### 2018 NABTEB MATHEMATICS QUESTIONS AND ANSWERS NOW AVAILABLE

(1a)

1 1 0 1 1 base 2

1 1 0 1 base 2

1 0 1 base 2

= 1 0 1 1 0 1 base 2

Hence,

1101 base 2 + 101 base 2 + 11011 base 2 = 101101 base 2

(1b)

Length = 3.0m

Width = 1.8m

Volume= 18,200 litres

Height= ?

But the volume of rectangle area = area ×height

18,200 = 3 × 1.8 ×height

18,200=5.4 × height

Height = 18,200/5.4

Height = 3370.37m

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(2ai)

Given that,

a = 2, b = 1

a²+b+3/√4a-b³

=(2)² + 1 + 3/√4(2) -(1)³ = 8/√7

= 8/√7 × √7/√7

=8√7/7

(2aii)

√(a+b)³

=√(2+1)³

=√3³ = √27

=3√3

(2b)

Let the original money be X

Amount spent on shop

=3/7 × X = 3x/7

Amount spent on school

=1/2 × (X – 3x/7)

=1/2(7x – 3x/7) = 4x/14

=2x/7

Amount left = #21

Hence,

X – (3x/7 + 2x/7) = 21

X – 5x/7 = 21

7x – 5x = 147

2x = 147

X = #73.50

i.e the original money

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(3a)

x+4 = y-5 …(1)

y-1=½(x+1) ….(2)

From equation 1

x-y = -5-4

x-y = -9 … (3)

From equation 2

y-1 = x/2+½

y-1/1 = x+1/2

Cross multiply

x+1=2y-2

x-2y = -3 …(2)

x-y = -9 …(1)

x = 9+y

Substitute -9+y for x in equation 2

-9+y-2y=3

-y=3+9

-y = 12

y= -12

To find; x substitute -12 for y in equation 1

x-y=-9

x-(-12)=-9

x+12=-9

x=-9-12

x=-21

Therefore x = -21, y = -12

(3b)

Radius = 7cm

Length of arc _ 10cm

π= 22/7

L= ∅/360×2πr

10 = ∅/360×2×22/7×2

3600=44∅

∅ =360/44 =81.82

∅ =81.82°

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(4a)

X/5=√y/Y-Z

By squaring both sides

x²/5² =y/y-z

x²/25 = y/y-z

25y = x²(y-z)

25y = x²y – x²z

25y – x²y = -x²z

y(25-x²) = -x²z

y(25-x²)/25-x² = -x²z/25-x²

y = x²z/25-x²

y = x²z/x²-25

(4b)

DRAW THE DIAGRAM

π=3.142

Area of the shaded portion = Area of square – Area of the circle

Area of square = L² = 20×20 = 400cm²

Area of circle = πr²

πr² = 3.142(10)²

= 3.142 x 100

= 314.2cm²

Area of the shaded portion = 400 – 314.2

= 85.8cm²

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(5a)

Draw the table

Hence the probability that the sum is 8 OR 10

Pr(8 or 10) = 5/36 + 3/36 = 8/36 = 2/9

(5b)

a = {2} b = {-1} and c ={0}

{1} {1 } {3}

a + kb = c

{ 2}+ k {-1}= {0}

{ 1} {1 } {3}

(2 – k) = (0) =

(1 + k) (3)

2 – K = 0 and

1 + k =3

K = 2

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(6a)

√4.842×1.872/0.0754²

Solution

[In a tabular form for “No” and “log”]

Under “No”

4.842

1.872

√4.842×1.872

0.0754²

Under “log”

0.6850

+

0.2723

=0.9573 ×1/2

= 0.4787 Numerator

bar2.8774×2

=bar3.7548 Denominator

0.4784

–

bar3.7548

=2.7239

Antilog of 2.7239 = 529.5

=529.5

(6bi)

[Draw the circle]

radius = 14cm

[Draw the circle]

