NABTEB 2017 MATHEMATICS Expo (May/June) Answers

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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
===================
MATHEMATICS ESSAY:
INSTRUCTION:ANSWER questions 1 to 5 and any other four questions
MATHS THEORY
10.30am-1:00p.m
======================
SECTION A(ANSWER ALL QUESTION)
QUESTION 1 -5)
======================
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
=====================
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
=====================
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
=======================
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
=====================
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
===============

ALSO READ:   NABTEB 2017 ENGLISH LANGUAGE Expo (May/June) Answers

==========================
SECTION B
ANSWER ONLY FOUR QUESIONS
QUESTION 8,9,10,11
POSTED
==========================
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
===================
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°
====================
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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