(1a)
Given A={2,4,6,8,…}
B={3,6,9,12,…}
C={1,2,3,6}
U= {1,2,3,4,5,6,7,8,9,10}
A’ = {1,3,5,7,9}
B’ = {1,2,4,5,7,8,10}
C’ = {4,5,7,8,9,10}
A’nB’nC’ = {5, 7}
(1b)
Cost of each premiere ticket = $18.50
At bulk purchase, cost of each = $80.00/50 = $16.00
Amount saved = $18.50 – $16.00
=$2.50
: 2ai).
p = (rk/Q -ms)2/3
p3/2 = (rk/Q -ms)2/8 * 3/2
p3/2 = rk/Q = ms
p3/2 + ms = rk/Q
q(p3/2 + ms ) = rk
Q =rk/p3/2 + ms
2aii)
Where :
r = 10, k = 4, s = 0.2
p = 3, m = 15
Q = 10 * 4/3^3/2 + (15 * 0.2)
Q = 40/5.2 + 3
Q = 40/8.2
Q = 4.88
. 🕵: 5a)
Using elimination method
2x: 2p-4q=4
3x: 4p+2q=9
Hence
4p-6q=8
9p+6 =27
Adding equation (i) to (ii)
Therefore
13p=35
Thus p=35/13
Substituting P=35/13 in equation (ii)
Hence
3p+2q=9……….. (ii)
3(35/13) + 2q= 9
=105/13+2q=9
Multiplying throughout by 13
~13~ *105/ ~13~ + ~13~*2q =13*9
Thus 105+26q=117
Collecting like terms
26q =117-105
26q=12
Q=12/26 or 6/13
*2bi* 2mn^2x/(3m-n)
But m= 3 n = -2 x= -1
= 2*3*(-2)^2*(-1)/3(3)-(-2)
= 2*3*4*-1/9+2
= -24/11
=2. 2/11
*2a.)*
P= (rk/Q – ms)⅔
Take the reciprocal of ⅔ by the power
P⅔= rk/Q – ms
rk/Q= P⅔ + ms
Q = rk/ P⅔+ ms
ii.) When P=3, m=15, s=0.2, k=4 and r=10
Q= rk/ P⅔ + ms
Q= 10(4)/(3)⅔ + (15)(0.2)
=40/5.196+3
=40/8.196
=4.88 (1dp)
(3a)
Diagram
CBD = CDB (base angles an scales D)
BCD+CBD+CDB=180° (Sum of < in a D)
2CDB+BCD=180°
2CDB+108°=180°
2CDB=180°-108°=72°
CDB=72/2=36°
BDE=90°(Angle in semi circle)
CDE=CDB+BDE
=36°+90
=126
(3b)
(Cosx)² – Sinx given
(Sinx)² + Cosx
Using Pythagoras theory thrid side of triangle
y²= 1²+√3
y²= 1+ 3=4
y=√4=2
(Cosx)² – sinx/(sinx)² + cosx
(1/2)² – √3/2/
(√3/2)² + 1/2 = 1/4 – √3/2 = 1-2√3/4
3/4+1/2 = 3+2/4
=1-2√3/4 * 4/5
=1-2√3/5
4a) Total Surface Area = 224πcm²
r:l = 2:5
r/l = 2/5
Cross multiply
2l/2 = 5r/2
L = 5r / 2
Total surface = πrl + πr²
= πr (l + r)
24π/π = πr (5r/2 + r )/ π
224 = 5r²/2 + r²/1
L.c.m = 2
448 = 5r² + 2r²
448 / 7= 7r²/7
r² = 64
r = √64 = 8cn
L = 5*8/2 = 20cm
4b)
Volume = 1/2πr²h
= 1/3 * 22/7 * 8 * 8 * 18.33
= 1228.98cm³
L² = h² + r ²
20² = h² + 8²
400 – 64 = h²
h² = 336
h = √ 336
h = 18.33cm
(5a)
Total income = 32+m+25+40+28+45
=170+m
PR(²)=m/170+m = 0.15/1
M=0.15(170+m)
M=25.5+0.15m
0.85m/0.85=25.5/0.85
M=30
(5b)
Total outcome = 170 + 30 = 200
(5c)
PR(even numbers) = 30+40+50/200
=115/200 = 23/40
(7a) Click here to view image
Using Pythagoras theorem, l²=48² + 14²
l²=2304 + 196
l²=2500
l=√2500
