14.2: Markup and Discount

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Solving Linear Equations. Commerce Problems

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Some problems here are simple. The solution can be worked by quick reasoning. The benefit of these problems is not to find the solution by reasoning, but by learning algebraic steps applicable to more challenging situations. Solving application problems is a skill. Cultivate it.

My approach is not to read a problem first till understood. Many of my colleagues will disagree with me. I propose creating a preamble in which you write a symbol or set of symbols for each element in the problem.

Look at the end of the problem where you usually find what to solve for. Let $x$ be the number we are looking for. Write this as the first step in the preamble (list of symbols in non-trivial exercises whose solution we seek).

Next start reading the problem from the beginning and develop the preamble step by step. Write the symbol(s) for each step on a new line.

Finish the preamble by converting all problem steps into symbols. Now look at your preamble and gain an overview of the your problem.

It should be easier to obtain an equation using preamble symbols. Then solve the equation by the method introduced this far for linear equations. (A linear equation has a variable to the first degree (exponent).

Markup and Discount

A retailer buys $20$ pairs of shoes from a manufacturer. The retailer has expenses. He needs to travel, probably pay rent, utilities, taxes, etc. To cover his/her expenses the retailer sells a pair of shoes at the cost increased by a fraction of the cost. This fraction of the cost is profit used to pay expenses. That fraction, usually a percent of the cost, is called mark up.

Now suppose the shoes don’t sell. Maybe the fashion changes, the economy worsens, the clientele migrates, etc. The retailer has to get rid of the shoes. As an incentive to entice customers to buy, the retailer takes off a fraction (a percent) of the before-discount price, that is the retail price or regular price or regular price. This fraction is called discount. The sales price is the price at which the shoes will sell after application of the discount. The discount is sometimes called a markdown.

The phrase“Markup “is" percent(decimal) “of" Cost, with emphasis on is and of, will be quite beneficial.

Examples

Example 1:

The Pedstore buys $16$ pairs of shoes at a cost of $\$27$ each. The markup rate is $40\%$. What is the markup and what is the sales price?

Solution:

Preamble:
Let $S$ be the sales price.

The cost is $\$27$ and the markup rate is $40\%=0.4$.

Equation:

$\begin{array}{rcl lll} \hbox{Mark up}&=&\hbox{percent(decimal) of Cost}\\ M&=&(0.4)(27)\\ &=&10.8\\ \end{array}$

The mark up is $\$10.80$. The sales price is $27.00+10.80=\$37.80$.

Note: The markup is $0.4$ for every dollar of cost. Then the sales price is $1.0+0.4=1.4$ dollars.

We could have found the sales price more quickly by multiplying $27$ by $1.4$.

Check: $1.4\cdot 27=37.80$
Example 2:

The Pedstore buys $16$ pairs of shoes at a cost of $\$35$ each. Each pair is on sale for $\$65$. Find the markup rate (as a percent). Round to the nearest percent.

Solution:

Preamble:
Let $x$ be the markup rate (as a decimal).

The cost is $\$35$, the sales price is $\$65$. The markup is
$65-35=\$30$

Equation:

$\begin{array}{rcl lll} \hbox{Mark up}&=&\hbox{percent(decimal) of Cost}\\ 30&=&(x)(35)\\[10pt] \displaystyle \frac{30}{35}&=&\displaystyle \frac{35}{35}x\\[15pt] \displaystyle \frac{6}{7}&=&x\\[15pt] x&=&0.8571\\[10pt] x&=&85.71\%\\[10pt] \end{array}$

The mark up rate is $86\%$.

Check: $1.86\cdot 35=65.1$ (the extra $10$ cents is the result of rounding $85.71$ to $86$.)
Example 3:

The Pedstore sells a pair of shoes for $\$89.99$. The mark up rate is $25\%$. What is the cost of a pair of shoes? (Round to the nearest cent.)

