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__100% VERIFIED MATHEMATICS THEORY QUESTIONS__

(Section A)

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(Section B)

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(Section C)

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# 👑 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

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

(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

(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)

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