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Thursday, 11 June 2015

NECO 2015 Mathematics Essay Questions and Answers

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Thursday 11th June

1022
Paper II: Essay – General Mathematics
2hrs 30mins
12:00noon – 2:30pm









Maths Theory
1a)
x:y:z*6:5:8
M=x/6=y/5=5/8
x:y:z*6m:5m:8m
1.8x=6m
y=3m
z=8m
12x-48/4y+2=12(6m)-8(8m)/4(5m)+8m
=72m-72m/20m+8m
=0m/28m
=0
1b)
un=ar^1
us=ar46
2/81=2r^4
81×2r^4/81×2=2/81×2
r^4=1/81
r=4rot1/81, r=1/3.
ii)sum=a(1-r^n)/1-r
=2(1-(1/3)5)/1-1/3
=2(1-1/243)/2/3
=(242/243)/2/3
484/243-2/3
484/243×3/2
=242/81
=============================
2a)
number of sides=12
radius of circle=10cm
area=?
nש2=360
12©2=360
©2=360/12=30°
©1+©2=180-30
©1=150
When ©1 and ©2 are interior and exterior angle of a polygon A sector has are.
Area of sector=©/360×rot8^2
=150/360×22/7×100/1
A=130-95cm^2
2b)
1/2(2x+1)-2/5(x-2)=3
2x+1/3-2x-4/3=3
10x+5-6x+12/15=3/1
Cross multiple
4x+17=45
4x/4=28/4
x=7.
=============================
3)
Apply 5m rule to find C P
t/sin T = P/sin P
t/sin 110 = 6/sin 40
t=6 * 0.9396/0.6428
=56376/6.6428
=87704
=877km
=============================
4)
Total Fruit = 80 + 60 = 140
(a)
(i) Pr one of each fruit is picked
(79/140 *60/139) + ( 59/140 * 80/139)
=4740/19,460 + 4720/19,460
=9460/19,460 = 0.486
4aii)
Pr one type of fruit is picked
(79/140 * 78/139) + (39/140 * 5p/139)
=6162/19,460 + 3422/19,460
= 9584/19,460 = 0.492
4b)
5X/8 - 1/6 ≤ X/3 + 7/24
Multiply through by 24 i:e
15X - 4≤ 8X + 7
15X - 8X ≤ 7 + 4
7X = 11
X ≤ 11/7 ===> X ≤ 1 4/9
=============================
5a)
3/X + 2 - 6/3X - 1
3(3X - 1) - 6 (X + 2)/ (X + 2) ( 3X - 1)
9X - 3 - 6X - 12/(X + 2)(3X - 1)
3X - 15/(X+2)(3X - 1)
5b)
C.I=P[1+r/100]^
=25000[1+12/100]^3
=25000[1+0.12]^3
=25000 * 1.4049
=35122.50
=N35,122.50
============================
6a)
X+-3/2
X=2/3 or X=2
(X + 3/2)^2 or (X -2)
(X + 3/2) (X - 2)
X (X - 2) + 3/2 (X - 2)
X^2 - 2X + 3 X/2 - 3
2X^2 - 4X + 3X - 6
2X^2 - X - 6
6b)
h/h+8 = 6/10
10h = 6h + 48
h = 12
H = h + 8
H = 12 + 8
H = 20
Volume = 1/3 A.h
= 1/3 (10*10) * 20 - 432/3
= 200/3 - 432/3
=1568/3
=522.67cm3
=============================
7a)
titan/360×2pie r cos t
d=55/360×2×22/7×640 cos 4
d=55×44×6400 cos 4/2520
d=55×44×6400×0496/2520
d=15,449.28/2520
d=6130.67
d~6130km.
ii)distance along gent circle
D=tita/360×2pie r
D=55/360×2/7×22/7×6400/1
D=55×44×6400/2520
D=15,488×6400/2520
D=6144.03
D=~6146km.
7b)
Length of sector tita/360×2pie r
L=120/360×2/1×22/7×42/1
L=120×44×42/2520
L=221760/2520
L=88cm
L=2pie r
Where r is the radius of circumference
88=2×22/7×r
88×7=44r
R=88×7/44
R=616/44
R=14cm.
Curved surface area
= pie rc
A=22/7×14×42/1
A=22×14×4^2/7
A=12936/7
A=1848cm^2.
=============================
8a)
X=60/t ---------->(i)
Y=180/t ---------->(ii)
T1=60/X
T2=100/Y
T1+T2=5
60/X+180/Y = 300 ---------->(i)
180/X + 60/Y = 260 ---------->(ii)
Let P =1/X
2 = 1/Y
60p + 180 Q = 300
180p + 6Q = 200
P + 3Q =5
9P + 3Q =13
Substract (i)from (ii)
8p = 8
P = 8/8 ÷ P = 1
Subtract P into (i)
P + 3Q =5
1 + 3Q =5
3Q =5-1
3Q =4
Q = 4/3
P = 1 ÷ 1 = 1/X ÷ X = 1
4/3 = 1/Y ÷ Y = 3/4
8b)
2001 --------- 25,700
2002 --------- 15/100 X 25,700 + 25,700 = 29,555
Amount of tax in 2002
= 29,555 * 12.5/100
= N3694.375
=N3690
8c)
Log25
Log16 25/100
Log16 2/4
Log4 6-1
-1/2
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