Neco 2017/2018 General Mathematics Free Expo Answers

Wednesday 14th June 2017 Paper III:OBJ - General Mathematics 10am - 11:45am COMPLETED -> OBJECTIVES 1-10:  CDAAEDEECE 11-20: EE...


Wednesday 14th June 2017
Paper III:OBJ - General Mathematics 10am - 11:45am

COMPLETED -> OBJECTIVES
1-10:  CDAAEDEECE
11-20: EEBABAABCC
21-30: BBCEDABECD
31-40: BCCCDDCADE
41-50: BDCBDCCECD
51-60: BCDC-EDCDB

a)
=1/2log25/4-2log4/5
+log320/125
=log(25/4)^1/2-log(4/5)^2
+log(320/125)
=log{sqroot(25/4)}-log
(16/25)+log(320/125)
=log(5/2)-log(320/125)-log(16/25)
=log[5/2*320/125/(16/25)
=log[5/2*320/125*25/16]
=log10
=1
(1b)
%Increment=20%
Grants per land=GH 15.00
The total population from 2003 to
2007=1.2*1.2*1.2*1.2*3000
=6220.8
Total grant=population * grant per head
=6220.8*15
=GH9331
Total grants=GH93312

(2a)
1/x+(1/x+3)=1/2
LCM=x(x+3)
(x+3+x)/x(x+3)=1/2
2(2x+3)=x(x+3)
4x+6=x^2+3x
x^2+3x=4x+6
x^2+3x-4x-6=0
(x^2-3x)+(2x-6)=0
x(x-3)+2(x-3)=0
(x+2)(x-3)=0
x=-2 or x=3
(2b)
Let the bag of rice be x
Let the bag of beans be y
x+y=17(eq1)
2250x+2400y=39600(eq2)
from (eq1)
x=17-y
substitute for x in eq2
2250(17-y)+2400y=39600
38250+150y=39600
y=(39600-3850)/150
y=9
therefore bags of beans=9
substitute for 9 in eq1
x+y=17
x+9=17
x=17-9
x=8

(3)
Area of garden=L^2
17=(L+2)*(L+L)
17=L^2+3L+2-17
L^2+3L+2-17=0
L^2+3L-15=0
-b+_sqroot(b^2-4ac)/2a
=-3+_sqroot(9-4*1*-15)/2*1
=-3+_sqroot69/2
=-3+_8.03/2
=11.307/2 or 5.307/2
=5.654 or 2.653
L=5.654 p=4L
p=4(5.654)
p=22.616m
(3b)
Area=L^2=5.654^2
=31.98m^2
Area of the path=L*b
=2*1
=2m^2

(4)
3^2+y^2=5^2
9+y^2=25
y^2=25-9
y^2=16
y=sqroot16
y=4
therefore (cosx+tanx)/sinx
=(4/5)+(3/4)/(3/5)
=(16+15/20)/(3/5)
=(31/20)/(3/5)
=31/20*5/3
=31/12
=2(7/12)
(4b)
From the diagram
200degrees+32degrees
+ydegrees=360degrees
(angles at a point)
ydegrees+232degrees=360degrees
ydegrees =360degrees-232degrees
y=128degrees
Ndegrees=128degrees(alternative angles)
xdegrees=128degrees+180degrees
xdegrees=308degrees

(5a)
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
(5b)
(i)Pr(sum of outcome is 8)=5/36
(ii)Pr(product of outcome 10)=17/36
(iii)Pr(outcome contain atleast a 3)
=32/36=8/9

(6a)
2basex(37basex)=75basex
(2*x^1)(3*x^1+7*x^0)=7*x+5*x^0
(2*1)(3x+7)=7x+5
2(3x+7)=7x+5
6x+14=7x+5
6x-7x=5-14
-x=-9
x=9
(6b)
let the number of boys=x no of girls=5+x
(x+5)/(x+2)=5/4
4(x+5)=5(x+20)
4x+20=5(x+2)
4x+20=5x+10
4x-5x=10-20
-x=-10
x=10
(i)No of girls=x+5
=10+5=15girls
(ii)Total No of pupils =x+x+5
=20+5=25pupils
(iii)probability of boy
=No of boy/total pupil
=10/25
=0.4

