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