Maths Assignment: Comprehensive Solutions for All Questions

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, dividen...Read More

Maths

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
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 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

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

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

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 = (40N, 15W) 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

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

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
= 15ab²
iii) 20xy² / 4x²y
= 5xy
7

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

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

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

Aug-2020 17 15,455.31 843.94 132.01 711.92 14,743.39
Sep-2020 18 14,743.39 843.94 125.93 718.01 14,025.38
Oct-2020 19 14,025.38 843.94 119.80 724.14 13,301.25
Nov-2020 20 13,301.25 843.94 113.61 730.32 12,570.92
Dec-2020 21 12,570.92 843.94 107.38 736.56 11,834.36
Jan-2021 22 11,834.36 843.94 101.09 742.85 11,091.50
Feb-2021 23 11,091.50 843.94 94.74 749.20 10,342.31
Mar-2021 24 10,342.31 843.94 88.34 755.60 9,586.71
Apr-2021 25 9,586.71 843.94 81.89 762.05 8,824.65
May-2021 26 8,824.65 843.94 75.38 768.56 8,056.09
Jun-2021 27 8,056.09 843.94 68.81 775.13 7,280.97
Jul-2021 28 7,280.97 843.94 62.19 781.75 6,499.22
Aug-2021 29 6,499.22 843.94 55.51 788.42 5,710.79
Sep-2021 30 5,710.79 843.94 48.78 795.16 4,915.64
Oct-2021 31 4,915.64 843.94 41.99 801.95 4,113.68
Nov-2021 32 4,113.68 843.94 35.14 808.80 3,304.88
Dec-2021 33 3,304.88 843.94 28.23 815.71 2,489.17
Jan-2022 34 2,489.17 843.94 21.26 822.68 1,666.50
Feb-2022 35 1,666.50 843.94 14.23 829.70 836.79
Mar-2022 36 836.79 843.94 7.15 836.79 -
B.
3 (7 – x) – 4 (x – 5)
= 21 – 3x – 4x + 20
= 21- 7x +20
7x = 41
x = (41/7) = 5.85
C.
11

The area of trapezium = h * [(b1 +b2)/2]
Where b1 and b2 are base of trapezium and h is the height.
Area of first trapezium
Height (h1) = 125 m
B1 = 95 m
B2 = 75 m
Area 1 = 125 * [(95 + 75) /2] = 10625 m
Area of second trapezium
Height (h2) = 125 m
B1 = 75 m
B2 = 63 m
Area 2 = 125 * [(75 + 63) /2] = 8625 m
Total area of the given field = 10625 + 8625 = 19250 m
SECTION C
Question 24
Question 24. a.
i. Equation of Income function.
The income function equation if n coffees are sold is as follows:
12

I = number of units sold * price of each unit
So, the equation of income function will be I = $5.00 n.
ii. Equation of Cost function of making coffee.
The cost function of making a coffee is as follows:
C(x) = FC + V(x)
where, C =total production cost, FC = total fixed cost, V = variable cost and x = number of units.
So, Equation of the cost function is C(x) = $900 + 2(x).
iii. Graph of Income function.
Revenue function
R = Mx
R = 5x
Profit function
P = R- C
P = 5 – 2
iv. Gradient of the cost functions.
C (x) = mx + b
v. Coordinates of Break even point
Particulars Figures Formula
Selling price per unit (SPU) $5
Variable cost per unit (VCPU) $2
Fixed cost $900
BEP (in units) Fixed cost / SPU - VCPU $900 / (5 – 2)
= 900 / 3
= 300
BEP (in $) BEP (in units) * SPU 300 * 5
= $1500
vi. Shading of profit zone on the graph.
13

vii. Cafe decided to buy a new machine:
1. The price of coffee machine – On buying new coffee machine, cafe has to pay a amount of
$8695.
2. Depreciation value of coffee machine -
Particulars Amount (in $)
Cost of coffee machine 8695
Useful Life in years 3
Rate of Depreciation 15.00%
Annual Depreciation =
Cost of coffee machine
*rate of depreciation *
Useful life
3913
B) i
Model 1:
Make: Ferrari
14

Model: Ferrari Portofino 2019 Australia
Fuel type: Petrol
Fuel consumption: 10.7 L / 100 km
Model 2:
Make: Audi
Model: Audi Q5 2018
Fuel type: Diesel
Fuel consumption: 5.3 L / 100 km
B.)ii
Audi Q5 2018 has more economical fuel consumption. For a distance of 3930 km from Sydney
to Perth the fuel consumption of the care will be:
3930 * (5.3/100)
208.29 litre
B.)iii
Distance d = 3930 km
Price of petrol = $1.49/litre
Fuel consumption of petrol based car = 10.7 L / 100 km
For given distance fuel consumption = 420.51 litre
Cost for the journey = 420.51*1.49 =626.55
Total cost of journey = $626.55
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