Maths Assignment: Comprehensive Solutions for All Questions
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Homework Assignment
AI Summary
This document presents a comprehensive solution to a maths assignment, covering various mathematical concepts and problem-solving techniques. The assignment includes multiple-choice questions, calculations involving formulas, and word problems. Topics such as calculating capacity, dividend yield, gradients, depreciation, speed, probability, and financial calculations are addressed. The solutions provide step-by-step explanations, formulas, and calculations to arrive at the correct answers. The assignment also covers topics like blood alcohol content, stamp duty calculations, algebraic manipulations, interquartile range, and geometric problems. The document is designed to help students understand the problem-solving process and improve their mathematical skills. The assignment is contributed by a student to be published on the website Desklib, which provides all the necessary AI based study tools for students.

Maths
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Table of Contents
SECTION A ....................................................................................................................................1
Question 1-20...............................................................................................................................1
SECTION B.....................................................................................................................................6
Question 21-23.............................................................................................................................6
SECTION C ..................................................................................................................................13
Question 24................................................................................................................................13
vi. Shading of profit zone on the graph......................................................................................14
SECTION A ....................................................................................................................................1
Question 1-20...............................................................................................................................1
SECTION B.....................................................................................................................................6
Question 21-23.............................................................................................................................6
SECTION C ..................................................................................................................................13
Question 24................................................................................................................................13
vi. Shading of profit zone on the graph......................................................................................14

SECTION A
Question 1-20
Question 1
Base radius (r) = 1.2 m
height (h)= 1.8
Capacity in litres = πr²h meter³ = πr²h *1000 liters
= π * 1.2² * 1.8 *1000 = Option C
Question 2. Dividend Yield for Vinh share.
Particulars Amount (in $)
Annual Dividend 15
Current Stock price 3
Dividend Yield = Annual
Dividend / Current Stock
Price
5
The current dividend yield for the shares of Vinh is 5%.
Question 3
Gradient can be calculated from the formula: y= mx +C
On comparing the above equation with the given equations:
A. x = 3
m = 1
B. y =3 + x
m = 1
C. y - 3x
m = 3
D. y = 10 – 3x
m = -3
Thus, equation C has gradient =3
1
Question 1-20
Question 1
Base radius (r) = 1.2 m
height (h)= 1.8
Capacity in litres = πr²h meter³ = πr²h *1000 liters
= π * 1.2² * 1.8 *1000 = Option C
Question 2. Dividend Yield for Vinh share.
Particulars Amount (in $)
Annual Dividend 15
Current Stock price 3
Dividend Yield = Annual
Dividend / Current Stock
Price
5
The current dividend yield for the shares of Vinh is 5%.
Question 3
Gradient can be calculated from the formula: y= mx +C
On comparing the above equation with the given equations:
A. x = 3
m = 1
B. y =3 + x
m = 1
C. y - 3x
m = 3
D. y = 10 – 3x
m = -3
Thus, equation C has gradient =3
1
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Question 4. Annual depreciation of car
Particulars Amount (in $)
Cost of Car 22990
Depreciated Value 18600
Useful Life 3
Annual Depreciation =
(Cost of car – Depreciated
value) / Useful life
1463.33
The correct annual depreciation of car is 1463.33
Answer: B
Question 5
Speed (s)= 75 km/h
Time (t) = 3 hours and 30 minutes = 3.5 hours
Distance (d)= s*t = 75*3.5
Distance (d)=262.5 km = Option D
Question 6
Systematic sampling Option B
Question 7
Wages per hour = $48.25
Additional allowance for toxic substance = $6.70
Total wage per hour for toxic handling toxic substance= $6.70 + $48.25 = $54.95
Total working hour of Todd = 35 hours (ordinary substance) + 5 hours (toxic)
Total wages for Todd = [35*48.25] + [5*54.95]
Total wages for Todd= $1963.5 Option B
Question 8
Total number of sections = 6
Number of sections containing number less than equal to 4 = 3
Probability = 3/6 = ½ Option C
2
Particulars Amount (in $)
Cost of Car 22990
Depreciated Value 18600
Useful Life 3
Annual Depreciation =
(Cost of car – Depreciated
value) / Useful life
1463.33
The correct annual depreciation of car is 1463.33
Answer: B
Question 5
Speed (s)= 75 km/h
Time (t) = 3 hours and 30 minutes = 3.5 hours
Distance (d)= s*t = 75*3.5
Distance (d)=262.5 km = Option D
Question 6
Systematic sampling Option B
Question 7
Wages per hour = $48.25
Additional allowance for toxic substance = $6.70
Total wage per hour for toxic handling toxic substance= $6.70 + $48.25 = $54.95
Total working hour of Todd = 35 hours (ordinary substance) + 5 hours (toxic)
Total wages for Todd = [35*48.25] + [5*54.95]
Total wages for Todd= $1963.5 Option B
Question 8
Total number of sections = 6
Number of sections containing number less than equal to 4 = 3
Probability = 3/6 = ½ Option C
2
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Question 9
A = (¾) VM²
V = 3.54
M= 1.07
A = (¾) * (3.54)*(1.07)²
A = 3.04 Option A
Question 10. Future value of investment.
