Download Financial Maths Solutions

Consumer arithmetic Test B
Name: ___________________________
Section A Multiple Choice
1
The simple interest on a loan of $16 000 B
at 9% p.a. over 4 years is:
A $6585.30
B $5760
C $1646.33
D $1440
2
Melissa has an outstanding credit card
balance of $2200. Interest is charged at
a rate of 15% p.a. The amount that
Melissa must pay to fully pay off the
card is:
A $27.50
B $330
C $2227.50
D $2500
3
A 600 g box of washing powder costs
D
$3.40. At the same rate, 1 kg of washing
powder would cost:
A
$1.76
B
$2.04
C
$2.94
D
$5.67
5
To buy Marmite, the option which
would give the best buy is:
A
250-g jar of Marmite for $2.75
B
550-g jar of Marmite for $5.95
C
850-g jar of Marmite for $9.40
D
975-g jar of Marmite for $10.50
6
A outdoor setting has a marked price of C
$980. The store offers a discount of 5%
to account customers and a further 5%
discount for accounts that are settled
within 7 days.
Calculate the price paid for the outdoor
setting by an account customer who
settles their account within 7 days.
A $735
B $882
C $884.45
D $931
7
Successive discounts of 15% and 7%
are equivalent to a single discount of:
A
20.95%
B
21%
C
21.5%
D
22%
C
The price of the lounge suite is $4350
B
but it can be purchased on the following
terms:
1

deposit
3
 interest rate 9% p.a.
 repayments to be made over 2 years.
The amount of each monthly repayment
would be:
A
$120.83
B
$142.58
C
$181.25
D
$213.88
Maths Quest 10 for New South Wales 5.3 pathway
4
Chapter 9
D
A
1
8
Terry invests $34 000 at 11% p.a. for 3
years with interest compounded
quarterly. The value of the investment
on maturity is (correct to the nearest
dollar):
A
$37 740
B
$45 220
C
$47 083
D
$51 614
9
A sum of money is to be invested for a C
6-year period. Which of the following
investments will give the greatest
return?
A 9% p.a. simple interest
B 8.5% p.a. interest compounded
annually
C 8.4% p.a. interest compounded sixmonthly
D 8.0% p.a. interest compounded
quarterly
10
The price of a new motorbike is
$22 500. The value of the motorbike
depreciates by 20% p.a. The value of
the motorbike after 4 years is:
A $9216
B $11 520
C $13 500
D $18 000
Maths Quest 10 for New South Wales 5.3 pathway
C
11
A mechanics garage has equipment that B
is valued at $250 000. The equipment
depreciates at a rate of 15% p.a. The
number of years that it will take for the
value of the equipment to fall below
$50 000 will be:
A 7 years
B 10 years
C 11 years
D 12 years
12
A $30 000 loan is taken out to be repaid C
over 4 years at a flat interest rate of
15% p.a. The total amount that is to be
repaid on the loan is:
A $18 000
B $22 470
C $48 000
D $52 570
A
Chapter 9
2
Consumer arithmetic Test B
Name: ___________________________
Section B Short/Extended answer
1
2
Find the simple interest payable on each of the
following investments.
(a) $7500 at 8% p.a. for 4 years
3
(a) I = PRT
= $7500  0.08  4
= $2400
(b) $14 800 at 9.6% p.a. for 3 years
(b) I = PRT
= $14 800  0.096  3
= $4262.40
(c) $19 000 at 6.5% p.a. for 9 months
(c) I = PRT
= $19 000  0.065  0.75
= $926.25
Gavin has an outstanding balance of $4620 on
his credit card at the end of April. The interest
rate charged on the credit card is 21% p.a.
Calculate:
(a) The interest that Gavin is charged on the
card for one month.
5
(a) I = PRT
= $4620  0.21 
= $80.85
1
12
(b) In May Gavin repays $1900 and makes
(b) Balance = $4620 + $80.85 – $1900 + $385
purchases of $385. Calculate the
= $3185.85
outstanding balance on Gavin’s credit card
at the end of May.
(c) Calculate the interest that is payable for
May.
Maths Quest 10 for New South Wales 5.3 pathway
Chapter 9
(c) I = PRT
= $3185.85  0.21 
= $55.75
1
12
3
3
4
A used car is purchased on terms. 10% deposit
is payable with the balance plus interest to be
repaid over 4 years in equal monthly
instalments. Interest is charged at a flat rate of
9% p.a. For a car with a marked price of
$12 500:
(a) Find the amount of the deposit.
5
(a) Deposit = 10% of $12 500
= $1250
(b) Find the balance owing.
(b) Balance = $12 500 – $1250
= $11 250
(c) Calculate the amount of interest payable.
(c) I = PRT
= $11 250  0.09  4
= $4050
(d) Calculate the total amount to be repaid.
(d) Total to be repaid = $11 250 + $4050
= $15 300
(e) Find the amount of each monthly
repayment.
