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Maths-Obj
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11ACBACCAADA
21DAAAABCDCB
31ABCDABCBCA
41DCABCCABBB

(1a)
S = ½n[a + L]
Where S = 130
S = n/2[a+a+(n-1)d]
2S = n[2a+(n-1)d]
2(130) = 10[2a + (10 - 1)d]
260 = 10[2a + 9d]
260 = 20a + 90d .....(I)
a + 4d = 3a...... (II)
4d = 3a - a
4d = 2a
2d = a...... (III)
Substitute a = 2d into Eqn (I)
260 = 20(2d) + 90d
260 = 40d + 90d
260 = 130d
260/130 = 130d/130
d = 2
Hence common difference d = 2

(1b)
First term a = 2d = 2(2) = 4

(1c)
L = a+(n-1)d
28 = 4+(ń - 1)2
28 = 4+2n - 2
2n = 28 - 2
2n = 26
2n/2 = 26/2
n = 13

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(4a)
(2y+x) + (6y-2x+1) + 4y = 28 ...(i)
6y-2x + 1 = 4y... (II)
2y+ x +6y - 2x + 1 +4 = 28
12y + x +6 -2x + 1 + 4 = 28
12y - x + 1 = 28
12y - x = 29... (III)
6y-2x + 1 = 4y, 6y-2x - 4y = 1
2y - 2x = -1... (iv)
24y - 2x = 54
2y - 2x = 1

22y/2w = 55/22 y = 2.5
12y - x = 27
12 (2.5) - x = 27
30 x 27 x=3

(b)
2y + x = 2 (2.5) + 3 = 5+3 = 8cm
6y - 2x + 1 = (2.5)-2 (3) + 1
= 15-6 + 1 = 10cm
4y = 4(2.5) = 10cm

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(5a)
5 - X > 1 9 + X >_ 8
5 - 1 > X X >_ 8 - 9
6 > X X >_ -1
Range is
-1_< X < 6 OR 6 >X >_ -1

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(5b)
PQR + PSR = 180(supplementary angles of a cyclic quad)
PQR + 56 = 180
PQR = 180 - 56
PQR = 124°
Next, join P to R
QRP = QPR(base angles of an isosceles)
PQR + 2QRP = 180(Sum of angles in a triangle)
124 + 2QRP = 180
2QRP = 56°
QRP = 56/2 = 28°
PRS = 90°(angle in a semi - circle)
= 28 + 90
= 118°



And 9-x>or equals to 8

9-8>or equals to X
1>or equals to x

Therefore 4>x and 1>or equals to x
Final answer.
4>x<or equals to 1

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(6ai)
The profit y = X²/8 + 5x
y = GHc20,000.00
Hence 20,000 = X²/8 + 5x
160,000 = X² + 40x
X² + 40x - 160,000 = 0
Since X is in thousands
X² + 40x - 160 = 0

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(6aii)
Using quadratic formula
X = -b±√b² - 4ac/2a
Where; a = 1, b = 40 & c = -160
X = -40±√40² - 4(1)(-160)/2(1)
X= -40 ±√1600 + 640/2
X = -40 ±√2240/2
X = -40 ± 47.32/2
X = -40±47.32/2
= 7.32/2
X = 3.66
X ≈ 4

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(6b)
Draw the diagram
Using ΔTOP
tan 28 = H/OP
OP = H/tan28

Then for ΔROP
tantita = H/2/OP
OP = H/2/tantita
Hence H/tan 28 = H/2/tantita
tantita = H/2 × tan 28/H
tantita = tan28/2
Hence Tita = 28/2 = 14.
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(
(I) Draw The Diagram

(II)
V = 1/3 Ah, = x r²
V = 1/3 xr²h
V = 4.158 liters
V = 4.158cm³, V = 1/3 xr²h
4158 = 1/3 x 22/7 x 21 x 21 x h
4158 x 3 = 22 x 63h
h = 4158/21 x 22 = 9cm
h = 9cm

(8b)
d = 28cm, r = d/2 = 28/2 = 14cm
V2 = 1/3 xr²h
V2 = 1/3 x 22/7 x 14 x 14 x 9
V2 = 1/2 x 22/1 x 2 X 14 x 3
= 1848cm³
V2 = 1.845 liters

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10a )
area of the farmland = 7200 m ^ 2
length X breath = 7200 m ^ 2.. . ( 1 )
perimeter = 360 m
2length +2breath = 360m . .. ( 2)
LXB = 7200 .. . ( 1 )
2L + 2B = 360 . .. ( 2)
i ) solving eqn ( 2) and ( 1)
L = 7200 / B .. .( 3)
put eqn ( 3) into ( 2)
2( 7200 / B ) +2B = 360
14400 / B + 2B / 1= 360
14400 +2B ^ 2/ B = 360
14400 +2B ^ 2= 360 B
2B ^ 2-360 B + 14400 = 0
B = 120 or 60
the maximum value is 120
hence B = 120 m
L = 7200 / B
L = 7200 / 120
L = 60 M
: . the length is 60m and the breath is 120m
or
the breath is 60 m and the length is 120m