Nabteb Gce Mathematics Answers – Nov/Dec 2018 Expo

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

 

 

 

 

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


NABTEB GCE 2018 MATHEMATICS Expo Answers


*NABTEB GCE 2018 O’LEVEL SOLUTIONS*

*O’level Nabteb Mathematics obj*

*1 DABBABBBAD*
*11 CDBDCCCBBA*
*21 CDABDACCCC*
*31 BCBCABABBC*
*41 DACCCCDBDB*

*SECTION A*
*_(Answer ALL QUESTIONS From This Section)_*

(1a)
6^n+1*9^n*4^2n/18^n*2^n*12^2n
=(2*3)^n+1*(3*3)^n*(2*2)^2n/(2*3*3)^n*2^n*(2*2*3)^2n

=2^n+1*2^n+1*3^n*3^n*2^2n*2^2n/2^n*3^n*3^n*2^n*2^n*2^n*3^2n

= 2^n*2^1*2^n*2^1*3^n*3^n*2^n*2^n*2^n*2^n/2^n*3^n*3^n*2^n*2^n*2^n*2^n*2^n*3^n*3^n

= 2^6n*3^2n*2*2/2^6n*3^4n

= (2^6n / 2^6n) *(3^2n / 3^4n)*2*2

=2^6n – 6n * 3^2n – 4n * 4

= 2^0 * 3^-2n * 4

= 4*3^-2n

= 4 * 1/3^2n

= 4/3^2n

(1b)
Log 243/log 27
= log 3^5/log 3^3
= 5log3/3log3 = 5/3 = 1 whole no 2/3

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

(2a)
3t – 2p = 8×2 …(I)
2t – 3p = 14×2…(II)

6t – 4p = 16 …(III)
6t – 9p = 42 …(IV)
9P – 4P = -42 + 16
5p = -26
P = -26/5 = -5 1/5

Put p = -26/5 into eqn (I)
3t – 2(-26/5) = 8
3t + 52/5 = 8
3t = 8 – 52/5
3t = 40 – 52/5 = -12/5
Divide both sides by 3
36/3 = -12/5 / 3
t = -12/5 × 1/3
= -12/15 = -4/5

(2b)
3/√3(2/√3 – √12/6)
2/√3 × 3/√3 – 3/√3 × √12/6
6/√9 – 3√12/6/√3
6/3 – 3√12/6√3
2 – √12/2√3
2 × 2√3 – √12/2√3
4√3 – (√4 × 3)/2√3
= 4√3 – 2√3/2√3
= 2√3/2√3 = 1

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

(3)
P(k) = 2/3
P(Y) = 5/8
P(I) = 3/4
P(k fail) = 1- 2/3 = 1/3
P(Y fail) 1 – 5/8 = 3/8
P(I fail) = 1 – 3/4 = 1/4

(a) 2/3*5/8*3/4 = 5/16
(b) 1/3*3/8*1/4 = 1/32
(c) 2/3*5/8*1*4 = 5/48

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

(4)
Draw the triangle
<GFN = 180 – ( 68+90)
Sum of angles in a Δ
= 180 – 158 = 22°
Sin 22/|GN| = Sin 68/8
|GN|= 8sin22/sin68
= 8×0.3746/0.9272
= 2.9969/0.9272
= 3.2322cm

|GT| = |TN| = 3.2322
= 1.616≈ 1.62cm
Let tita = <FTG

(i) tan tita = 8/1.6161 = 4.9502
tita = tan^-¹(4.9502)
tita = 78.6°

FTN = 180 – tita
= 180 – 78.6
(sum of angles on a straight line) = 101.4

(ii) >TFN
= 180 – (68+101.4)
Sum of angles in a Δ
180 – 169.4
=10.6°

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

(5a)
4 1/2 – 3(y – 2) = 2y + 1/3
= 9/2 – 3y + 6 = 2y + 1/3
= -3y + 6 + 9/2 = 2y + 1/3
6 + 9/2 = 2y + 1/3 + 3y
12 + 9/2 = 2y + 1/3 + 3y
21/2 = 2y + 1 + 9y/3
21 × 3 = 2(2y + 1 + 9y)
63 = 4y + 2 + 18y
63 = 22y + 2
22y = 63 – 2
22y = 61
y = 61/22 = 2 17/22

