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Neco Gce Mathematics Answers – Nov/Dec 2018 Expo


Here is the Real NECO Gce 2018 MATHEMATICS Expo Answer, 2018/2019 NECO GCE MATHEMATICS ESSAY AND OBJECTIVE,

 

 

 

 

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Here Is The Verified NECO GCE 2018 MATHEMATICS Expo Answers therory /objective Questions And Answers – Nov/Dec Exam Expo Runz 

NECO GCE 2018 MATHEMATICS Expo Answers

NECO VERIFIED MATHS OBJ
1 CCBEAEBECE
11 AADCEECABB
21 CBBAACEBAD
31 CAABEEBCAD
41 ECCECEDDAC
51 ACABCCCCCB



(1)
TABULATE:
No| Log
3081| 3.4887
0.775 | 1.8893
0.456 | 1. 6589
| UNDER Log
| 3.4887
| 1. 5482
| 3.9405
| 0.9851
Square root of 4/3081/0.775.0456
Antilog = 9663
= 9.663
= 2.9

(2a)
1101₂ = 2x + 1
1 x 2² + 0 x 2¹+ 1 x 2º = 2x + 1
8 + 4 + 1 = 2x +1
13 – 2x
X = 12/2 = 6

(2b)
Sin x = 12/13 = 0.9231
Z = sin-¹ 0.9231 = 67.4º
Therefore 3 sin x + ½ cos x
= 3 sin 67.4 + ½ cos 67.4
= 3 x 0.9232 + ½ x 0.3843
= 7.7696 + 0.1921


(3a)
PQ x 90º [angle in a semicircle]
QPO = 90º – 62 [ angle in a triangle]
= 28º
Therefore POZ = 28º [alternative angles]
(ii)
PXZ = ½ x 28º = 14º [angle at center is twice angle at cirumfeence]

(3b)
6/5 + 3/x+3 – 9/5(x+3)
= 6 (x+3) + 3 (5) – 9/5(x+3)
= 6x + 18 + 15 – 9/5 (x+3)
= 6x + 24/5(x+3)
= 6(x+4)/5(x+3)


(4a)
1/3(y-1)+2>1/2(2y-1)+1
2y-2+12>6y-3+1
2y-6y>-2-10
-4y>-12
y<3

(4bi)
M=y2-y1/X2-X1
=-1-2/2-3=-3/-1=3

(4bii)
2X+1=X+3
2X-X=3-1
X=2


(6)
126y = 86
1 x y² 2 x y¹ + 6 x yº = 86
y² + 2y + 6 = 86
y² + 2y + 6 – 86 = 0
y² + 2y – 80 = 0
y + 10y – 8y – 80 = 0
y (y+10)-8y (y+10) = 0
y – 8 = 0 or y + 10 = 0
the positive value of y

(6b) Area of triangle = ½ x b x h
Let the Acheal Area = x
Area = ½ x b x h
Base = x – 9x/100 = 93/100
Height = 9x/100 + x = 107x/100
963x

(6c)
p/100 + 2p + 7 = 11.02 x 100
p + 200p + 700 = 1102
201p = 1102 – 700
201p = 402
P = 402/201
P = N2
P = 200K


(8)
S²₁ (3x – 1) (x+2) dx
By expansion’
S²₁ 3x² + 6x – x – 2 dx
S²₁ 3x² + 5x – 2 dx
By integrating using d formula
Xn+1/n+1
Therefore 3x ²+¹/2+1 + 5x¹+¹/1+1
= 2xº+¹/0+1 + c
Therefore 3x³/3 + 5x²/2 + 2x/1 + c
Therefore x³ + 5x²/2 + 2x
But x = 2 at higher and are at lower by substituting 2 in x than of value of 1
(2)³ + 5/2(2)² + 2(2) – ((1)³ + 3/2(1) + 2 (1)
8 + 5/2 x 4 + 4 – (1 + 5/2 + 2)
8 + 10 + 4 – (3 + 5/2)
22 – 11/2 = = 22-11/2
= 44- 11/2
= 33/2

(8bi)
T = 2π square l/g
Dividing both side by 2π
T/2π = square root of L/g
By squaring both side
(T/2 π)² = (L/g)²
T²/4 π² = l/g
Cross multiplication
gT²/T² = 4π²L/T²
g = 4π²L/T²

(8bii)
T = (0.4)1/2 = square root of 0.4
L = 0.04 T1 = 3.14
G = 4π²L/T2
= 4 x (3.14)² x 0.04/(0.4)²
= 4 x 9.8596 x 0.04/0.4
= 1.578/0.4
g = 3.94


(9a)
2p-q=10…….(1)
3p+q^2=22……..(2)
Eq(1) x3 and eq(2) x2
6p-3q=30
6p+2q=44
Subtracting eq(1) from eq(2)
-3q-2q= -14
2q^2+3q= 14
2q^2+3q-14=0
2q^2+7q-4q-14=0
q(2q+7)-2(2q+7)=0
(q-2)(2q+7)=0
q-2=0 or 2q+7=0
q=2 or q= -7/2
Substitute q into eq(1)
2p-q=10
2p-2=10
2p=10+2
2p=12
p=12/2=6
When q= -7/2
2p-q=10
2p-(-7/2)=10
2p+7/2=10
4p+7=10
4p=10-7
4p=3
p=3/4

(9b)
z^2 -25/z^2-9z+20
If Z is undefined
Z^2-9z+20=0
Using factorization method
-4z and -5z
Z^2 -4z-5z+20=0
Z(z-4) – 5(z-4)=0
(z-4)(z-5)=0
Z-4=0 or Z-5=0
Z=4 or Z=5
Z=4 or 5
Z is undefined when it is equal to 4 or 5


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