The manner of writing the expression of future value of the investment is part A I.e. 40
000 x (1+0.05)3.
Answer: C
Question 11
(3- x)/2 = 5
(3- x) = 10
x = 3-10
x = -7 Option A
Question 12
3.39 m Option B
Question 13
Radius of outer circle= 0.35 + 0.1 = 0.45 m
Radius of inner circle= 0.35 m
Area of outer circle= πr² = π (0.45)² = 0.635
Area of inner circle= π (0.35)² = 0.384
Area of the shaded annulus = Area of outer circle - Area of inner circle
0.635-0.384
Area of the shaded annulus = 0.25 m ² Option B
Question 14
MHR = 205.8 – 0.685a
a = age
3
A = (¾) VM²
V = 3.54
M= 1.07
A = (¾) * (3.54)*(1.07)²
A = 3.04 Option A
Question 10. Future value of investment.
The manner of writing the expression of future value of the investment is part A I.e. 40
000 x (1+0.05)3.
Answer: C
Question 11
(3- x)/2 = 5
(3- x) = 10
x = 3-10
x = -7 Option A
Question 12
3.39 m Option B
Question 13
Radius of outer circle= 0.35 + 0.1 = 0.45 m
Radius of inner circle= 0.35 m
Area of outer circle= πr² = π (0.45)² = 0.635
Area of inner circle= π (0.35)² = 0.384
Area of the shaded annulus = Area of outer circle - Area of inner circle
0.635-0.384
Area of the shaded annulus = 0.25 m ² Option B
Question 14
MHR = 205.8 – 0.685a
a = age
3

MHR = 190 beats per minute
a = ?
0.685a = 205.8 – MHR
0.685a = 205.8 – 190
a = 23 Option B
Question 15. Percentage of depreciation
Particulars Amount (in $)
Cost of Car 28500
Depreciated Value 24000
Useful Life 1
Annual Depreciation = (Cost of
car – Depreciated value) /
Useful life
4500.00
Percentage of depreciation
=(Annual depreciation * 100) /
cost of asset
15.79
The depreciation percentage for car is 15.79% or 16%.
Answer: A
Question 16
Speed = 90 km/h
Reaction time = 1.8 seconds
Braking distance = 30 m
Stopping distance = Reaction distance + braking distance
Reaction distance = Reaction time * speed
= 1.8*90*(5/18) meters
Reaction distance = 45 m
Stopping distance = 45 + 30
Stopping distance = 75 m Option D
4
a = ?
0.685a = 205.8 – MHR
0.685a = 205.8 – 190
a = 23 Option B
Question 15. Percentage of depreciation
Particulars Amount (in $)
Cost of Car 28500
Depreciated Value 24000
Useful Life 1
Annual Depreciation = (Cost of
car – Depreciated value) /
Useful life
4500.00
Percentage of depreciation
=(Annual depreciation * 100) /
cost of asset
15.79
The depreciation percentage for car is 15.79% or 16%.
Answer: A
Question 16
Speed = 90 km/h
Reaction time = 1.8 seconds
Braking distance = 30 m
Stopping distance = Reaction distance + braking distance
Reaction distance = Reaction time * speed
= 1.8*90*(5/18) meters
Reaction distance = 45 m
Stopping distance = 45 + 30
Stopping distance = 75 m Option D
4
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Question 17
Number of text
message sent (X)
Frequency
(f)
Frequncy * no. of text message sent (X*f)
1 3 3
2 5 10
3 5 15
4 8 32
5 9 45
SUM 30 105
Mean = Ʃ (X*f) / Number of data values
Mean = 105/30
Mean = 3.5 Option B
Question 18
y = (40N, 15W) Option C
Question 19
Power = 15 W
time = 4 hours per day for 365 days
Energy = power * time = 15*4*365
Energy= 21.9 kWh Option D
Question 20. Peta's total earnings for the year.
5
Number of text
message sent (X)
Frequency
(f)
Frequncy * no. of text message sent (X*f)
1 3 3
2 5 10
3 5 15
4 8 32
5 9 45
SUM 30 105
Mean = Ʃ (X*f) / Number of data values
Mean = 105/30
Mean = 3.5 Option B
Question 18
y = (40N, 15W) Option C
Question 19
Power = 15 W
time = 4 hours per day for 365 days
Energy = power * time = 15*4*365
Energy= 21.9 kWh Option D
Question 20. Peta's total earnings for the year.
