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Maths-obj
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21BCADBCCAAB
31CDCACDBACB
41BDABCDDCAD
SECTION A ANS ALL
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1 a )
( y - 1 ) log 4 ^ 10 = ylog 16 ^ 10
log 4 ^ 10 ( y - 1 )= log 16 ^ y 10
4 ^ ( y - 1 )= 16 y
4 ^ y - 1 = 4 ^ 2 y
y - 1 = 2 y
- 1 = 2 y = y
- 1 = y
y = - y
1 b )
let the actual time for 5 km /hr be t
for 4 km /hr = 30 mint + t
4 km /hr = 0 . 5 + t
distance = 4 ( 0 . 5 + t)
= 2 * 4 t
for 5 km /hr , time = t
distance = 5 t
1 + 4 t= 5 t
t= 2 hrs
actual distance = 5 * 2 = 10 km
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2 a )
2 /3 ( 3 x - 5 ) - 3 /5 ( 2 x - 3 )= 3 /1
L C M = 15
10 ( 3 x - 5 ) - 9 ( 2 x - 3 )= 45
30 x - 50 - 18 x + 27 = 45
30 x - 18 x = 45 + 50 - 27
12 x - 23 = 45
12 x = 45 + 23
12 x = 68
x = 68 /12
x = 34 /6
x = 17 /3
2 b )
U ' aS = 180 - ( n + 88 )
= 180 - n - 88 = 92 - n
also , u ' TQ = 18 m
80 degree + 92 - n + 180 - m= 180 degree
80 + 92 + 180 - n - m= 180 degree
352 - n - m= 180 degree
- n - m= 180 - 352
- n - m= - 172
+ ( n + m)= + 172
m+ n = 172 dgree
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3 a )
Tan 23 . 6 ° = h /50
Cross multiply
Tan 23 . 6 ° x h / 50
h = 50 tan 23 . 6 °
= 21 . 844 m
3 b )
Area of < TRU = 45 cm ^ 2 ( Note: This ^ means Raise to power )
A = 1 /2 bh
45 = 1 /2 x 10 x h
45 = 5 h
h = 9 cm
Area of < QTUS = 1 /2 ( QT + US ) h
= 1 /2 ( 6 + 16 ) 9
= 99 cm^ 2
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4 a )
T6 = 37
T6 = a + ( 6 - 1 ) d
T6 = a + 5 d
a + 5 d = 37 - - - - - ( eq 1 )
s 6 = 147
sn= n /2 ( 2 a + ( n - 1 ) d )
147 = 3 ( 2 a + 5 d )
49 = 2 a + 5 d
2 a + 5 d = 49 - - - - ( eq 2 )
a + 5 d = 37 - - - ( eq 1 )
2 a + 5 d = 49 - - - ( eq 2 )
a = 12
4 b )
S15 = 15 / 2 ( 2 ( 12 ) + 14 d )
S15 = 15 / 2 ( 24 + 14 d )
from ( 1 )
a + 5 d = 37
12 + 5 d = 37
5 d = 37 - 12
5 d = 25
d = 5
S15 = 15 /2 ( 24 + 14 ( 15 )
S15 = 15 /2 ( 24 + 70 )
S15 = 15 / 2 * 94
S15 = 15 * 42
S15 = 630
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5 a )
draw
U = 20
B = y - 45
S= y - 34
B = bag
S= shoe
let n ( B )= y
n ( S )= y + 11
for bag only y - 45
for shoe only y - 11 - 45 = y - 34
5 b )
y - 45 + 45 + y - 34 = 120
2 y - 34 = 120
2 y = 154
y = 154 / 2
y = 77
number of customers who bought shoe = y + 11
77 + 11 = 88
5 c )
n ( bag )= 77 customers
probability = 77 /120
= 0 . 642
SECTION B ANS 5 QUESTIONS ONLY