Solution:

Preamble:

Let $C$ be the cost.
Then the mark up is $M=0.25C$
and the sales price is $S=C+0.25C=1.25C$.

Special note: Mark up is a percent of cost. Be careful if the saes price is given and the cost is to be solved for.

Equation:

$\begin{array}{rcl lll} \hbox{Sales price}&=&\hbox{Sales price}\\[5pt] 1.25C&=&89.99\\[10pt] \displaystyle \frac{1.25}{1.25}C&=&\displaystyle \frac{89.99}{1.25}x\\[15pt] C&=&71.99\\[10pt] C&=&\$72\\[10pt] \end{array}$

Check: $M=0.25(72)=\$18$, $S=18+72=90$.
Example 4:

A $\$159.99$ (Regular price) dress has been sitting in Elegante Fashion window for over a year. The merchant encourages you to buy the dress after applying a $15\%$ discount. How much are you paying for the dress?

Solution:

Preamble:

Let $S$ be the sales price. The discount is $15\%$ of the regular price.

Equation:

$\begin{array}{rcl lll} \hbox{Discount}&=&\hbox{discount rate of regular price}\\[5pt] D&=&(0.15)(159.99)\\[5pt] &=&24 \end{array}$

The sales price $S=159.99-24=\$135.99$

Example 5:

The regular price of a hat in Elegante Fashion window is $34.99$. It is discounted to sell for $27.99$. What is the discount rate (in percent)? Round to the nearest percent.

Solution:

Preamble:

Let $x$ be the discount rate (as a decimal). The discount is
$D=34.99-27.99=7.00$.

Equation:

$\begin{array}{rcl lll} \hbox{Discount}&=&\hbox{discount rate of regular price}\\[5pt] 7.00&=&(x)(34.99)\\[15pt] \displaystyle \frac{7.00}{34.99}&=&\displaystyle \frac{34.99}{34.99}x\\[15pt] 0.20&=&x\\[5pt] x&=&20\% \end{array}$

The discount rate is $20\%$.
Example 6:

Find the regular price of a mattress that sells for $\$469.99$ if the discount rate is $33.33\%$

Solution:

Preamble:

Let $R$ be the regular price.
Then the discount is $D=0.3333R$ and the sales price is
$S=R-0.3333R=0.6667R$.

Equation:

$\begin{array}{rcl lll} \hbox{Regular price}&=&\hbox{Regular price}\\[5pt] 0.6667R&=&469.99\\[15pt] \displaystyle \frac{0.6667}{0.6667}R&=&\displaystyle \frac{469.99}{0.6667} =704.95 \end{array}$

The regular price is $\$704.95$

Check: $704.95-0.3333(704.95)=469.99$
Example 7:

A retailer buys a pair of boots for $\$200$. The markup is $10\%$ or $\$20$. The sales price is $200+20=220.$

A year later, the previous sales price is now called the regular price and the boots are discounted at a discount rate of $10\%$.

Since the mark up rate equals the discount rate, has the retailer made a profit, broken even, or lost money?

Solution:

Preamble:

$D=(0.1)(220)=\$22$

Equation:

$S=R-D=220-22=198$.

The retailer lost $\$2$.

Note that mark up is computed on $\$200$ while discount is calculated on he larger figure of $\$220$.

Exercise 14

The Golden Nest buys a wedding ring at a cost of $\$1,395$. The Markup rate is $65\%$. What is the markup and what is the sales price?
The Golden Nest buys a necklace at a cost of $\$235$. It is on sale for $\$675$. Find the markup rate (as a percent). Round to the nearest percent.
The Golden Nest sells a bracelet for $\$59.99$. The mark up rate is $35\%$. What is the cost of the bracelet? (Round to the nearest cent.)
A $\$1,250$ (Regular price) computer has been sitting in at TechTech store for over a year. The merchant encourages you to buy the computer after applying a $33\%$ discount. How much are you buying the computer for?
The regular price of a phone at CallMe is $89.99$. It is discounted to sell for $\$67.99$. What is the discount rate (in percent)? Round to the nearest percent.
Find the regular price of a TV set that sells for $\$345$ if the discount rate is $21\%$

Solution:
Preamble:

Let $S$ be the sales price.

