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# Nabteb 2018 Mathematics Answers (Obj & Theory) Question And Answers Expo  Nabteb 2018 Mathematics Answers, How To Get Real 2018/2019 NABTEB Mathematics Obj And Theory Answers, 2018 NABTEB Maths questions and Answers, Expo/Runz

## Nabteb 2018 Mathematics Answers Now Out

Here Is The Verified Nabteb 2018 Mathematics Questions Obj And Theory Answers – May/June Expo

MATHS OBJ:
1-10: DBDCDABBCD
11-20: DCDBDCDAAB
21-30: BCDAABCCCD
31-40: CABABCBCCD

===

(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

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

(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)
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

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

(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]

[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,
=616 – 254.64
=361.36cm²

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

(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

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

We Wishes You Best Of Luck.

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