## 2018 Nabteb Gce Mathematics Expo Answers

NABTEB GCE MATHEMATICS WELL ARRANGE VERSION FROM NO. 1 – 15

(1a)

6n+¹ x 9n x 4²n/18n x 2n x 12²n

= 6n+¹ x 3³n x 4²n/(6×3)n x 2n x (4×3)²n

= 6n+¹ x 3³n x 4²n/6n x 3n x 2n x 3²n x 4²n

= 6n+¹/6n x 6n = 6n+¹/6²n

= 6n+¹ – ²n

= 6¹-n

(1b)

Log243/log27

= log3⁵/log3³

= 5log⁵/5log³

= 5/3

(2a)

3t – 2p = 8 ——(i) x2

2t – 3p = 14 —- (ii) x3

6t – 4p = 16 ——–(iii)

6t – 9p = 42 ——–(iv)

Subtract equation (iv) from equation (iii)

– 4p + 9p = 16 – 42

5p = 26

P = -26/5

Therefore p = -5 1/5

From the equator (ii) above

3t – 2p = 8 substitute -26/5 for p

3t – 2 (-26/5) = 8

3t + 52/5 = 8

3t = 8 -52/5

3t = 40 – 53/5

3t = – 12/5

t = -12/5 ÷ 3/1

t = 12/5 x 1/3

t = -4/5

therefore t = -4/5, p = -5 1/5

(2b)

3√3(2/√3 – √12/6)

= 3√3 (2√3 – 2√3/6)

3√3 (2√3 – 3√3)

= 6/3 – 1

= 6 – 3/3

= 3/3 = 1

(3)

Given that;

Pr (Rabiru will pass the examination) = 2/3

Pr (Yahana will pass the examination) = 5/8

Pr (Ibrahim will pass the examination) = ¾

And the Pr (Kabra will fail) = 1 -2/3 = 1/3

Pr (Yahana will fail) = 1 -5/8 = 3/8

Pr (Ibrahim will fail) = 1 -3/4 = ¼

(a) Pr (three will pass) = 2/3 x 5/8 x ¾

(b) Pr (non of them will pass) = 1/3 x 3/8 x ¼ = 1/32

(c) pr (Kabiru and Yahana only will pass)

= Pr (K, Y, I)

= 2/3 x 5/8 x ¼

= 5/48

(4)

DRAW THE DIAGRAM

Let |GT| = |TB| = x

|GN| = |GT| = |TN| = x + x = 2x

|GN| = 2x

From ∆ FGN

Tan 68 = 8/|GN|

|GN| = 8/tan68 = 8/2.4751

|GN| = 3.23m

|GN| = 2 |GT|

|GT| = 3.23/2 = 1.616m

(4i) from the ∆ FGT

tanØ = 8/1.616

tanØ = 4.95

Ø = tan-¹ (4.95) = 78.58º

< FTG = 78.58 = 79º (correct to the nearest degree)

(4ii)

< TFN = 〆

〆 + β- = y

Y = 90 – 68 = 22

β + 0 = 90

β = 90 – 0 = 90 – 79 = 11º

hence, 〆 = y – β

〆 = 22 – 11

〆 = 11º

Therefore <TFN = 11º

(5a)

4 ½ – 3 (y – 2) = 2y +1/3

9/2 – 3 (y-2) = 2y+1/3

Multiply both sides by 3

27/2 – 9 (y-2) = 2y + 1

27 – 18 (y-2) = 2(2+1)

27 – 18y + 36 = 4y + 2

Collect the like terms

-18y – 4y = 2 – 36 – 27

– 22y = 61

Y = -61/-22

Y = 2 17/22 or 2.7727

(5bi)

7 – 2y ≥ 16

(5bii)

7 – 2y ≥ 16

-2y ≥ 16-7

-2y ≥ 9

Y ≤ -9/2

Therefore y ≤ -4 ½

(5biii)

The greatest possible value of y that satisfy the equation = -4 ½

(6)

Let the cost of water melon be x

And let the cost of mangoes be y

12x + 24y = 432 ——-(1) x2

24x + 12y = 360 ——-(ii) x1

24x + 42y = 864 ——–(iv)

Subtract equation (iv) from equation (iii)

36y = 504

Y = 504/36

Y = N14

From equation (i) above

12x + 24y = 432, substitute 14 for y

12x + 24(14) = 432

12x = 432 – 336

12x = 96

Therefore x = N8

(6ai)

The cost of water melon per kg is N8

(6aii) The cost of mangoes per kg is N14

(6aiii)

3x + 2y

3(8) + 2(14)

24 + 28 = 52

The cost of 3kg of water melon and 2kg of mangoes is N52

(6b)

123x = 83 base 10

123ºx = 83

1 x x² + 2x x+3 x xº = 83

x² + 2x + 3 = 83

x² + 2x – 80 = 0

x² + 10x – 8x – 80 = 0

x (x+10) – 8 (x+10) = 0

(x-8) (x+10) = 0

X = 8 or -10

(7a)

