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NABTEB 2018 - MATHEMATICS ANSWERS

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NABTEB 2018 - MATHEMATICS ANSWERS

Monday 21/05/2018
Mathematics obj & essay
9.00am – 10.30am
12.00pm – 2.30pm

ICT
3.00pm – 5.40pm
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COMPLETED
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MATHS OBJ:
1-10: DBDCDABBCD
11-20: DCDBDCDAAB
21-30: BCDAABCCCD
31-40: CABABCBCCD
41-50: AABBAABDAD



NOTE: Nos 12-15 is for only SECRETARIAL & BUSINESS CANDIDATES[Answers two]
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Click here to view Image Version 1
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Click here to view Image Version 2 with SECRETARIAL AND BUSINESS CANDIDATE'S MATHS


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(1a)
1 1 0 1 1 base 2
1 1 0 1 base 2
1 0 1 base 2
____
101101base2
____

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
|1 | 2| 3| 4| 5| 6|
|1 | 2| 3| 5| 6| 7|
|2 | 3| 4| 6| 7| 8|
|3 | 4| 5| 7| 8| 9|
|4 | 5| 6| 8| 9| 10|
|5 | 6| 7| 9| 10| 11|
|6 | 7| 8| 10| 11| 12|

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

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(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°

(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
=1680km(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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FOR SECRETARIAL & BUSINESS CANDIDATES:
(You are asked to answer two from nos 12-15)
Click here



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