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Obj

01-10: CBBCBDDACC

11-20: ABDBABDBCB

21-30: CABBCBAACD

31-40: BCABDABBDB

41-50: CBCBACCBAA

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

Theory

**SOLUTION IN PICTURES **

- https://i.imgur.com/YTOBt5S.jpg

https://i.imgur.com/iXS9no4.jpg

https://i.imgur.com/ee2MPD0.jpg

https://i.imgur.com/vCRhP9P.jpg

https://i.imgur.com/ShNAMNU.jpg

https://i.imgur.com/X5UWuSq.jpg

https://i.imgur.com/IpocDqR.jpg

https://i.imgur.com/gCw6i0y.jpg

https://i.imgur.com/cxJazvA.jpg

https://i.imgur.com/poKm15j.jpg

https://i.imgur.com/vqmxfJe.jpg

https://i.imgur.com/Qw7g43v.jpg

https://i.imgur.com/sgY6bbW.jpg

(1a)

214base5 and 23base5

214base5 = 2×5² + 1×5¹ + 4×5⁰ = 2×25 + 5 + 4

= 59base10

23base5 = 2×5¹ + 3×5⁰ = 10 + 3 = 13base10

59base10 × 13base10 = 767

Convert to base Five

5| 767

5| 153 > 2

5| 30 > 3

5| 6 > 0

5| 1 > 1

5| 0 > 1

11032base5

(1b)

20+3 – x/3 = x+7

20+3-7= x+x/3 = 3x+x/3

16 = 4x/3

x =16×3/4

x=12

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

(2a)

Draw an equilateral triangle

A

/\

m+n / \ (3m-5n)

/ \

B ——– C

(m-2n+3)

Since the triangle is equilateral

m+n=3m-5n=m-2n+3

m+n=m-2n+3

Collect Like terms

m-m+n+2n=3

3n=3

n=3/3

n=1

Also,

m+n=3m-5n

Collect like terms

n+5n=3m-m

6n=2m

6(1)=2m

2m=6

m=6/2

m=3

Length=m+n

=3+1

=4

(2b)

Perimeter=3L

L=m+n=3m-5n=m-2n+3

L=4

P=3L

P=3×4

:. Perimeter=12 unit

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

(3a)

√0.81×10-⁵/√2.25×10⁷

= √81×10-²×10-⁵/√225×10⁷×10-²

= √81×10-⁵/225×10⁷

= √81/225 × √10-⁵-⁷

= √81/√225 × √10-¹²

= (9²)½/(15²)½ × (10-¹²)½

= 9/15 ×10-⁶

Multiply through by 3

9/15×10-⁶

3/5×10-⁶

0.6×10-¹×10-⁶

= 6×10-⁷

(3b)

3/√3(2/√3 – 12/√6)

= 3/√3(2/√3) – 3/√3(12/√6)

=6/√9 – 36/√18 = 6/3 – 36/√9×√2

= 6/3 – 36/3√2

= 2 – 12√2 = 2-12√2 × √2/√2

= 2 – 12√2/2

= 2-6√2

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

(4a)

U=80

n(P)=40

n(C)=45

n(B)=30

n(PΠC)=20

n(PΠB)=12

n(P’ΠC’ΠB’)=3

n(PΠCΠB)=x

n(BΠC)=y

Draw a Venn diagram

[img]https://i.imgur.com/MgLuFIb.jpg[/img]

Number of student offering physics only = 40–(20 – x + x + 12 – x)

=8 + x

Number of student offering Chemistry only = 45 –(20 – x + x + y – x)

= 25 + x – y

Number of student offering Biology only = 30 –(12 – x + x + y – x)

=18 + x – y

:. 8 + x + 25 + x – y + 18 + x – y + 20 – x + 12 – x + y – x + x + 3 = 80

x = y – 6

(ii)

Students offering one subject = 8 + y – 6 + 25 + y – 6 – y + 18 + y – 6 – y = 33 + y

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

(5a)