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10 a )
Sin x = 5 /13
Using pythagoras rule
M^ 2 = 13 ^ 2 - 5 ^ 2 (^ means Raise to power )
M^ 2 = 169 - 25
M ^ 2 = 144
M = √144
M = 12
Hence :
Cos x - 2 sin x / 2 tan x
12 /13 - 2 ( 5 /13 ) / 2 ( 5 /12 )
= 12 /13 - 10 /23 / 5 /6
FIND LCM
= 12 - 10 /13 / 5 /6
= 12 /65
10 bi)
Considering < LMB
/MB /^ 2 . = 12 ^ 2 - 9 . 6 ^ 2
/MB /^ 2 = 51 . 84
/MB / = √ 51 . 84
/MB / = 7 . 2 m
From < AML
/LA /^ 2 = 2 . 8 ^ 2 + 9 . 6 ^ 2
/LA / ^ 2 = 100
/LA / = √100
/LA / = 10 m
10 bii )
Let the angle be. θ
From < AML
Tanθ = 9 . 6 /2 . 8
Tan θ = 3 . 4288
θ = Tan ^ - 1 ( 3 . 4288 )
= 73 . 74 °
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13 ai)
given
x (*)y = x + y /2
i ) 3 (*) 2 / 5 = 3 + 2 /5 /2
= ( 15 + 2 /5 )* 1 /2
= 17 /5 * 1 /2
= 17 /10 = 1 ,7 /10
13 aii )
8 (*) y = 8 ^ 1 /4
= 8 + y / 2 = 33 /4
32 + 4 y = 66
4 y = 66 - 32
4 y = 34
y = 34 /4
y = 17 /2
y = 8 ^ 1 / 2
13 b )
given DABC
AB= (^ - 4 /6 ) and AC = ( 3 /^ - 8 )
so AP = 1 / 2 (^ - 4 /6 )
AP= (^ - 2 /3 )
hence
CP = CA + AP
CP= - ( 3 / ^ 8 ) + (^ - 2 / 3 )
CP = (^ - 5 / 11 )
12a)
3y^2-5y+2=0
y^2 - 5/3y + 2/3=0
y^2-5/3y=-2/3
y^2-5/3y+(^-5/6)^2=(-^5/6)^2-2/3
(y-5/6)^2=25/36-2/3
(y-5/6)^2=25/-24/36
(y-5/6)^2=1/36
(y-5/6)=+sqr1/36
y=5/6+1/6
y=5+1/6 or 5-1/6
y=6/6 or 2/3
y=1 or 2/3
12b)
given
M N = [2,3 1,4]
hence
[1,4 2,3] * [m,n x,y] =[2,3 1,4]
[m+2n, x*2y]
[4m+3n, 4x+5y] = [2,3, 1,4]
therefore
m+2n=2------(i)
4m+3n=3------(ii)
from ------(i)
m=2-2n
4(2-2n)+3n=3
8-8n+3n=3
8-5n=3
8-3=5n
5=5n
n=1
hence
m=2-2(1)
M=0
also
x+2y=1------(i)
4x+3y=4------(ii)
from ------(iii)
x=1-2y
4(1-2y)+3y=4
4-8y+3y=4
y=0
therefore x=1-2(0)
x=1
this N=[i i]
11 a )
8 students finished
12 tanks in 2 /3 ( 60 ) mins
= 40 mins
4 student wil finish
X tanks in 1 /3 ( 60 ) min
= 20 mins
X = 4 x 20 x 12 /8 x 40
= 3 tanks
11 b )
L ( AB ) = 200 m | ON | = 12 cm
r 2 = ( AN ) 2 + ( ON ) 2
r 2 = ( 10 ) 2 + ( 12 ) 2
r 2 = 100 + 144
r 2 = 244
r = Sqr 244
r = 15 . 6 CM
11 bii )
L ( AB ) = 2 r sin 0 /2
20 = 2 ( 15 . 6 ) sin 0 /2
20 = 31 . 2 sin 0 / 2
sin 0 /2 = 20 /31 . 2
sin 0 /2 = 0 . 6410
0 /2 = sin - 1 ( 0 . 6410 )
0 /2 = 39 . 87
0 = 2 ( 39 . 87 )
0 = 79 . 74
= 79 . 7 ' ( 1 d . p )
11 bii )
p 2 r + 0 /360 x 2 TTr
= 2 ( 15 . 6 ) + 79 . 7 /360 x 2 x 3 x 42 x 15 . 6
= 31 . 2 + 21 . 7
= 52 . 9 cm
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Completed
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