WAEC GCE MATHEMATICS EXPO (OBJ/THEORY) ANSWERS 2016
1a )
6whole 1 / 2 – 3 whole 2/ 5 / 2whole 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
= 3whole 1 / 6
1b)
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
3x = 27
X = 27/3
X = 9 students
========
2a)
m/m-y+2 = r/y+r-1
r(m-y+2)=m(y+r-1)
rm-ry+2r = my+mr-m
rm+2r-m+mr=my+ry
y(m+r)=rm+2r-m+mr
y=2mr+2r-m/m+r
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
=======
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
SCROL DOWN FOR 3B
========
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*Ï€r^2 – (½abSinc)
=$90/360*22/7*7^2) – ½*7*7*sin90`
=154/4 – 42/2
=(754-98)/4
=56/4
=14cm^2
Cost of painting it =14*750
=N10500
========
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(0.2588)
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
========
8a)
Drawing
8bi)
Using Pythagoras theory
|xz|² =550²+320²
= 302500+102400=404900
xz=√404999
=636km(3 s.f)
Total distance = 320+550+636
= 1506km
8bii)
Using SOHCAHTOA
Tan z = 320/550
Z= Tan-¹ (320/550)
Z= Tan-¹(0.5818)
Z = 30°
From the diagram in 8a
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
2Ï€(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=Ï€r²h
Volume= 22/7 * (23)² * 8
22/7 * 529/1 * 8/1
= 93104/7 = 13,300.57
13301m³ 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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6whole 1 / 2 – 3 whole 2/ 5 / 2whole 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
= 3whole 1 / 6
1b)
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
3x = 27
X = 27/3
X = 9 students
========
2a)
m/m-y+2 = r/y+r-1
r(m-y+2)=m(y+r-1)
rm-ry+2r = my+mr-m
rm+2r-m+mr=my+ry
y(m+r)=rm+2r-m+mr
y=2mr+2r-m/m+r
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
=======
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
SCROL DOWN FOR 3B
========
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*Ï€r^2 – (½abSinc)
=$90/360*22/7*7^2) – ½*7*7*sin90`
=154/4 – 42/2
=(754-98)/4
=56/4
=14cm^2
Cost of painting it =14*750
=N10500
========
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(0.2588)
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
========
8a)
Drawing
8bi)
Using Pythagoras theory
|xz|² =550²+320²
= 302500+102400=404900
xz=√404999
=636km(3 s.f)
Total distance = 320+550+636
= 1506km
8bii)
Using SOHCAHTOA
Tan z = 320/550
Z= Tan-¹ (320/550)
Z= Tan-¹(0.5818)
Z = 30°
From the diagram in 8a
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
2Ï€(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=Ï€r²h
Volume= 22/7 * (23)² * 8
22/7 * 529/1 * 8/1
= 93104/7 = 13,300.57
13301m³ 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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