(5b)
Let the number be y
7 – 2y >_ 16
-2y >_16 – 7
-2y >_ 9
y _< _9/2 = 4 1/2
The greatest possible values of y are; -4, -3, -2, -1, 0

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

*SECTION B*
_(All Candidates should Answer FOUR while SECRETARIAL & BUSINESS CANDIDATES Should Answers Only TWO from this Section)_

(6a)
Let the water melon be x
Let the mango be y

12x + 24y = 432 …(i) × 24
24x + 12y = 360…(ii) × 12

288x + 576y = 10368 …(iii)
288x + 144y = 4320 … (iv)
-144y + 576y = -4320+10368
432y = 6048
y = 6018/432 = 14

Put y = 14 into eqn (i)

12x + 24(14) = 432
12x + 336 = 432
12x = 432 – 336
12x = 96
X = 96/12 = 8

(i) Water melon per kg = 8
(ii) Mango per kg = 14
(iii) (3 × 8) + (2 × 14)
= 24 + 28
= 52

(6b)
123x = 83ten
1*X² + 2*X¹ + 3*X^0
= 83

X² + 2x + 3 = 83
X² + 2x + 3 – 83 = 0
X² + 2x – 80 = 0
X² + 10x -8x – 80 = 0
X(x+10) – 8(x+10) = 0
Then either
X – 8 = 0 or X + 10 = 0
X = 8 or X = -10

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

(8a)
Distance /xy/ = tita/360 × 2πr
tita = 45° – 15° = 30°
r = Rcosα
r = 6400 × cos40°
r = 6400 × 0.7600
r = 4902.4km
Distance /xy/ = 30/360 × 2 × 22/7 × 4902.4
= 30 × 2 × 22 × 4902.4/2520
=6471168/2520 = 2567.9238
= 2600km

(8bi)
Time taken to fly from X to y
Speed = Total distance/Total time taken
850 = 2600/t
t = 2600/850 = 3.0588hrs
t = 3hours

Time taken to fly to Y from X
Time = Distance/Speed
= 6700/850 = 7.8824hrs = 8hrs
Total time = 3hrs + 8hrs = 11hrs

(8bii)
Latitude of Z
Distance /yz/ = tita/360 × 2πr
6700 = Z-40/360×2×22/7×6400
6700×360×7 =(Z – 40)281600
16884000 = 281600Z – 11264000
281600Z = 16884000 + 11264000
281600Z = 28148000
Z = 28148000/281600
Z = 99.9574 = 100°

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

(9a)
DRAW THE TRIANGLE DAIGRAM
|YX| = x + 620
From ∆ YFP
Tan56 = R/620
R = 620 x tan56
= 620 x 1.4826
h = 919.19m
To find x
Tan 20 = h/x
X = h/tan20 = 919.19/0.3639
X = | XF | = 2525.45m
|YX| = |XF| + |FY|
|YX| = 2525.45 + 620
= 3145.5.45m
Hence the value of |YX| correct to the four significant figure is 3145m
|XY| = 3145m

(9b)
DRAW THE DAIGRAM
|YX|² = |OY|² + |OX|² – 2|OY| |OX| cosØ
(6)² = (5)² + (5)² – 2 (5) (5) cosØ
36 = 25 +25 – 50cosØ
36 = 50 – 50cos Ø
50cosØ = 50 – 36
50cos = 14
cosØ = 14/50 = 0.28
Ø = cos-¹ (0.028) = 73.34º
Therefore <XZY = ½ (73.74) (Angle at the center twice the angle at the circumference)
<XZY = ½ (73.74) = 36.87º
<XZY = 37º (nearest degree)

