VIKTUZ

Tuzer 1 a)
6 whole 1 / 2 - 3 whole 2 / 5 / 2 whole 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
= 3 whole 1 / 6
1 b)
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
3 x = 27
X = 27/3
X = 9 students
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2 a)
m / m -y+ 2 = r /y+ r - 1
r ( m -y +2 )=m ( y+ r -1 )
rm- ry + 2 r = my +mr -m
rm+ 2 r -m +mr = my + ry
y( m + r )=rm + 2r - m + mr
y= 2 mr+ 2 r -m / m + r
2 b)
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
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3 a)
c= - 1, y= -3 , z = -4 and w= - 7
x^ 2 - y^2 / 2w - z
( -1 )^3 - ( -3 )^2 / 2 ( - 7) -( - 4)
= -1 - 9/ + 4 +4 = -10/ -10
= 1
3 b)
<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
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4 a)
2 /3 ( 1 -4 x) -1 /2 ( 5 -3 x) less/ equal to 1 /4 ( 7 + 9 x) -1 / 3
multiply through by 12
8 ( 1- 4 x) - 6( 5 - 3x ) less/ equal to 3 ( 7 + 9x ) -4
8 -32x- 30+18x less/ equal to 21+ 27x- 4
-32x+ 18x - 27x less /equal to 21- 4 +30- 8
-41x less/ equal to 39
x less /equal to -39/ 41
4 b)
DRAW THE ANGLE
from DTMR /
Tan65degree = TR /3
TR = 3Tan 65
= 3 x 22.1445
From DRMB
Tan20 = RB/3
RB = 3Tan 20
= 3 x 0 .3640
= 1 .092 m
H = TR +RB
= 6 .4335 + 1 .092
= 7 .5255
= 7 .53m
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5 )
Area of shed segment = Area of Sector - Area of Triangle
= Φ/ 360* πr^ 2 - (½abSinc)
= $90/ 360 *22/ 7 *7 ^ 2 ) - ½* 7 *7 * sin 90`
= 154/ 4 - 42/ 2
= ( 754- 9 /4
= 56/4
= 14cm ^ 2
Cost of painting it = 14* 750
= N 10500
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6 a)
Taxable income = x
25/100 x x/ 1
= 14, 000
25x= 1400000
X = 1400000 /25
X = 560000
Total income =
56, 000/ 4 x 5
= 70,000
6 b)
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 month
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8 a)
Drawing
8 bi)
Using Pythagoras theory
| xz | ² = 550² + 320²
= 302500+ 102400= 404900
xz=√404999
= 636km ( 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 .581
Z = 30°
From the diagram in 8 a
Bearing of X from Z = 55+ 30
= 085 ° or N 85°E
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9 a)
3 log 10^ 2 - 2log 10^3 = 1 + log ( 1/ x)
log 10^ ( 2 )^3 - log 10^ ( 3)^2 - log 10^( 1 / x) = 1
Log10^ ( 8 / 9 ) / ( 1 / x) = log 10^10
8 / 9 * X / 1 = 10
8 x = 9 *10
8 x = 90
X = 90/8
= 11.25
9 b)
Distance = speed * time
Distance = 3 km / h * 3 mins
= 3000m / 60min * 3 min
= 50*3 = 150 min.
9 bi)
Circumference= 150 m
2 π( 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
9 bii)
Volume of cylinder = πr² h
Volume = 22/ 7 * ( 23) ² * 8
22/7 * 529/ 1 * 8 / 1
= 93104 /7 = 13, 300. 57
13301 m ³ to the nearest whole number
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7 a)
diagram
7 b)
Angula difference in long ( tita)=42- 12
tita= 30 degree
7 bi)
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 )
7 bii)
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 .258
alpha/ 2= 14. 999
alpha= 14.999 * 2
alpha= 29. 998
alpha= 30. 0 degree ( to 1 dp)
7 biii)
XY bar = tita/ 360 * 2 pie R cos lat
= 30 degeree/ 360 * 2 *3 . 142 * 6400* cos 60
XY bar = 30/ 360 * 2 * 3. 142 * 6400* 0. 5
XY bar = 1675.73 km
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2016-09-10 09:53 ·

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