2018 Neco Gce Mathematics Obj And Theory Answer

2018 NECO GCE Mathematics Obj And Theory ANSWERS

Refresh Every 10mins Starting From 2hrs To Exam


1 CCBEAEBECE
11 AADCEECABB
21 CBBAACEBAD
31 CAABEEBCAD
41 ECCECEDDAC
51 ACABCCCCCB
===============

*NECO GCE MATHS*

(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)COMPLETED
(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
Area = ½ x 93x/100 x 107x/100
=963x/2000
% error = actual Area – wrong/actual area x 100
= x – 963x/2000 /x 100
= 20000x – 963x/20000 x 100
= 19037/20000 x 100
= 95.185%

(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

(10)
Draw the triangle

Hence we have A+B+C = 180
50+80+C = 180
130+C = 180
C = 180 – 130
C = 50

(i)
The bearing of B from C = 90 – 50 = 40°

(ii)
Bearing of A from B
90+50(alternate angle to A)
= 140°

(iii)
Distance between B and C
Using sine rule
C/sinC = a/SinA
40/sin50 = a/sin50
Cross multiply
asin50 = 40sin50
a = 40km
Hence distance between B and C = 40km

(iv)
Using cosine rule
b²= a²+c² – 2acCosB
where a = 40, c= 40 and B=80
b² = 40²+40² -2*40*40cos80
b² = 1600+1600 – 3200(0.1736)
b² = 3200 – 555.52
b² = 2644.48
b = √2644.48
b = 51.42km

(v)
Height of ΔABC
Draw the triangle

Area of the triangle
=1/2acSinB
=1/2*40*40sin80
=800(0.9848)
=787.84km²
Hence 1/2bh=787.84
bh=2*787.84
h=2*787.84/b
h=2*787.84/51.42
h=30.64
=31

(12)
TABULATE:
Marks | frq | c.f | boundary
11-20 | 8 | 8 | 10.5 – 20.5
21-30 | 6 | 14 | 20.5 – 30. 5
31-40 | 10 | 24 | 30.5 – 40.5
41-50 | 12 | 36 | 40.5 – 50.5
51-60 | 8 | 44 | 50.5-60.5
61-70 | 6 | 50 | 60.5 – 70.5

(12b)
DRAW THE CUMULATIVE FREQUENCE CURVE

(12ci)
Medium mark = N + ½
= 50 + 1/2 = 51/2
= 25. 5
= 48

(12cii)
Lower quartile Q1
= ¼ x 50
= 12.5
= 28

(12ciii)
Upper quartile Q2
= ¾ x 50
= 37.5
=53

(12civ)
Both percentile
70/100 x 50
= 35
= 50

More loading…


CLICK HERE TO JOIN OUR WHATSAPP GROUP

For CRS or IRS Each

#500 WhatsApp

#400 Pin

Send as MTN Cards to +17036594269 (via WhatsApp) or CLICK HERE

SERIEZLOADED REMAINS THE BEST

CLICK HERE TO LEARN HOW TO EARN $300 DAILY AT HOME!!!

2019 Scholarship
2019 Scholarship
SHARE THIS POST WITH FRIENDS BELOW Click Here To Comment
Check Also  2018 Neco Gce Government Obj And Theory Answer

JOIN SERIEZLOADED.COM.NG TODAY!!!

Promote your Music's /Videos on Seriezloaded, WhatsApp 08077722280 or Send Them a Mail To[email protected]

     

About SL EXAM BOSS 24 Articles
100% Sure Expo Boss, Contact Me on : [email protected] I deliver Asap

Be the first to comment

Leave a Reply

Your email address will not be published.