MATHEMATICS ANSWERS
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 15 x
15 x + 45 / x + 1 = 18
15 x + 45 = 18 ( x + 1 )
15 x + 45 = 18 x + 18
18 x – 15 x = 45 – 18
3 x = 27
X = 27 / 3
X = 9 students
==================================
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 + 2 r – 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
==================================
3 a )
c = – 1 , y = -3 , z = -4 and w = – 7
x ^ 2 – y ^ 2 / 2 w – 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 ==================================
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 -3 x ) less / equal to 3 ( 7 + 9 x ) -4 8 -32 x -30 + 18 x less / equal to 21 + 27 x - 4 -32 x + 18 x - 27 x less / equal to 21 - 4 + 30 - 8 -41 x less / equal to 39 x less / equal to - 39 / 41 4 b ) DRAW THE ANGLE from DTMR / Tan 65 degree = TR / 3 TR = 3 Tan 65 = 3 x 22 . 1445 From DRMB Tan 20 = RB / 3 RB = 3 Tan 20 = 3 x 0 . 3640 = 1 . 092 m H = TR + RB = 6 . 4335 + 1 . 092 = 7 . 5255 = 7 . 53 m ==================================
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 - 98 )/ 4 = 56 / 4 = 14 cm ^ 2 Cost of painting it = 14 * 750 = N 10500 ==================================
6 a ) Taxable income = x 25 / 100 x x / 1 = 14 ,000 25 x = 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 ==================================
(7a) diagram (7b) Angula difference in long(tita)=42-12 tita=30 degree
(7bi) 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)
(7bii) 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(0.2588) alpha/2=14.999 alpha=14.999 *2 alpha= 29.998 alpha= 30.0 degree (to 1 dp)
(7biii) XY bar= tita/360 * 2pie R cos lat =30 degeree/360 * 2*3.142 * 6400* cos60 XY bar= 30/360 * 2* 3.142* 6400* 0.5 XY bar=1675.73 km ==================================
8 a ) Drawing 8 bi ) Using Pythagoras theory | xz | ² = 550² + 320² = 302500 + 102400 = 404900 xz =√ 404999 = 636 km ( 3 s .f ) Total distance = 320 + 550 + 636 = 1506 km 8 bii ) Using SOHCAHTOA Tan z = 320 / 550 Z = Tan - ¹ ( 320 / 550 ) Z = Tan - ¹( 0 . 5818 ) Z = 30 ° From the diagram in 8 a Bearing of X from Z = 55 + 30 = 085 ° or N 85 ° E ==================================
9 a ) 3 log 10 ^ 2 - 2 log 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 = 3000 m / 60 min * 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 = 23 min
9 bii ) Volume of cylinder = πr ² 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
10a)
GIVEN: y = 2X^2- 3X – 1
When x = -3
Y= 2(-3)^2- 3(-3) – 1 = 18 + 9 – 1 = 26
When x= -1
Y = 2(-1)^2 – 3(-1)-1 = 2+3-1 = 4
When x = 1
Y = 2(-1)^2 – 3(1)-1 = 2 – 3 – 1 = -2
When x = 2
Y=2(2)^2-3(2)-1= 8 – 6 – 1 =1
When x = 3
Y =2(3)3-3(3)-1 = 18-9-1 = 8
When x = 4
Y = 2(4)^3 – 3(4)-1 = 32-12-1 =19
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 15 x
15 x + 45 / x + 1 = 18
15 x + 45 = 18 ( x + 1 )
15 x + 45 = 18 x + 18
18 x – 15 x = 45 – 18
3 x = 27
X = 27 / 3
X = 9 students
==================================
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 + 2 r – 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
==================================
3 a )
c = – 1 , y = -3 , z = -4 and w = – 7
x ^ 2 – y ^ 2 / 2 w – 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 ==================================
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 -3 x ) less / equal to 3 ( 7 + 9 x ) -4 8 -32 x -30 + 18 x less / equal to 21 + 27 x - 4 -32 x + 18 x - 27 x less / equal to 21 - 4 + 30 - 8 -41 x less / equal to 39 x less / equal to - 39 / 41 4 b ) DRAW THE ANGLE from DTMR / Tan 65 degree = TR / 3 TR = 3 Tan 65 = 3 x 22 . 1445 From DRMB Tan 20 = RB / 3 RB = 3 Tan 20 = 3 x 0 . 3640 = 1 . 092 m H = TR + RB = 6 . 4335 + 1 . 092 = 7 . 5255 = 7 . 53 m ==================================
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 - 98 )/ 4 = 56 / 4 = 14 cm ^ 2 Cost of painting it = 14 * 750 = N 10500 ==================================
6 a ) Taxable income = x 25 / 100 x x / 1 = 14 ,000 25 x = 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 ==================================
(7a) diagram (7b) Angula difference in long(tita)=42-12 tita=30 degree
(7bi) 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)
(7bii) 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(0.2588) alpha/2=14.999 alpha=14.999 *2 alpha= 29.998 alpha= 30.0 degree (to 1 dp)
(7biii) XY bar= tita/360 * 2pie R cos lat =30 degeree/360 * 2*3.142 * 6400* cos60 XY bar= 30/360 * 2* 3.142* 6400* 0.5 XY bar=1675.73 km ==================================
8 a ) Drawing 8 bi ) Using Pythagoras theory | xz | ² = 550² + 320² = 302500 + 102400 = 404900 xz =√ 404999 = 636 km ( 3 s .f ) Total distance = 320 + 550 + 636 = 1506 km 8 bii ) Using SOHCAHTOA Tan z = 320 / 550 Z = Tan - ¹ ( 320 / 550 ) Z = Tan - ¹( 0 . 5818 ) Z = 30 ° From the diagram in 8 a Bearing of X from Z = 55 + 30 = 085 ° or N 85 ° E ==================================
9 a ) 3 log 10 ^ 2 - 2 log 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 = 3000 m / 60 min * 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 = 23 min
9 bii ) Volume of cylinder = πr ² 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
10a)
GIVEN: y = 2X^2- 3X – 1
When x = -3
Y= 2(-3)^2- 3(-3) – 1 = 18 + 9 – 1 = 26
When x= -1
Y = 2(-1)^2 – 3(-1)-1 = 2+3-1 = 4
When x = 1
Y = 2(-1)^2 – 3(1)-1 = 2 – 3 – 1 = -2
When x = 2
Y=2(2)^2-3(2)-1= 8 – 6 – 1 =1
When x = 3
Y =2(3)3-3(3)-1 = 18-9-1 = 8
When x = 4
Y = 2(4)^3 – 3(4)-1 = 32-12-1 =19
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