5
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4000 Shares bought at price of $3.00 per share.
Annual dividend receive during the year @ 5%.
At year end 4000 shares sold for $3.30 per share.
1. Profit on selling of 4000 shares = $3.30 - $3.00 = $0.30 per share
4000 shares x $0.30 per share = $1200
2. Dividend receive = (4000 x $3.00) x 5% = $600
So, total profit earned = $1200 + $600 = $1800
Particulars Amount (in $)
Current Stock price 3.3
Old stock price 3
Profit per share 0.3
Number of share sold 4000
Profit earned 1200
Dividend Rate 5.00%
Dividend receive during
year 600
Total profit earned 1800
Answer: B
SECTION B
Question 21-23
Question 21
A.
Mass in Kg (M) = 75
Number of hours drinking (H) = 2
Number of standard drinks consumed (N) = 1.4 * 3 = 4.2
Blood alcohol content= BAC(male) = (10N – 7.5H) / 6.8 M
= (10*4.2 – 7.5*2) / 6.8*75 = 0.052
BAC(male) = 0.052
6
Annual dividend receive during the year @ 5%.
At year end 4000 shares sold for $3.30 per share.
1. Profit on selling of 4000 shares = $3.30 - $3.00 = $0.30 per share
4000 shares x $0.30 per share = $1200
2. Dividend receive = (4000 x $3.00) x 5% = $600
So, total profit earned = $1200 + $600 = $1800
Particulars Amount (in $)
Current Stock price 3.3
Old stock price 3
Profit per share 0.3
Number of share sold 4000
Profit earned 1200
Dividend Rate 5.00%
Dividend receive during
year 600
Total profit earned 1800
Answer: B
SECTION B
Question 21-23
Question 21
A.
Mass in Kg (M) = 75
Number of hours drinking (H) = 2
Number of standard drinks consumed (N) = 1.4 * 3 = 4.2
Blood alcohol content= BAC(male) = (10N – 7.5H) / 6.8 M
= (10*4.2 – 7.5*2) / 6.8*75 = 0.052
BAC(male) = 0.052
6

B.) i
Pine apple : Ginger : Orange juice = 6:1:3
Total punch = 250 ml
Amount of Ginger in 250 ml = 250*(1/10) = 25 ml
B.) ii
Amount of orange juice = 270 ml
Amount of punch containing 270 ml orange juice = 270* (10/3) = 900 ml
c. Stamp duty calculation
For first $45000 Stamp duty @ 3% & after $45000 the Stamp duty applicable will be @ 5%
Particulars
Amount (in
$)
Total Car price 52000
Car price 45000
Stamp duty Rate 3.00%
Car price 7000
Stamp duty Rate 5.00%
Stamp duty calculation 1700
Total amount of stamp duty payable is $1700.
D.
i) 7x + 3y -3x -8y
4x = 5y
x/y = 5/4
ii) 3a²b * 5ab
= 15ab²
iii) 20xy² / 4x²y
= 5xy
7
Pine apple : Ginger : Orange juice = 6:1:3
Total punch = 250 ml
Amount of Ginger in 250 ml = 250*(1/10) = 25 ml
B.) ii
Amount of orange juice = 270 ml
Amount of punch containing 270 ml orange juice = 270* (10/3) = 900 ml
c. Stamp duty calculation
For first $45000 Stamp duty @ 3% & after $45000 the Stamp duty applicable will be @ 5%
Particulars
Amount (in
$)
Total Car price 52000
Car price 45000
Stamp duty Rate 3.00%
Car price 7000
Stamp duty Rate 5.00%
Stamp duty calculation 1700
Total amount of stamp duty payable is $1700.
D.
i) 7x + 3y -3x -8y
4x = 5y
x/y = 5/4
ii) 3a²b * 5ab
= 15ab²
iii) 20xy² / 4x²y
= 5xy
7
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e.
Interquartile range: Q3 – Q1
= (5*5*6) – (0*3*6)
= 0
F.
Using Pythagoras theorem:
? = 31.54
BD = 29.5
Question 22
A.
F = (9C/5) +32
C = -25
8
Interquartile range: Q3 – Q1
= (5*5*6) – (0*3*6)
= 0
F.
Using Pythagoras theorem:
? = 31.54
BD = 29.5
Question 22
A.
F = (9C/5) +32
C = -25
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F = (9(-25)/5) +32
F = -13 Fahrenheit
B.
4 = 9 + (1/m)
(9 m +1) / m = 4
(9 m +1) = 4m \
5m = -1
m = -(1/5)
C.
V = IR – E
IR = V +E
I = (V +E)/R
D. Calculation of amount of Income tax of Sam.
Particulars Amount Tax Rate Tax Amount
18200 18200 Nil 0
$18201 - $37000 18799 19.00% 3572
$37001 - $56920 19919 32.50% 6474
Tax amount payable 10045
Thus, the amount of income tax payable is $10 045 over income of $56 920.
