2019 NABTEB GCE Mathematics Questions and Answers – For 0′ Level

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 NABTEB GCE Mathematics Questions and Answers

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100% VERIFIED MATHEMATICS THEORY QUESTIONS

(Section A)
CLICK HERE

(Section B)
CLICK HERE and HERE

(Section C)
CLICK HERE

👑 2019 NABTEB GCE MATHEMATICS ANSWERS👑

100% VERIFIED MATHEMATICS OBJ ANSWERS

1-10: CDCDACCCDA
11-20: BDACDDBBCD
21-30: DBCBDDDBCC
31-40: CCCBBDDBAB
41-50: BBCBBCBBCB

100% VERIFIED 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)
Draw the diagram [ CLICK HERE]

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

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(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)
Draw the diagram [ CLICK HERE]

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)
Draw the diagram [ CLICK HERE]

(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 [ CLICK HERE]

(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

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(8d)
The minimum value of the function = -10
========================

(9a)
Draw the diagram [ CLICK HERE]
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 [ CLICK HERE]
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)

SNAPPED ANSWERS

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