Nabteb 2017 Mathematics (Essay & Obj) Answers – May/June Expo

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VERIFIED NABTEB MATHS
OBJ:
1-10=DBDCDABBCD
11-20=DCDBDCDAAB
21-30=BCDAADCACD
31-40=DABABCBDCD
41-50=ABBBAABDAD

COMPLETED

MATHS-THEORY ANSWERS

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

SECTION B
(ANSWER ONLY FOUR QUESIONS)

QUESTION 8,9,10,11

(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

GOODLUCK

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