The cost is $\$1,395$ and the markup rate is $65\%=0.65$.

Equation:

$\begin{array}{rcl lll} \hbox{Markup}&=&\hbox{percent(decimal) of Cost}\\[5pt] M&=&(0.65)(1,395)\\[5pt] &=&906.75\\[5pt] \end{array}$

The markup is $\$906.75$.
The sales price is $1,395+906.75=\$2,301.75$.

Note: The mark up is $0.65$ for every dollar of cost. Then the sales price is $1.0+0.65=1.65$ dollars.

We could have found the sales price more quickly by multiplying $1,395$ by $1.65$.

Check: $1.65\cdot 1,395=2301.75$
Solution:
Preamble:

Let $x$ be the mark up rate (as a decimal).

The cost is $\$235$, the sales price is $\$675$. The mark up is $675-235=\$440$

Equation:

$\begin{array}{rcl lll} \hbox{Mark up}&=&\hbox{percent(decimal) of Cost}\\ 440&=&(x)(235)\\[15pt] \displaystyle \frac{440}{235}&=&\displaystyle \frac{235}{235}x\\[15pt] \displaystyle \frac{88}{47}&=&x\\[10pt] x&=&1.8723\\[10pt] x&=&187.23\%\\[10pt] \end{array}$

The mark up rate is $187\%$.

Check: $(1+1.87)\cdot 235=674.45$ (the $55$ cents discrepancy is the result of rounding.)
Solution:
Preamble:

Let $C$ be the cost. Then the markup is $M=0.35C$ and the sales price is $S=C+0.35C=1.35C$.

Equation:

$\begin{array}{rcl lll} \hbox{Sales price}&=&\hbox{Sales price}\\ 1.35C&=&59.99\\[15pt] \displaystyle \frac{1.35}{1.35}C&=&\displaystyle \frac{59.99}{1.35}x\\[15pt] C&=&44.44\\[15pt] \end{array}$

Check: $M=0.35(44.44)=\$15.55$,
$S=44.44+15.55=59.99$.
Solution:
Preamble:

Let $S$ be the sales price. The discount is $33\%$ of the regular price.

Equation:

$\begin{array}{rcl lll} \hbox{Discount}&=&\hbox{discount rate of regular price}\\[5pt] D&=&(0.33)(1,250)\\[5pt] &=&412.50\\[5pt] \end{array}$

The sales price $S=1,250-412.50=\$837.50$

Check: $1,250*0.67=837.50$
Solution:
Preamble:

Let $x$ be the discount rate (as a decimal). The discount is $D=89.99-67.99=22.00$.

Equation:

$\begin{array}{rcl lll} \hbox{Discount}&=&\hbox{discount rate of regular price}\\[5pt] 22&=&(x)(89.99)\\[15pt] \displaystyle \frac{22}{89.99}&=&\displaystyle \frac{89.99}{89.99}x\\[15pt] 0.2447&=&x\\[15pt] x&=&24.47\% \end{array}$

The discount rate is $24\%$.
Solution:
Preamble:

Let $R$ be the regular price.

Then the discount is $D=0.21R$ and the sales price is $S=R-0.21R=0.79R$.

Equation:

$\begin{array}{rcl lll} \hbox{Regular price}&=&\hbox{Regular price}\\[5pt] 0.79R&=&345\\[15pt] \displaystyle \frac{0.79}{0.79}R&=&\displaystyle \frac{345}{0.79}\\[15pt] &=&436.71 \end{array}$

The regular price is $\$436.71$

Check: $436.71-0.21(436.71)=345$

Solving Linear Equations. Commerce Problems

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Markup and Discount

Examples

Exercise 14

Supplementary Homework