2018 NABTEB MATHEMATICS QUESTIONS AND ANSWERS NOW AVAILABLE

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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

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

(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°

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

(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

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

(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

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

(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)

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

(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

 

2018 NABTEB MATHEMATICS QUESTIONS AND ANSWERS NOW AVAILABLE

2018 NABTEB MATHEMATICS QUESTIONS AND ANSWERS NOW AVAILABLE

2018 NABTEB MATHEMATICS QUESTIONS AND ANSWERS NOW AVAILABLE

2018 NABTEB MATHEMATICS QUESTIONS AND ANSWERS NOW AVAILABLE
2018 NABTEB MATHEMATICS QUESTIONS AND ANSWERS NOW AVAILABLE
2018 NABTEB MATHEMATICS QUESTIONS AND ANSWERS NOW AVAILABLE

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