Question

A convenience store sell pre-paid mobile phones. It purchases them for $12 each and uses a mark up rate of 250%.

a. The markup rate is the constant of proportionality in the equation y= kx. Write an equation that can be used to find the amount of markup on the cost x of a phone


b. Use the equation to calculate the markup amount. What is the retail price of a phone?


c. Next week the store is going to reduce the price to $31. 50. Calculate what percent the sale price is of the retail price.


d. Based on your answer to part C, what is an equation in the form y = kx that relates the sale price to the retail price x?

Answer 1

a. y = 2.5x

b. markup: $30; retail price: $42

c. 75%

d. y = 0.75x

Explanation:

You want equations relating markup, retail price, and sale price on a phone that costs $12, is marked up 250%, then put on sale for $31.50.

a. Markup equation

The problem statement tells you the markup is 250% of the cost. You could write the equation as ...

y = (250%)x

Or you can simplify it to a decimal number:

y = 2.5x

b. Markup amount

The markup amount is ...

y = 2.5(12) = 30

The markup amount is $30.

That is added to the cost to get the retail price:

retail = $12 +30 = $42

The retail price of the phone is $42.

c. Sale price

The sale price as a percent of the retail price is ...

percent = (sale price)/(retail price) × 100%

percent = 31.50/42 × 100% = 75%

The sale price is 75% of the retail price.

d. Sale price equation

As in part (a), we could write this as ...

y = (75%)x

or, in decimal, as ...

y = 0.75x

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