100% REAL NABTEB 2018 MAY/JUN MATHEMATICS OBJ AND THEORY QUESTIONS AND ANSWERS

(1a)
1101 base2 + 101 base2 + 11011base2
11011
1101
101base2
____
101101base2
____

Hence 1101base2 + 101base2 + 11011base2
= 101101base2

(1b)
DRAW THE DIAGRAM
L = 3.0m
B = 1.8m
H = ?
Volume = 18200 liters
= 18.2m
Therefore volume = L x B x H
18.2 = 3 x 1.8 x H
H = 18.2/5.4
H = 3.37m

(2a)
Given that,
a = 2, b = 1
(i) a + b + 3/√4a(2) – b
= (2) + 1 + 3/√4(2) – (1) = 8/√7
= 8/√7 x √7/ √7
= 8 √7/7

(2aii)
√ (a + b)
= √ (2 + 1)
= √3 = √27
= 3 √3

(2b)
Let the original money be x
Amount spent in shop
= 3/7 x = 3x/7
Amount spent on school
= ½ x (x – 3x/7)
= ½ (7x – 3x)/7 = 4x/14
= 2x/7
Amount left = N 21
Hence x – (3x/7 + 2x/7) = 21
x – 5x/7 = 21
7x – 5x = 147
2x = 147
x = N 73.50 s the original money

(3a)
x + 4 = y – 5———– (1)
y – 1 = ½ (x+1) ——— (2)
x – y = -9 ———— (3)
-x +2y = 3 = ———- (4)
Solving (3) and (4) simultaneously we have
x – y – x +2y = -9 + 3
y = -6
from (3)
x – y = -9
x + 6 = -9
x = -9 – 6 = -15
hence,
x = – 15 and y = -6

(3b)
DRAW THE DIAGRAM
Radius = 7cm
Length of aec = 10cm
π = 22/7
L = ϴ /360 x 2 πr
10 = ϴ /360 x 2 x 22/7 x 7
3600 = 440
ϴ = 3600/44 = 81.82ᵒ
ϴ = 81.82ᵒ

(4a)
x/5 = square root of y/y-z
by squaring both sides
x raise power2/5² = y/y-z
x² (y-z) = 5² y
x raise power y – xᵒz = 5² y
(x² – 5²) y = x² z
Y = x² z/x² – 5²
= x² z/ x²- 5²

(4b)
DRAW THE DIAGRAM
π = 3.142
Area of the shaded portion
= area of the square – area of the circle
Area of the square = L²
= 20 x 20 = 400cm²
Area of the circle
= πr²

(5a)
TABULATE
|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 = (2/1), b = (-1/1) and c = (0/3)
a + kb = c
(2/1) + k (-1/1) = (0/3)
==> (2-k/1+k) = (0/3)
==> 2 – k = 0 and 1 + k = 3
k = 2

(6a)
√4.842 x 1.872/0.0754²
TABULATE
|No| Log|
|4.842 | 0.6850
|1.872| + 0.2723/0.9573 x ½
|√4.842 x 1.872| = 0.4787 Numerator 1
|0.0754²| bar2 . 8774 x 2 = bar3.7548 Denominator
0.4787-bar3.7548/­2.7239
= 529.5
Antilog = 529.5

(6bi)
DRAW THE DIAGRAM
Radius = 14cm
The length of the chord AC
1 = AC = 2r sin ϴ/2
= 2 x 14sin 120/2
= 28 x 0.866
= 24.25cm

(6bii)
Area of the shaded portion
= area of the circle – area of triangle
Area of the circle = πr²
= 22/7 x (14)²
= 22 x 196/7 = 616cm²
Area of equilateral triangle = ½ ab sin c
= ½ x 24.25 x 24.25 sin60
= 294.03 x sin60
= 254.64cm²
Hence area of the shaded portion
= 616 – 254.64
= 361.36cm²

(7a)
4x² – 4x
4(x2 – x)
Add and subtract the square of ½
4(x2 bar x + (1/2)² – (1/2)²
4 (x² – x + ¼ – ¼)
= 4 [(x-1/2)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 base 8 10 + 2 = 4log 6 base 8 10
2 (logy 8 10 + 1) = 4 log 6 base 10
Log y base 10 + log 10 base 10 = 2 log 6 base 10
Log 10y base 10 = log 36 base 10
10y = 36
y = 36/10 = 3.6
y approximate = 4

(7c)
Sum of interior angle of a regular polygon = (2n – 4) 90
n = 8
sum = 2 (n-2) 90
= 2 (8-2) 90
= 2 x 6 x 90 = 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 42√15/2
= 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 rectangle + 2 (area of small of semi circle) + 2 (area of big semi circle) area of rectangle = 20 x 16
= 320cm²
Area of small semi circle
= πr2/2 = π(8)² 2/2 = 64π/2
Area of big semi circle
= πr2/2 = π(10)²/2 = 100 π/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
(i) The length of the distance XO
|OX|raise power2 = |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
tanϴ = 4.5/6.5
ϴ = tan-1 (0.6923)
= 34.69ᵒ
Approximate 35ᵒ

(10b)
DRAW THE DIAGRAM
X (30ᵒN, 40ᵒW)
Y (15ᵒN, 40ᵒW)
Angle XY = ϴ/360 x 2πR
= 15 x 2 x 3.142 x 6400/360
= 1675. 7km
Approximate 168km (3sf)

(11a)
DRAW THE DIAGRAM
r = 6cm
= 0.6m
H = 8m
The curved surface area of the a cone = πr l
L = √(8)2 + (0.06)2
= √64 + 0.0036
= √64. 0036
L = 8.0002m
Curved surface area = πrl
= 22/7 x 8 x 0.06
= 1.51m²

(11bi)
T base3 = ar² = -1/4
T base5 = ar⁴ = -1/16
ar² = -1/­4————-eq(1)
ar⁴ = -1/16 ———–eq(2)
divide (2) by (1)
ar²/ar² = -1/16/-1/4
r2 = ¼
r = √1/4 = ½
common ratio = ½ 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 term
Sn = a (1 – rn)/1 – r
= (-1) (1 – (1/2) raise to power 6 /1-1/2
= (-1) (1 – 1/64)/1/2
= -63/64 ÷ ½ = -63/64 x 2/1
Sum of 1st term = -63/32