(TABULATE AND DRAW A TABLE) p = 3+5t – 2t²

t | -2 | -1 | 0 | 3 | 4 | 5 |

t² | 4 | 1 | 0 | 9 | 16 | 25 |

3 | 3 | 3 | 3 | 3 | 3 | 3 |

5t | -10 | -5 | 5 | 15 | 20| 25 |

-2t² | -8 | – 2 | -2 | -18 | -32 | -50 |

P | -15 | -4 | 6 | 0 | -9 | -22 |

(7b)

DRAW THE GRAPH

(7c)

The root of the equation are -1/2 and 3

(7d)

(TABULATE AND DRAW A TABLE) p = 3 – 2t

t | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |

3 | 3 | 3 | 3 | 3 | 3 | 3 | 3 | 3|

-2t | 4 | 2 | 0 | -2| -4 | -6 | -8 | -10 |

P | 7 | 5 | 3 | 1 | -1 | -3 | -5 | -7 |

P = 3 – 2t

P = 3 + 5t – 2t²

3 + 5t – 2t² > 3 – 2t

5t – 2t² > -2t

7t – 2t² > 0

t (7 – 2t) > 0

t > 0 or 7 – 2t > 0

t > 0 or -2t > -9 or t <7/2

the range is 0 < t 7/2

0 < t < 3.5

(8a)

X (40ºS, 15 ºW)

Y (40ºS, 45ºW)

(DRAW THE CIRCLE)

Distance along the parallel of latitude = Ø/360 x 2πr

ˉXY = Ø/360 x 2πR cos 〆

= (45-15)/360 x 2 x 22/7 x 6400 x cos 40

= 30/360 x 44 x 6400 x 0.7660/7

= 30 x 44 x 6400 x 0.7660/2520

= 6471.168/2520 = 2567.9km

Distance ˉXY = 257km to the nearest 100km

(8bi)

Z (xºN, 45ºW)

Y (40ºS, 45ºW)

ˉYZ = 6700km

Speed = total distance/total time

850 = (2567.9 + 6700/total time

Total time = 9267.9/850 = 10.9hr

Total time = 10.9hrs approximate 11hrs

(8bii)

Latitude Z

Distance ˉ YZ

ˉYZ = Ø/360 x 2πR

6700 = (x + 40)/360 x 2 x 22/7 x 6400

6700 x 2520 = 281600 (x+40)

x + 40 = 16884000/281600

x + 40 = 59.96

x = 59.96 – 40

x = 19.96º

hence, the distance of Z to the nearest degree = 20ºN

(9)

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)

TABULATE

Scores x | frequency (f) | YF

2 | 2 | 4

3 | 4 | 12

4 | 5 | 20

5 | 3 | 15

6 | 4 | 24

7 | 2 | 14

– | 20 | 89

(11ai)

Mode is the most frequently occurring number from an observation

Therefore the modal score is 4

(11aii)

The median score = 4 + 4/2 = 8/2 = 4

Therefore the median score is 4

(11aiii)

Mean = Epx/Ef

= 3a/20

Therefore mean = 4.45

(11b)

A = P(1+r/100)n

= 53,000(1+7/100)⁵

= 53,000 (1.07) ⁵

= 53,000 (1.402551731)

A = 74,335.24

Compound interest = 74,335.24 – 53,000

Therefore compound interest = 21,335.24

(12a)

Monthly salary – 6,000.00

Commission = (5% x 8,400 x 52) – 21,840.00

Take home income – 27,840.00

(12b)

Gross pay – N125,018.35

Less:

Income tax (5% x 125,018.35) – N 6250.92

Union dues (2% x 125,018.35) – N 2,500.37

Housing fund (1% x 125,018.35) – N 1,250.18

Pension schemes (7.5% x 125,018.35) – N 3,125.46

Cooperative contribution (10% x 125,018.35) – N 12,5018.84 (23,754.15)

Net pay – N 101,264.20

(14a)

N 1 shares at N 1.10 divided 11%

Yield

Divided 11% x 1 – N 0.11

Less premium – (N 0.10)

Yield – N 0.01

N 1 shares at N 2.50 divided at N 20%

Yield

Divided 20% x 1 – N 0.20

Premium (2.50 – 1) – (N 1.50)

Loss (N 1.30)

N share at N 1.10 divided at 11% gives the highest yield.

(14b)

A = p[(1+r)n-1]/0.04

= 60,000 [(1.04)⁴-1]/0.04

= 60,000 [1.04)⁴-1]/0.04

= 60,000 (1.1698585 – 1)/0.04

A = N 254,787.84

(15a)

Annual tax free allowance:

Rent (15% x 365,500 – 54,825

Personal (12% x 365,500) – 43,680

= 98,685

Therefore annual tax free allowance = N98,685

(15b)

Calculation of annual income taxi

Gross salary – 365,500

Less tax free allowance – 98,685

Taxable income – 266,815

Tax payable:

Taxable income – 266,815

1st 65,000 (a) 0% – (64,000)

Balance after 1st scale – 201,815

Next 100,000 (a) 10% – (100,000)/101,815 10,000

Balance (a) 201% (1,815) 363

Total tax = 25,363

His annual income tax is N25,363

(15c)

Percentage of salary paid as tax

= 25,363/364.500 x 100/1

Therefore percentage of his salary paid as tax approximately 6.9%