20pencils = ₦55

440pencils= X

X=55×440/20 = ₦1210

Unsaleable = 10/100 ×440 = 44pencils

Saleable = 440 – 44= 396pencils

Total expenses = 1210+150+200= ₦1560 = c.p

Sold

12pencils = ₦60

326pencils = X

X= 60×396/12= ₦1980 = s.p

Profit = s.p – c.p = 1980-1560 = ₦420

(5b)

% Profit = Profit/c.p ×100%

= 420/1560 ×100

= 26.92%

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

(6)

Area of triangle FBD =1/2 × base × height

= 1/2 × 8 × 3

=12cm²

Area of ABCD = L × B

=8×6

=48cm²

Area of CDIH = L×B

=12 × 6

=72cm²

Area of FDGH= L × B

=12 × 5

=60cm²

:. Total surface Area = 2(12) + 2(48) + 2(72) + 2(60)

=24 + 96 + 144 + 120

=384cm²

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

(7a)

y= 5x – 2x² , –2 ≤ x ≤ 4

TABULATE

x | -2 | -1 | 0 | 1 | 2 | 3 | 4 |

x² | 4 | 1 | 0 | 1 | 4 | 9 | 16 |

5x | -10 | -5 | 0 | 5 | 10 | 15 | 20 |

-2x² | -8 | -2 | 0 | -2 | -8 | -18 | -32 |

y | -18 | -7 | 0 | 3 | 2 | -3 | -12 |

(7b)

View Image ➡ [https://i.imgur.com/oG4h0WO.jpg]

(7ci)

The line of symmetry is at point (1,3).

:. The lube is x=1

(7cii)

The solution of the equation 5 + 5x – 2x² = 0, From the graph can be obtained as follow;

y=5x – 2x²

y=5 + 5x – 2x²

5x – 2x² = 5 + 5x – 2x²

y= -5.

y= -5 is traced from the graph and the values are -0.75 and 3.25

(7ciii)

The maximum turning point is y=3

(7civ)

Gradient at point x=1

Gradient=∆y/∆x

=4 – 2.2 ÷ 2.3 – 0

=1.8/2.3

=0.78

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

(8a)

TABULATE

Class interval | Class mark | Frequency | Tally | Cumulative frequency | Class Boundaries |

UNDER Class interval

21-30

31-40

41-50

51-60

61-70

71-80

81-90

91-100

UNDER Class mark

25.5

35.5

45.5

55.5

65.5

75.5

85.5

95.5

UNDER Frequency

2

5

7

9

11

8

5

3

UNDER Tally

//

~////~

~////~ //

~////~ ////

~////~ ~////~ /

~////~ ///

~////~

///

UNDER Cummulative Frequency

2

7

14

23

34

42

47

50

UNDER Class boundaries

20.5 – 30.5

30.5 – 40.5

40.5 – 50.5

50.5 – 60.5

60.5 – 70.5

70.5 – 80.5

80.5 – 90.5

90.5 – 100.5

(8bi)

Median = 1/2 of N

=N/2

=50/2

25th Cummulative Frequency

Median = 62.5

(8bii)

Semi interquantite Range = Q3 – Q1 ÷ 2

Q3= 3N/4

=30 × 50 ÷ 4

37.5th Cummulative Frequency

Q3= 74.5

Q1= N/4

=50/4

12.5th Cummulative Frequency

Q1=48.5

Semi Interquantite Range= 74.5 – 48.5 ÷ 2

=26/2

=13

(8biii)

View Graph ➡ [https://i.imgur.com/DfExmsk.jpg]

(8c)

The percentage of the students that passed the examination. If 45% is the passed mark = 7 + 9 + 11 + 8 + 5 + 3 × 100 ÷ 50

= 43/50 × 100

=86%

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

(9)

t∝v

t∝1/p

t∝v/p1= >t=kv/p

P=5,t=10minutes

V=20

10=20k/5

10*5=20k

50=20k

K=50/20=5/2

(9ai)

t=5/2v/p

t=5v/2p

(9aii)

V=50,t=?,p=2

t=5/2*50/2

=125/2 = 62.5minutes

(9aiii)

V=40, t=20minutes, p=?