Question 23
A.
i. Balance owing after deposit paid.
Particulars Amount (in $)
Car price 29000
Deposit rate 10.00%
Amount deposited 2900
9
F = -13 Fahrenheit
B.
4 = 9 + (1/m)
(9 m +1) / m = 4
(9 m +1) = 4m \
5m = -1
m = -(1/5)
C.
V = IR – E
IR = V +E
I = (V +E)/R
D. Calculation of amount of Income tax of Sam.
Particulars Amount Tax Rate Tax Amount
18200 18200 Nil 0
$18201 - $37000 18799 19.00% 3572
$37001 - $56920 19919 32.50% 6474
Tax amount payable 10045
Thus, the amount of income tax payable is $10 045 over income of $56 920.
Question 23
A.
i. Balance owing after deposit paid.
Particulars Amount (in $)
Car price 29000
Deposit rate 10.00%
Amount deposited 2900
9

Interest rate 6.20%
Interest paid 1618
Balance owing after
deposit paid 24482
ii. Monthly repayment
Car’s cash price: $29000
10% deposit: 29000 * 105
= 2900
Loan amount: 29000 – 2900
= 26100
Loan Principle Amount 26,100.00
Annual Interest Rate 6.20%
Loan Period (in months) 36.00
Original Repayment Amount 796.38
Loan Start Date 4/27/2019
Repayment Type End
Residual Value -
Month
Repayment
Number
Opening
Balance
Loan
Repayment
Interest
Charged
Capital
Repaid
Closing
Balance
Apr-2019 1 26,100.00 796.38 134.85 661.53 25,438.47
May-2019 2 25,438.47 843.94 217.29 626.65 24,811.82
Jun-2019 3 24,811.82 843.94 211.93 632.00 24,179.81
Jul-2019 4 24,179.81 843.94 206.54 637.40 23,542.41
Aug-2019 5 23,542.41 843.94 201.09 642.85 22,899.56
Sep-2019 6 22,899.56 843.94 195.60 648.34 22,251.22
Oct-2019 7 22,251.22 843.94 190.06 653.88 21,597.35
Nov-2019 8 21,597.35 843.94 184.48 659.46 20,937.89
Dec-2019 9 20,937.89 843.94 178.84 665.09 20,272.79
Jan-2020 10 20,272.79 843.94 173.16 670.78 19,602.02
Feb-2020 11 19,602.02 843.94 167.43 676.51 18,925.51
Mar-2020 12 18,925.51 843.94 161.66 682.28 18,243.23
Apr-2020 13 18,243.23 843.94 155.83 688.11 17,555.12
May-2020 14 17,555.12 843.94 149.95 693.99 16,861.13
Jun-2020 15 16,861.13 843.94 144.02 699.92 16,161.21
Jul-2020 16 16,161.21 843.94 138.04 705.90 15,455.31
10
Interest paid 1618
Balance owing after
deposit paid 24482
ii. Monthly repayment
Car’s cash price: $29000
10% deposit: 29000 * 105
= 2900
Loan amount: 29000 – 2900
= 26100
Loan Principle Amount 26,100.00
Annual Interest Rate 6.20%
Loan Period (in months) 36.00
Original Repayment Amount 796.38
Loan Start Date 4/27/2019
Repayment Type End
Residual Value -
Month
Repayment
Number
Opening
Balance
Loan
Repayment
Interest
Charged
Capital
Repaid
Closing
Balance
Apr-2019 1 26,100.00 796.38 134.85 661.53 25,438.47
May-2019 2 25,438.47 843.94 217.29 626.65 24,811.82
Jun-2019 3 24,811.82 843.94 211.93 632.00 24,179.81
Jul-2019 4 24,179.81 843.94 206.54 637.40 23,542.41
Aug-2019 5 23,542.41 843.94 201.09 642.85 22,899.56
Sep-2019 6 22,899.56 843.94 195.60 648.34 22,251.22
Oct-2019 7 22,251.22 843.94 190.06 653.88 21,597.35
Nov-2019 8 21,597.35 843.94 184.48 659.46 20,937.89
Dec-2019 9 20,937.89 843.94 178.84 665.09 20,272.79
Jan-2020 10 20,272.79 843.94 173.16 670.78 19,602.02
Feb-2020 11 19,602.02 843.94 167.43 676.51 18,925.51
Mar-2020 12 18,925.51 843.94 161.66 682.28 18,243.23
Apr-2020 13 18,243.23 843.94 155.83 688.11 17,555.12
May-2020 14 17,555.12 843.94 149.95 693.99 16,861.13
Jun-2020 15 16,861.13 843.94 144.02 699.92 16,161.21
Jul-2020 16 16,161.21 843.94 138.04 705.90 15,455.31
10
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