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» NECO GCE » 2017 NECO GCE Mathematics Expo Answer Obj And Essay Solution Answer – Nov/Dec Free Expo

Monday, November 6, 2017

2017 NECO GCE Mathematics Expo Answer Obj And Essay Solution Answer – Nov/Dec Free Expo

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2017  NECO GCE Mathematics Expo Answer Obj And Essay Solution Answer – Nov/Dec Free Expo


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NOTE::: BELOW ANSWER IS OUR 2016 NECO GCE MATHEMATICS ANSWERS..
OUR 2016 NECO GCE MATHEMATICS WENT WELL..  


UPDATED NUMBERS: 1,2,3,4,5,7,8,9,10,11…
FULL ANSWERS COMPLETED
MATHS OBJ
1EAACBAECBE
11CDABECBCCA
21AEDCBACEDE
31ADADCCDABD
41BCEBDCEABE
51DBAECBECBE
==========
SECTION A(ANSWER ALL QUESTIONS 1-5)
1a)
(6y²)³ / 2+y^6
= 6³y^6 / 24y^6
=6³*y^6/24*y^6
= 216/24
=9
1b)
Log10(2m²+5m-2)=1
2m²+5m-2=10
2m²+5m-2-10=0
2m²+5m-12=0
Using m=-b+-√b²-4ac/2a
a=2
b=5
c=-12
m=-5+-√5²-4*2*(-12)/2*2
m=-5+-√2+96/4
m=-5+-√98/4
m=-5+-9.9/4
m=-5+9.9/4 or m=-5-9.9/4
M=1.225 or -3.725
==========
2a)
w*3√D
W=k 3√D
10=k*3√27
10/3 =k*3/3 k=10/3
W=10/3*3√64 = 10/3*4*40/3=13 1/3
2b)
X/2 + y/6 = 2/3
L.C.M = 6
Rx + y = 4 —–(1)
X/3 – y/2 = 5/3
L.C.M = 6
2x – 3y = 10 —–(2)
From eq (1)
3x + y = 4
Y= 4 – 3x ——(3)
Sub for y in eq (2)
2x – 3(4-3x) = 10
2x – 12 + 9x = 10
11x = 22
X=2
Sub for x in eq (3)
Y = 4-3(1)
Y = 4-3-
Y = 1
===========
3a)
Sn = a(r^n – 1)
a = 5
r = 3
n= 8
5(3^8 – 1)/ (3-1) = 5(6561-1)/2
= 32800/2 = 164000
3b
Cos30°=12/P
PCos30°=12
Divide through by cos 30°
PCos30°/Cos30°=12/Cos30°
P=12/cos30°
P=12/0.8660
P=13.8568
==========
4a)V=2/3Πr
1527.4=2/3*22/7*r
1527.4=44r/21
32075.4=44rb
r=32075.4/44
r=728.98
=729cm
4b)
3^3y+1 = 81^-y
3^3y= 3^-4y
3y+1=-4y
1=-4y-3y
1=-7y
Y= – 1/7
===========
5a)
Red=4
Green = 6
Black= 10
Total=20
1)p(Red) = 4/20 = 1/5
2)p(green)= 6/20 = 3/10
3) p (black)= 10/20= 1/2
5b)
( X-1)(X-8)+12=0
X²-8x-X+8+12=0
X²-9x+20=0
X= -b +√b²-4ai
X= 9+/-√-9²-4×1×20÷ 2×1
X=9+/-√81-80÷2
X=9+/-√1÷2
X= 9+1/2 or 9-1/2
X= 10/2 or 8/2
X= 5 or 4
============
+++++++++++++
SECTION B(ANSWER 5 QUESTIONS 7-11)
+++++++++++++
7a)
lets k and p be our constant
R=k+P3√5
S = speed
k + P3√64 = 182…(i)
k + P3√343 = 200…(ii)
subtitute eqn (i) from (ii)
k+4p=182
k+7p=200
3p/3 = 18/3
p=6
subtitute 6 in eqn(i)
k+4p=182
k+4(6)=182
k=182-24
k=158
recall R=k+p3√5
R=158+63√5
(a)find S when R is 188
188=158+6×3√s
188-158 = 6 3√s
30/s = 6 3√s/s
√s=30
(s) = 30^3
s = 30^3
s= 27,000k/h
7b)
find R when S = 27
R=158+6 3√27
R=158+6×3
R=158+18
R=176 newtons
============
(8a)
4a^2 b*3ab^-2/4a^3*(3ab)
=12 a^4 b^1/12a^4 b
=6^-1/b^1
=b^-1-1 =b^2
6^-2 or 1/b^2
8b)
Let the pocket money be x
2x/5 =stations
X/4=school levies
X/6= food
X-(24x/5+x/4+x/6)=55
X-(24x+15x+10x)/60 =55
X-49x/60 =55
60*x-60-49x/60 = 55*60
60x-49x=#3,300
11x/11=3300/11
X=#300
The pocket money =#300
8c)
16^x-4 = 64^x+4/8^x
24^x-16 = 2^6x+24/2^3x
2^4x-16 = 2^6x+24-3x
2^4x-16 = 2^3x+24
4x-16 = 3x + 24
4x – 3x = 16+24
x = 40
============
9a)
solve using cosine rule
m^2 = 70^2+40^2 – (2 x 70 x 40 x cos135)
m^2 = 4900 + 1600 – (5600 x cos135)
m^2 = 6500 – (5600 x cos135)
m√6500-(5600 x cos135)
m = √6500+5578.2
m = √12078.16
m = 109.9006
m == 109.9
9b)
lets find tita (θ)
using sine rule
70/sinθ = 109.9/sin135
70sin135/109.9 = 109.9sinθ/109.9
sinθ = 70 x sin135/109.9
sinθ = 0.45038
θ = sin^-1(0.45038)
θ = 26.76935
θ == 27
:- Bearing of k from R
180° + 85° + 27°
= 292°
==============
10a)
1/27^(x-2)=9^(2-x)
3^3(x-2)-1=3^2(2-x)
3-1(x-2)=4-2x
2x-4=4-2x
2x+2x=4+4
4x=8
X=8/4
X=2
10b
5x+y/x+5y= 3/1
5x+y= 3(x+5y)
5x+y=3x+15y
5x-3x=15y-y
2x=14y
X/y= 14/2 =7, y/x= 1/7
X2-y2/4xy
Divide all through by my
(X²- y²/) ÷(4xy/xy)
Xy- xy
(X/y- y/x) ÷4
(7- 1/7)÷4 ~~~49-1/7×1/4
48/7× 1/4= 24/7
= 4
============
11a)
A=(1 7,3 2) and B=(2 5,4 6)
i)A^-1=1/det(2 -7,-3 1)
Det= 2 -21=-19
A^-1=1/-19(2 -7,-3 1)
-1/19(2 -7,-3 1)
ii)B^T =(2 -4,5 6)
iii)AB =(1 7,3 2) (2 5,4 6)
(2 -28,6-8, 5+46,15+12)
(-26,-2, 51,27,)
11b)
7x^2 + 12x + 5 = 0
7x^2/7 + 12x/7 = -5/7
X^2 + 12x/7 = -5/7
X^2 12x/7 + (6/7)^2 – -5/7 + (6/7)2
(X+6/7)^2= -5/7 + 36/49
(X+6/7)^2 = -35+36/49
(X +6/7)^2 = 1/49
(X +6/7) = + and – √1/19
X+ 6/7 = + and – 1/7
X = -6/7 + and – 1/7
X = -6+1/7 or -6-1/7
X= -5/7 or -7/7
X = -5/7 or -1
============
GOODLUCK
============

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