20=5/2*40/p

20p=100

P=100/20=5

(9bi)

A=P(1+r/100)^n

(A/P)^1/n =(1+r/100)^n*1/n

(A=P)^1/n =1+r/100

(A=P)^1/n – 1 =r/100

100[(A/P)]^1/n – 1 ] =r

(9bii)

100[(506.19/450.0)^⅓-1]=r

r=100(1.040-1)

r=100(0.04)

=4

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

(10ai)

X²-10/X²+4x-5. =0

X²-10=0

X=±√10

X=+√10 or

X=-√10

(10aii)

X²+4X-5=0

X²-X+5X-5=0

X(X-1)+5(X-1)=0

X+5=0. or X-1=0

X=-5 or X=1

(10b)

(i)r=a+b

=7i + 2j-k

(ii)r=a+b+c

=2c+3j

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

(11ai)

Given

x² – 10 / x² + 4x – 5

The value of x for which the above fraction is zero is

x² – 10 = 0

x² = 10

x² = ±√10

(11aii)

The value of x for which the above fraction is undefined is

x² + 4x – 5 = 0

(x² – x) + (5x – 5) = 0

x(x–1) +5(x–1)=0

(x+5)(x–1)=0

x+5=0 or x–1=0

x=–5 or x=1

(11bi)

The resultant of the vectors

a=3i + j + 2k

And

b=4i + j – 3k

Resultant = | a + b |

a + b = (3i + j + 2k) + (4i + j – 3k)

=7i + 2j – k

:. The resultant

R=√(7)² + (2)² + (-1)²

R=√49+4+2

R=√54

R=7.35 units.

(11bii)

a=3i, b= -2i – j

c= i + 4j

Resultant = | a + b + c |

a + b + c = 3i + ( -2i – j) + i + 4j

=3i – i + 3j

=2i – 3j

Resultant = √(2)² + (3)²

=√4+9

=√13

=3.6 units

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

(12a)

Five montly moving average

0.3 +0.3+2.8 + 8.6+20.3/5= 32.8/5 = 6.46

0.3+2.8 +8.6+20.3+22.6/5 = 54.6/5 = 10.92

2.8+8.6+20.3+22.6+33.0/5 = 87.3/5 = 17.46

8.6+20.3+22.6+33.0 +29.2/5 = 113.7/5 = 22.74

(12b)

Deposite = ₦10,250

Instalment payment = ₦3600 per week

Total numbers of Week = 6*4 = 24 Weeks

Total instalment payment=24*3600 = ₦86,400

Total cost of motor bike = Intial deposit +

Total instalment = 10,250 +86,400= ₦96,650

(12c)

Worth of goods 7,000,000

custom duty = 25/100*7000,000 = 1750,000
profit made = 35/100*7000,000 = 2,450,000

selling price = Worth of goods & Custom dutyt

proft made 7,000,000 + 1,750,000 + 2,450,000= ₦11,200,000

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

(14)

Flat fee = ₦1000

Distance charge = ₦250 per km

Neight charge = ₦100 per gram

If weight =₦75g

Weight Charge = 75*100 = ₦7,500
If distance = 900KM
distance Charge = 900*250 = ₦225,000

(14ai)

Company charge = flat feet distance Changet

Weight charge =

1000 + 7500 +225000= ₦233,500

(14aii)

Customer’s bill = company charge + VAT

Stree VAT=5/100 * 233,500=₦11,675

Customer’s bill = 233,500 +11675

= ₦235,175

(14bi)

Personal allowance = ₦18,000

(14bii)

Sponse allowance = ₦5000

(14biii)

children allowance = ₦4,000 per child

(14biv)

Dependent relative = ₦6000 each

(14bv)

Given gross per annum= ₦1,020,000

INHIS insurance = 1/100 * 1,020,000 = ₦10,200

Union dues = 2/100*1,020,000 = ₦20,400
pension scheme = 7.5/100 * 1,020,000 =₦76,500
Tax paid =10/100*1020,000 = ₦102,000

(14ci)

monthly tax; 116700/12 ₦9,725.50

(14cii)

Total monthly pay =1167000/12= 97,250.00

Monthly net pay = 97,250 -9725=₦87,525.00

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