Sample Test 1 Solutions MCD 2140 PDF

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Test 1 sample paper answers of business maths of monash college. MCD 2140...

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INTRODUCTORY MATHS FOR BUSINESS (MCD1550) AND BUSINESS MATHEMATICS (MCD2140)

SOLUTIONS Sample Test 1

2

Section A: Short Answer Questions

Question 1 There are two types of tickets on a train: an adult ticket and a child ticket. The cost of buying 2 adult tickets and 3 child tickets would be $20. The cost of buying one of each type would be $8.50.

a.

Write down the simultaneous linear equations for this question. x = Price of an adult ticket y = Price of a child ticket 2 x + 3 y = 20 x + y = 8.5

b.

( 1) (2 )

Solve the above simultaneous equations using substitution method and work out the cost of a single adult ticket and a single child ticket. Eq. 2: x + y = 8.5 x = 8.5 − y

Substitute x into Eq. 1: 2(8.5 − y) + 3 y = 20 17 − 2 y + 3 y = 20

y =3

Substitute y in x = 8.5 − y

x = 8.5 − 3 = 5.5 Price of an adult ticket = $5.50 Price of a child ticket = $3.00

Note: The prices must have the unit ($).

3

Question 2

A balloon vendor at Southbank sells novelty blown up balloons for $5 each.

a.

Write down a function which gives the revenue R( x) obtained when x balloons are sold.

R ( x) = 5x

b.

If each balloon costs the vendor $2, and he pays $100 for a cylinder of gas used to blow up the balloons, write down the cost function C ( x) for getting x balloons blown up and ready for sale.

C ( x ) = 100 + 2x

c.

Sketch the graphs of R( x) and C ( x) on the axes below.

y 240 220 200 180 160 140 120 100 80 60 40 20 O

d.

5

10

15

20

25

30

35

40

45

50 x

The profit obtained when x balloons are sold is given by the function P(x). Write down the function for P(x) in simplest terms.

P ( x ) = R (x ) − C (x ) = 5x − (100+ 2x )

P ( x) = 3 x −100

4

e.

Find the minimum number of balloons needed to be sold in order to make a profit.

P ( x ) = 3x − 100  0

100 3 x  33.33 x

Need to sell 34 balloons

5

Question 3

A hybrid function is defined by for x  0 2  y = 1− 4x for 0  x  0.5 4 x − 3 for x  0.5. 

Sketch the graph of the hybrid function on the axes below.

y 3

2

1

– 1

O

1

– 1

– 2

6

x

Question 4 A rule of the form y = kx + c has been used to partially complete the following table of values. 2

2 0

y x

14 2

50 4

110 6

194 8

50 4 16

110 6 36

194 8 64

x2

a

Complete the table of values above. y x x2

2 0 0

14 2 4

Plot the graph of y against x 2 .

b

y

200 160 120 80 40

10 c

20

30

40

50

60

70

Calculate the values of k and c by reference to your answers in part a.

y = kx2 + c when x 2 = 0, y = 2 c = 2 when x 2 = 4, y = 14  14 = 4 k + 2 k =3 7

80

Question 5

a.

On the axes below shade the feasible region given by: 0 x  5 0  y 3 y

1 x 2

y 3

2

1

O

b.

1

2

3

4

5

6

x

State the coordinates of the four corner points of the feasible region shaded in part a.

(0, 0 ), (0, 3 ), (5, 2.5 ), (5, 3 ) c.

Find the maximum value of the objective function P = 2 x + 3 y over the shaded feasible region in part a. State also the values of x and y which produce this maximum.

P( 0,0 ) = 2  0 + 3  0 = 0 P( 0,3) = 2  0 + 3  3 = 9 P( 5, 2.5) = 2 5 + 3 2.5 =17.5 P ( 5, 3) = 2  5 + 3 3 = 19 Maximum P occurs at (5, 3)

8

Question 6 A share loses 15% of its opening price over the course of a day’s trading, and closes at a value of $34. a.

Find the opening price of the share.

original price = new price

b.

100 100 100 = 34 = 34 = $40 85 ( 100 − r) ( 100 − 15)

The following day the share rises to a value of $45 from its closing price of $34 in the previous day. Find the percentage rise in value of the share over the following day. Give your answer correct to one decimal place.

45 − 34 100 = 32.4% 34

9

Question 7

An investor whose annual income is $70 000 makes a capital gain of $6 000 by selling shares in the last financial year. Use the table below to answer the following questions.

a.

Taxable income ($)

Marginal rate (cents in each $)

0-6 000

0

6 001-30 000

15

30 001-75 000

30

75 001-150 000

40

150 000+

45

Calculate the income tax for this person.

Taxable income ($)

Income Tax

6 0000 – 0 = 6 000

6 000  0

= $0

30 000 – 6 000 = 24 000

24 000  0.15

= $3 600

70 000 – 30 000 = 40 000

40 000  0.30

= $12 000 = $15 600

Total tax on income

b.

Calculate the net monthly income for this person. $70 000 - $15 600 = $54 400

$54400 = $4533.33 12

c.

Given that this person pays tax at the marginal rate on the capital gain, calculate the capital gains tax paid for the share sale. Total income = 70 000 + 6 000 = $ 76 000 Capital gains tax to be paid ((76 000 − 75 000) × 0.40) = $400 ((75 000 − 70 000) × 0.30) = $1 500 Total capital gains tax payable = $1 500 + $400 = $1900 10

Question 8

A car worth $20 000 can be bought using a hire purchase contract for a deposit of 12.5% of the price of the car, with the balance being paid off over 5 years via weekly repayments. The simple interest rate applying to the contract is 8% p.a. a.

Find the balance (principal) still owed after the deposit has been paid.

Balance ( Principal) = 20 000 - (20 000

b.

12.5 ) = $17500 100

Find the interest which will be paid on the deal.

V 0 nr 17500 5 8 = 100 100 D = 7000

D=

c.

Find the weekly repayment amount. Total repaid = 17 500 + 7000 = 24 500 24500 Weekly repayment = = $94.23 5 52

d.

Find the effective interest rate applicable to this contract, correct to two decimal places.

2t 2  260  rf =  8% (260 + 1) ( t + 1) = 15.94% p.a.

Effective rate = re =

11

Section B: Analysis Questions Answer each question in the space provided. Show all working and calculations.

Question 1

The following bank statement shows savings account transactions over the month of May. Date 1 May 6 May 16 May 3 June

a.

Debit ($)

Credit ($) 75

?

Balance ($) 1 850 1 925 1 650 1 650

Find the value of the debit transaction on 16 May.

Debit on 16th May = 1925 – 1650 = $275

b.

If interest is calculated on the minimum daily balance at 2.75% per annum, find the interest due for May. Give your answer correct to two decimal places.

D=

V0rn 100 May 1 – 6:

1850 2.75 6 / 365 = $0.84 100

1925 2.75 9 / 365 = $1.31 100 1650 2.75 16 / 365 May 16 – 31: = $1.99 100

May 7 – 15:

Total interest = 0.84 + 1.31+1.99 = $4.14

c.

Calculate the interest earned to the nearest cent, which pays 2.80% p.a. simple interest on the minimum monthly balance for May.

Interest earned D =

V0 rn 1650  2.80 1 = = $3.85 100 100 12 12

d.

What would be the better offer for this customer and Why?

Daily minimum balance is better as it earns extra interest of $0.29 ($4.14 – $3.85) compared to monthly minimum balance.

End of Test 1

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