2018 NECO GCE Mathematics Questions and Answers

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👑 2018 NECO GCE MATHEMATICS ANSWERS👑

100% VERIFIED MATHEMATICS OBJ ANSWERS

1-10: CCDEAEBDCE
11-20: AAACEECACE
21-30: CBBAACEBAD
31-40: CDABEEBCAD
41-50: ECCECEDEAC
51-60: EBABCCCCCB

100% VERIFIED MATHEMATICS THEORY ANSWERS

SNAPPED ANSWERS

5. CLICK HERE

6. CLICK HERE and HERE

7. CLICK HERE and HERE

10. CLICK HERE and HERE

TYPED ANSWERS

(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)
1101base2 = 2x+1
1 * 2^3 + 1 * 2^2 + 0 * 2^1 + 1 * 2^0 = 2x + 1
8+4+1=2x+1
13 = 2x+1
13-1=2x
12=2x
x = 12/2=6

(2b)
sinx=12/13= 0.9231
x=sin^-1 0.9231=67.4degree
.: 3sinx+1/2cosx
= 3sin67.4 + 1/2cos67.4
= 3*0.9232 + 1/2*0.3843
=2.7696+0.1921
=2.9617
=2.96
===================================

(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
=====================================

(5) [ CLICK HERE ]
Tabulate
X|8,9,10,12,14,17
F|6,4,8,5,4,3,(30)
FX|48,36,80,60,56,51,(331)

(i)
Mean = ƩFX/ƩF=331/30=11
Tabulate
d=x-x̅| -3.03,-2.03,-1.03,-0.97,-2.97,-5.97
d^2| 9.18,4.12,1.06,0.94,8.82,35.64
fd^2| 50.08,16.48,10.60,11.28,123.48,605.88,(8178)

(ii)
Standard deviation
=√Ʃfd^2/Ʃf
=√817.8/30
=27.26
====================================

(6a) [ CLICK HERE ]

126y = 86
Convert all to base 10
(1*y²)+(2*y¹)+(6*y°)=86
y²+2y+6=86
y²+2y+6-86=0
y²+2y – 80 = 0
Using -b±√b² – 4ac/2a
Where a = 1, b=2, c=-80
-2±√2²-4*1*-80/2*1
y = -2±√4+320 = -2±√324/2
y = -2±18 = -2+18/2 = 16/2 = 8
Hence the positive value of y is 8

(6b) [ HERE ]
Error in Base = 93%
Error in height = 100 + 5
Hence error in Area = 1/2bh
= 1/2(93)×(105) = 9765/2 = 4883
Actual Base = 100
Actual error = 100
Actual area = 1/2×100×100 = 5000
Hence % Error in Area =
Error in Area/Actual Area × 100/1
=9765/10000×100/1 = 97.65%
OR
4883/5000 × 100/1 = 97.66%

(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
================================
(7)
name
name

================================

(8a)
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 [ CLICK HERE ]

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 [ HERE ]

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
=================================

(12a)
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 FREQUENCY 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

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