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**NABTEB GCE Mathematics Questions and Answers**, NABTEB GCE 2018 Mathematics Questions and Answers, NABTEB GCE 2018 Mathematics Expo, NABTEB GCE 2018 Mathematics Runs, NABTEB GCE 2018 Mathematics Choke, NABTEB GCE 2018 Mathematics Dubs, NABTEB GCE 2018 Mathematics Link, **NABTEB GCE Mathematics Questions and Answers**, NABTEB GCE Mathematics Expo, NABTEB GCE Mathematics Runs, NABTEB GCE Mathematics Choke, NABTEB GCE Mathematics Dubs, NABTEB GCE Mathematics Link

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# 👑 2018 NABTEB GCE MATHEMATICS ANSWERS👑

__100% VERIFIED MATHEMATICS OBJ ANSWERS__

1-10: DCDBABBAAD

11-20: CDBDCCCBDA

21-30: CDABDACACC

31-40: BABCABABBC

41-50: DCDCCCDBDA

__100% VERIFIED MATHEMATICS THEORY ANSWERS__

__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

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(4)

Draw the triangle

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

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(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

(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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__IMAGE SOLUTIONS:__

(1a & 3)

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(1b & 4)

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(2)

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(5)

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(6)

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(7)

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(8)

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(9)

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(11)

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(12)

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(13)

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(14)

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(15)

**======COMPLETED======**

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