NABTEB Mathematics Questions 2022 Objective and Theory Update
– NABTEB Mathematics Questions –
NABTEB Mathematics NABTEB Expo Questions is out now on our website. In this article, I will show you past NABTEB Mathematics random repeated questions for free. All you need to do is to stay focused and follow this guide.
About NABTEB
The National Business and Technical Examinations Board were founded in 1992 to domesticate craft level examinations previously conducted by City & Guilds, Pittman’s, and the Royal Society of Arts throughout the United Kingdom in conformity with the principles of the National Policy on Education.
Its formation was the culmination of a 15-year evolutionary process in which FOUR Government Panels were established at various periods to assess the place and structure of public examinations in our educational system.
Mission
Develop as a globally recognized assessment body for Craftsmen and Technicians.
Vision
NABTEB is a vision-led, mission-driven public institution with the following vision: To be a Globally Recognised Assessment and Certification Body Preparing Candidates for the Workplace and Academic/Professional Excellence.
The management of the National Business and Technical Examinations Board (NABTEB) is yet to release the 2022 November/December NBC/NTC and ANBC/ANTC examinations results.
Candidates who intend to take part in the 2022 NABTEB GCE can learn how to check their results online.
NABTEB is one of the examination bodies set up by the Federal Government in 1992 to reduce the burden of conducting examinations, which involves a lot of technical and trade-related practicals.
The board conducts the National Technical Examination (NTC), National Business Certificate (NBC) and their respective advance level examinations (ANBC and ANTC).
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NABTEB Mathematics Questions and Answers
MATHEMATICS OBJ ANSWERS:
1-10: CDCDACCCDA
11-20: BDACDDBBCD
21-30: DBCBDDDBCC
31-40: CCCBBDDBAB
41-50: BBCBBCBBCB
MATHEMATICS THEORY ANSWERS:
(1a)
4/5 – (1/16 + 1/4 of 3¾)
4/5 – (1/16 + 1/4 × 15/4)
4/5 – (1/16 + 15/16)
4/5 – (1+15/16)
4/5 – 16/16
4/5 – 1
= 4 – 5/5 = -1/5
(1b)
√75 + 2√48 – √108
Solution
√25×3 + 2√16×3 – √36×3
5√3 + 28√3 – 6√3
(5 + 28 – 6)√3
27√3
(2a)
0.25×0.000081/0.09×0.08
=2.5×10^-¹ × 8.1×10^-5/9×10^-2 × 8×10^-2
=2.5×8.1×10^-6/9×8×10^-4
=2.5×81×10^-8/9×8×10^-4
=25×9/8 × 10^-8 × 10^4
=28.125×10^-4
=2.8125×10^-3
(2b)
1101₂ × 101₂
1101₂
× 101₂
1101
0000
1101
1000001₂
Therefore 1101₂ × 101₂ = 1000001₂
(3a)
Total surface area = Area of rectangle + Area of triangle
Area of rectangle = L×B
= 12×6 = 72cm²
From ΔACB
|AB|² = |BC|²+|AC|²
(13)² = x² + (12)²
169 = x² + 144
x² = 169 – 144
x = √25
x = 5cm
Area of triangle = 1/2bh
= 1/2×5×12
=30cm²
Total surface area=30+72
= 102cm²
(3b)
Volume of the prism
= base area × length
= area of triangle × length
= 1/2bh × L
= 1/2×5×12 × 6
= 30 × 6
= 180cm³
(4a)
16^x – 1/4 = 128
2^4(x-1/4) = 2^7
4(x-1/4) = 7
4x – 1 = 7
4x = 7 + 1
4x = 8
x = 8/4 = 2
x = 2
(4b)
Given that
I = #49,500
P = #310,000
T = 5years
R = ?