(e) Monthly repayments = $15 300  48
= $318.75
A brand of dishwashing liquid is sold in three
different sizes
Bottle A – 400 mL for $1.40
Bottle B – 750 mL for $2.60
Bottle C – 1.25 L for $4.40
(a) Find the cost of each bottle as a rate per
litre.
(b) Which bottle of dishwashing liquid is the
cheapest?
Maths Quest 10 for New South Wales 5.3 pathway
Chapter 9
4
(a) Bottle A = $1.40  400  1000
= $3.50/L
Bottle B = $2.60  750  1000
= $3.47/L
Bottle C = $4.40  1.25
= $3.52/L
(b) Bottle B is the cheapest.
4
5
A hardware store has a sale in which all goods
are discounted by 15%.
(a) Find the price paid for a set of taps that
have a marked price of $320.
3
(a) Price = 85% of $320
= 0.85  $320
= $272
(b) Account customers are then given a further (b) Price = 98% of $272
discount of 2%. What would an account
= 0.98  $272
customer pay for the taps?
= $266.56
6
A timber yard offers its account customers a
5% discount. Carpenters are then given a trade
discount of 12.5%.
(a) Find the amount that a carpenter would pay (a) After 1st discount price = 95% of $1500
to settle an account of $1500.
= 0.95  $1500
= $1425
After 2nd discount price = 87.5% of $1425
= 0.875  $1425
= $1246.88
4
(b) What single discount is equivalent to
successive discounts of 5% and 12.5%?
(b) Discount = $1500 – $1246.88
= $253.13
$253.13
Percentage discount =
 100%
$1500
= 16.875%
7
Find the single discount equivalent to
successive discounts of 12% and 4%.
Percentage paid = 88% of 96%
= 0.88  0.96
= 0.8448
= 84.48%
Percentage discount = 100% – 84.48%
= 15.52%
3
8
Find the amount to which an investment of
$32 000 will grow if invested at 8% p.a. for 4
years with interest compounded annually.
A = P(1 + R)n
= $32 000(1.08)4
= $43 535.65
2
Maths Quest 10 for New South Wales 5.3 pathway
Chapter 9
5
9
Marilyn invests $50 000 at 6% p.a. for a period
of 5 years. Calculate the amount of compound
interest earned if interest is compounded:
(a) annually
(a) A = P(1 + R)n
= $50 000(1.06)5
= $66 911.28
Interest = $66 911.28– $50 000
= $16 911.28
(b) six-monthly.
(b) A = P(1 + R)n
= $50 000(1.03)10
= $67 195.82
Interest = $67 195.82 – $50 000
= $17 195.82
10
A sound system that is purchased new for
$3600 depreciates at the rate of 12% p.a. Find
the value of the sound system after 5 years.
Give your answer correct to the nearest dollar.
A = P(1 – R)n
= $3600(0.88)5
= $1900
11
A tractor purchased for $450 000 depreciates at Test n = 5
the rate of 17.5% p.a. The tractor can be written A = P(1 – R)n
off for tax purposes when the value falls below
= $450 000(0.825)5
$50 000. Use a trial and error method to find
= $171 982
how long it will take for the value to first fall
below $50 000.
Test n = 10
A = P(1 – R)n
= $450 000(0.825)10
= $65 728
4
2
3
Test n = 11
A = P(1 – R)n
= $450 000(0.825)11
= $54 226
Test n = 12
A = P(1 – R)n
= $450 000(0.825)12
= $44 736
It will take 12 years for the tractor to be written
off.
Maths Quest 10 for New South Wales 5.3 pathway
Chapter 9
6
12
Calculate to total cost of repaying a loan of
$80 000 to be repaid over 10 years at a flat
interest rate of 12% p.a.
13
A loan of $25 000 is to be repaid over 3 years
with interest charged at 9% p.a.
(a) Find the amount of interest that is paid if
the annual repayment made is $10 000.
(b) Calculate the saving made if the annual
repayment is increased to $12 500.
Maths Quest 10 for New South Wales 5.3 pathway
Chapter 9
I = PRT
= $80 000  0.12  10
= $96 000
Total repayments = $80 000 + $96 000
= $176 000
2
4
(a) 1st year interest = 9% of $25 000
= $2250
2nd year balance = $25 000 + $2250 – $10 000
= $17 250
2nd year interest = 9% of $17 250
= $1552.50
3rd year balance
= $17 250 + $1552.50 – $10 000
= $8802.50
3rd year interest = 9% of $8802.50
= $792.23
Total interest = $2250 + $1552.50 + $792.23
= $4594.73
(b) 1st year interest = 9% of $25 000
= $2250
2nd year balance = $25 000 + $2250 – $12 500
= $14 750
2nd year interest = 9% of $14 750
= $1327.50
3rd year balance
= $14 750 + $1327.50 – $12 500
= $3577.50
3rd year interest = 9% of $3577.50
= $321.98
Total interest = $2250 + $1327.50 + $321.98
= $3899.48
Saving = $4594.73 – $3899.48
= $695.25
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