2016 maths GCE answers
1a)
6whole1/2 - 3whole2/5 / 2whole1/2 - 1whole3/5
25/4-17/5, / 5/2-8/5
125-68/,20,/25-16,
/10
==> 57/20/9/10
=57/20+9/10
=57/20x10/9
=57/18
=======
1b) let x be total number of students
Y is number of sample.
x/y = 15
x= 15y.....(I)
x+ 45/y+1= 18.
x+45=18y-45.
x-18y=-27......(ii)
Substitute for x in equation 2.
15y-18y=-27.
-3y=-27.
Therefore y= 9.
Substitute y back into equation 1.
x=15y.
x= 15*9
x=135.
Therefore the number of student in the class is 9.
3a)
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
3b)< means angle.
<MQO= 90°= angles at tangent.
<MNO=90°= angles at tangent of the circle.
<MQN+ <NQO=90° = angles at tangent.
<MQN+46= 90°.
<MQN= 44°.
ii) Let <MQN=<MNQ (isosceles)
But, <MQN=44°,
So, x=<MNQ= 44°.
ii) Considering <MNQ
y+x+<MQN=180° ( sum of angles in a triangle)
y+ 44+ 44 = 180°.
Y= 180- 88= 92°
4a)
2/3(1-4x) -1/2(5-3x) less/equal to 1/4(7+9x)-1/3
multiply through by 12
8(1-4x) -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
4b)
DRAW THE ANGLE
from DTMR/
Tan65degree = TR/3
TR = 3Tan65
= 3 x 22.1445
From DRMB
Tan20 = RB/3
RB = 3Tan20
= 3 x 0.3640
= 1.092m
H = TR+RB
= 6.4335 + 1.092
= 7.5255
= 7.53m
==================================
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 * sin90 `
=154 /4 - 42 /2
=( 754 - 98 ) /4
=56 / 4
=14 cm^ 2
Cost of painting it =14 * 750
=N 10500
CLICK HERE FOR NUMBER 6 ANSWER
6a)
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
6b)
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 years
(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(0.258
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
= 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 N85 °E
9a)
3log10^2 - 2log10^3 = 1 + log(1/x)
log10^(2)^3 - log10^(3)^2 - log10^(1/x) = 1
Log10^( 8 /9 ) / (1 / x) = log10^10
8 / 9 * X / 1 = 10
8x = 9 *10
8x = 90
X = 90/8
= 11.25
9b.
Distance= speed * time
Distance= 3km/h * 3mins
= 3000m/60min * 3min
=50*3 =150min.
9bi.
Circumference= 150m
2π(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
9bii.
Volume of cylinder=πr²h
Volume= 22/7 * (23)² * 8
22/7 * 529/1 * 8/1
= 93104/7 = 13,300.57
13301m³ to the nearest whole number
25/4-17/5, / 5/2-8/5
125-68/,20,/25-16,
/10
==> 57/20/9/10
=57/20+9/10
=57/20x10/9
=57/18
=======
1b) let x be total number of students
Y is number of sample.
x/y = 15
x= 15y.....(I)
x+ 45/y+1= 18.
x+45=18y-45.
x-18y=-27......(ii)
Substitute for x in equation 2.
15y-18y=-27.
-3y=-27.
Therefore y= 9.
Substitute y back into equation 1.
x=15y.
x= 15*9
x=135.
Therefore the number of student in the class is 9.
3a)
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
3b)< means angle.
<MQO= 90°= angles at tangent.
<MNO=90°= angles at tangent of the circle.
<MQN+ <NQO=90° = angles at tangent.
<MQN+46= 90°.
<MQN= 44°.
ii) Let <MQN=<MNQ (isosceles)
But, <MQN=44°,
So, x=<MNQ= 44°.
ii) Considering <MNQ
y+x+<MQN=180° ( sum of angles in a triangle)
y+ 44+ 44 = 180°.
Y= 180- 88= 92°
4a)
2/3(1-4x) -1/2(5-3x) less/equal to 1/4(7+9x)-1/3
multiply through by 12
8(1-4x) -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
4b)
DRAW THE ANGLE
from DTMR/
Tan65degree = TR/3
TR = 3Tan65
= 3 x 22.1445
From DRMB
Tan20 = RB/3
RB = 3Tan20
= 3 x 0.3640
= 1.092m
H = TR+RB
= 6.4335 + 1.092
= 7.5255
= 7.53m
==================================
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 * sin90 `
=154 /4 - 42 /2
=( 754 - 98 ) /4
=56 / 4
=14 cm^ 2
Cost of painting it =14 * 750
=N 10500
CLICK HERE FOR NUMBER 6 ANSWER
6a)
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
6b)
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 years
(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(0.258
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
= 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 N85 °E
9a)
3log10^2 - 2log10^3 = 1 + log(1/x)
log10^(2)^3 - log10^(3)^2 - log10^(1/x) = 1
Log10^( 8 /9 ) / (1 / x) = log10^10
8 / 9 * X / 1 = 10
8x = 9 *10
8x = 90
X = 90/8
= 11.25
9b.
Distance= speed * time
Distance= 3km/h * 3mins
= 3000m/60min * 3min
=50*3 =150min.
9bi.
Circumference= 150m
2π(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
9bii.
Volume of cylinder=πr²h
Volume= 22/7 * (23)² * 8
22/7 * 529/1 * 8/1
= 93104/7 = 13,300.57
13301m³ to the nearest whole number
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