(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

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

(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)
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

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

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

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

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

(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

: (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

08133702199 OJCA🔥*

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

I h = 18.33cm

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