CORRECT: 2015/2016 WAEC GCE MATHEMATICS OBJ Essay Theory Questions And Answer Expo/ Runs

MATHEMATICS OBJECTIVE
1-10 ABCDBCBADB
11-20 DCCCCACDBA
21-30 BDBAABDACA
31-40 CCACBDCDBD
41-50 DACBDABBBC
Completed

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
••••••••••••••••••••••••••••••••••

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

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
==================================
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
••••••••••••••••••••••••••••••••••

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)
<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
••••••••••••••••••••••••••••••••••

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
=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 year
••••••••••••••••••••••••••••••••••

(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
••••••••••••••••••••••••••••••••••
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
••••••••••••••••••••••••••••••••••
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

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
••••••••••••••••••••••••••••••••••
10a)
GIVEN : y = 2 X ^ 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

Post a Comment

0 Comments