9.1 markup on cost selling price: price for product offered to public markup, margin, or gross...
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9.1 Markup on Cost
• Selling Price: price for product offered to public• Markup, margin, or gross profit: difference
between the cost and the selling price• Basic formula: Cost + Markup = Selling Price
(in this section markup is based on cost)
SMC
9.1 Markup on Cost• Example: A coffee maker is purchased for $15 and
sold for $18.75. Find the percent of markup based on cost.
Markup = M = $18.75 - $15 = $3.75Percent equation: P = partB = baseR = rate = percent %2525.0
15$
75.3$
15$75.3$
R
R
BRP
9.1 Markup on Cost• Example: A baseball glove is sold for $42,
which is 140% of cost. How much is the store’s cost?
Selling price= 140% of cost so the markup is 40% of cost (cost is 100% of itself)
30$4.1
42$
42$4.14.0
C
CCC
SMC
9.1 Markup on Cost• Example: North American Coins priced a proof
coin at $868, which was 112% of cost. Find (a) the cost, (b) the markup as a percent of cost, and (c) the markup.
Selling price= 112% of cost so the markup is 12% of cost
93$775$868$
775$12.1
868$
868$12.112.0
M
C
CCC
SMC
9.2 Markup on Selling Price
• Sometimes markup is based on the selling price rather than cost. The same basic formula applies:
• The difference is that markup is now considered a percent of the selling price rather than cost
SMC
9.2 Markup on Selling Price• Example: An auto parts dealer pays $7.14 per 12
gallons of windshield washer fluid and the markup is 50% on selling price. Find the selling price.Markup = 50% of the selling price
gallonperS
C
SSSC
SSCSofMandSMC
19.1$5.
595.0$
595.0$12
14.7$
5.05.0
5.0%50
9.2 Markup on Selling Price• Example: A retailer purchases silk flowers for
$31.56 per dozen and sells them for $4.78 each. Find the percent markup on selling price and the equivalent percent markup on cost.
%7.81817.
%45450.
15.2$63.2$78.4$
63.2$12
56.31$
63.215.2
78.415.2
CM
SM
M
C
CSMSMC
9.2 Markup on Selling Price• Converting percent markup on cost to
percent markup on selling price:
• Converting percent markup on selling price to percent markup on cost:
S
SC M
MM
%100
C
CS M
MM
%100
9.2 Markup on Selling Price• Example: Convert a markup of 20% on
selling price to its equivalent markup on cost.
%2525.08.0
2.0
%80
%20
%20%100
%20
%100
S
SC M
MM
9.3 Markup with Spoilage
• Markup with spoilage: Some items may not be fit for sale or will go bad. Sometimes they can be sold for a reduced price. Sometimes they are a total loss. The selling price has to be higher to make up for this loss.
9.3 Markup with Spoilage• Example: The cost for 36 items is $540. If 6 items
cannot be sold, what is the selling price per item for a % markup of 25% on selling price?
itemper
S
SSS
SSCSMC
24$30
720$
636
720$
720$75.
540$
75.540$25.540$
%25
9.3 Markup with Spoilage• The cost for 120 items is $360. If 10% are sold at
a reduced price of $2, what is the selling price per item for a markup of 20% on cost?
10812120#
121201.%)10(#
432$)360($2.1
%20
priceregularat
pricereducedatsold
S
SCCSMC
78.3$108
408
408$24$432$
24$2$12
itemperprice
salespriceregular
salespricereduced
10.1 Markdown• When merchandise does not sell at the original
price the price must be reduced. The basic formula for markdown is:
• Example: What is the reduced price if the original price was $960 and the markdown is 25%?
720$240$960$
240$960$25.0
pricereduced
markdown
markdownpriceoriginalpricereduced
10.1 Markdown• Example: Given an original price of $240 and a
markdown of $96, what is the percent markdown and the reduced price?
144$96$240$
%404.0240
96
96$240$
pricereduced
R
RPBR
10.1 Markdown• Markdown equations:
Break-even point = Cost + Operating expenses
Operating Loss = Break-even point – Reduced selling price
Absolute loss = Cost – Reduced selling price
10.1 Markdown• Given a cost of $25, operating expense of $8, and
reduced price of $22, what is the break-even point, the operating loss, and the absolute loss?
3$22$25$
11$22$33$
33$8$25$
lossabsolute
lossoperating
ptevenbreak
11.1 Basics of Simple Interest
• Simple Interest Formula:I = interest, P = principal, R = rate of interest per year, T = time in years
• Example: Given an investment of $9500 invested at 12% interest for 1½ years, find the simple interest.
PRTI
1710$
5.112.09500$
I
PRTI
11.1 Basics of Simple Interest• Example: If money invested at 10% interest for 7
months yields $84, find the principal.
1440$0583.
84$
0583.12
710.084$
P
PP
PRTI
11.1 Basics of Simple Interest• Example: If $2600 is invested for 7 months and
yields simple interest of $144.08, what is the interest rate?
%5.9095.67.1516
08.144
67.151612
72600$08.144$
R
RR
PRTI
11.2 Simple Interest for a Given Number of Days
• To find the exact number of days between two dates (2 methods):
1. Get the number corresponding to each date (Julian date) from table 11.1 and subtract
2. Add the number of days in between the two dates going month by month using the number of days in each month
11.2 Simple Interest for a Given Number of Days
• Find the number of days from April 24 to July 7:
(1) Using table 11.1, April 24 = day 114July 7 = day 188, # days = 188 – 114 = 74
(2) # days left in April = 6# days in May = 31# days in June = 30# days in July = 7Total days = 6 + 31 + 30 + 7 = 74
11.2 Simple Interest for a Given Number of Days
• Exact interest:• Ordinary or banker’s interest:• Example: Given an investment of $2600
invested at 10.5% interest for 180 days, find the ordinary interest.