The length of chord AC

1 = AC = 2rsin∅/2

= 2×14sin120/2

=28×0.866

= 24.25cm

(6bii)

Area of the shaded portion = Area of the circle – Area of triangle

Area of the circle = πr²

= 22/7 * (14)²

= 22*196/7 = 616cm²

Area of equilateral triangle = ½ a.b sin C

=1/2*24.25*24.25sin60

=294.03*sin60

=254.64cm²

Hence,

Area of the shaded portion

=616 – 254.64

=361.36cm²

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(7a)

4(4x²-x)

Add and subtract the square √ of ½

i.e 4(x²-x +(½)²-(½)²

=4[(x²-x+¼-¼]

=4[(x-½)²-¼]

=4[(x-½)²-1

The first term is a perfect square and 1 must be added to have

4(x-½)²=4x²-4x+1

(7b)

2log^y base10 + 2 = 4log^6 base10

2log^y base10 + 1 = 4log^6 base10

Log^y base10 + log^10 base10 = 2log^6 base10

Log10y base10 = log^36 base10

10y=36

Y=36/10=3.6

Y = 4

(7c)

Sum of interior angle of a regukar polygon = (2n-4)90°

n = 8

Sum 2(n-2)90

= 2(8-2)90

= 2x6x90 = 1080

Sum of interior angle = 1080

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(9a)

√5/√5 – √3+ √3/√5 + √3

=√5(√5+√3)+√3(√5 – √3)/(√5 – √3)(√5 + √3)

=5 + √5 + √15 – 3/5 -3

=2 + 2√15/2

= 1 + 1√15

Then, we compare with a+b√c

Where,

a=1, b=1 and c=15

(9b)

[Draw the diagram]

π = 3.142

The total surface Area = Area of the rectangle + 2(area of small semi circle)+ 2(area of big semicircle)

Area of rectangle= 20×16

=320cm2

Area of small semi circle

=πr²/2 =π(8)²/2 = 64π/2

Area of big semi circle

=πR²/2 =π(10)²/2 = 100π/2

Hence, the total surface Area

=320 + 2(64π/2)+2(100π/2)

=320+64π+100π

=320+164π

=320+164(3.142)

=835.29cm2

==================================================

(10ai)

Draw the diagram

The length of the distance XO

|OX|² = |OY|² + |XY|²

|OX|² = (4.5)² + (6.5)²

=20.25+42.25

=62.5

|OX| = √62.5

=7.9m

(10aii)

Bearing of O from X

tan∅ = 4.5/6.5

∅ = tan-1(0.693)

=34.69°

=35°s.

(10b)

X(30°N, 40°W)

Y(15°N, 40°W)

Draw the diagram

barXY = ∅/360 * 2πR

=(30- 15)*2*3.142*6400/360

=15*2*3.142*6400/360

=1675.7km

=168km(3 s.f)

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(11)

Draw the diagram

r = 6cm = 0.6m

H = 8m

The curved surface Area of a cone = πrL

L = √(8)² + (0.06)²

= √64 + 0.0036

= √64.0036

L = 8.0002m

Curved surface Area =πrl

=22/7*8*0.06

=1.51m²

(11bi)

T3 = ar² = -1/4

T5 = ar^4 = -1/16

ar² = -1/4 —–(1)

ar^4 = -1/16—-(2)

Divide 2 by 1

ar^4/ar² = -1/16/-1/4

r² = 1/4

r = √1/4 = 1/2

Common ratio = 1/2 and the first term is;

From (1)

ar² = -1/4

a[1/2]² = -1/4

a/4 = -1/4

a = -1

The first term = -1

(11bii)

Sum of the 1st six terms

Sn = a(1 – rn)/1-r

=(-1)[1-(½)^6]/1 – ½

=(-1)(1 – 1/64)/½

= -63/64 divided by ½ = -63/64*2/1

Sum of 1st six term = -63/32

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