(7a)
PQ=(5-x)^2+x^2
PQ=25+x^2-10x+x^2
therefore Area of the square=2x^2-10x
+25
If the area of PQRS=3/5
2x^2-10x+25=3/5*25
2x^2-10x+25=15
2x^2-10x=15-25
2x^2-10x+10=0
divide through by 2
x^2-5x+5
Using formular==-b+_sqroot(b^2-4ac)/2a
=5+_sqroot(25-4*1*5)/2*1
=5+_sqroot(25-20)/2
=5+_sqroot4/2
=5+_2/2
=5+2/2 or 5-2/2
=7/2 or 3/2
=3.5 or 1.5
(7b)
(1+a)/(n-1)=d
1+a=dn-d
a=d(n-1)-L
2s=n(a+L)
s=n(d(n-1)+L)-L/2
s=n(dn-d+L)-L/2

(8a)
diagram
(8+x)^2 = x^2+32
64+16x+x^2= x^2 + 1024
16x=1024-64=960
therefore 960/16= 60
x=960/16
=60
therefore the radius = 60+8
=68cm
(8b)
diagram
(i)volume of a pyramid
=1/3 AH
2601= 1/3 * A * 27
A=7803/27
=289cm^3
Area of square =289
t^2= 289
t= sqr rut(289)
l=17cm
(8bii)
AC^2 = 17^2 +17^2
AC^2 = 289 +289
Ac^2 =578
AC =sqr root (578)
AC=24.04cm
for the triangele COP
CO= 1/2 AC
=1/2 * 24.04
VC^2= 27^2 + 12.07
VC^2= 929 +144.49
VC= SQR root (873.48)
=29.55cm
cos tita = ADJ/hyp
cos x= 8.5/29.55
cos x=0.2877
x=cos^-1 0.2877
=73.66 degree

(9a)
CBP=128-x(sum of angle in a triangle)
CBA=180-(128-x)
sum of angle on a straight line
CBA=52+x
ADC=180-(128-x)
=52+x
Also BCD=180-x(angle on a straight line)
DCQ=180-(180-x)
DCQ=180-180+x
DCQ=x
x+52+x+76=180
2x=180-52-76
2x/2=52/2
x=26degrees

10a)
YX/XZ=XM/MZ
W/10=8/15
15W=10*8
W=5.33cm
10bi)
q^2=p^2+r^2-2prcos tita
q^2=20^2+15-2*20*15 c0s 90
q^2=400+225-0
q^2=625
q=sqroot625
q=25km
10bii)
p/sinP=q/sinQ=r/sinR
25/sin90=15/sinR
sinR=15*1/25
sinR=0.6
R=sin^-1(0.6)
R=36.86degrees
The bearing of p from R
=90+90+90+alpha
alpha=45-36.86
=90+90+90+8.14
=278.14
=278degrees
The bearing of p from
R=278degrees
--++++++++++++++---++++++


1a)
(2x+1)/(3-4x)=2/3
3(2x+1)=2(3-4x)
6x+3=6-8x
6x+8x=6-3
14x/14=3/14
x=3/14
1bi)
E=MV^2/2
2E/M =MV^2/M
V^2=2E/M
V=sqr2E/M
1bii)
Vsqr2E/M
Vsqr2*64/2
Vsqr64
V=8

2 a )
number of sides =12
radius of circle =10 cm
area =?
n ש 2= 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 × rot 8 ^2
= 150 / 360 ×22 / 7× 100 / 1
A =130 – 95 cm^2

2 b )
1 /2 ( 2x + 1 ) – 2 /5 ( x – 2 )= 3
2 x +1/ 3 – 2 x – 4/ 3= 3
10 x +5 – 6x + 12 / 15 =3 / 1
Cross multiple
4 x +17 =45
4 x /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
= 877 km



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

4 aii )
Pr one type of fruit is picked
( 79 / 140 * 78 / 139 ) + ( 39 / 140 * 5 p/ 139 )
= 6162 / 19, 460 + 3422 / 19 , 460
= 9584 / 19 , 460 = 0.492

4 b )
5 X / 8 – 1 / 6 ≤ X / 3 + 7 /24
Multiply through by 24 i : e
15 X – 4≤ 8X + 7
15 X – 8X ≤ 7 + 4
7 X = 11
X ≤ 11 / 7 ===> X ≤ 1 4 / 9
=============================

5 a )
3 /X + 2 – 6 /3 X – 1
3 ( 3 X – 1) – 6 ( X + 2) / ( X + 2) ( 3X – 1)
9 X – 3 – 6X – 12 /( X + 2) ( 3X – 1)
3 X – 15 / ( X + 2) ( 3X – 1)