365
# daysexactT
360
# daysexactT
50.136$360
180105.02600$
I
PRTI
11.2 Simple Interest for a Given Number of Days
• Example: Bella missed an income tax payment. The payment was due on June 15 and was paid September 7. The penalty was 14% simple interest on the unpaid tax of $4600. Find the penalty using exact interest.#days = 15 + 31 + 31 + 7 = 84 days
21.148$365
8414.04600$
I
PRTI
11.3 Maturity Value• Maturity Value = amount loaned + interest
• Maturity Date = date the loan is paid off
• Example: A $12,200 loan is borrowed at 9.5% for 10 months. Find the interest and maturity value.
)1( RTPM
PRTPIPM
83.165,13$83.965$200,12$
83.965$12
10095.200,12$
M
I
PRTI
11.3 Maturity Value• Find the Time: If a loan of $7400 is borrowed at
9.5% has interest of $292.92, find the time in days and the maturity value
92.7692$92.292$7400$
150
3604167.04167.0095.7400$
92.292$
095.7400$92.292$
M
daysT
yearsT
TPRTI
11.3 Maturity Value• Find the Principal and Rate: If a loan is borrowed
with interest of $300 for 120 days with a maturity value of $7800, find the principal and interest rate.
%1212.02500$
300$
2500$360
1207500$300$
7500$
300$7800$
R
RRPRTI
P
PIPM
11.4 Inflation and the Time Value of Money
• Inflation: continuing rise in the general price level of goods and services
• Consumer Price Index (CPI): one way to measure inflation. The CPI reflects the average change in prices from one year to the next.
• Time Value of Money: the idea that loaning money has value and that value is repaid by returning interest in addition to principal.
11.4 Inflation and the Time Value of Money
• Present value: principal amount that must be invested today to produce a given future value.
• Future value: amount that a present value grows to; also called the maturity amount.
RT
MPorRTPM
1)1(
11.4 Inflation and the Time Value of Money
• Time Value of Money – with simple interest of 5% per year.
2000 2010 2020
)10)05(.1(1000$
)1(1000$
RT
10)05(.1
1000$
1
1000$
RT1000$
11.4 Inflation and the Time Value of Money
• Example: If the present value = $8000 at 8.5% for 140 days, what is the future value?
44.8264$
)18
7085.01(8000$)1(
18
7
360
140
M
RTPM
T
11.4 Inflation and the Time Value of Money
• Example: If the future value = $1985.50 at 9% for 180 days, what is the present value?
1900$5.009.01
50.1985$
1
5.0360
180
PRT
MP
T
12.1 Simple Interest Notes• Promissory note: Legal note in which a person
agrees to pay a certain amount of money at a stated time and interest rate to another person
• Face value of note: principal (P)• Maturity value: M = P + I = P + PRT = P(1 + RT)• Term of the note: T – often given in days (convert
to years for formulas)
12.1 Simple Interest Notes• Example: For a promissory note with face value of
$9500, term of 200 days, rate of 10%, and date made of March 18, find the due date and the maturity value.Using table 11.1, March 18 = day 7777 + 200 = day 277 = October 4 (due date)
78.027,10$
)9
510.01(9500$)1(
9
5
360
200
M
RTPM
T
12.1 Simple Interest Notes• Example: For a simple interest note with maturity
value of $7632, term of 240 days, and rate of 9%, find the principal.
7200$3
209.01
7632$
1
3
2
360
240
P
RT
MP
T
12.2 Simple Discount Notes
• Simple discount note: interest is deducted in advance from the face value written on the note.
• M = face value = maturity value (not the principal)• B = bank discount (similar to interest)• D = discount rate (similar to rate of interest)• T = time in years
PRTItosimilarMDTB
12.2 Simple Discount Notes• Maturity for simple interest:
• Maturity for discount notes:(similar but you subtract the discount from the maturity)
)1( RTPM
PRTPIPM
)1( DTMP
MDTMBMP
12.2 Simple Discount Notes• Example: For a simple discount note with a
maturity value of $6800, discount rate of 10%, and time of 180 days, find the discount and the proceeds.
6460340$6800$
340$5.010.06800$
5.0360
180
BMP
MDTB
T
12.2 Simple Discount Notes• Example: For a simple discount note with a
maturity value of $8200, discount of $205, and date made of 2/9, due date of 5/10, find the discount rate, time in days, and the proceeds.
7995$205$8200$
%1010.025.08200$
205$
205$25.08200$
25.0)(
9010303119)(
36090
BMP
D
DMDTB
yearsT
daysT
12.3 Comparing Simple Interest and Simple Discount
• Similarities between simple interest notes and simple discount notes:
1. Borrower receives money at the beginning of each note.
2. Both notes are repaid with a single payment at the end of the period.
3. Length of time is generally less than 1 year.
12.3 Comparing Simple Interest and Simple Discount
• Differences between simple interest notes and simple discount notes:
1. Formulas• Discount notes:
• Interest notes:
2. A simple interest rate 12% (relative to present value) is not the same as a simple discount rate of 12% (relative to maturity value.
)1( RTPM
PRTPIPM
)1( DTMP
MDTMBMP
12.3 Comparing Simple Interest and Simple Discount
1. Converting interest rate to discount rate
2. Converting discount rate to interest rate
DT
DR
1
RT
RD
1
12.3 Comparing Simple Interest and Simple Discount
• Example: Given an interest rate of 8% and a time period of 240 days, find the corresponding simple discount rate:
%59.7
0759.008.01
08.0
1
3
2
360
240)(
32
D
RT
RD
yearsT