Corey brought $54.50 to the art supply store. He bought a brush, a sketchbook, and a paint set. The brush was 1 6 as much as the sketchbook, and the sketchbook cost 3 4 the cost of the paint set. Corey had $2.00 left over after buying these items. What was the cost of each item?

Answer: The items cost as follows;

(a) Brush = $3.50, (b) Sketchbook = $21 and (c) Paint set = $28

Step-by-step explanation: First of all we shall assign letters to the unknown variables, hence we shall call the brush x, the sketchbook y, and the paint set we shall call z.

The total amount spent by Corey was $52.50 because after buying all he needed, he had $2.00 left out of a total of $54.50. This means his transaction can be expressed as follows;

x + y + z = 52.50

However, we know that the brush was 1/6 as much as the sketchbook, that is, if the sketchbook is y, then the brush which is x shall be 1/6 of y. Simply put, x = 1/6y or x = y/6.

Similarly, the sketchbook cost 3/4 of the paint set. If the paint set is z, that means y = 3/4z or y = 3z/4.

Note however that, if y = 3z/4, then by cross multiplication,

4y = 3z

4y/3 = z

Having derived that x = y/6 and z = 4y/3, we can substitute for the values of x and z into the original equation

x + y + z = 52.50

y/6 + y + 4y/3 = 52.5

using a common denominator which is 6, you now have;

(y + 6y + 8y) / 6 = 52.5

By cross multiplication, you now have

15y = 52.5*6

15y = 315

Divide both sides of the equation by 15

y = 21

Therefore, if the sketchbook is y, then the sketchbook costs $21.

Also if the brush is x, then the brush costs

x = y/6

x = 21/6

x = 3.50 (The brush costs $3.50)

Also if the sketchbook costs 3/4 of the paint set, then

y = 3z/4

21 = 3z/4

21*4 = 3z

84 = 3z

Divide both sides of the equation by 3

28 = z (The paint set costs $28)