Question

For the class of 2013's prom Norman's dress shop sold cheaper dresses for $90 each and more expensive dresses for $140 each. They took in $5250 and sold 20 more of the cheaper dresses than the expensive dresses. How many of each kind did they sell? Simply need help setting up the equations. I can solve them on my own.

Answer 1

Answer:

Number of cheaper dresses sold  is 35

Number of expensive  dresses sold  is 15

Step-by-step explanation:

Given:

Cost of cheaper dresses = $90

Cost of  expensive dresses = $140

Total cost of  the dresses = $5250

To Find:

Number of cheaper dress = ?

Number of expensive  dress = ?

Solution:

Let

The number of  cheaper dresses be x

The number of  expensive dresses be y

(Number of cheaper dresses X cost of cheap dress) +  (Number of Expensive dresses X cost of  expensive dress)  =  $5250

$x \times90 +y \times 140 = 5250$=  $5250

It is given that the 20 more of the cheaper dresses than the expensive dresses is sold

So,

number of cheaper dress  =  20  +  number of expensive dress

x = 20 + y---------------------------------------(1)

$(20+y) \times90 +y \times 140 = 5250 = 5250$

$(20 \times 90 +y\times 90) +y \times 140= 5250$

$1800 + 90y+ 140y = 5250$

$1800 + 230y = 5250$

$230y =5250 -1800$

$230y = 3450$

$y = \frac{3450}{230}$

y = 15

Substituting y in (1)

x = 20 +15

x= 35