5 b )
C .I= P [1 +r / 100 ]^
= 25000 [ 1+ 12/ 100 ]^ 3
= 25000 [ 1+ 0. 12 ]^ 3
= 25000 * 1 . 4049
= 35122 . 50
= N 35 ,122 . 50
============================

6 a )
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
2 X ^2 – 4X + 3X – 6
2 X ^2 – X – 6

6 b )
h /h +8 = 6 / 10
10 h = 6 h + 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 . 67 cm3

7 a )
titan / 360 ×2 pie 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 ~ 6130 km .

ii ) distance along gent circle
D = tita / 360 ×2 pie r
D = 55 / 360 × 2/ 7× 22 /7 × 6400/ 1
D = 55 × 44 ×6400 / 2520
D = 15 , 488 ×6400 / 2520
D = 6144 .03
D =~ 6146 km .

7 b )
Length of sector tita/ 360 × 2 pie 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 = 44 r
R =88 ×7 /44
R =616 / 44
R =14 cm.

Curved surface area
= pie rc
A =22 / 7 ×14 × 42 / 1
A =22 ×14 ×4 ^2 /7
A =12936 /7
A =1848 cm^ 2.

8 a )
X =60 / t — – – – – – – – – > ( i )
Y = 180 / t – – – – – — – – – > ( ii )
T 1=60 / X
T 2=100 /Y
T 1+T 2= 5
60 / X + 180 / Y = 300 – — – – – – – – – > ( i )
180 / X + 60 / Y = 260 – – – – – – – – — > ( ii )
Let P =1 / X
2 = 1 / Y
60 p + 180 Q = 300
180 p + 6Q = 200
P + 3 Q =5
9 P + 3Q = 13
Substract ( i ) from ( ii )
8 p = 8
P = 8 / 8 ÷ P = 1
Subtract P into ( i )
P + 3 Q =5
1 + 3 Q = 5
3 Q =5 – 1
3 Q =4
Q = 4/ 3
P = 1 ÷ 1 = 1/ X ÷ X = 1
4 /3 = 1/ Y ÷ Y = 3/ 4

8 b )
2001 – – – – – — – – 25 , 700
2002 – – – – – — – – 15 / 100 X 25, 700 + 25,
700 = 29 , 555
Amount of tax in 2002
= 29 , 555 * 12 .5 /100
= N 3694 .375
= N 3690

8 c )
Log 25
Log 16 25 / 100
Log 16 2/ 4
Log 4 6 – 1
– 1/ 2 Log 4^4
– 1/ 2

9 ai)
W= K +C / 2
24 = k + C / 16
384 = 16 K + C – — – – – – ( i )
18 = K + C / 4
72 + 4K + C – – – – – – – — ( ii )
16 K + C = 384
4 K + C = 72
Substact ( ii ) from ( i )
12 k / 12 = 312 /12
K = 26
Substract K into ( i )
16 k + C = 384
C = 384 – 416
C = – 32
W= k +C / t 2
W= 26- 32 / t 2

( 9aii )
When W= – 46, t =?
– 46 = 26 – 32 / t 2
t 2 = – 32 / -72
t = Sqrt 16 / 36 = 4 /6
= 2 / 3

9 b )
V = Pie r 2 . d = 14 , r = 7cm
1232 = 22 / 7 * 7 ^2 * h
h = 7 * 1232/ 22* 49
h = 8624 /1074
h = 8 cm

11 a)
y ^ 1 =x ^2 ( 3x +1 )^ 2
v = ( 2x + 1 )^ 2
v = m ^2
dm / dx =2
dv / dm =2 m
dy /dx = dv / dm ×dm / dx
= 2 m× 2
= 4 m
dy /dx = 4( 2x + 1)
dy /dx = udv /dx +v whole no . dy / dx
= x ^2 4( xx + 1)^ 2, × 3x
= 4 x ^2 ( 2x +1) + 2x ( 2x + 1)^ 2.

11 b)
[ 3 3 -1 ] [ 1 0 2] [ 3 – 2 3] + 2 [0 – ( – 4) – 3[ 63 ] + – 1 ( – 2)
8 + 9+ 2
= 19 .

11 c )
m= y 2- y 1 /x 2- x 1
y – y 1 = m ( x – x 2)
m= 4- 3/ – 1- 2
m= – 1/ 3
y 1 -y 2=- 1 /3 ( x – x 2)
y – 3= 1/ 3 ( x – 2)
y – 3= – 4/ 3 + 2/ 3
3 y =- x + 11
y = -1 / 3x + 11/ 3 .




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