(9bii)
Let |XZ| = |YZ| = y
|XY|² = |YZ|² + |XZ|² – 2 |YZ| |XZ| Cos〆
(6)² = y² + y² – 2y² cos 36.87
36 = 0.4y²
y² = 36/0.4 = 60
y = √60 = 7.75m
hence, |XZ| = 7.8m (1dp)

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

(11a)
In a tabular form

Under scores(x)
2, 3, 4, 5, 6, 7

Under frequency(f)
2, 4, 5, 3, 4, 2 Ef, = 20

Under fx
4, 12, 20, 15, 24, 14 Efx = 89

(i) Modal score = 4
(ii) Median = n1+n2 = 4+4/2 = 8/2 = 4
(iii) Mean X = Efx/Ef = 89/20 = 4.45

(11b)
A = P(1+r/100)n
A = amount in compound interest.
P = principal, r = rates, n = no of years the compound interest is charged.
A = 53000(1+7/100)^5
= 53000(1+0.07)^5 = 53000(1.07)^5

In a tabular form
No | Log
53000| 4.7243 = 4.7243
| 0.0294×5 = 0.1469
4.8712
Antilog of 4.8712 = ₦74436.14
Compound interest = 74436.14 – 53000 = ₦21336.14

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

*SECTION C*
_(Only For SECRETARIAL & BUSINESS Candidates. They shoud answers only TWO from this section)_

(12a)
If passengers paid ₦8,400.00 per trip
Commission on each trip = 5/100 × 8400 = ₦420.00

For 52 trips, his commission would be 52 × ₦420 = ₦21,840.00

His take home for the month = ₦6000.00 + ₦21,840.00 = ₦27,840.00

(12bi)
Gross pay = ₦125018.35 per month.
5% income tax = 5/100 × 125018.35 = 6250.9175

2% Union dues = 2/100 × 125018.35 = 2500.3670

1% housing fund = 1/100 × 125018.35 = 1250.1835

7.5% pension scheme = 7.5/100 × 125018.35 = 9376.3763

2.5% health insurance scheme = 2.5/100 × 125018.35 = 3125.4588

10% cooperative contribution = 10/100 × 125018.35 = 12501.8350

Total deductions = 6250.9175+2500.3670+1250 1835+9376.3763+3125.4588+12501.8350 = 35005.14

(12bii)
His net pay = ₦125018.35 – 35005.14 = ₦90013.21

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

(13a)
4 – point moving average.
16, 18, 20, 27, 28, 31, 32

16+11+20+27/4 = 81/4 = 20.25

18+20+27+28/4 = 93/4 = 23.25

20+27+28+31/4 = 106/4 = 23.25

20+27+28+31/4 = 106/4 = 26.5

27+28+31+32/4 = 118/4 =
29.5

(13b)
First 20weeks, she earns ₦400×20 = ₦8000

The next 20weeks she earns ₦720×20 = ₦14400

Total earns for 40weeks = ₦8000 + ₦14400 = ₦22400

Total earns for the whole
year = ₦52×600 = ₦31200

Amount earns for the remaining 12months = ₦31200 – ₦22400 = ₦8800

Average weekly earns for the last 12 months = 8800/12 = ₦733.33

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

(15a)
Rent = 15/100 * ₦365,500 = ₦54,825
Personal allowance = 12/100 ×₦365,500 = ₦43,860
Annual tax free allowance
= ₦54825 + ₦43860 = ₦98685

(15b)
Taxable income = ₦(365500 – 98685) = ₦266,815
Tax calculation
The first ₦65,000 = 0% = ₦0

Next ₦100,000 = 10/100 × 100,000 = ₦10,000

Next 100,000 = 15/100 × 100,000 = ₦15,000

The remaining ₦1815 = 20/100 × 1815 = ₦363

Annual income tax = ₦(0 + 10,000 + 15,000 + 363)
=₦25,363

(15c)
Percentage of his salary paid as tax = 25,363/365,500 × 100 = 6.94%

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

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