I = PRT/100
R = 100I/PT
R = 100×49500/310,000×5
R = 495/31×5
R = 495/155
R = 3.19%
(5)
Given that
U = {1,2,3,4,…,15}
B = {1,3,5,7,9,11}
C = {3,6,9,12,15}
D = {2,3,5,7,11,13}
(B U C)^c n D
B U C = {1,3,5,6,7,9,11,12,15}
(B U C)^c = {2,4,8,10,13,14}
(B U C)^c n D = {2,13}
(6a)
√21.81² × 37.2/92.63
Solution
No | Log
21.81² |1.3387 x 2
|=2.6774
37.2 |1.5705
| 2.6774
|+1.5705
| 4.2479
92.63 |-1.9667
| = 2.2812
|2.2812÷2
| =1.1406
(6b)
3sin²Φ – 4sinΦ, given that cosΦ = 1/√2
But, sin²Φ + cos²Φ = 1
sin²Φ + (1/√2)² =1
sin²Φ + 1/2 = 1
sin²Φ = 1 – 1/2 = 1/2
sin²Φ = 1/2
and sin Φ = 1/√2
3sin²Φ – 4sinΦ = 3×1/2 – 4/√2
= 3/2 – 4√2/2 = 3/2 – 2√2
or 3-4√2/2
Antilog of 1.1406 = 13.82
(7a)
The bearing of Q from O = 50 + Φ
Sin P/p = sin Q/t
p/Sin P = t/sinα
250/sin115 = t/sinα
sin α = 100×sin115/250
=90.63/250 = 0.3625
α = sin-¹(0.3625) = 21.26°
α = 21.26°
α+Φ+115 = 180(sum of angles in a triangle)
Φ + 21.26 + 115 = 180
Φ = 180 – 136.26
Φ = 43.74°
Therefore the bearing of Q from O = 50 + 43.74°
= 93.74°
(7aii)
The distance PQ
Using sine rule
P/sin P = 0/sinΦ
P/sin P = y/sinΦ
y = P×sinΦ/sinP
= 250×sin43.74/sin115
y = 172.846/0.906
y = 190.77
y = 190.8km
The distance PQ = 190.8km
(7b)
(i)
Area of the curved surface = 2πrh
=2×3.142×3.5×15.5
=340.907cm²
(ii)
Total surface area
= 2πr(r+h)
= 2×3.142×3.5(3.5+15.5)
=6.284(19)
= 119.4cm²
(8a)
Given that y = 2x² – x – 10
-3 ≤ x ≤ 3
X | -3 | -2 | -1 | 0 | 1 | 2 | 3
x² | 9 | 4 | 1 | 0 | 1 | 4 | 9
2x²|18| 8 | 2 | 0 | 2 | 8 | 18
-X | 3 | 2 | 1 | 0 | -1| -2 | -3
-10|-10|-10|-10|-10|-10|-10|-10
y | 11 | 0 | -7 | -10 | -9 | -4 | -5
(8b)
Plot the graph
(8ci)
The roots of the equation y=2x² – x – 10 = 0 are -2 and 2.5
(8cii)
The root of the equation
2x² – x – 3 = 0 are -1 and 1.4
This can be obtained from the graph as
2x² – x – 3 = 2x² – x – 10
The difference is 7
y = -7
A parallel line is drawn to cross the parabola through -7. The roots are -1 and 1.4
(8d)
The minimum value of the function = -10
(9a)
Draw the diagram
AB = Φ/360 × 2πRcosα
AB = 1100km
1100 = (70-50)/360×2×22/7×640×cosα
1100 = 20/360× 281600×cosα/7
1100 = 5632000cosα/2520
cosα = 1100×2520/5632000
cosα = 2772000/5632000
cosα = 0.4922
α = cos-¹(0.4922)
α = 60.52°
The parallel latitude to the nearest degree = 61°
(9b)
Draw the diagram
Perimeter = 2(L+B)
2(L+B) = 34
L+B = 17 ——(1)
Area = L×B
LB = 72 ——(2)
Solving (1) and (2) simultaneously
L + B = 17——(1)
LB = 72——(2)
From (1) L = 17 – B —–(3)
Putting (3) into (2)
LB = 72
(17 – B) B = 72
17B – B² = 72
B² – 17B + 72 = 0
Solving quadratically
(B² – 8B) -(9B + 72) = 0
B(B – 8) -9(B – 8) = 0
(B – 8)(B – 9) = 0
B – 8 = 0 or B – 9 = 0
B = 8cm or B = 9cm
Breadth = 8cm or 9cm
From (3)
L = 17 – B
When B = 8cm
L = 17 – 8 = 9cm
When B = 9cm
L = 17 – 9 = 8cm
Lengths = 9cm or 8cm
(11ai)
Draw the diagram
Length of arc AB = 10.5cm
L = Φ/360 × 2πr
10.5 = 65/360×2×3.142r
65×6.284r = 360×10.5
r = 360×10.5/65×6.284
=3780/408.46 = 9.25cm
radius = 9.3cm
(11aii)
Length of major arc
= Φ/360 × 2πr
Φ = 360 – 65 = 295°
L = 295/360×2×3.142×9.3
L = 17240.154/360
L = 47.9cm
Length of major arc = 48cm(2s.f)
(11b)
Draw the diagram
Perimeter of a Sector
=Length of arc + 2r
= 2r + Φ/360 × 2πr
= 2r[1 + 0.6109]
= 40(1.6109)
= 64.436cm
Therefore the perimeter of a Sector is 64.436cm²
P = 64.4cm(1d.p)
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