 93104/7 = 13,300.57
13301m³ to the nearest whole number
(7a)
diagram
(7b)
Angula difference in long(tita)=42-12
tita=30 degree
(7bi)
lenght of chord Xy=2Rsin tita/2
XY=2*6400*Sin 30/2
=2*6400*sin15 degree
=12800*0.2588
=3.312.64km
=3312.64km
=3310km(to the nearest 10km)
(7bii)
let the angle that the chord xy substends at the
centre of the earth be alpha degree
diagram
sin alpha/2= opp/hyp=/NY/ /6400
/NY/= 1/2 /XY/= 1/2 * 3312.64
=1656.32km
sin alpha/2= 1656.32/6400
sin alpha/2= 0.25888
alpha/2= sin^-1(0.2588)
alpha/2=14.999
alpha=14.999 *2
alpha= 29.998
alpha= 30.0 degree (to 1 dp)
(7biii)
XY bar= tita/360 * 2pie R cos lat
=30 degeree/360 * 2*3.142 * 6400* cos60
XY bar= 30/360 * 2* 3.142* 6400* 0.5
XY bar=1675.73 km
2a)
m/m-y+2 = r/y+r-1
r(m-y+2)=m(y+r-1)
rm-ry+2r = my+mr-m
rm+2r-m+mr=my+ry
y(m+r)=rm+2r-m+mr
y=2mr+2r-m/m+r
2b)
p[one of them owns a bicycle]
[60/100+70/100] – [40/100 + 30/100]
Because % of boys that don’t have bicycle =
40/100
% of girls without bicycle = 30/100
=130/100 – 70/100
=60/100
=0.6