Good Day, The National Business and Technical

Examination board has commenced

and Galantwap Team, collaborating with

some NABTEB invigilators has

promised to make this years NABTEB a fruitful one. We serve ANSWERS, EXPOS, RUNS to

our subscribers two hours to the

commencement of the examination,

we wouldn’t want to lie that its any

earlier than that like other sites do,

we send our EXPOS orderly, complete and at once. Our SUBSCRIBERS have nothing to

worry about as we assure you that

the best will be served and your

result would be a TESTIMONY.

MATHS Obj:

1-10=DBDCDABBCD

11-20=DCDBDCDAAB

21-30=BCDAADCACD

31-40=DABABCBDCD

41-50=ABBBAABDAD

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MATHEMATICS ESSAY:

INSTRUCTION:ANSWER questions 1 to 5 and any other four questions

MATHS THEORY

10.30am-1:00p.m

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SECTION A(ANSWER ALL QUESTION)

QUESTION 1 -5)

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1a)

1 4/5 × 2 1/3 / 3 3/4 – 4/5 × 2/3

=9/5 × 7/3 / 18/5 – 4/5 × 2/3

=21/5 / 18 – 4/5 × 2/3

=(21/5 ÷ 14/5) × 2/3

=21/5 × 5/14 × 2/3

=1

1b)

x^2 +5x – 6= 0

(x^2 + 6x) – ( x– 6) = 0

x(x+6) –1 ( x+6)= 0

(x–1) (x +6)= 0

x – 1 =0 or x+6 = 0

x=1 or x =– 6

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2a)

U ={2,3 ,4 ,5, 6, 7,8 ,9}

A ={2,3 ,5 , 7}

B ={3,6,9 }

i)AUB ={2,3,5,6,7,9}

ii)

A’nB’

A’ ={4,6, 8,9,}

B’ ={2,4,5,7,8,}

A’nB’ {4,8}

2b)

161n = 32less down 5

1×n^2+6×n^1+1×n° = 3× 5^1+2×5°

n^2 + 6n + 1 = 15 + 2

n^2 +6n + 1 =17

n^2 + 6n – 16 = 0

(n^2 +8n) – (2n – 16) = 0

n(n + 8) –2 (n + 8) = 0

(n – 2)(n + 8) = 0

n – 2 = 0 or n + 8 =0

n = 2 or n = –8

hence,

n = 2

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3a)

1/2logy^8 = 2

logy^√81 =2

logy^9 =2

y^2 = 9

y = √9=3

y = 3

3b)

0.016 × 0.048 / 0.64

=16 ×10^-3 / 64 × 10^-2

=16 × 48 × 10^-6 / 64 × 10^-2

=768 / 64 × 10^-6 × 10^2

=12 × 10^-4

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4)

h/8 = h + 20/12

12h = 8h + 160

4h =160

h = 160/4

h =40cm

Hence ,H =h +20=40 +20 =60cm

H=60cm

Volume of bucket =

1/3πR^2H – 1/3πr^2h

=1/3π(R^2H – r^2h)

=1/3(3.142)( (12)^2(60) – (8)^2(40))

=1/3(3.142)(8640 – 2560)

=1/3 × 3.142 × 6080

volume is =6367.7cm^3

since ,

1litre = 1000cm^3

Capacity =6367.7 / 1000

=6.3677litre

=6.4litre

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5a)

2/3(x – 2) – x –1/x–2

=2–3(x –1)/3(x–2)

=2 – 3x + 3/5x – 6

=5–3x/3x–6

5b)

x^2 + 3x + 2/ x^2 – 4

=(x^2 + x) + (2x + 2)/(x+2)+(x –2)

=x(x+1) +2(x+1)/(x+2) +(x – 2)

= (x+2) (x – 2)/(x+2)(x –2)

= x+1/x –2

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SECTION B

ANSWER ONLY FOUR QUESIONS

QUESTION 8,9,10,11

POSTED

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8a)

if (x – 6), 2x and (8x – 20) are consecutive terms of G.P

the common ratio is

r = 2x / x– 2 ——–(1)

r = 8x + 20 / 2x ——–(2)

Equating (1) and (2)

2x/x – 6 = 8x + 20/2x

4x^2 = (x – 6) (8x + 20)

4x^2 = 8x^2 + 20x – 48x – 120

4x^2 – 28x – 120 = 0

x^2 – 7x – 30 = 0

Solving quadratically,

(x^2 + 3x) – (10x – 30) = 0

x(x + 3) –10 (x + 3) = 0

(x – 20) (x + 3) = 0

x – 10 =0 or x + 3 = 0

x = 10 or x = -3

8b)

√72 × 3√18 × 14√6 / 2√24 ×√12 √36×2 × √9×3 × 24√6 /2√4×6 × √4×3

6√2 x 9√2 × 14√16 / 4√6 × 2√3

6 × 9 × 2 × 14 √6 / 4√6 × 2√3

3 × 9 × 9 / √3

by rationalizing the denomenator

= 189/√3 × √3/√3

=189√3 / 3

=63√3

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9a)

x =30°

y =θ = 180 – 60

θ=120° , r = 5cm

the length of the chord AC

L =2rsinθ/2

=2 × 5 × sin120/2

=10sin60

=10 × 0.866

Length of chord is =8.66cm

9aii)

Area of shaded segment = area of sector – Area ot triangel

θ/360 × πr^2 – 1/2 (5)^2sinθ

=120/360 × 3.142 × (5)^2sin120

=120 × 3.142 × 25 /360 – 25 × sin120/2

=26.18 – 10.825

=15.358cm^2

9b)

Speed = Distance / time

Distance = 500 × 2

=1000km

x^2 =(1000)^2 + (450)^2 -2(1000)(450)cos120

=1000000 + 202500 +450,000

x^2 =1652500

x =√1652500

x = 1285.5km

The bearing the airport (θ)

450/sinθ = 1285.5/sin120

sinθ = 450 × sin120 / 1285.5

sinθ =0.3032

θ = sin(0.3032)= 17.6°

=18°

10a)

blue marble = 3

white marble = 2

Red marble = 4/9

i)

Pr (both of them will be red)

first drawn = 4/9

second drawn = 3/8

Pr (both red) =4/9 × 3/8

=1/8

ii)

Pr (the two are of the same color)

= RR or BB or WW

=(4/9 × 3/8)+ (3/9 × 2/8)+ (2/9×1/8)

=12/72 + 6/72 + 2/72

=12+6+2 / 72

=20/72

=5/18

10b)

sine(2θ – 30°) and (3θ – 45)are supplementary that there sum is 180°

2θ – 30 + 3θ–45=180

5θ –75 =180

5θ=180 + 75

5θ =225

θ=51°

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11a)

hence, volume of hemispherical portion is half of the volume of the cone

Volume of hemisphere=2/3πr^3

Volume of cone =1/3πr^3

2/3πr^3 = 1/2(1/3πr^2 h)

2/3πr^3 = 1/6 πr2 h

h = 2 × 6 × πr^3 / 3πr^2 = 4r

h = 4r = 4 × 4 = 16cm

hence the vertical angle is

tanθ = r/h

tanθ = 4/6 = 1/4

θ =tan(0.25) = 14.036°

θ =14°(correct to nearest degree)

11b)

Total volume of solid

=1/3πr^2h (h + 2r)

=1/3 × 22/7 × (4)^2 (16 + 8)

=1/3 × 22/7 × 16 × 24

= 22 × 16 × 24 / 21

= 402.285cm^3

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