MAB122 MAP Revision - Full Solutions[1]

8/4/2019 MAB122 MAP Revision - Full Solutions[1]

1/7

Year 12 Mathematics BTerm 3 MAP Revision

SolutionsQuestion 1A graphic design for a cats eye emblem is madefrom two ellipses and a circle. The design of bothellipses is such that the ratio of the shortestdimension to the longest dimension is 1:2. Theouter ellipse is 10 unit high and 20 units long.

Write a set of equations (using the origin at thecentre of the design) that will create the emblem.

For the outer ellipse, a circle of radius 1 has been dilated by a factor of 10 horizontally andby a factor of 5 vertically. The centre is at the origin.

i.e. its equation is 1510

22

=

+

yxor 1

25100

22

=+yx

The circle has a radius of 5, so its equation is 2522 =+ yx or 12525

22

=+yx

The inner ellipse is a circle of radius 1 which has been dilated by a factor of 2.5horizontally and by a factor of 5 vertically. The centre is at the origin.

i.e. its equation is 155.2

22

=

+

yxor 1

2525.6

22

=+yx

Question 2

Colleen has to repay a debt which is currently $ 18 500 and is growing with interest at 6%pa compounded semi-annually. She plans to repay this debt with semi-annual paymentsof $1900 each, the first due in 6 months, together with a final payment if necessary.

Verify that the amount she still owes just after the eighth payment is in fact $6539.82.

A = 18500; i= 0.03; P= 1900; to find the amount repaid after 8 payments:

( )

( )

42.1333703.0

)03.01(11900

)1(1

=

+=

+=

n

n

i

iPA

Amount of original loan unpaid = 18500 13337.42= 5162.58

Future value of unpaid portion = 5162.58 1.038= 6539.81

i.e. Colleen still owes $6539.81


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Question 3Judy Judge buys a house in West End taking out a $95 000 mortgage over 25 years. Shetakes a fixed interest loan with the NAB at 5.95% pa. At the end of the 5 years, her loanchanges to a fixed interest loan at 6.75%pa.

Assuming the rate remains at 6.75% pa for the term of the loan, find what her repayments

should increase to in order to maintain the loan at 25 years.

Let Rbe the monthly repayments for the first 5 years, 3004958.012

0595.0==i

( )

( )

( )19.609)3004958.01(1

3004958.095000

)1(195000

)1(1

300

=+

=

+=

+=

R

i

iR

i

iRP

n

n

Let Qbe the remaining debt after 5 years (i.e. with 20 years ofloan remaning),

( )

59.85374

3004958.0

)3004958.1(119.609240

=

=

Q

Let Xbe the monthly repayments required to repay this, where 005625.012

0675.0

==i ,

( )

( )

95.648

005625.11

005625.059.85374

005625.0

)005625.01(159.85374

)1(1

240

240

=

=

+=

+=

R

R

i

iRP

n

i.e Judge Judy will need to increase her monthly repayments to $648.95


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Question 4An investor wants to have a holiday in 5 years. He estimates that in current prices it willcost $5000.If he assumes that inflation will average 3% over the next 5 years, but his pay will increaseby 10% each year, how much should he invest now, so that in five years (after four morepayments, indexed to reflect his annual pay) he will have sufficient to pay for the holiday?

Assuming that his investments earn interest at 6% p.a. compounded annually,

The anticipated (inflated) price of the holiday in 5 years, A, is

37.5796

)03.01(5000

)1(

5

=

+=

+=n

iPA

If the first payment is P, then the subsequent payments are: )1.01( +P , 2)1.01( +P ,3)1.01( +P and 4)1.01( +P

Including interest, the future value of each of the five payments will be: 5)06.01( +P ,4)06.01)(1.01( ++P , 32 )06.01()1.01( ++P , 23 )06.01()1.01( ++P and )06.01()1.01( 4 ++P .

These amounts form a G.P. with a first term of5)06.01( += Pa and a common

ratio of 0377.106.1

1.1=r .

So using the sum of a G.P. formula:

32.803

106.1

1.1)06.01(

106.1

1.137.5796

106.1

1.1

106.1

1.1)06.01(

37.5796

1

1

5

5

5

5

+

=

+

=

=

P

P

ir

raS

n

n

i.e. His first payment will have to be $803.32.


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Question 5Use the classpad to solve this, as we are dealing with different P/Yand C/Y values.

Question 6Equation of smaller circle

Another point on this smaller circle is (-3,-4), this is the centre of the larger circle

Equation of larger circle

Question 7First, work out how much will need to be saved up.

= $114848.16


5/7

Question 8

Given centre at origin, equation is . We know that the foci occur at Given

that the equation for foci is

Equation for eccentricity is

Vertices occur at and . The four vertices are and .

Question 9

Jack

Given A=-250000, i=0.072, k=12 and R=1000n = 153.17 months

Jill

Given A=250000, i=0.073, k=1 and R=-1000n = 13.12 years, n = 157.47 months

Jills inheritance lasts 4.3 months longer than Jacks.


6/7

Question 10

Assume a 25 year loan.

Therefore, if we move to fortnightly payments, . Given this, solve for n.

Total paid (fortnightly) = 530.93 x 539.69= $286,537.61

Total paid (monthly) = 300 x 1079.38= $323,814

Fortnightly is cheaper by $37276.39

Question 11

Where R=1000, i=0.075, k=12, n=10x12

This now becomes our Present Value, where we now need to solve n.


7/7

Question 12Romeos carInitial value = $10,000. After 3 years it has appreciated in value to $15,000.

Juliets carJuliets car depreciates by 15% each year, and was initially worth $35,000.

Find when the car prices are equal

Therefore, the cars will be worth the same amount after 4.21 years

Question 13Period Start Balance Interest Payment Close Balance1 1,000,000 1,080,000 30,000 1,050,0002 1,050,000 1,134,000 31,200 1,102,8003 1,102,800 1,191,024 32,448 1,158,576

As it can be seen here, the investment will increase in value, even with his withdrawal.

MAB122 MAP Revision - Full Solutions[1]

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