l=50m
Area of Cone(Curved) =πrl
Area of hemisphere=2πr²
Total area of structure =πrl + 2πr²
=πr(l + 2r)
=22/7 * 14 [50 + 2(14)]
=22/7 * 14 * 78
=3432cm²
~3430cm² (3 S.F)
(b) let the percentage of Musa be x
Let the percentage of sesay be y
x + y=100 ——————-1
(x – 5)=2(y – 5)
x – 5=2y – 10
x – 2y=-5 ——————-2
Equ (1) minus equ (2)
y – (-2y)=100 – (-5)
3y=105
y=105/3
y=35
Sesay’s present age is 35years
8a) Cost price = Ghd
150
Let the number of plates
be X profit per one = 1
GHd
Total profit = 6 GHd
Remaining plates = x – 4
Total profit = (x – 4)
1Ghd
6 = x – 4
X = 6 + 4
X = 10
The number of plates is
10
(8a)
Let Ms Maureen’s Income = Nx
1/4x = shopping mall
1/3x = at an open market
Hence shopping mall and open market = 1/4x + 1/3x
= 3x + 4x/12 = 7/12x
Hence the remaining amount
= X-7/12x = 12x-7x/12 =5x/12
Then 2/5(5x/12) = mechanic workshop
= 2x/12 = x/6
Amount left = N225,000
Total expenses
= 7/12x + X/6 + 225000
= Nx
7x+2x+2,700,000/12 =Nx
9x + 2,700,000 = 12x
2,700,000 = 12x – 9x
2,700,000/3 = 3x/3
X = N900,000
(ii) Amount spent on open market = 1/3X
= 1/3 × 900,000
= N300,000
(8b)
T3 = a + 2d = 4m – 2n
T9 = a + 8d = 2m – 8n
-6d = 4m – 2m – 2n + 8n
-6d = 2m + 6n
-6d/-6 = 2m+6n/-6
d = -m/3 – n
d = -1/3m – n
(9a)
Draw the triangle
(9b)
(i)Using cosine formulae
q² = x² + y² – 2xycosQ
q² = 9² + 5² – 2×9×5cos90°
q² = 81 + 25 – 90 × 0
q² = 106
q = square root 106
q = 10.30 = 10km/h
Distance = 10 × 2 = 20km
(ii)
Using sine formula
y/sin Y = q/sin Q
5/sin Y = 10.30/sin 90°
Sin Y = 5 × sin90°/10.30
Sin Y = 5 × 1/10.30
Sin Y = 0.4854
Y = sin‐¹(0.4854), Y = 29.04
Bearing of cyclist X from y
= 90° + 19.96°
= 109.96° = 110°
(9c)
Speed = 20/4, average speed = 5km/h
11b) 8y = – 4x + 24
Y = – 4/8 x + 24/8
Y = – 1/2 x + 3
Gradient = m = – 1/2
:. Y- y1 = m(x-x)
Y- 12 = 1/2 (x + 8)
2(y-12) = 1 (x+8)
2y – 24 = -x-8
X + 2y – 24 + 8 = 0
X + 2y – 24 + 8 = 0
X + 2y – 16 = 0
Is the line above
=====================
✍️✍️✍️✍️✍️✍️✍️
[8/17, 12:58 PM] Aro Okeigbo Blog: 11b) 8y = – 4x + 24
Y = – 4/8 x + 24/8
Y = – 1/2 x + 3
Gradient = m = – 1/2
:. Y- y1 = m(x-x)
Y- 12 = 1/2 (x + 8)
2(y-12) = 1 (x+8)
2y – 24 = -x-8
X + 2y – 24 + 8 = 0
X + 2y – 24 + 8 = 0
X + 2y – 16 = 0
Is the line above
=====================
MATHS-OBJ
1CBCDACDCCD
11AADBDACBBC
21BDDABDADAD
31CDACCCCCDA
41BBBCDC
==================
Share On:
CLICK HERE TO PROMOTE YOUR MUSIC/VIDEO