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

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